Added June 22, 2019
Solve the heat equation \[ u_t = k u_{xx} \] For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &= A \\ u(L,t) &= B \end {align*}
Initial condition is \(u(x,0)=f(x)\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}]; bc = {u[0, t] == A, u[L, t] == B}; ic = u[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t},Assumptions->{k>0,L>0,t>0}], 60*10]]; sol = sol /. K[1] -> n;
\[\left \{\left \{u(x,t)\to \underset {n=1}{\overset {\infty }{\sum }}\frac {2 e^{-\frac {k n^2 \pi ^2 t}{L^2}} \left (\int _0^L \left (-A+\frac {(A-B) x}{L}+f(x)\right ) \sin \left (\frac {n \pi x}{L}\right ) \, dx\right ) \sin \left (\frac {n \pi x}{L}\right )}{L}+\frac {x (B-A)}{L}+A\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,t),t)=k*diff(u(x,t),x$2); ic := u(x,0)=f(x); bc := u(0,t)=A, u(L,t)=B; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t)) assuming k>0,L>0,t>0),output='realtime'));
\[u \left (x , t\right ) = \frac {A L -2 L \left (\sum _{n=1}^{\infty }\frac {\left (\int _{0}^{L}\left (B x -L f \left (x \right )+\left (L -x \right ) A \right ) \sin \left (\frac {\pi n x}{L}\right )d x \right ) {\mathrm e}^{-\frac {\pi ^{2} k n^{2} t}{L^{2}}} \sin \left (\frac {\pi n x}{L}\right )}{L^{2}}\right )-\left (A -B \right ) x}{L}\]
Hand solution
Solve\begin {equation} u_{t}=ku_{xx}\qquad t>0,0<x<L \tag {1} \end {equation} BC are\begin {align*} u\left ( 0,t\right ) & =A\\ u\left ( L,t\right ) & =B \end {align*}
Initial conditions\[ u\left ( x,0\right ) =f\left ( x\right ) \] Solution
Since boundary conditions are nonhomogeneous, then the first step is to reduce the problem to one with homogeneous B.C. to be able to use separation of variables (separation of variables can only be done on a PDE with homogeneous B.C.)
This is done by using steady state solution. Let the total solution be\begin {equation} u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x\right ) \tag {2} \end {equation} Where \(v\left ( x,t\right ) \) is the transient solution which satisfies homogeneous version of the B.C. and \(r\left ( x\right ) \) is the steady state solution which do not depend on time and just needs to satisfy the nonhomogeneous BC. Since \(r\left ( x\right ) \) is the steady state solution, then the PDE becomes an ODE\begin {align*} 0 & =kr^{\prime \prime }\left ( x\right ) \\ r\left ( 0\right ) & =A\\ r\left ( L\right ) & =B \end {align*}
This has the solution \(r\left ( x\right ) =c_{1}x+c_{2}\). Using BC at \(x=0\) leads to \(A=c_{2}\). Therefore the solution is \(r\left ( x\right ) =c_{1}x+A\). Using BC at \(x=L\) gives \(B=c_{1}L+A\) or \(c_{1}=\frac {B-A}{L}\). Hence\[ r\left ( x\right ) =\left ( \frac {B-A}{L}\right ) x+A \] Therefore (1) becomes\[ u\left ( x,t\right ) =v\left ( x,t\right ) +\left ( \frac {B-A}{L}\right ) x+A \] Substituting the above back in the original PDE \(u_{t}=ku_{xx}\) gives\begin {align} v_{t} & =kv_{xx}\nonumber \\ v\left ( 0\right ) & =0\nonumber \\ v\left ( L\right ) & =0\nonumber \\ v\left ( x,0\right ) & =F\left ( x\right ) \nonumber \\ & =u\left ( x,0\right ) -r\left ( x\right ) \nonumber \\ & =f\left ( x\right ) -\left ( \frac {B-A}{L}x+A\right ) \tag {3} \end {align}
The above PDE was solved in problem 3.1.1.1 on page 395 and the solution is\begin {align} v\left ( x,t\right ) & =\sum _{n=1}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}F\left ( s\right ) \sin \left ( \sqrt {\lambda _{n}}s\right ) ds\right ) e^{-k\lambda _{n}t}\sin \left ( \sqrt {\lambda _{n}}x\right ) \tag {4}\\ \lambda _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \nonumber \end {align}
Substituting (3) into (4) gives\[ v\left ( x,t\right ) =\frac {2}{L}\sum _{n=1}^{\infty }\left ( \int _{0}^{L}\left ( f\left ( s\right ) -\left ( \frac {B-A}{L}s+A\right ) \right ) \sin \left ( \sqrt {\lambda _{n}}s\right ) ds\right ) e^{-k\lambda _{n}t}\sin \left ( \sqrt {\lambda _{n}}x\right ) \] From (2)\begin {align*} u\left ( x,t\right ) & =v\left ( x,t\right ) +r\left ( x\right ) \\ & =A+\left ( \frac {B-A}{L}\right ) x+\frac {2}{L}\sum _{n=1}^{\infty }\left ( \int _{0}^{L}\left ( f\left ( s\right ) -\left ( \frac {B-A}{L}s+A\right ) \right ) \sin \left ( \sqrt {\lambda _{n}}s\right ) ds\right ) e^{-k\lambda _{n}t}\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ & =A+\left ( \frac {B-A}{L}\right ) x+\frac {2}{L}\sum _{n=1}^{\infty }\left ( \int _{0}^{L}\left ( f\left ( s\right ) -\left ( \frac {B-A}{L}s+A\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \end {align*}
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Taken from Maple PDE help pages
Solve the heat equation \[ u_t = u_{xx} \] For \(0<x<1\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &= 20 \\ u(1,t) &= 50 \end {align*}
Initial condition is \(u(x,0)=0\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == D[u[x, t], {x, 2}]; bc = {u[0, t] == 20, u[1, t] == 50}; ic = u[x, 0] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], x, t], 60*10]]; sol = sol /. K[1] -> n;
\[\left \{\left \{u(x,t)\to -\frac {2 \underset {n=1}{\overset {\infty }{\sum }}\frac {\left (20-50 (-1)^n\right ) e^{-n^2 \pi ^2 t} \sin (n \pi x)}{n}}{\pi }+30 x+20\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,t),t)=diff(u(x,t),x$2); ic := u(x,0)=0; bc := u(0,t)=20, u(1,t)=50; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t))),output='realtime'));
\[u \left (x , t\right ) = 30 x +20 \left (\sum _{n=1}^{\infty }\frac {\left (5 \left (-1\right )^{n}-2\right ) {\mathrm e}^{-\pi ^{2} n^{2} t} \sin \left (\pi n x \right )}{\pi n}\right )+20\]
Hand solution
The general solution to \begin {equation} u_{t}=ku_{xx}\qquad t>0,0<x<L\tag {1} \end {equation} BC are\begin {align*} u\left ( 0,t\right ) & =A\\ u\left ( L,t\right ) & =B \end {align*}
Initial conditions\[ u\left ( x,0\right ) =f\left ( x\right ) \] Is given in problem 3.1.5.1 on page 688 as\[ u\left ( x,t\right ) =A+\left ( \frac {B-A}{L}\right ) x+\frac {2}{L}\sum _{n=1}^{\infty }\left ( \int _{0}^{L}\left ( f\left ( x\right ) -\left ( \frac {B-A}{L}x+A\right ) \right ) \sin \left ( \sqrt {\lambda _{n}}x\right ) dx\right ) e^{-k\lambda _{n}t}\sin \left ( \sqrt {\lambda _{n}}x\right ) \] Where \(\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2},n=1,2,3,\cdots \). Substituting \(A=20,B=50,L=1,k=1,f\left ( x\right ) =0\) gives the solution as\[ u\left ( x,t\right ) =20+30x+2\sum _{n=1}^{\infty }\left ( \int _{0}^{1}-\left ( 30x+20\right ) \sin \left ( n\pi x\right ) dx\right ) e^{-kn^{2}\pi ^{2}t}\sin \left ( n\pi x\right ) \] But \(\int _{0}^{1}-\left ( 30x+20\right ) \sin \left ( n\pi x\right ) dx=\frac {50\left ( -1\right ) ^{n}-20}{n\pi }\) and the above becomes\begin {align*} u\left ( x,t\right ) & =20+30x+2\sum _{n=1}^{\infty }\left ( \frac {50\left ( -1\right ) ^{n}-20}{n\pi }\right ) e^{-kn^{2}\pi ^{2}t}\sin \left ( n\pi x\right ) \\ & =20+30x+\sum _{n=1}^{\infty }\left ( \frac {100\left ( -1\right ) ^{n}-40}{n\pi }\right ) e^{-kn^{2}\pi ^{2}t}\sin \left ( n\pi x\right ) \end {align*}
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Added December 20, 2018.
Example 8.4.1 from Partial differential equations and boundary value problems with Maple by George A. Articolo, 2nd ed.
Solve the heat equation for \(u(x,t)\) \[ u_t= k u_{xx} \] For \(0<x<1\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t)&=10 \\ u(1,t) &= 20\\ \end {align*}
Initial condition is \(u(x,0)=60 x - 50 x^2+10\) and \(k=\frac {1}{20}\)
Mathematica ✓
ClearAll["Global`*"]; k = 1/20; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}]; bc = {u[0, t] == 10, u[1, t] == 20}; ic = u[x, 0] == 60*x - 50*x^2 + 10; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], x, t], 60*10]]; sol = sol /. K[1] -> n;
\[\left \{\left \{u(x,t)\to \underset {n=1}{\overset {\infty }{\sum }}-\frac {200 \left (-1+(-1)^n\right ) e^{-\frac {1}{20} n^2 \pi ^2 t} \sin (n \pi x)}{n^3 \pi ^3}+10 (x+1)\right \}\right \}\]
Maple ✓
restart; k := 1/20; pde := diff(u(x,t),t)= k*diff(u(x,t),x$2); bc := u(0, t) = 10, u(1, t) = 20; ic := u(x, 0) = 60*x - 50*x^2 + 10; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde, ic, bc], u(x, t))),output='realtime'));
\[u \left (x , t\right ) = 10 x -200 \left (\sum _{n=1}^{\infty }\frac {\left (\left (-1\right )^{n}-1\right ) {\mathrm e}^{-\frac {\pi ^{2} n^{2} t}{20}} \sin \left (\pi n x \right )}{\pi ^{3} n^{3}}\right )+10\]
Hand solution
The general solution to \begin {equation} u_{t}=ku_{xx}\qquad t>0,0<x<L\tag {1} \end {equation} BC are\begin {align*} u\left ( 0,t\right ) & =A\\ u\left ( L,t\right ) & =B \end {align*}
with Initial conditions\[ u\left ( x,0\right ) =f\left ( x\right ) \] Is given in problem 3.1.5.1 on page 688 as\[ u\left ( x,t\right ) =A+\left ( \frac {B-A}{L}\right ) x+\frac {2}{L}\sum _{n=1}^{\infty }\left ( \int _{0}^{L}\left ( f\left ( x\right ) -\left ( \frac {B-A}{L}x+A\right ) \right ) \sin \left ( \sqrt {\lambda _{n}}x\right ) dx\right ) e^{-k\lambda _{n}t}\sin \left ( \sqrt {\lambda _{n}}x\right ) \] Where \(\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2},n=1,2,3,\cdots \). Substituting \(A=10,B=20,L=1,k=\frac {1}{20},f\left ( x\right ) =60x-50x^{2}+10\) gives the solution as\begin {align*} u\left ( x,t\right ) & =10+10x+2\sum _{n=1}^{\infty }\left ( \int _{0}^{1}\left ( 60x-50x^{2}+10-\left ( 10x+10\right ) \right ) \sin \left ( n\pi xx\right ) dx\right ) e^{-\frac {1}{20}n^{2}\pi ^{2}t}\sin \left ( n\pi xx\right ) \\ & =10+10x+2\sum _{n=1}^{\infty }\left ( \int _{0}^{1}\left ( 50x-50x^{2}\right ) \sin \left ( n\pi xx\right ) dx\right ) e^{-\frac {1}{20}n^{2}\pi ^{2}t}\sin \left ( n\pi xx\right ) \end {align*}
But \(\int _{0}^{1}\left ( 50x-50x^{2}\right ) \sin \left ( n\pi xx\right ) dx=\frac {100-100\left ( -1\right ) ^{n}}{n^{3}\pi ^{3}}\) and the above becomes\begin {align*} u\left ( x,t\right ) & =10+10x+2\sum _{n=1}^{\infty }100\left ( \frac {1-\left ( -1\right ) ^{n}}{n^{3}\pi ^{3}}\right ) e^{-\frac {1}{20}n^{2}\pi ^{2}t}\sin \left ( n\pi xx\right ) \\ & =10+10x-\sum _{n=1}^{\infty }200\left ( \frac {\left ( -1\right ) ^{n}-1}{n^{3}\pi ^{3}}\right ) e^{-\frac {1}{20}n^{2}\pi ^{2}t}\sin \left ( n\pi xx\right ) \end {align*}
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Added July 6,2019
Solve the heat equation for \(u(x,t)\) \[ u_t= k u_{xx} + Q(x) \] For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &=A \\ u(1,t) &=B\\ \end {align*}
Initial condition is \(u(x,0)=f(x)\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}] + Q[x]; bc = {u[0, t] == A, u[L, t] == B}; ic = u[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t},Assumptions->{k>0,L>0}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} \left (\frac {\left (1-e^{-\frac {k \pi ^2 t K[1]^2}{L^2}}\right ) \left (\int _0^L \frac {\sqrt {2} Q(x) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}} \, dx\right ) L^2}{k \pi ^2 K[1]^2}+e^{-\frac {k \pi ^2 t K[1]^2}{L^2}} \int _0^L \frac {\sqrt {2} (-B x+A (x-L)+L f(x)) \sin \left (\frac {\pi x K[1]}{L}\right )}{L^{3/2}} \, dx\right ) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}}+\frac {x (B-A)}{L}+A\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,t),t)= k*diff(u(x,t),x$2)+Q(x); bc := u(0, t) = A, u(L, t) = B; ic := u(x, 0) = f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde, ic, bc], u(x, t)) assuming L>0,k>0),output='realtime'));
\[u \left (x , t\right ) = \frac {-2 L k \left (\sum _{n=1}^{\infty }\frac {\left (\int _{0}^{L}-\left (L \left (\int _{0}^{\mathit {\_a}}\int _{0}^{\mathit {\_z1}}Q \left (\mathit {\_z1} \right )d \mathit {\_z1} d \mathit {\_z1} \right )-\mathit {\_a} \left (\int _{0}^{L}\int _{0}^{\mathit {\_z1}}Q \left (\mathit {\_z1} \right )d \mathit {\_z1} d \mathit {\_z1} \right )+\left (-A L +L f \left (\mathit {\_a} \right )+\left (A -B \right ) \mathit {\_a} \right ) k \right ) \sin \left (\frac {\pi \mathit {\_a} n}{L}\right )d \mathit {\_a} \right ) {\mathrm e}^{-\frac {\pi ^{2} k n^{2} t}{L^{2}}} \sin \left (\frac {\pi n x}{L}\right )}{L^{2} k}\right )-L \left (\int _{0}^{x}\int _{0}^{\mathit {\_a}}Q \left (\mathit {\_a} \right )d \mathit {\_a} d \mathit {\_a} \right )+x \left (\int _{0}^{L}\int _{0}^{\mathit {\_a}}Q \left (\mathit {\_a} \right )d \mathit {\_a} d \mathit {\_a} \right )+\left (A L -\left (A -B \right ) x \right ) k}{L k}\]
Hand solution
Solving\begin {equation} u_{t}=ku_{xx}+Q\left ( x\right ) \tag {1} \end {equation} With initial conditions \(u\left ( x,0\right ) =f\left ( x\right ) \) and boundary conditions \(u\left ( 0,t\right ) =A,u\left ( L,t\right ) =B\) with \(0<x<L,t>0\)
Since boundary conditions are nonhomogeneous, the first step is to reduce the problem to one with homogeneous B.C. to be able to use separation of variables. This is done by using steady state solution. Let the total solution be\begin {equation} u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x\right ) \tag {2} \end {equation} Where \(v\left ( x,t\right ) \) is the transient solution which satisfies the homogeneous B.C. and \(r\left ( x\right ) \) is the steady state solution which do not depend on time and just needs to satisfy the nonhomogeneous BC. Since \(r\left ( x\right ) \) is the steady state solution, then the PDE becomes an ODE\begin {align*} 0 & =kr^{\prime \prime }\left ( x\right ) \\ r\left ( 0\right ) & =A\\ r\left ( L\right ) & =B \end {align*}
This has the solution \(r\left ( x\right ) =c_{1}x+c_{2}\). Using BC at \(x=0\) leads to \(A=c_{2}\). Therefore the solution is \(r\left ( x\right ) =c_{1}x+A\). Using BC at \(x=L\) gives \(B=c_{1}L+A\) or \(c_{1}=\frac {B-A}{L}\). Hence\begin {equation} r\left ( x\right ) =\left ( \frac {B-A}{L}\right ) x+A\tag {3} \end {equation} Substituting (2) back in the original PDE \(u_{t}=ku_{xx}+Q\left ( x\right ) \) gives\begin {align} v_{t} & =kv_{xx}+Q\left ( x\right ) \tag {4}\\ v\left ( 0,t\right ) & =0\nonumber \\ v\left ( L,t\right ) & =0\nonumber \end {align}
The initial conditions are\begin {align*} v\left ( x,0\right ) & =F\left ( x\right ) \\ & =u\left ( x,0\right ) -r\left ( x\right ) \\ & =f\left ( x\right ) -\left ( \left ( \frac {B-A}{L}\right ) x+A\right ) \end {align*}
The general solution to (4) was solved in 3.1.1.11 on page 443 and the solution is\begin {align*} v\left ( x,t\right ) & =\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\Phi _{n}\left ( x\right ) \left ( \frac {2}{L}\int _{0}^{L}F\left ( s\right ) \Phi _{n}\left ( s\right ) ds\right ) +\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\Phi _{n}\left ( x\right ) \left ( \int _{0}^{t}\frac {2}{L}e^{k\lambda _{n}\tau }\left ( \int _{0}^{L}Q\left ( s\right ) \Phi _{n}\left ( s\right ) ds\right ) d\tau \right ) \\ \Phi _{n}\left ( x\right ) & =\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end {align*}
But \(F\left ( x\right ) =f\left ( x\right ) -\left ( \left ( \frac {B-A}{L}\right ) x+A\right ) \) in this case, hence the solution becomes\begin {align*} v\left ( x,t\right ) & =\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \frac {2}{L}\int _{0}^{L}\left ( f\left ( s\right ) -\left ( \left ( \frac {B-A}{L}\right ) s+A\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}\frac {2}{L}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}Q\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
Since \(u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x\right ) \), then the final solution is\begin {align*} u\left ( x,t\right ) & =\left ( \frac {B-A}{L}\right ) x+A+\\ & \sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \frac {2}{L}\int _{0}^{L}\left ( f\left ( s\right ) -\left ( \left ( \frac {B-A}{L}\right ) s+A\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}\frac {2}{L}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}Q\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
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Added July 6,2019
Solve the heat equation for \(u(x,t)\) \[ u_t= k u_{xx} + Q(x) \] For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &=A \\ u(1,t) &=B\\ \end {align*}
Initial condition is \(u(x,0)=f(x)\), Using the following values \begin {align*} A &=20 \\ B &=50\\ f(x)&=60-2 x\\ L&=30\\ k&=\frac {1}{10}\\ Q(x) &=\frac {x}{10} \end {align*}
Mathematica ✓
ClearAll["Global`*"]; A=20; B=50; f=60-2*x; L=30; k=1/10; Q=x/10; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}] + Q; bc = {u[0, t] == A, u[L, t] == B}; ic = u[x, 0] == f; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], x, t], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}-\frac {20 e^{-\frac {\pi ^2 t K[1]^2}{9000}} \left (-\left (\left (4+5 (-1)^{K[1]}\right ) \pi ^2 K[1]^2\right )-2700 (-1)^{K[1]}+2700 (-1)^{K[1]} e^{\frac {\pi ^2 t K[1]^2}{9000}}\right ) \sin \left (\frac {1}{30} \pi x K[1]\right )}{\pi ^3 K[1]^3}+x+20\right \}\right \}\]
Maple ✓
restart; A:=20; B:=50; f:=60-2*x; L:=30; k:=1/10; Q:=x/10; pde := diff(u(x,t),t)= k*diff(u(x,t),x$2)+Q; bc := u(0, t) = A, u(L, t) = B; ic := u(x, 0) = f; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde, ic, bc], u(x, t))),output='realtime'));
\[u \left (x , t\right ) = -\frac {x^{3}}{6}+151 x +20 \left (\sum _{n=1}^{\infty }\frac {\left (5 \pi ^{2} n^{2} \left (-1\right )^{n}+4 \pi ^{2} n^{2}+2700 \left (-1\right )^{n}\right ) {\mathrm e}^{-\frac {\pi ^{2} n^{2} t}{9000}} \sin \left (\frac {\pi n x}{30}\right )}{\pi ^{3} n^{3}}\right )+20\]
Hand solution
Solving\begin {align} u_{t} & =kv_{xx}+Q\left ( x\right ) \tag {1}\\ u\left ( 0,t\right ) & =A\nonumber \\ u\left ( L,t\right ) & =B\nonumber \\ u\left ( x,0\right ) & =f\left ( x\right ) \nonumber \end {align}
Where \(A=20,B=50,f\left ( x\right ) =60-2x,Q\left ( x\right ) =\frac {x}{10},k=\frac {1}{10},L=30\).
The general solution to above PDE was solved in 3.1.5.4 on page 698 and the solution is\begin {align*} u\left ( x,t\right ) & =\left ( \frac {B-A}{L}\right ) x+A+\\ & \sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \frac {2}{L}\int _{0}^{L}\left ( f\left ( s\right ) -\left ( \left ( \frac {B-A}{L}\right ) s+A\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}\frac {2}{L}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}Q\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
Replacing the specific values, the solution becomes\begin {align*} u\left ( x,t\right ) & =x+20+\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \frac {2}{30}\int _{0}^{30}\left ( 60-2s-\left ( s+20\right ) \right ) \sin \left ( \frac {n\pi }{30}s\right ) ds\right ) \\ & +\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}\frac {2}{30}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{30}\frac {s}{10}\sin \left ( \frac {n\pi }{30}s\right ) ds\right ) d\tau \right ) \end {align*}
Or\begin {align*} u\left ( x,t\right ) & =x+20+\frac {1}{15}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{30}\left ( 40-3s\right ) \sin \left ( \frac {n\pi }{30}s\right ) ds\right ) \\ & +\frac {1}{150}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{30}s\sin \left ( \frac {n\pi }{30}s\right ) ds\right ) d\tau \right ) \end {align*}
But \(\int _{0}^{30}\left ( 40-3s\right ) \sin \left ( \frac {n\pi }{30}s\right ) ds=\frac {1500\left ( -1\right ) ^{n}+1200}{n\pi }\) and \(\int _{0}^{30}s\sin \left ( \frac {n\pi }{30}s\right ) ds=\frac {-900\left ( -1\right ) ^{n}}{n\pi }\), hence the above becomes\begin {align*} u\left ( x,t\right ) & =x+20+\frac {1}{15}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \frac {1500\left ( -1\right ) ^{n}+1200}{n\pi }\right ) \\ & +\frac {1}{150}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \frac {-900\left ( -1\right ) ^{n}}{n\pi }\right ) d\tau \right ) \end {align*}
Or\begin {align*} u\left ( x,t\right ) & =x+20+\frac {300}{15}\sum _{n=1}^{\infty }\left ( \frac {5\left ( -1\right ) ^{n}+4}{n\pi }\right ) e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \\ & -\frac {900}{150}\sum _{n=1}^{\infty }\frac {\left ( -1\right ) ^{n}}{n\pi }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }d\tau \right ) \end {align*}
But \(\int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }d\tau =\left [ \frac {e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }}{k\left ( \frac {n\pi }{L}\right ) ^{2}}\right ] _{0}^{t}=\frac {e^{k\left ( \frac {n\pi }{L}\right ) ^{2}t}-1}{k\left ( \frac {n\pi }{L}\right ) ^{2}}\), therefore the above becomes\begin {align*} u\left ( x,t\right ) & =x+20+\frac {300}{15}\sum _{n=1}^{\infty }\left ( \frac {5\left ( -1\right ) ^{n}+4}{n\pi }\right ) e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \\ & -\frac {900}{150}\sum _{n=1}^{\infty }\frac {\left ( -1\right ) ^{n}}{n\pi }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \frac {e^{k\left ( \frac {n\pi }{L}\right ) ^{2}t}-1}{k\left ( \frac {n\pi }{L}\right ) ^{2}}\right ) \end {align*}
Or\begin {align*} u\left ( x,t\right ) =x+20+20 & \sum _{n=1}^{\infty }\left ( \frac {5\left ( -1\right ) ^{n}+4}{n\pi }\right ) e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \\ & -\frac {900}{150}\sum _{n=1}^{\infty }\frac {\left ( -1\right ) ^{n}}{n\pi k\left ( \frac {n\pi }{L}\right ) ^{2}}\sin \left ( \frac {n\pi }{L}x\right ) \left ( 1-e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\right ) \end {align*}
Finally replacing the remaining variables in the above for \(L,k\,\) gives\begin {align*} u\left ( x,t\right ) =x+20+20 & \sum _{n=1}^{\infty }\left ( \frac {5\left ( -1\right ) ^{n}+4}{n\pi }\right ) e^{-\frac {1}{10}\left ( \frac {n\pi }{30}\right ) ^{2}t}\sin \left ( \frac {n\pi }{30}x\right ) \\ & -\frac {900}{150}\sum _{n=1}^{\infty }\frac {9000\left ( -1\right ) ^{n}}{n^{3}\pi ^{3}}\sin \left ( \frac {n\pi }{30}x\right ) \left ( 1-e^{-\frac {1}{10}\left ( \frac {n\pi }{30}\right ) ^{2}t}\right ) \end {align*}
Or\begin {align*} u\left ( x,t\right ) & =x+20+20\sum _{n=1}^{\infty }\left ( \frac {5\left ( -1\right ) ^{n}+4}{n\pi }\right ) e^{-\frac {1}{10}\left ( \frac {n\pi }{30}\right ) ^{2}t}\sin \left ( \frac {n\pi }{30}x\right ) \\ & -54000\sum _{n=1}^{\infty }\frac {\left ( -1\right ) ^{n}}{n^{3}\pi ^{3}}\sin \left ( \frac {n\pi }{30}x\right ) \left ( 1-e^{-\frac {1}{10}\left ( \frac {n\pi }{30}\right ) ^{2}t}\right ) \end {align*}
Or\[ u\left ( x,t\right ) =\left ( x+20\right ) +20\sum _{n=1}^{\infty }\left ( \frac {5\left ( -1\right ) ^{n}+4}{n\pi }\right ) e^{-\frac {\pi ^{2}n^{2}}{9000}t}\sin \left ( \frac {n\pi }{30}x\right ) -54000\sum _{n=1}^{\infty }\frac {\left ( -1\right ) ^{n}}{n^{3}\pi ^{3}}\sin \left ( \frac {n\pi }{30}x\right ) \left ( 1-e^{-\frac {\pi ^{2}n^{2}}{9000}t}\right ) \] Animation is below
Source code used for the above
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Added July 6,2019
Solve the heat equation for \(u(x,t)\) \[ u_t= k u_{xx} + Q(x,t) \] For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &=A \\ u(1,t) &=B\\ \end {align*}
Initial condition is \(u(x,0)=f(x)\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}] + Q[x,t]; bc = {u[0, t] == A, u[L, t] == B}; ic = u[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t},Assumptions->{k>0,L>0}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} \left (\int _0^t e^{-\frac {k \pi ^2 K[1]^2 (t-K[2])}{L^2}} \text {Integrate}\left [\frac {\sqrt {2} Q(x,K[2]) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}},\{x,0,L\},\text {Assumptions}\to k>0\land L>0\right ] \, dK[2]+e^{-\frac {k \pi ^2 t K[1]^2}{L^2}} \int _0^L \frac {\sqrt {2} (-A L+f(x) L+A x-B x) \sin \left (\frac {\pi x K[1]}{L}\right )}{L^{3/2}} \, dx\right ) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}}+\frac {x (B-A)}{L}+A\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,t),t)= k*diff(u(x,t),x$2)+Q(x,t); bc := u(0, t) = A, u(L, t) = B; ic := u(x, 0) = f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde, ic, bc], u(x, t)) assuming L>0,k>0),output='realtime'));
\[u \left (x , t\right ) = \frac {A L +L \left (\int _{0}^{t}\left (\sum _{n=1}^{\infty }\frac {2 \left (\int _{0}^{L}Q \left (x , \tau \right ) \sin \left (\frac {\pi n x}{L}\right )d x \right ) {\mathrm e}^{-\frac {\pi ^{2} \left (t -\tau \right ) k n^{2}}{L^{2}}} \sin \left (\frac {\pi n x}{L}\right )}{L}\right )d \tau \right )-2 L \left (\sum _{n=1}^{\infty }\frac {\left (\int _{0}^{L}\left (A L -L f \left (\tau \right )+\left (-A +B \right ) \tau \right ) \sin \left (\frac {\pi n \tau }{L}\right )d \tau \right ) {\mathrm e}^{-\frac {\pi ^{2} k n^{2} t}{L^{2}}} \sin \left (\frac {\pi n x}{L}\right )}{L^{2}}\right )-\left (A -B \right ) x}{L}\]
Hand solution
Solving\begin {equation} u_{t}=ku_{xx}+Q\left ( x,t\right ) \tag {1} \end {equation} With initial conditions \(u\left ( x,0\right ) =f\left ( x\right ) \) and boundary conditions \(u\left ( 0,t\right ) =A,u\left ( L,t\right ) =B\) with \(0<x<L,t>0\)
Since boundary conditions are nonhomogeneous, the first step is to reduce the problem to one with homogeneous B.C. to be able to use separation of variables. This is done by using steady state solution. Let the total solution be\begin {equation} u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x\right ) \tag {2} \end {equation} Where \(v\left ( x,t\right ) \) is the transient solution which satisfies the homogeneous B.C. and \(r\left ( x\right ) \) is the steady state solution which do not depend on time and just needs to satisfy the nonhomogeneous BC. Since \(r\left ( x\right ) \) is the steady state solution, then the PDE becomes an ODE\begin {align*} 0 & =kr^{\prime \prime }\left ( x\right ) \\ r\left ( 0\right ) & =A\\ r\left ( L\right ) & =B \end {align*}
This has the solution \(r\left ( x\right ) =c_{1}x+c_{2}\). Using BC at \(x=0\) leads to \(A=c_{2}\). Therefore the solution is \(r\left ( x\right ) =c_{1}x+A\). Using BC at \(x=L\) gives \(B=c_{1}L+A\) or \(c_{1}=\frac {B-A}{L}\). Hence\begin {equation} r\left ( x\right ) =\left ( \frac {B-A}{L}\right ) x+A\tag {3} \end {equation} Substituting (2) back in the original PDE \(u_{t}=ku_{xx}+Q\left ( x,t\right ) \) gives\begin {align} v_{t} & =kv_{xx}+Q\left ( x,t\right ) \tag {4}\\ v\left ( 0,t\right ) & =0\nonumber \\ v\left ( L,t\right ) & =0\nonumber \end {align}
The initial conditions are\begin {align*} v\left ( x,0\right ) & =F\left ( x\right ) \\ & =u\left ( x,0\right ) -r\left ( x\right ) \\ & =f\left ( x\right ) -\left ( \left ( \frac {B-A}{L}\right ) x+A\right ) \end {align*}
The general solution to (4) was solved in 3.1.6.4 on page 796 and the solution is
\begin {align*} v\left ( x,t\right ) & =\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\Phi _{n}\left ( x\right ) \left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \Phi _{n}\left ( s\right ) ds\right ) +\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\Phi _{n}\left ( x\right ) \left ( \int _{0}^{t}\frac {2}{L}e^{k\lambda _{n}\tau }\left ( \int _{0}^{L}Q\left ( s,\tau \right ) \Phi _{n}\left ( s\right ) ds\right ) d\tau \right ) \\ \Phi _{n}\left ( x\right ) & =\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end {align*}
But \(F\left ( x\right ) =f\left ( x\right ) -\left ( \left ( \frac {B-A}{L}\right ) x+A\right ) \) in this case, hence the solution becomes\begin {align*} v\left ( x,t\right ) & =\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \frac {2}{L}\int _{0}^{L}\left ( f\left ( s\right ) -\left ( \left ( \frac {B-A}{L}\right ) s+A\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}\frac {2}{L}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}Q\left ( s,\tau \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
Since \(u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x\right ) \), then the final solution is\begin {align*} u\left ( x,t\right ) & =\left ( \frac {B-A}{L}\right ) x+A+\\ & \sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \frac {2}{L}\int _{0}^{L}\left ( f\left ( s\right ) -\left ( \left ( \frac {B-A}{L}\right ) s+A\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}\frac {2}{L}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}Q\left ( s,\tau \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
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Added June 23, 2019
Solve the heat equation for \(u(x,t)\) \[ u_t= k u_{xx} \] For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t)&=A(t) \\ u(L,t) &= B(t)\\ \end {align*}
Initial condition is \(u(x,0)=f(x)\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}]; bc = {u[0, t] == A[t], u[L, t] == B[t]}; ic = u[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t},Assumptions->{k>0,L>0}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} \left (e^{-\frac {k \pi ^2 t K[1]^2}{L^2}} \int _0^L \frac {\sqrt {2} (-L A(0)+x A(0)-x B(0)+L f(x)) \sin \left (\frac {\pi x K[1]}{L}\right )}{L^{3/2}} \, dx+\int _0^t \frac {\sqrt {2} e^{-\frac {k \pi ^2 K[1]^2 (t-K[2])}{L^2}} \sqrt {L} \left ((-1)^{K[1]} B'(K[2])-A'(K[2])\right )}{\pi K[1]} \, dK[2]\right ) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}}+\frac {x (B(t)-A(t))}{L}+A(t)\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,t),t)= k*diff(u(x,t),x$2); bc := u(0, t) = A(t), u(L, t) = B(t); ic := u(x, 0) = f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde, ic, bc], u(x, t)) assuming L>0,k>0),output='realtime'));
\[u \left (x , t\right ) = \frac {L \left (\int _{0}^{t}\left (\sum _{n=1}^{\infty }-\frac {2 \left (-\left (-1\right )^{n} \left (\frac {d}{d \tau }B \left (\tau \right )\right )+\frac {d}{d \tau }A \left (\tau \right )\right ) {\mathrm e}^{-\frac {\pi ^{2} \left (t -\tau \right ) k n^{2}}{L^{2}}} \sin \left (\frac {\pi n x}{L}\right )}{\pi n}\right )d \tau \right )+L \left (\sum _{n=1}^{\infty }\frac {2 \left (\int _{0}^{L}\frac {\left (L f \left (x \right )-B \left (0\right ) x +\left (-L +x \right ) A \left (0\right )\right ) \sin \left (\frac {\pi n x}{L}\right )}{L}d x \right ) {\mathrm e}^{-\frac {\pi ^{2} k n^{2} t}{L^{2}}} \sin \left (\frac {\pi n x}{L}\right )}{L}\right )+x B \left (t \right )+\left (L -x \right ) A \left (t \right )}{L}\]
Hand solution
\begin {align} u_{t} & =ku_{xx}\tag {1}\\ u\left ( 0,t\right ) & =A\left ( t\right ) \nonumber \\ u\left ( L,t\right ) & =B\left ( t\right ) \nonumber \\ u\left ( x,0\right ) & =f\left ( x\right ) \nonumber \end {align}
Since boundary conditions are nonhomogeneous, the first step is to reduce the problem to one with homogeneous B.C. to be able to use separation of variables. This is done by using a reference solution \(r\left ( x,t\right ) \) which only needs to satisfy the B.C. Let the total solution be\begin {equation} u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x,t\right ) \tag {2} \end {equation} Where \(v\left ( x,t\right ) \) is the transient solution which satisfies the homogeneous B.C. We see that\begin {align} r\left ( x,t\right ) & =A\left ( t\right ) +\frac {B\left ( t\right ) -A\left ( t\right ) }{L}x\nonumber \\ & =A\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B\left ( t\right ) }{L}x\tag {3} \end {align}
Satisfies the nonhomogeneous. Substituting (2) back into the original PDE (1) gives\begin {align*} \frac {\partial }{\partial t}\left ( v\left ( x,t\right ) +r\left ( x,t\right ) \right ) & =k\frac {\partial ^{2}}{\partial x^{2}}\left ( v\left ( x,t\right ) +r\left ( x,t\right ) \right ) \\ v_{t}\left ( x,t\right ) +r_{t}\left ( x,t\right ) & =kv_{xx}\left ( x,t\right ) +kr_{xx}\left ( x,t\right ) \end {align*}
But \(r_{xx}\left ( x,t\right ) =0\) and \(r_{t}=A^{\prime }\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B^{\prime }\left ( t\right ) }{L}x\) and the above PDE becomes\[ v_{t}\left ( x,t\right ) =kv_{xx}\left ( x,t\right ) -r_{t}\left ( x,t\right ) \] Let \begin {align*} Q\left ( x,t\right ) & =-r_{t}\left ( x,t\right ) \\ & =-\left ( A^{\prime }\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B^{\prime }\left ( t\right ) }{L}x\right ) \end {align*}
Therefore the problem has been transformed to\begin {align*} v_{t} & =kv_{xx}+Q\left ( x,t\right ) \\ v\left ( 0,t\right ) & =0\\ v\left ( L,t\right ) & =0\\ v\left ( 0,x\right ) & =F\left ( x\right ) \\ & =u\left ( x,0\right ) -r\left ( x,0\right ) \\ & =f\left ( x\right ) -\left ( A\left ( 0\right ) \left ( \frac {L-x}{L}\right ) +\frac {B\left ( 0\right ) }{L}x\right ) \end {align*}
The above problem was solved in 3.1.6.4 on page 796. The solution is\begin {equation} v\left ( x,t\right ) =\sum _{n=1}^{\infty }\left [ \left ( \frac {2}{L}\int _{0}^{L}F\left ( s\right ) \Phi _{n}\left ( s\right ) ds\right ) e^{-k\lambda _{n}t}+e^{-k\lambda _{n}t}\int _{0}^{t}\frac {2}{L}e^{k\lambda _{n}\tau }\left ( \int _{0}^{L}Q\left ( s,\tau \right ) \Phi _{n}\left ( s\right ) dx\right ) d\tau \right ] \Phi _{n}\left ( x\right ) \tag {4} \end {equation} Where \begin {align*} \Phi _{n}\left ( x\right ) & =\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end {align*}
Hence (4) becomes
\begin {align*} v\left ( x,t\right ) & =\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}F\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\sin \left ( \frac {n\pi }{L}x\right ) \int _{0}^{t}e^{k\lambda _{n}\tau }\left ( \int _{0}^{L}Q\left ( s,\tau \right ) \sin \left ( \frac {n\pi }{L}s\right ) dx\right ) d\tau \end {align*}
Since \(u\left ( x,t\right ) =r\left ( x,t\right ) +v\left ( x,t\right ) \) then the final solution becomes
\begin {align*} u\left ( x,t\right ) & =\left ( A\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B\left ( t\right ) }{L}x\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}F\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\sin \left ( \frac {n\pi }{L}x\right ) \int _{0}^{t}e^{k\lambda _{n}\tau }\left ( \int _{0}^{L}Q\left ( s,\tau \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \end {align*}
Or
\begin {align*} u\left ( x,t\right ) & =\left ( A\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B\left ( t\right ) }{L}x\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}\left ( f\left ( s\right ) -\left ( A\left ( 0\right ) \left ( \frac {L-s}{L}\right ) +\frac {B\left ( 0\right ) }{L}s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}Q\left ( s,\tau \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
Or
\begin {align*} u\left ( x,t\right ) & =\left ( A\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B\left ( t\right ) }{L}x\right ) \\ & +\frac {2}{L^{2}}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}\left ( f\left ( s\right ) L-\left ( A\left ( 0\right ) \left ( L-s\right ) +B\left ( 0\right ) s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}Q\left ( s,\tau \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
Or, since here \(Q\left ( x,t\right ) =-\left ( A^{\prime }\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B^{\prime }\left ( t\right ) }{L}x\right ) \) the above becomes\begin {align*} u\left ( x,t\right ) & =\left ( A\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B\left ( t\right ) }{L}x\right ) \\ & +\frac {2}{L^{2}}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}\left ( f\left ( s\right ) L-\left ( A\left ( 0\right ) \left ( L-s\right ) +B\left ( 0\right ) s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}\left ( \frac {B^{\prime }\left ( \tau \right ) }{L}s-A^{\prime }\left ( \tau \right ) \left ( \frac {L-s}{L}\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
Or
\begin {align*} u\left ( x,t\right ) & =\left ( A\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B\left ( t\right ) }{L}x\right ) \\ & +\frac {2}{L^{2}}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}f\left ( s\right ) L\sin \left ( \frac {n\pi }{L}s\right ) -A\left ( 0\right ) \left ( L-s\right ) \sin \left ( \frac {n\pi }{L}s\right ) -B\left ( 0\right ) s\sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}\frac {B^{\prime }\left ( \tau \right ) }{L}s\sin \left ( \frac {n\pi }{L}s\right ) -A^{\prime }\left ( \tau \right ) \left ( \frac {L-s}{L}\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
Or
\begin {align} u\left ( x,t\right ) & =\left ( A\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B\left ( t\right ) }{L}x\right ) \tag {5}\\ & +\frac {2}{L^{2}}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}f\left ( s\right ) L\sin \left ( \frac {n\pi }{L}s\right ) -A\left ( 0\right ) \left ( L-s\right ) \sin \left ( \frac {n\pi }{L}s\right ) -B\left ( 0\right ) s\sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \nonumber \\ & +\frac {2}{L^{2}}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}B^{\prime }\left ( \tau \right ) s\sin \left ( \frac {n\pi }{L}s\right ) -A^{\prime }\left ( \tau \right ) \left ( L-s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \nonumber \end {align}
But \begin {multline*} \int _{0}^{L}f\left ( s\right ) L\sin \left ( \frac {n\pi }{L}s\right ) -A\left ( 0\right ) \left ( L-s\right ) \sin \left ( \frac {n\pi }{L}s\right ) -B\left ( 0\right ) s\sin \left ( \frac {n\pi }{L}s\right ) ds=\\ L\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds-A\left ( 0\right ) \int _{0}^{L}\left ( L-s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds-B\left ( 0\right ) \int _{0}^{L}s\sin \left ( \frac {n\pi }{L}s\right ) ds \end {multline*}
Since \(\int _{0}^{L}\left ( L-s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds=\frac {L^{2}}{n\pi }\) and \(\int _{0}^{L}s\sin \left ( \frac {n\pi }{L}s\right ) dx=-\frac {L^{2}\left ( -1\right ) ^{n}}{n\pi }\), then the above becomes\begin {multline} \int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) -A\left ( 0\right ) \left ( \frac {L-s}{L}\right ) \sin \left ( \frac {n\pi }{L}s\right ) -\frac {B\left ( 0\right ) }{L}s\sin \left ( \frac {n\pi }{L}s\right ) ds=\tag {6}\\ L\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds-A\left ( 0\right ) \frac {L^{2}}{n\pi }+B\left ( 0\right ) \frac {L^{2}\left ( -1\right ) ^{n}}{n\pi }\nonumber \end {multline}
And \begin {align} \int _{0}^{L}B^{\prime }\left ( \tau \right ) s\sin \left ( \frac {n\pi }{L}s\right ) -A^{\prime }\left ( \tau \right ) \left ( L-s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds & =-A^{\prime }\left ( \tau \right ) \int _{0}^{L}\left ( L-s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds+B^{\prime }\left ( \tau \right ) \int _{0}^{L}s\sin \left ( \frac {n\pi }{L}s\right ) ds\tag {7}\\ & =-A^{\prime }\left ( \tau \right ) \frac {L^{2}}{n\pi }+B^{\prime }\left ( \tau \right ) \frac {L^{2}\left ( -1\right ) ^{n}}{n\pi }\nonumber \end {align}
Substituting (6,7) back into (5) gives
\begin {align*} u\left ( x,t\right ) & =\left ( A\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B\left ( t\right ) }{L}x\right ) \\ & +\frac {2}{L^{2}}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( L\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds-A\left ( 0\right ) \frac {L^{2}}{n\pi }+B\left ( 0\right ) \frac {L^{2}\left ( -1\right ) ^{n}}{n\pi }\right ) \\ & -\frac {2}{L^{2}}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( -B^{\prime }\left ( \tau \right ) \frac {L^{2}\left ( -1\right ) ^{n}}{n\pi }+A^{\prime }\left ( \tau \right ) \frac {L^{2}}{n\pi }\right ) d\tau \right ) \end {align*}
Or
\begin {align*} u\left ( x,t\right ) & =\left ( A\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B\left ( t\right ) }{L}x\right ) \\ & +\frac {2}{L^{2}}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}f\left ( s\right ) L\sin \left ( \frac {n\pi }{L}s\right ) ds-A\left ( 0\right ) \frac {L^{2}}{n\pi }+B\left ( 0\right ) \frac {L^{2}\left ( -1\right ) ^{n}}{n\pi }\right ) \\ & -2\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \frac {1}{n\pi }\int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( -\left ( -1\right ) ^{n}B^{\prime }\left ( \tau \right ) +A^{\prime }\left ( \tau \right ) \right ) d\tau \right ) \end {align*}
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Added June 26, 2019
Solve the heat equation for \(u(x,t)\) \[ u_t= k u_{xx} \] For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &=A(t) \\ u(L,t) &= B(t)\\ \end {align*}
Initial condition is \(u(x,0)=f(x)\) using the following values \begin {align*} L&=2 \\ k&=\frac {1}{10}\\ f(x)&= x \\ A(t)&= \sin (t)\\ B(t)&= 2 \cos (t) \end {align*}
Mathematica ✓
ClearAll["Global`*"]; f=x; L=2; k=1/10; A=Sin[t]; B=2*Cos[t]; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}]; bc = {u[0, t] == A, u[L, t] == B}; ic = u[x, 0] == f; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {80 \left (e^{-\frac {1}{40} \pi ^2 t K[1]^2} \pi ^2 K[1]^2-80 (-1)^{K[1]} e^{-\frac {1}{40} \pi ^2 t K[1]^2}+\cos (t) \left (80 (-1)^{K[1]}-\pi ^2 K[1]^2\right )-2 \left ((-1)^{K[1]} \pi ^2 K[1]^2+20\right ) \sin (t)\right ) \sin \left (\frac {1}{2} \pi x K[1]\right )}{\pi K[1] \left (\pi ^4 K[1]^4+1600\right )}-\frac {1}{2} x \sin (t)+x \cos (t)+\sin (t)\right \}\right \}\]
Maple ✓
restart; f:=x; L:=2; k:=1/10; A:=sin(t); B:=2*cos(t); pde := diff(u(x,t),t)= k*diff(u(x,t),x$2); bc := u(0, t) = A, u(L, t) = B; ic := u(x, 0) = f; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde, ic, bc], u(x, t)) ),output='realtime'));
\[u \left (x , t\right ) = x \cos \left (t \right )-\frac {x \sin \left (t \right )}{2}-80 \left (\sum _{n=1}^{\infty }\frac {\left (\pi ^{2} n^{2} \cos \left (t \right )+\left (2 \pi ^{2} n^{2} \sin \left (t \right )-80 \cos \left (t \right )\right ) \left (-1\right )^{n}+\left (-\pi ^{2} n^{2}+80 \left (-1\right )^{n}\right ) {\mathrm e}^{-\frac {\pi ^{2} n^{2} t}{40}}+40 \sin \left (t \right )\right ) \sin \left (\frac {\pi n x}{2}\right )}{\pi \left (\pi ^{4} n^{4}+1600\right ) n}\right )+\sin \left (t \right )\]
Hand solution
The basic solution for this type of PDE was already given in problem 3.1.5.7 on page 716 as
\begin {align*} u\left ( x,t\right ) & =\left ( A\left ( t\right ) \left ( \frac {L-x}{L}\right ) +\frac {B\left ( t\right ) }{L}x\right ) \\ & +\frac {2}{L^{2}}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}f\left ( s\right ) L\sin \left ( \frac {n\pi }{L}s\right ) ds-A\left ( 0\right ) \frac {L^{2}}{n\pi }+B\left ( 0\right ) \frac {L^{2}\left ( -1\right ) ^{n}}{n\pi }\right ) \\ & -2\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \frac {1}{n\pi }\int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( -\left ( -1\right ) ^{n}B^{\prime }\left ( \tau \right ) +A^{\prime }\left ( \tau \right ) \right ) d\tau \right ) \end {align*}
In this problem we have \begin {align*} L & =2\\ k & =\frac {1}{10}\\ f\left ( x\right ) & =x\\ A\left ( t\right ) & =\sin \left ( t\right ) \\ B\left ( t\right ) & =2\cos \left ( t\right ) \end {align*}
This is animation of the above solution using these specific values for \(20\) seconds. (Animation will only show in the HTML version)
Source code used for the above
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Added January 18, 2019.
Solve the heat equation for \(u(x,t)\) \[ u_t = u_{xx} \] For \(0<x<\pi \) and \(t>0\). The boundary conditions are \begin {align*} u_x(0,t) &= 1\\ u_x(1,t) &= -1 \end {align*}
Initial condition is \(u(x,0)=\sin (x)\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == D[u[x, t], {x, 2}] + (x - (1*x^2)/Pi); ic = u[x, 0] == Sin[x]; bc = {Derivative[1, 0][u][0, t] == 1, Derivative[1, 0][u][Pi, t] == -1}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {e^{-t K[1]^2} \cos (x K[1]) \left (\sqrt {2 \pi } K[1]^4 \left (\begin {array}{cc} \{ & \begin {array}{cc} 0 & K[1]=1 \\ -\frac {\left (1+(-1)^{K[1]}\right ) \sqrt {\frac {2}{\pi }}}{K[1]^2 \left (K[1]^2-1\right )} & \text {True} \\\end {array} \\\end {array}\right )-2 \left (1+(-1)^{K[1]}\right ) \left (-1+e^{t K[1]^2}\right )\right )}{\pi K[1]^4}-\frac {2 t+x^2-2}{\pi }+\frac {1}{6} \pi (t-1)+x\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x, t), t) = diff(u(x, t), x$2): ic := u(x, 0) = sin(x): bc := eval(diff(u(x,t),x),x=0)=1, eval( diff(u(x,t),x),x=Pi)=-1: cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve({pde, ic, bc}, u(x, t))),output='realtime'));
\[u \left (x , t\right ) = -\frac {x^{2}}{\pi }-\frac {2 t}{\pi }+x +\left (\sum _{n=2}^{\infty }-\frac {2 \left (\left (-1\right )^{n}+1\right ) \cos \left (n x \right ) {\mathrm e}^{-n^{2} t}}{\left (n^{2}-1\right ) \pi n^{2}}\right )+\frac {2-\frac {\pi ^{2}}{6}}{\pi }\]
Hand solution
Since the boundary conditions are not homogeneous, we can’t use separation of variables. Let the solution be
\[ u=v\left ( x,t\right ) +r\left ( x\right ) \] Where \(v\left ( x,t\right ) \) is the solution to \(v_{t}=v_{xx}\) and homogenous B.C. \(v_{x}\left ( 0,t\right ) =0,v_{x}\left ( \pi ,t\right ) =0\) and \(r\left ( x\right ) \) is any reference solution which only needs to satisfy the nonhomogeneous boundary conditions: \(r^{\prime }\left ( 0\right ) =1,r^{\prime }\left ( \pi \right ) =-1\). By guessing, let \(r\left ( x\right ) =Ax+Bx^{2}\). Let see if this satisfies the boundary conditions. \(r^{\prime }=A+2Bx\). At \(x=0\) this implies \(1=A\). Hence \(r=x+Bx^{2}\). Now \(r^{\prime }=1+2Bx\). At \(x=\pi \) this gives \(-1=1+2B\pi \) or \(B=-\frac {1}{\pi }\). Therefore \[ r\left ( x\right ) =x-\frac {1}{\pi }x^{2}\] Substituting \(u=v\left ( x,t\right ) +r\left ( x\right ) \) into the PDE \(u_{t}=u_{xx}\) and noting that \(r^{\prime \prime }\left ( x\right ) =-\frac {2}{\pi }\) gives\begin {equation} v_{t}=v_{xx}-\frac {2}{\pi }\tag {1} \end {equation} PDE (1) is now solved using eigenfunction expansion. We need to find eigenfunctions and eigenvalues of \(v_{t}=v_{xx}\) with \(v_{x}\left ( 0,t\right ) =0,v_{x}\left ( \pi ,t\right ) =0\). This is known PDE and have eigenfunctions and eigenvalues as follows. For zero eigenvalue, the eigenfunction is an arbitrary constant. Say \(\beta \). let \(\beta =1\) since scale is not important.\[ \Phi _{0}\left ( x\right ) =1 \] And for \(n=1,2,3,\cdots \) \begin {align*} \Phi _{n}\left ( x\right ) & =\cos \left ( \sqrt {\lambda _{n}}x\right ) \\ & =\cos \left ( nx\right ) \end {align*}
with eigenvalues \(\lambda _{n}=n^{2}\) for \(n=1,2,3,\cdots \). Now we can eigenfunction expansion and assume the solution to (1) is \begin {equation} v\left ( x,t\right ) =\sum _{n=0}^{\infty }A_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \tag {2} \end {equation} Plugging this into the PDE (1) gives\[ \sum _{n=0}^{\infty }A_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) =\sum _{n=0}^{\infty }A_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) -\frac {2}{\pi }\] But \(\Phi _{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\Phi _{n}\left ( x\right ) \) and the above simplifies to\[ \sum _{n=0}^{\infty }A_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) =-\sum _{n=0}^{\infty }A_{n}\left ( t\right ) \lambda _{n}\Phi _{n}\left ( x\right ) -\frac {2}{\pi }\] Since eigenfunctions are complete, we can expand \(\frac {2}{\pi }\) using them and the above becomes\begin {align} \sum _{n=0}^{\infty }A_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) & =-\sum _{n=0}^{\infty }A_{n}\left ( t\right ) \lambda _{n}\Phi _{n}\left ( x\right ) -\sum _{n=0}^{\infty }C_{n}\Phi _{n}\left ( x\right ) \nonumber \\ A_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) +A_{n}\left ( t\right ) \lambda _{n}\Phi _{n}\left ( x\right ) & =-C_{n}\Phi _{n}\left ( x\right ) \nonumber \\ A_{n}^{\prime }\left ( t\right ) +A_{n}\left ( t\right ) \lambda _{n} & =-C_{n}\tag {3} \end {align}
To find \(C_{n}\)\[ \sum _{n=0}^{\infty }C_{n}\Phi _{n}\left ( x\right ) =\frac {2}{\pi }\] For \(n=0\)\[ C_{0}\Phi _{0}\left ( x\right ) =\frac {2}{\pi }\] But \(\Phi _{0}\left ( x\right ) =1\), hence \[ C_{0}=\frac {2}{\pi }\] All other \(C_{m}\,\ \) for \(m>0\) are zero. Hence (3) becomes, for \(n=0\) (since \(\lambda _{0}=0\))\begin {align*} A_{0}^{\prime }\left ( t\right ) & =-\frac {2}{\pi }\\ A_{0}\left ( t\right ) & =-\frac {2}{\pi }t+B_{0} \end {align*}
Where \(B_{0}\) is integration constant. For \(n>0\) (3) becomes\[ A_{n}^{\prime }\left ( t\right ) +A_{n}\left ( t\right ) n^{2}=0 \] This has the solution\[ A_{n}\left ( t\right ) =B_{n}e^{-n^{2}t}\] Where \(B_{n}\) is constant of integration. Hence from (2)\begin {align*} v\left ( x,t\right ) & =\sum _{n=0}^{\infty }A_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \\ & =A_{0}\left ( t\right ) +\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \\ & =-\frac {2}{\pi }t+B_{0}+\sum _{n=1}^{\infty }B_{n}e^{-n^{2}t}\cos \left ( nx\right ) \end {align*}
Since \(u=v\left ( x,t\right ) +r\left ( x\right ) \) then the solution becomes\begin {equation} u\left ( x,t\right ) =\left ( x-\frac {1}{\pi }x^{2}\right ) -\frac {2}{\pi }t+B_{0}+\sum _{n=1}^{\infty }B_{n}e^{-n^{2}t}\cos \left ( nx\right ) \tag {4} \end {equation} At \(t=0\)\begin {equation} \sin \left ( x\right ) =\left ( x-\frac {1}{\pi }x^{2}\right ) +B_{0}+\sum _{n=1}^{\infty }B_{n}\cos \left ( nx\right ) \tag {5} \end {equation} case \(n=0\)\[ \int _{0}^{\pi }\sin \left ( x\right ) \cos \left ( \sqrt {\lambda _{0}}x\right ) dx=\int _{0}^{\pi }\left ( x-\frac {1}{\pi }x^{2}\right ) \cos \left ( \sqrt {\lambda _{0}}x\right ) dx+\int _{0}^{\pi }B_{0}\cos \left ( \sqrt {\lambda _{0}}x\right ) dx \] But \(\lambda _{0}=0\) hence\begin {align*} \int _{0}^{\pi }\sin \left ( x\right ) dx & =\int _{0}^{\pi }\left ( x-\frac {1}{\pi }x^{2}\right ) dx+\int _{0}^{\pi }B_{0}dx\\ 2 & =\frac {\pi ^{2}}{6}+B_{0}\pi \\ B_{0} & =\frac {2}{\pi }-\frac {\pi }{6} \end {align*}
For \(n>0\), Multiplying both sides of (5) by \(\cos \left ( mx\right ) \) and integrating\[ \int _{0}^{\pi }\sin \left ( x\right ) \cos \left ( mx\right ) dx=\int _{0}^{\pi }\left ( x-\frac {1}{\pi }x^{2}\right ) \cos \left ( mx\right ) dx+\sum _{n=1}^{\infty }B_{n}\int _{0}^{\pi }\cos \left ( nx\right ) \cos \left ( mx\right ) dx \] For \(m=1\)\begin {align*} 0 & =0+B_{1}\frac {\pi }{2}\\ B_{1} & =0 \end {align*}
For \(m>1\)\begin {align*} -\frac {1+\left ( -1\right ) ^{m}}{m^{2}\left ( -1+m^{2}\right ) } & =\frac {\pi }{2}B_{m}\\ B_{m} & =\frac {-2}{\pi }\left ( \frac {1}{m^{2}}\frac {\left ( -1\right ) ^{m}+1}{m^{2}-1}\right ) \end {align*}
Hence solution (4) becomes\begin {align*} u\left ( x,t\right ) & =\left ( x-\frac {1}{\pi }x^{2}\right ) -\frac {2}{\pi }t-\frac {\pi }{6}+\frac {2}{\pi }+\sum _{n=1}^{\infty }B_{n}e^{-n^{2}t}\cos \left ( nx\right ) \\ u\left ( x,t\right ) & =\left ( x-\frac {1}{\pi }x^{2}\right ) -\frac {2}{\pi }t-\frac {\pi }{6}+\frac {2}{\pi }+\sum _{n=2}^{\infty }\frac {-2}{\pi }\left ( \frac {1}{n^{2}}\frac {\left ( -1\right ) ^{n}+1}{n^{2}-1}\right ) e^{-n^{2}t}\cos \left ( nx\right ) \end {align*}
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Added Nov 27, 2018
This is problem 8.2.1 part(a) from Richard Haberman applied partial differential equations 5th edition.
Solve the heat equation
\[ \frac { \partial u}{\partial t} = k \frac { \partial ^2 u}{\partial x^2} \]
For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &= A \\ \frac { \partial u}{\partial x}(L,t) &= B \end {align*}
Initial condition is \(u(x,0)=f(x)\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}]; bc = {u[0, t] == A, Derivative[1, 0][u][L, t] == B}; ic = u[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], x, t, Assumptions -> L > 0], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} e^{-\frac {k \pi ^2 t (1-2 K[1])^2}{4 L^2}} \left (\int _0^L -\frac {\sqrt {2} (A+B x-f(x)) \sin \left (\frac {\pi x (2 K[1]-1)}{2 L}\right )}{\sqrt {L}} \, dx\right ) \sin \left (\frac {\pi x (2 K[1]-1)}{2 L}\right )}{\sqrt {L}}+A+B x\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t)=k*diff(u(x,t),x$2); ic := u(x,0)=f(x); bc := u(0,t)=A, eval( diff(u(x,t),x),x=L)=B; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t)) assuming L>0),output='realtime'));
\[u \left (x , t\right ) = B x +A -2 \left (\sum _{n=0}^{\infty }\frac {\left (\int _{0}^{L}\left (B x +A -f \left (x \right )\right ) \sin \left (\frac {\left (2 n +1\right ) \pi x}{2 L}\right )d x \right ) {\mathrm e}^{-\frac {\pi ^{2} \left (2 n +1\right )^{2} k t}{4 L^{2}}} \sin \left (\frac {\left (2 n +1\right ) \pi x}{2 L}\right )}{L}\right )\]
Hand solution
Let \begin {equation} u\left ( x,t\right ) =v\left ( x,t\right ) +u_{E}\left ( x\right ) \tag {1} \end {equation} We can look for \(u_{E}\left ( x\right ) \) which is the steady state solution that satisfies the non-homogenous boundary conditions. In (1) \(v\left ( x,t\right ) \) satisfies the PDE itself but with homogenous boundary conditions. The first step is to find \(u_{E}\left ( x\right ) \). We use the equilibrium solution in this case. At equilibrium \(\frac {\partial u_{E}\left ( x,t\right ) }{\partial t}=0\) and hence the solution is given \(\frac {d^{2}u_{E}}{\partial x^{2}}=0\) or \[ u_{E}\left ( x\right ) =c_{1}x+c_{2}\] At \(x=0,u_{E}\left ( x\right ) =A\), Hence \[ c_{2}=A \] And solution becomes \(u_{E}\left ( x\right ) =c_{1}x+A\). at \(x=L,\frac {\partial u_{E}\left ( x\right ) }{\partial x}=c_{1}=B\), Therefore\[ u_{E}\left ( x\right ) =Bx+A \] Now we plug-in (1) into the original PDE, this gives\[ \frac {\partial v\left ( x,t\right ) }{\partial t}=k\left ( \frac {\partial ^{2}v\left ( x,t\right ) }{\partial x}+\frac {\partial ^{2}u_{E}\left ( x\right ) }{\partial x}\right ) \] But \(\frac {\partial ^{2}u_{E}\left ( x\right ) }{\partial x}=0\), hence we need to solve \[ \frac {\partial v\left ( x,t\right ) }{\partial t}=k\frac {\partial ^{2}v\left ( x,t\right ) }{\partial x}\] for \(v\left ( x,t\right ) =u\left ( x,t\right ) -u_{E}\left ( x\right ) \) with homogenous boundary conditions \(v\left ( 0,t\right ) =0,\frac {\partial v\left ( L,t\right ) }{\partial t}=0\) and initial conditions \begin {align*} v\left ( x,0\right ) & =u\left ( x,0\right ) -u_{E}\left ( x\right ) \\ & =f\left ( x\right ) -\left ( Bx+A\right ) \end {align*}
This PDE we already solved before and we know that it has the following solution\begin {align} v\left ( x,t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }b_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) e^{-k\lambda _{n}t}\nonumber \\ \lambda _{n} & =\left ( \frac {n\pi }{2L}\right ) ^{2}\qquad n=1,3,5,\cdots \tag {2} \end {align}
With \(b_{n}\) found from orthogonality using initial conditions \(v\left ( x,0\right ) =f\left ( x\right ) -\left ( Bx+A\right ) \)\begin {align*} v\left ( x,0\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }b_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ \int _{0}^{L}\left ( f\left ( x\right ) -\left ( Bx+A\right ) \right ) \sin \left ( \sqrt {\lambda _{m}}x\right ) dx & =\int _{0}^{L}\sum _{n=1,3,5,\cdots }^{\infty }b_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) \sin \left ( \sqrt {\lambda _{m}}x\right ) dx\\ \int _{0}^{L}\left ( f\left ( x\right ) -\left ( Bx+A\right ) \right ) \sin \left ( \sqrt {\lambda _{m}}x\right ) dx & =b_{m}\frac {L}{2} \end {align*}
Hence\begin {equation} b_{n}=\frac {2}{L}\int _{0}^{L}\left ( f\left ( x\right ) -\left ( Bx+A\right ) \right ) \sin \left ( \sqrt {\lambda _{n}}x\right ) dx\qquad n=1,3,5,\cdots \tag {3} \end {equation} Therefore, from (1) the solution is\begin {align*} u\left ( x,t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }b_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) e^{-k\lambda _{n}t}+\overset {u_{E}\left ( x\right ) }{\overbrace {Bx+A}}\\ & =Bx+A+\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac {2}{L}\int _{0}^{L}\left ( f\left ( x\right ) -\left ( Bx+A\right ) \right ) \sin \left ( \sqrt {\frac {n\pi }{L}}x\right ) dx\right ) \sin \left ( \sqrt {\frac {n\pi }{L}}x\right ) e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t} \end {align*}
Or
\[ u\left ( x,t\right ) =Bx+A+\sum _{n=0}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}\left ( f\left ( x\right ) -\left ( Bx+A\right ) \right ) \sin \left ( \sqrt {\frac {\left ( 2n+1\right ) \pi }{2L}}x\right ) dx\right ) \sin \left ( \sqrt {\frac {\left ( 2n+1\right ) \pi }{2L}}x\right ) e^{-k\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}t}\]
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This is problem 8.2.1 part(d) from Richard Haberman applied partial differential equations 5th edition.
Solve the heat equation \[ \frac { \partial u}{\partial t} = k \frac { \partial ^2 u}{\partial x^2} + k \] For \(0<x<1\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &= A \\ u(L,t) &= B \end {align*}
Initial condition is \(u(x,0)=f(x)\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}] + k; bc = {u[0, t] == A0, u[L0, t] == B0}; ic = u[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], x, t], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}-\frac {e^{-\frac {k \pi ^2 t K[1]^2}{\text {L0}^2}} \left (2 \left (-1+(-1)^{K[1]}\right ) \left (-1+e^{\frac {k \pi ^2 t K[1]^2}{\text {L0}^2}}\right ) \text {L0}^2-\sqrt {2} \sqrt {\frac {1}{\text {L0}}} \pi ^3 \left (\int _0^{\text {L0}} \sqrt {2} \left (\frac {1}{\text {L0}}\right )^{3/2} (-\text {A0} \text {L0}+f(x) \text {L0}+\text {A0} x-\text {B0} x) \sin \left (\frac {\pi x K[1]}{\text {L0}}\right ) \, dx\right ) K[1]^3\right ) \sin \left (\frac {\pi x K[1]}{\text {L0}}\right )}{\pi ^3 K[1]^3}+\frac {x (\text {B0}-\text {A0})}{\text {L0}}+\text {A0}\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t)=k*diff(u(x,t),x$2)+k; ic := u(x,0)=f(x); bc := u(0,t)=A, u(L,t)=B; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t))),output='realtime'));
\[u \left (x , t\right ) = -\frac {x^{2}}{2}+A +\sum _{n=1}^{\infty }\left (-\frac {\left (\int _{0}^{L}2 \left (\frac {L^{2} x}{2}-L f \left (x \right )+\left (-\frac {x^{2}}{2}+A \right ) L -\left (A -B \right ) x \right ) \sin \left (\frac {\pi n x}{L}\right )d x \right ) {\mathrm e}^{-\frac {\pi ^{2} k n^{2} t}{L^{2}}} \sin \left (\frac {\pi n x}{L}\right )}{L^{2}}\right )+\frac {\left (L^{2}-2 A +2 B \right ) x}{2 L}\]
Hand solution
Let \begin {equation} u\left ( x,t\right ) =v\left ( x,t\right ) +u_{E}\left ( x\right ) \tag {1} \end {equation} Where \(u_{E}\left ( x\right ) \) is the equilibrium solution which needs to satisfy only the nonhomogeneous B.C. And \(v\left ( x,t\right ) \) is transient solution to heat PDE with homogeneous B.C.
At equilibrium, \(u_{t}=ku_{xx}+Q\left ( x\right ) \) becomes\begin {align*} 0 & =ku_{E}^{\prime \prime }+Q\left ( x\right ) \\ & =ku_{E}^{\prime \prime }+k\\ & =k\left ( u_{E}^{\prime \prime }+1\right ) \end {align*}
Hence\[ u_{E}^{\prime \prime }=-1 \] The solution to this ODE is \[ u_{E}=c_{1}x+c_{2}-\frac {1}{2}x^{2}\] At \(x=0\), the above gives\[ A=c_{2}\] And at \(x=L\)\begin {align*} B & =c_{1}L+A-\frac {1}{2}L^{2}\\ c_{1} & =\frac {B-A+\frac {1}{2}L^{2}}{L}\\ & =\frac {B}{L}-\frac {A}{L}+\frac {1}{2}L \end {align*}
Hence \[ u_{E}=\left ( \frac {B}{L}-\frac {A}{L}+\frac {1}{2}L\right ) x+A-\frac {1}{2}x^{2}\] Hence from (1)\begin {align} u\left ( x,t\right ) & =v\left ( x,t\right ) +u_{E}\tag {1A}\\ & =v\left ( x,t\right ) +\left ( \frac {B}{L}-\frac {A}{L}+\frac {1}{2}L\right ) x+A-\frac {1}{2}x^{2}\nonumber \end {align}
Substituting this in \(u_{t}=ku_{xx}+k\) gives\begin {align} v_{t} & =k\left ( v_{xx}-1\right ) +k\nonumber \\ & =kv_{xx}\tag {2} \end {align}
We need to solve the above for \(v\left ( x,t\right ) \), but with homogeneous B.C. \(v\left ( 0,t\right ) =0,v\left ( L,t\right ) =0\). The eigenvalues for the homogeneous PDE \(v_{t}=kv_{xx}\) with these boundary conditions is known to be \(\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2}\,\), for \(n=1,2,\cdots \) and the corresponding eigenfunctions are \(X_{n}\left ( x\right ) =\sin \left ( \sqrt {\lambda _{n}}x\right ) \). Now, using eigenfunction expansion, let\begin {equation} v\left ( x,t\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) X_{n}\left ( x\right ) \tag {3} \end {equation} Substituting (3) into (2) gives\[ \sum _{n=1}^{\infty }b_{n}^{\prime }\left ( t\right ) X_{n}\left ( x\right ) =k\sum _{n=1}^{\infty }b_{n}\left ( t\right ) X_{n}^{\prime \prime }\left ( x\right ) \] But \(X_{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}X_{n}\left ( x\right ) \), therefore the above becomes\[ \sum _{n=1}^{\infty }b_{n}^{\prime }\left ( t\right ) X_{n}\left ( x\right ) +k\sum _{n=1}^{\infty }\lambda _{n}b_{n}\left ( t\right ) X_{n}\left ( x\right ) =0 \] Since the above is true for each \(n\) and since eigenfunctions can not be zero, the above simplifies to\begin {equation} b_{n}^{\prime }\left ( t\right ) +k\lambda _{n}b_{n}\left ( t\right ) =0\tag {4} \end {equation} This is linear in \(b\left ( t\right ) \). The solution using integrating factor is\[ b_{n}\left ( t\right ) =b_{0}\left ( 0\right ) e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\] Therefore (3) becomes\begin {align*} v\left ( x,t\right ) & =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) X_{n}\left ( x\right ) \\ & =\sum _{n=1}^{\infty }b_{0}\left ( 0\right ) e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \end {align*}
And from (1)\begin {align} u\left ( x,t\right ) & =v\left ( x,t\right ) +u_{E}\left ( x\right ) \nonumber \\ & =\overset {u_{E}}{\overbrace {\left ( \frac {B}{L}-\frac {A}{L}+\frac {1}{2}L\right ) x+A-\frac {1}{2}x^{2}}}+\sum _{n=1}^{\infty }b_{0}\left ( 0\right ) e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \tag {5} \end {align}
At \(t=0\) the above becomes\[ f\left ( x\right ) =\frac {Bx}{L}-\frac {Ax}{L}+\frac {1}{2}Lx+A-\frac {1}{2}x^{2}+\sum _{n=1}^{\infty }b_{0}\left ( 0\right ) \sin \left ( \frac {n\pi }{L}x\right ) \] For \(n>0\), and applying orthogonality\[ \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx=\int _{0}^{L}\left ( \frac {Bx}{L}-\frac {Ax}{L}+\frac {1}{2}Lx+A-\frac {1}{2}x^{2}\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx+\int _{0}^{L}b_{0}\left ( 0\right ) \sin ^{2}\left ( \frac {n\pi }{L}x\right ) dx \] Hence\[ \int _{0}^{L}\left ( f\left ( x\right ) -\left ( \frac {Bx}{L}-\frac {Ax}{L}+\frac {1}{2}Lx+A-\frac {1}{2}x^{2}\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) dx=\frac {L}{2}b_{0}\left ( 0\right ) \] Therefore\[ b_{0}\left ( 0\right ) =\frac {2}{L}\int _{0}^{L}\left ( f\left ( x\right ) -\left ( \frac {Bx}{L}-\frac {Ax}{L}+\frac {1}{2}Lx+A-\frac {1}{2}x^{2}\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) dx \] Substituting the above in (5) gives\begin {align*} u\left ( x,t\right ) & =\left ( \frac {Bx}{L}-\frac {Ax}{L}+\frac {1}{2}Lx+A-\frac {1}{2}x^{2}\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}\left ( f\left ( x\right ) -\left ( \frac {Bx}{L}-\frac {Ax}{L}+\frac {1}{2}Lx+A-\frac {1}{2}x^{2}\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) dx\right ) e^{-k\frac {n\pi }{L}t}\sin \left ( \frac {n\pi }{L}x\right ) \end {align*}
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Added July 3, 2019
Solve the heat equation \[ u_t = k u_{xx} + Q(x) \] For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &= A(t) \\ u(L,t) &= B(t) \end {align*}
Initial condition is \(u(x,0)=f(x)\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}] + Q[x]; bc = {u[0, t] == A[t], u[L, t] == B[t]}; ic = u[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t},Assumptions->{L>0,k>0}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} \left (\int _0^t e^{-\frac {k \pi ^2 K[1]^2 (t-K[2])}{L^2}} \text {Integrate}\left [\frac {\sqrt {2} \sin \left (\frac {\pi x K[1]}{L}\right ) \left (L Q(x)+(x-L) A'(K[2])-x B'(K[2])\right )}{L^{3/2}},\{x,0,L\},\text {Assumptions}\to k>0\land L>0\right ] \, dK[2]+e^{-\frac {k \pi ^2 t K[1]^2}{L^2}} \int _0^L \frac {\sqrt {2} (-L A(0)+x A(0)-x B(0)+L f(x)) \sin \left (\frac {\pi x K[1]}{L}\right )}{L^{3/2}} \, dx\right ) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}}+\frac {x (B(t)-A(t))}{L}+A(t)\right \}\right \}\]
Maple ✗
restart; interface(showassumed=0); pde := diff(u(x,t),t)=k*diff(u(x,t),x$2)+Q(x); ic := u(x,0)=f(x); bc := u(0,t)=A(t), u(L,t)=B(t); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t)) assuming L>0,k>0),output='realtime'));
time expired Possible bug. Maple can solve with \(Q(x,t)\) source but not with \(Q(x)\)
Hand solution
Solve \begin {align*} u_{t} & =ku_{xx}+Q\left ( x\right ) \\ u\left ( 0,t\right ) & =A\left ( t\right ) \\ u\left ( L,0\right ) & =B\left ( t\right ) \\ u\left ( x,0\right ) & =f\left ( x\right ) \end {align*}
Since boundary conditions are nonhomogeneous, the first step is to reduce the problem to one with homogeneous B.C. to be able to use separation of variables. This is done by using a reference solution \(r\left ( x,t\right ) \) which only needs to satisfy the B.C. Let the total solution be\begin {equation} u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x,t\right ) \tag {2} \end {equation} Where \(v\left ( x,t\right ) \) is the transient solution which satisfies the homogeneous B.C. One can easily see that the reference function is\begin {equation} r\left ( x,t\right ) =A\left ( t\right ) +\frac {B\left ( t\right ) -A\left ( t\right ) }{L}x\tag {3} \end {equation} Substituting (1) back into the original PDE gives\begin {align*} \frac {\partial }{\partial t}\left ( v\left ( x,t\right ) +r\left ( x,t\right ) \right ) & =k\frac {\partial ^{2}}{\partial x^{2}}\left ( v\left ( x,t\right ) +r\left ( x,t\right ) \right ) +Q\left ( x\right ) \\ v_{t}\left ( x,t\right ) +r_{t}\left ( x,t\right ) & =kv_{xx}\left ( x,t\right ) +kr_{xx}\left ( x,t\right ) +Q\left ( x\right ) \end {align*}
But \(r_{xx}\left ( x,t\right ) =0\) and \(r_{t}=A^{\prime }\left ( t\right ) +\frac {B^{\prime }\left ( t\right ) -A^{\prime }\left ( t\right ) }{L}x\) and PDE becomes\[ v_{t}\left ( x,t\right ) =kv_{xx}\left ( x,t\right ) -r_{t}\left ( x,t\right ) +Q\left ( x\right ) \] Let \begin {align} \tilde {Q}\left ( x\right ) & =Q\left ( x\right ) -r_{t}\left ( x,t\right ) \nonumber \\ & =Q\left ( x\right ) -\left ( A^{\prime }\left ( t\right ) +\frac {B^{\prime }\left ( t\right ) -A^{\prime }\left ( t\right ) }{L}x\right ) \tag {4} \end {align}
Therefore the problem has been transformed to\begin {align*} v_{t} & =kv_{xx}+\tilde {Q}\left ( x\right ) \\ v\left ( 0,t\right ) & =0\\ v\left ( L,t\right ) & =0\\ v\left ( 0,x\right ) & =F\left ( x\right ) \\ & =u\left ( x,0\right ) -r\left ( x,0\right ) \\ & =f\left ( x\right ) -\left ( A\left ( 0\right ) +\frac {B\left ( 0\right ) -A\left ( 0\right ) }{L}x\right ) \end {align*}
The basic solution for this type of PDE was already given in problem 3.1.1.11 on page 443 and the solution is
\begin {align*} v\left ( x,t\right ) & =\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\Phi _{n}\left ( x\right ) \left ( \frac {2}{L}\int _{0}^{L}F\left ( s\right ) \Phi _{n}\left ( s\right ) ds\right ) \\ & +\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\Phi _{n}\left ( x\right ) \left ( \int _{0}^{t}\frac {2}{L}e^{k\lambda _{n}\tau }\left ( \int _{0}^{L}Q\left ( s\right ) \Phi _{n}\left ( s\right ) ds\right ) d\tau \right ) \end {align*}
Where \begin {align*} \Phi _{n}\left ( x\right ) & =\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end {align*}
Hence, using our \(\tilde {Q}\left ( x\right ) \) and \(F\left ( x\right ) \) found above into this solution gives
\begin {align*} v\left ( x,t\right ) & =\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}\left ( f\left ( s\right ) -\left ( A\left ( 0\right ) +\frac {B\left ( 0\right ) -A\left ( 0\right ) }{L}s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}\left ( Q\left ( s\right ) -\left ( A^{\prime }\left ( \tau \right ) +\frac {B^{\prime }\left ( \tau \right ) -A^{\prime }\left ( \tau \right ) }{L}s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
Since \(u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x,t\right ) \) then the final solution is\begin {align*} u\left ( x,t\right ) & =A\left ( t\right ) +\frac {B\left ( t\right ) -A\left ( t\right ) }{L}x\\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}\left ( f\left ( s\right ) -\left ( A\left ( 0\right ) +\frac {B\left ( 0\right ) -A\left ( 0\right ) }{L}s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}\left ( Q\left ( s\right ) -\left ( A^{\prime }\left ( \tau \right ) +\frac {B^{\prime }\left ( \tau \right ) -A^{\prime }\left ( \tau \right ) }{L}s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
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Added July 4, 2019
Solve the heat equation
\[ u_t = k u_{xx} + Q(x) \]
For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &= A(t) \\ u(L,t) &= B(t) \end {align*}
Initial condition is \(u(x,0)=f(x)\) using these values \begin {align*} L&=2\\ k&=\frac {1}{10}\\ A(t)&=\sin (t)\\ B(t)&=2 \cos (t)\\ Q(x)&= x \\ f(x)&=x \end {align*}
Mathematica ✓
ClearAll["Global`*"]; L=2; k=1/10; A=Sin[t]; B=Cos[t]; f=x; Q=x; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}] + Q; bc = {u[0, t] == A, u[L, t] == B}; ic = u[x, 0] == f; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\left (-\frac {80 \left (\pi ^4 \cos (t) K[1]^4+40 \pi ^2 \sin (t) K[1]^2-e^{-\frac {1}{40} \pi ^2 t K[1]^2} \left (\pi ^4 K[1]^4+2 (-1)^{K[1]} \left (\pi ^4 K[1]^4-20 \pi ^2 K[1]^2+1600\right )\right )+(-1)^{K[1]} \left (\pi ^4 (\sin (t)+2) K[1]^4-40 \pi ^2 \cos (t) K[1]^2+3200\right )\right )}{\pi ^3 K[1]^3 \left (\pi ^4 K[1]^4+1600\right )}-\frac {2 (-1)^{K[1]} e^{-\frac {1}{40} \pi ^2 t K[1]^2}}{\pi K[1]}\right ) \sin \left (\frac {1}{2} \pi x K[1]\right )+\frac {1}{2} x (\cos (t)-\sin (t))+\sin (t)\right \}\right \}\]
Maple ✓
restart; L:=2; k:=1/10; A:=sin(t); B:=cos(t); f:=x; Q:=x; pde := diff(u(x,t),t)=k*diff(u(x,t),x$2)+Q; ic := u(x,0)=f; bc := u(0,t)=A, u(L,t)=B; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t)) ),output='realtime'));
\[u \left (x , t\right ) = -\frac {5 x^{3}}{3}-\frac {\left (-3 \cos \left (t \right )+3 \sin \left (t \right )-40\right ) x}{6}+\left (\sum _{n=1}^{\infty }-\frac {2 \left (40 \pi ^{2} \left (\pi ^{2} n^{2} \cos \left (t \right )+\left (\pi ^{2} n^{2} \sin \left (t \right )-40 \cos \left (t \right )\right ) \left (-1\right )^{n}+40 \sin \left (t \right )\right ) n^{2}+\left (-40 \pi ^{4} n^{4}+\left (\pi ^{6} n^{6}-80 \pi ^{4} n^{4}+3200 \pi ^{2} n^{2}-128000\right ) \left (-1\right )^{n}\right ) {\mathrm e}^{-\frac {\pi ^{2} n^{2} t}{40}}\right ) \sin \left (\frac {\pi n x}{2}\right )}{\pi ^{3} \left (\pi ^{4} n^{4}+1600\right ) n^{3}}\right )+\sin \left (t \right )\]
Hand solution
Solve \begin {align*} u_{t} & =ku_{xx}+Q\left ( x\right ) \\ u\left ( 0,t\right ) & =A\left ( t\right ) \\ u\left ( L,0\right ) & =B\left ( t\right ) \\ u\left ( x,0\right ) & =f\left ( x\right ) \end {align*}
With \(k=\frac {1}{10},L=2,f\left ( x\right ) =x,Q\left ( x\right ) =x,A\left ( t\right ) =\sin \left ( t\right ) ,B\left ( t\right ) =2\cos \left ( t\right ) \).
The general problem above was solved in 3.1.5.12 on page 741 and the solution is
\begin {align*} u\left ( x,t\right ) & =A\left ( t\right ) +\frac {B\left ( t\right ) -A\left ( t\right ) }{L}x\\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}\left ( f\left ( s\right ) -\left ( A\left ( 0\right ) +\frac {B\left ( 0\right ) -A\left ( 0\right ) }{L}s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}\left ( Q\left ( s\right ) -\left ( A^{\prime }\left ( \tau \right ) +\frac {B^{\prime }\left ( \tau \right ) -A^{\prime }\left ( \tau \right ) }{L}s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
Substituting the specific values given above into this solution gives\begin {align*} u\left ( x,t\right ) & =\sin \left ( t\right ) +\frac {2\cos \left ( t\right ) -\sin \left ( t\right ) }{2}x\\ & +\sum _{n=1}^{\infty }e^{-\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}t}\sin \left ( \frac {n\pi }{2}x\right ) \left ( \int _{0}^{2}\left ( s-\left ( A\left ( 0\right ) +\frac {B\left ( 0\right ) -A\left ( 0\right ) }{2}s\right ) \right ) \sin \left ( \frac {n\pi }{2}s\right ) ds\right ) \\ & +\sum _{n=1}^{\infty }e^{-\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}t}\sin \left ( \frac {n\pi }{2}x\right ) \int _{0}^{t}e^{\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}\tau }\left ( \int _{0}^{2}\left ( s-\left ( A^{\prime }\left ( \tau \right ) +\frac {B^{\prime }\left ( \tau \right ) -A^{\prime }\left ( \tau \right ) }{2}s\right ) \right ) \sin \left ( \frac {n\pi }{2}s\right ) ds\right ) d\tau \end {align*}
But \(A\left ( 0\right ) =0,B\left ( 0\right ) =2,A^{\prime }\left ( t\right ) =\cos \left ( t\right ) ,B^{\prime }\left ( t\right ) =-2\sin \left ( t\right ) \) and the above becomes\begin {align*} u\left ( x,t\right ) & =\sin \left ( t\right ) +\frac {2\cos \left ( t\right ) -\sin \left ( t\right ) }{2}x\\ & +\sum _{n=1}^{\infty }e^{-\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}t}\sin \left ( \frac {n\pi }{2}x\right ) \left ( \int _{0}^{2}\left ( s-\left ( 0+\frac {2-0}{2}s\right ) \right ) \sin \left ( \frac {n\pi }{2}s\right ) ds\right ) \\ & +\sum _{n=1}^{\infty }e^{-\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}t}\sin \left ( \frac {n\pi }{2}x\right ) \int _{0}^{t}e^{\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}\tau }\left ( \int _{0}^{2}\left ( s-\left ( \cos \left ( \tau \right ) +\left ( \frac {-2\sin \left ( \tau \right ) -\cos \left ( \tau \right ) }{2}\right ) s\right ) \right ) \sin \left ( \frac {n\pi }{2}s\right ) ds\right ) d\tau \end {align*}
Or\begin {align*} u\left ( x,t\right ) & =\sin \left ( t\right ) +\frac {2\cos \left ( t\right ) -\sin \left ( t\right ) }{2}x\\ & +\sum _{n=1}^{\infty }e^{-\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}t}\sin \left ( \frac {n\pi }{2}x\right ) \int _{0}^{t}e^{\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}\tau }\left ( \int _{0}^{2}\left ( s-\left ( \cos \left ( \tau \right ) -\left ( \frac {2\sin \left ( \tau \right ) +\cos \left ( \tau \right ) }{2}\right ) s\right ) \right ) \sin \left ( \frac {n\pi }{2}s\right ) ds\right ) d\tau \end {align*}
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Source code used for the above
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Added June 27, 2019
Solve the heat equation \[ u_t = k u_{xx} + Q(x,t) \] For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &= A(t) \\ u(L,t) &= B(t) \end {align*}
Initial condition is \(u(x,0)=f(x)\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}] + Q[x,t]; bc = {u[0, t] == A[t], u[L, t] == B[t]}; ic = u[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t},Assumptions->{L>0,k>0}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} \left (\int _0^t e^{-\frac {k \pi ^2 K[1]^2 (t-K[2])}{L^2}} \text {Integrate}\left [\frac {\sqrt {2} \sin \left (\frac {\pi x K[1]}{L}\right ) \left (L Q(x,K[2])+(x-L) A'(K[2])-x B'(K[2])\right )}{L^{3/2}},\{x,0,L\},\text {Assumptions}\to k>0\land L>0\right ] \, dK[2]+e^{-\frac {k \pi ^2 t K[1]^2}{L^2}} \int _0^L \frac {\sqrt {2} (-L A(0)+x A(0)-x B(0)+L f(x)) \sin \left (\frac {\pi x K[1]}{L}\right )}{L^{3/2}} \, dx\right ) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}}+\frac {x (B(t)-A(t))}{L}+A(t)\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t)=k*diff(u(x,t),x$2)+Q(x,t); ic := u(x,0)=f(x); bc := u(0,t)=A(t), u(L,t)=B(t); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t)) assuming L>0,k>0),output='realtime'));
\[u \left (x , t\right ) = \frac {L \left (\int _{0}^{t}\left (\sum _{n=1}^{\infty }-\frac {2 \left (\int _{0}^{L}\left (-L Q \left (x , \tau \right )+x \left (\frac {d}{d \tau }B \left (\tau \right )\right )+\left (L -x \right ) \left (\frac {d}{d \tau }A \left (\tau \right )\right )\right ) \sin \left (\frac {\pi n x}{L}\right )d x \right ) {\mathrm e}^{-\frac {\pi ^{2} \left (t -\tau \right ) k n^{2}}{L^{2}}} \sin \left (\frac {\pi n x}{L}\right )}{L^{2}}\right )d \tau \right )-2 L \left (\sum _{n=1}^{\infty }\frac {\left (\int _{0}^{L}\left (-L f \left (\tau \right )+B \left (0\right ) \tau +\left (L -\tau \right ) A \left (0\right )\right ) \sin \left (\frac {\pi n \tau }{L}\right )d \tau \right ) {\mathrm e}^{-\frac {\pi ^{2} k n^{2} t}{L^{2}}} \sin \left (\frac {\pi n x}{L}\right )}{L^{2}}\right )+x B \left (t \right )+\left (L -x \right ) A \left (t \right )}{L}\]
Hand solution
Solve \begin {align*} u_{t} & =ku_{xx}+Q\left ( x,t\right ) \\ u\left ( 0,t\right ) & =A\left ( t\right ) \\ u\left ( L,0\right ) & =B\left ( t\right ) \\ u\left ( x,0\right ) & =f\left ( x\right ) \end {align*}
Since boundary conditions are nonhomogeneous, the first step is to reduce the problem to one with homogeneous B.C. to be able to use separation of variables. This is done by using a reference solution \(r\left ( x,t\right ) \) which only needs to satisfy the B.C. Let the total solution be\begin {equation} u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x,t\right ) \tag {2} \end {equation} Where \(v\left ( x,t\right ) \) is the transient solution which satisfies the homogeneous B.C. One can easily see that the reference function is\begin {equation} r\left ( x,t\right ) =A\left ( t\right ) +\frac {B\left ( t\right ) -A\left ( t\right ) }{L}x\tag {3} \end {equation} Substituting (1) back into the original PDE gives\begin {align*} \frac {\partial }{\partial t}\left ( v\left ( x,t\right ) +r\left ( x,t\right ) \right ) & =k\frac {\partial ^{2}}{\partial x^{2}}\left ( v\left ( x,t\right ) +r\left ( x,t\right ) \right ) +Q\left ( x,t\right ) \\ v_{t}\left ( x,t\right ) +r_{t}\left ( x,t\right ) & =kv_{xx}\left ( x,t\right ) +kr_{xx}\left ( x,t\right ) +Q\left ( x,t\right ) \end {align*}
But \(r_{xx}\left ( x,t\right ) =0\) and \(r_{t}=A^{\prime }\left ( t\right ) +\frac {B^{\prime }\left ( t\right ) -A^{\prime }\left ( t\right ) }{L}x\) and PDE becomes\[ v_{t}\left ( x,t\right ) =kv_{xx}\left ( x,t\right ) -r_{t}\left ( x,t\right ) +Q\left ( x,t\right ) \] Let \begin {align} \tilde {Q}\left ( x,t\right ) & =Q\left ( x,t\right ) -r_{t}\left ( x,t\right ) \nonumber \\ & =Q\left ( x,t\right ) -\left ( A^{\prime }\left ( t\right ) +\frac {B^{\prime }\left ( t\right ) -A^{\prime }\left ( t\right ) }{L}x\right ) \tag {4} \end {align}
Therefore the problem has been transformed to\begin {align*} v_{t} & =kv_{xx}+\tilde {Q}\left ( x,t\right ) \\ v\left ( 0,t\right ) & =0\\ v\left ( L,t\right ) & =0\\ v\left ( 0,x\right ) & =F\left ( x\right ) \\ & =u\left ( x,0\right ) -r\left ( x,0\right ) \\ & =f\left ( x\right ) -\left ( A\left ( 0\right ) +\frac {B\left ( 0\right ) -A\left ( 0\right ) }{L}x\right ) \end {align*}
The basic solution for this type of PDE was already given in problem 3.1.6.4 on page 796 and the solution is\begin {align*} v\left ( x,t\right ) & =\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\Phi _{n}\left ( x\right ) \left ( \frac {2}{L}\int _{0}^{L}F\left ( s\right ) \Phi _{n}\left ( s\right ) ds\right ) \\ & +\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\Phi _{n}\left ( x\right ) \left ( \int _{0}^{t}\frac {2}{L}e^{k\lambda _{n}\tau }\left ( \int _{0}^{L}\tilde {Q}\left ( s,\tau \right ) \Phi _{n}\left ( s\right ) ds\right ) d\tau \right ) \end {align*}
Where \begin {align*} \Phi _{n}\left ( x\right ) & =\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end {align*}
Hence, using our \(\tilde {Q}\left ( x,\tau \right ) \) and \(F\left ( x\right ) \) found above into this solution gives
\begin {align*} v\left ( x,t\right ) & =\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}\left ( f\left ( s\right ) -\left ( A\left ( 0\right ) +\frac {B\left ( 0\right ) -A\left ( 0\right ) }{L}s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}\left ( Q\left ( s,\tau \right ) -\left ( A^{\prime }\left ( \tau \right ) +\frac {B^{\prime }\left ( \tau \right ) -A^{\prime }\left ( \tau \right ) }{L}s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
Since \(u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x,t\right ) \) then the final solution is\begin {align*} u\left ( x,t\right ) & =A\left ( t\right ) +\frac {B\left ( t\right ) -A\left ( t\right ) }{L}x\\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}\left ( f\left ( s\right ) -\left ( A\left ( 0\right ) +\frac {B\left ( 0\right ) -A\left ( 0\right ) }{L}s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}\left ( Q\left ( s,\tau \right ) -\left ( A^{\prime }\left ( \tau \right ) +\frac {B^{\prime }\left ( \tau \right ) -A^{\prime }\left ( \tau \right ) }{L}s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \right ) \end {align*}
____________________________________________________________________________________
Added June 27, 2019
Solve the heat equation \[ u_t = k u_{xx} + Q(x,t) \] For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &= A(t) \\ u(L,t) &= B(t) \end {align*}
Initial condition is \(u(x,0)=f(x)\) using these values \begin {align*} L&=2\\ k&=\frac {1}{10}\\ A(t)&=\sin (t)\\ B(t)&=2 \cos (t)\\ Q(x,t)&= x t e^{-t} \cos (t) \\ f(x)&=x \end {align*}
Mathematica ✓
ClearAll["Global`*"]; L=2; k=1/10; A=Sin[t]; B=Cos[t]; f=x; Q=x*t*Exp[-t]*Cos[t]; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}] + Q; bc = {u[0, t] == A, u[L, t] == B}; ic = u[x, 0] == f; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {2 e^{-\frac {1}{40} \pi ^2 t K[1]^2} \left (\frac {40 \left (-80 (-1)^{K[1]} \pi ^2 \left (\pi ^2 K[1]^2-80\right ) \left (\pi ^4 K[1]^4+1600\right ) K[1]^2-\left (40 (-1)^{K[1]}-\pi ^2 K[1]^2\right ) \left (\pi ^4 K[1]^4-80 \pi ^2 K[1]^2+3200\right )^2-e^{\frac {1}{40} t \left (\pi ^2 K[1]^2-40\right )} \left (\cos (t) \left (2 (-1)^{K[1]} \left (\pi ^4 K[1]^4+1600\right ) \left (t \left (\pi ^6 K[1]^6-120 \pi ^4 K[1]^4+6400 \pi ^2 K[1]^2-128000\right )-40 \pi ^2 K[1]^2 \left (\pi ^2 K[1]^2-80\right )\right )-e^t \left (40 (-1)^{K[1]}-\pi ^2 K[1]^2\right ) \left (\pi ^4 K[1]^4-80 \pi ^2 K[1]^2+3200\right )^2\right )+\left (e^t \left ((-1)^{K[1]} \pi ^2 K[1]^2+40\right ) \left (\pi ^4 K[1]^4-80 \pi ^2 K[1]^2+3200\right )^2+80 (-1)^{K[1]} \left (\pi ^4 K[1]^4+1600\right ) \left (t \left (\pi ^4 K[1]^4-80 \pi ^2 K[1]^2+3200\right )-80 \left (\pi ^2 K[1]^2-40\right )\right )\right ) \sin (t)\right )\right )}{\left (\pi ^4 K[1]^4+1600\right ) \left (\pi ^4 K[1]^4-80 \pi ^2 K[1]^2+3200\right )^2}+(-1)^{K[1]+1}\right ) \sin \left (\frac {1}{2} \pi x K[1]\right )}{\pi K[1]}+\frac {1}{2} x (\cos (t)-\sin (t))+\sin (t)\right \}\right \}\]
Maple ✓
restart; L:=2; k:=1/10; A:=sin(t); B:=cos(t); f:=x; Q:=x*t*exp(-t)*cos(t); pde := diff(u(x,t),t)=k*diff(u(x,t),x$2)+Q; ic := u(x,0)=f; bc := u(0,t)=A, u(L,t)=B; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t)) ),output='realtime'));
\[u \left (x , t\right ) = \frac {x \cos \left (t \right )}{2}+\left (\sum _{n=1}^{\infty }-\frac {2 \left (80 \left (\pi ^{4} n^{4}+1600\right ) \left (\left (\pi ^{6} n^{6} t -120 \pi ^{4} \left (t +\frac {1}{3}\right ) n^{4}+6400 \left (t +\frac {1}{2}\right ) \pi ^{2} n^{2}-128000 t \right ) \cos \left (t \right )+40 \left (\pi ^{4} n^{4} t -80 \pi ^{2} \left (t +1\right ) n^{2}+3200 t +3200\right ) \sin \left (t \right )\right ) \left (-1\right )^{n} {\mathrm e}^{-t}+\left (-40 \pi ^{2} \left (\pi ^{4} n^{4}-80 \pi ^{2} n^{2}+3200\right )^{2} n^{2}+\left (\pi ^{12} n^{12}-160 \pi ^{10} n^{10}+19200 \pi ^{8} n^{8}-1280000 \pi ^{6} n^{6}+56320000 \pi ^{4} n^{4}-2048000000 \pi ^{2} n^{2}+32768000000\right ) \left (-1\right )^{n}\right ) {\mathrm e}^{-\frac {\pi ^{2} n^{2} t}{40}}+40 \left (\pi ^{2} n^{2} \cos \left (t \right )+\left (\pi ^{2} n^{2} \sin \left (t \right )-40 \cos \left (t \right )\right ) \left (-1\right )^{n}+40 \sin \left (t \right )\right ) \left (\pi ^{4} n^{4}-80 \pi ^{2} n^{2}+3200\right )^{2}\right ) \sin \left (\frac {\pi n x}{2}\right )}{\pi \left (\pi ^{4} n^{4}+1600\right ) \left (\pi ^{4} n^{4}-80 \pi ^{2} n^{2}+3200\right )^{2} n}\right )-\frac {\left (x -2\right ) \sin \left (t \right )}{2}\]
Hand solution
Solve \begin {align*} u_{t} & =ku_{xx}+Q\left ( x,t\right ) \\ u\left ( 0,t\right ) & =A\left ( t\right ) \\ u\left ( L,0\right ) & =B\left ( t\right ) \\ u\left ( x,0\right ) & =f\left ( x\right ) \end {align*}
With \(k=\frac {1}{10},L=2,f\left ( x\right ) =x,Q\left ( x,t\right ) =xte^{-t}\cos \left ( t\right ) ,A\left ( t\right ) =\sin \left ( t\right ) ,B\left ( t\right ) =2\cos \left ( t\right ) \).
The general problem above was solved in 3.1.5.14 on page 754 and the solution is\begin {align*} u\left ( x,t\right ) & =A\left ( t\right ) +\frac {B\left ( t\right ) -A\left ( t\right ) }{L}x\\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \left ( \int _{0}^{L}\left ( f\left ( s\right ) -\left ( A\left ( 0\right ) +\frac {B\left ( 0\right ) -A\left ( 0\right ) }{L}s\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \\ & +\frac {2}{L}\sum _{n=1}^{\infty }e^{-k\left ( \frac {n\pi }{L}\right ) ^{2}t}\sin \left ( \frac {n\pi }{L}x\right ) \int _{0}^{t}e^{k\left ( \frac {n\pi }{L}\right ) ^{2}\tau }\left ( \int _{0}^{L}\left ( Q\left ( s,\tau \right ) -\left ( A^{\prime }\left ( \tau \right ) +\frac {B^{\prime }\left ( \tau \right ) -A^{\prime }\left ( \tau \right ) }{L}x\right ) \right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) d\tau \end {align*}
Substituting the specific values given above into this solution gives\begin {align*} u\left ( x,t\right ) & =\sin \left ( t\right ) +\frac {2\cos \left ( t\right ) -\sin \left ( t\right ) }{2}x\\ & +\sum _{n=1}^{\infty }e^{-\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}t}\sin \left ( \frac {n\pi }{2}x\right ) \left ( \int _{0}^{2}\left ( s-\left ( A\left ( 0\right ) +\frac {B\left ( 0\right ) -A\left ( 0\right ) }{2}s\right ) \right ) \sin \left ( \frac {n\pi }{2}s\right ) ds\right ) \\ & +\sum _{n=1}^{\infty }e^{-\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}t}\sin \left ( \frac {n\pi }{2}x\right ) \int _{0}^{t}e^{\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}\tau }\left ( \int _{0}^{2}\left ( s\tau e^{-\tau }\cos \left ( \tau \right ) -\left ( A^{\prime }\left ( \tau \right ) +\frac {B^{\prime }\left ( \tau \right ) -A^{\prime }\left ( \tau \right ) }{2}s\right ) \right ) \sin \left ( \frac {n\pi }{2}s\right ) ds\right ) d\tau \end {align*}
But \(A\left ( 0\right ) =0,B\left ( 0\right ) =2,A^{\prime }\left ( t\right ) =\cos \left ( t\right ) ,B^{\prime }\left ( t\right ) =-2\sin \left ( t\right ) \) and the above becomes\begin {align*} u\left ( x,t\right ) & =\sin \left ( t\right ) +\frac {2\cos \left ( t\right ) -\sin \left ( t\right ) }{2}x\\ & +\sum _{n=1}^{\infty }e^{-\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}t}\sin \left ( \frac {n\pi }{2}x\right ) \left ( \int _{0}^{2}\left ( s-\left ( 0+\frac {2-0}{2}s\right ) \right ) \sin \left ( \frac {n\pi }{2}s\right ) ds\right ) \\ & +\sum _{n=1}^{\infty }e^{-\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}t}\sin \left ( \frac {n\pi }{2}x\right ) \int _{0}^{t}e^{\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}\tau }\left ( \int _{0}^{2}\left ( s\tau e^{-\tau }\cos \left ( \tau \right ) -\left ( \cos \left ( \tau \right ) +\left ( \frac {-2\sin \left ( \tau \right ) -\cos \left ( \tau \right ) }{2}\right ) s\right ) \right ) \sin \left ( \frac {n\pi }{2}s\right ) ds\right ) d\tau \end {align*}
Or\begin {align*} u\left ( x,t\right ) & =\sin \left ( t\right ) +\frac {2\cos \left ( t\right ) -\sin \left ( t\right ) }{2}x\\ & +\sum _{n=1}^{\infty }e^{-\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}t}\sin \left ( \frac {n\pi }{2}x\right ) \int _{0}^{t}e^{\frac {1}{10}\left ( \frac {n\pi }{2}\right ) ^{2}\tau }\left ( \int _{0}^{2}\left ( s\tau e^{-\tau }\cos \left ( \tau \right ) -\left ( \cos \left ( \tau \right ) -\left ( \frac {2\sin \left ( \tau \right ) +\cos \left ( \tau \right ) }{2}\right ) s\right ) \right ) \sin \left ( \frac {n\pi }{2}s\right ) ds\right ) d\tau \end {align*}
Animation is below
Source code used for the above
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Added July 2, 2018.
Pinchover and Rubinstein’s exercise 6.17. Taken from Maple document for new improvements in Maple 2018.1
Solve the heat equation
\[ {\frac {\partial }{\partial t}}u \left ( x,t \right ) -{\frac {\partial ^{2}}{\partial {x}^{2}}}u \left ( x,t \right ) =1+x\cos \left ( t\right ) \]
For \(0<x<1\) and \(t>0\). The boundary conditions are \begin {align*} \frac {\partial u}{\partial x}(0,t) &=\sin (t) \\ \frac {\partial u}{\partial x}(1,t) &=\sin (t) \end {align*}
Initial condition is \(u (x,0) =1+ \cos ( 2 \pi x)\).
Mathematica ✗
ClearAll["Global`*"]; pde = D[u[x, t], x] == D[u[x, t], {x, 2}] + 1 + x*Cos[t]; bc = {Derivative[1, 0][u][0, t] == Sin[t], Derivative[1, 0][u][1, t] == Sin[t]}; ic = u[x, 0] == 1 + Cos[2*Pi*x]; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, ic, bc}, u[x, t], x, t]], 60*10]];
Failed
Maple ✓
restart; pde := diff(u(x, t), t)= (diff(u(x, t), x, x)) + 1+x*cos(t); bc := (D[1](u))(0, t) = sin(t), (D[1](u))(1, t) = sin(t); ic := u(x, 0) = 1+cos(2*Pi*x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde, ic, bc],u(x,t))),output='realtime'));
\[u \left (x , t\right ) = x \sin \left (t \right )+\cos \left (2 \pi x \right ) {\mathrm e}^{-4 \pi ^{2} t}+t +1\]
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Added March 28, 2018. A problem from my PDE animation page.
Solve the heat equation
\[ u_t = k u_{xx} + x \]
For \(0<x<\pi \) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &= \frac {t \sin t}{5} \\ u(\pi ,t) &= \frac {t \cos t}{10} \\ \end {align*}
Initial condition is \(u(x,0)=60 - 20 x\).
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == D[u[x, t], {x, 2}] + x; bc = {u[0, t] == (t*Sin[t])/5, u[Pi, t] == (t*Cos[t])/10}; ic = u[x, 0] == 60 - 2*x; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], x, t, Assumptions -> {t > 0, x > 0}], 60*10]];
\[\left \{\left \{u(x,t)\to \frac {10 \pi \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {e^{-t K[1]^2} \left (20 \left (30+(-1)^{K[1]} (-30+\pi )\right ) K[1]^2+\frac {(-1)^{K[1]+1} K[1]^8+10 (-1)^{K[1]} \pi K[1]^8-10 (-1)^{K[1]} e^{t K[1]^2} \pi K[1]^8-4 K[1]^6+(-1)^{K[1]} K[1]^4+20 (-1)^{K[1]} \pi K[1]^4-20 (-1)^{K[1]} e^{t K[1]^2} \pi K[1]^4+e^{t K[1]^2} \cos (t) \left (\left ((-1)^{K[1]} K[1]^4+4 K[1]^2+(-1)^{K[1]+1}\right ) K[1]^2+t \left ((-1)^{K[1]}-2 K[1]^2\right ) \left (K[1]^4+1\right )\right ) K[1]^2-e^{t K[1]^2} \left (2 \left (K[1]^4+(-1)^{K[1]+1} K[1]^2-1\right ) K[1]^2+t \left ((-1)^{K[1]} K[1]^2+2\right ) \left (K[1]^4+1\right )\right ) \sin (t) K[1]^2+10 (-1)^{K[1]} \pi -10 (-1)^{K[1]} e^{t K[1]^2} \pi }{\left (K[1]^4+1\right )^2}\right ) \sin (x K[1])}{5 \pi K[1]^3}+2 t (\pi -x) \sin (t)+t x \cos (t)}{10 \pi }\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t)=diff(u(x,t),x$2)+x; ic := u(x,0)=(60-2*x); bc := u(0,t)=t/5*sin(t), u(Pi,t)=t/10*cos(t); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t)) assuming t>0,x>0),output='realtime'));
\[u \left (x , t\right ) = \frac {\left (x \cos \left (t \right )+\left (-2 x +2 \pi \right ) \sin \left (t \right )\right ) t +2 \pi \left (\sum _{n=1}^{\infty }\frac {20 \left (-\frac {\left (\left (n^{4} t -2 n^{2}+t \right ) n^{2} \cos \left (t \right )+\left (n^{6}+n^{4} t -n^{2}+t \right ) \sin \left (t \right )\right ) n^{2}}{10}+\left (-\frac {\left (n^{4} t -2 n^{2}+t \right ) n^{4} \sin \left (t \right )}{20}+\frac {\left (n^{6}+n^{4} t -n^{2}+t \right ) n^{2} \cos \left (t \right )}{20}-\frac {\pi \left (n^{4}+1\right )^{2}}{2}\right ) \left (-1\right )^{n}+\left (30 n^{10}+\frac {299 n^{6}}{5}+30 n^{2}+\left (\left (\pi -30\right ) n^{10}+\left (-\frac {1}{20}+\frac {\pi }{2}\right ) n^{8}+\left (2 \pi -60\right ) n^{6}+\left (\frac {1}{20}+\pi \right ) n^{4}+\left (\pi -30\right ) n^{2}+\frac {\pi }{2}\right ) \left (-1\right )^{n}\right ) {\mathrm e}^{-n^{2} t}\right ) \sin \left (n x \right )}{\pi \left (n^{4}+1\right )^{2} n^{3}}\right )}{10 \pi }\]
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Added Nov 27, 2018.
Problem 8.2.2 part(a) from Richard Haberman applied partial differential equations book, 5th edition
Solve the heat equation for \(u(x,t)\)
\[ u_t=u_{xx} + Q(x,t) \]
For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} \frac { \partial u}{\partial x}(0,t) &= A(t) \\ \frac { \partial u}{\partial x}(L,t) &= B(t)\\ \end {align*}
Initial condition is \(u(x,0)=f(x)\)
For hand solution see my HW9, Math 322, UW Madison. The text does not actually asks to solve this PDE but only to reduce the problem to one with homogeneous B.C.
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}] + Q[x, t]; bc = {Derivative[1, 0][u][0, t] == A[t], Derivative[1, 0][u][L, t] == B[t]}; ic = u[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], x, t, Assumptions -> L > 0], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} \cos \left (\frac {\pi x K[1]}{L}\right ) \left (e^{-\frac {k \pi ^2 t K[1]^2}{L^2}} \int _0^L \frac {\cos \left (\frac {\pi x K[1]}{L}\right ) (x (x (A(0)-B(0))-2 L A(0))+2 L f(x))}{\sqrt {2} L^{3/2}} \, dx+\int _0^t e^{-\frac {k \pi ^2 K[1]^2 (t-K[2])}{L^2}} \text {Integrate}\left [\frac {\cos \left (\frac {\pi x K[1]}{L}\right ) \left (A'(K[2]) x^2-B'(K[2]) x^2-2 L A'(K[2]) x-2 k A(K[2])+2 k B(K[2])+2 L Q(x,K[2])\right )}{\sqrt {2} L^{3/2}},\{x,0,L\},\text {Assumptions}\to L>0\right ] \, dK[2]\right )}{\sqrt {L}}+\frac {\int _0^t \text {Integrate}\left [\frac {2 L Q(x,K[2])-2 L x A'(K[2])+x^2 A'(K[2])-2 k A(K[2])-x^2 B'(K[2])+2 k B(K[2])}{2 L^{3/2}},\{x,0,L\},\text {Assumptions}\to L>0\right ] \, dK[2]+\int _0^L \frac {-2 A(0) L x+A(0) x^2-B(0) x^2+2 L f(x)}{2 L^{3/2}} \, dx}{\sqrt {L}}+\frac {x^2 (B(t)-A(t))}{2 L}+x A(t)\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t)+k*diff(u(x,t),x$2)+Q(x,t); ic := u(x,0)=f(x); bc := eval( diff(u(x,t),x),x=0)=A(t), eval( diff(u(x,t),x),x=L)=B(t); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t)) assuming L>0),output='realtime'));
\[u \left (x , t\right ) = \frac {x^{2} B \left (t \right )}{2 L}+\left (-\frac {x^{2}}{2 L}+x \right ) A \left (t \right )+\int _{0}^{t}\left (\left (\sum _{n=1}^{\infty }\frac {2 \left (\int _{0}^{L}-\frac {\left (\frac {x^{2} \left (\frac {d}{d \tau }B \left (\tau \right )\right )}{2}+L Q \left (x , \tau \right )+\left (L -\frac {x}{2}\right ) x \left (\frac {d}{d \tau }A \left (\tau \right )\right )-\left (A \left (\tau \right )-B \left (\tau \right )\right ) k \right ) \cos \left (\frac {\pi n x}{L}\right )}{L}d x \right ) \cos \left (\frac {\pi n x}{L}\right ) {\mathrm e}^{\frac {\pi ^{2} \left (t -\tau \right ) k n^{2}}{L^{2}}}}{L}\right )+\frac {\int _{0}^{L}\frac {-x^{2} \left (\frac {d}{d \tau }B \left (\tau \right )\right )-2 L Q \left (x , \tau \right )+2 \left (A \left (\tau \right )-B \left (\tau \right )\right ) k +\left (-2 L x +x^{2}\right ) \left (\frac {d}{d \tau }A \left (\tau \right )\right )}{2 L}d x}{L}\right )d \tau +\left (\sum _{n=1}^{\infty }\frac {2 \left (\int _{0}^{L}\frac {\left (2 L f \left (x \right )-B \left (0\right ) x^{2}+\left (-2 L x +x^{2}\right ) A \left (0\right )\right ) \cos \left (\frac {\pi n x}{L}\right )}{2 L}d x \right ) \cos \left (\frac {\pi n x}{L}\right ) {\mathrm e}^{\frac {\pi ^{2} k n^{2} t}{L^{2}}}}{L}\right )+\frac {\int _{0}^{L}\frac {2 L f \left (x \right )-B \left (0\right ) x^{2}+\left (-2 L x +x^{2}\right ) A \left (0\right )}{2 L}d x}{L}\]
Hand solution
Solve \begin {align*} u_{t} & =ku_{xx}+Q\left ( x,t\right ) \\ u_{x}\left ( 0,t\right ) & =A\left ( t\right ) \\ u_{x}\left ( L,0\right ) & =B\left ( t\right ) \\ u\left ( x,0\right ) & =f\left ( x\right ) \end {align*}
Let \begin {equation} u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x,t\right ) \tag {1} \end {equation} Since the problem has time dependent source function \(Q\left ( x,t\right ) \) then \(r\left ( x,t\right ) \) is now a reference function that only needs to satisfy the non-homogenous boundary conditions which in this problem are at both ends and \(v\left ( x,t\right ) \) has homogenous boundary conditions. The first step is to find \(r\left ( x,t\right ) \). Let\[ r\left ( x,t\right ) =c_{1}\left ( t\right ) x+c_{2}\left ( t\right ) x^{2}\] Then\[ \frac {\partial r\left ( x,t\right ) }{\partial x}=c_{1}\left ( t\right ) +2c_{2}\left ( t\right ) x \] At \(x=0\)\[ A\left ( t\right ) =c_{1}\left ( t\right ) \] And at \(x=L\)\begin {align*} B\left ( t\right ) & =c_{1}\left ( t\right ) +2c_{2}\left ( t\right ) L\\ c_{2}\left ( t\right ) & =\frac {B\left ( t\right ) -c_{1}\left ( t\right ) }{2L} \end {align*}
Solving for \(c_{1},c_{2}\) gives\begin {equation} r\left ( x,t\right ) =A\left ( t\right ) x+\left ( \frac {B\left ( t\right ) -A\left ( t\right ) }{2L}\right ) x^{2} \tag {2} \end {equation} Replacing (1) into the original PDE \(u_{t}=ku_{xx}+Q\left ( x,t\right ) \) gives\begin {align*} \frac {\partial }{\partial t}\left ( v\left ( x,t\right ) -r\left ( x,t\right ) \right ) & =k\frac {\partial ^{2}}{\partial x}\left ( v\left ( x,t\right ) -r\left ( x,t\right ) \right ) +Q\left ( x,t\right ) \\ \frac {\partial v}{\partial t}-\frac {\partial r}{\partial t} & =k\frac {\partial ^{2}v}{\partial x^{2}}-k\frac {\partial ^{2}r}{\partial x^{2}}+Q\left ( x,t\right ) \end {align*}
But \(r_{xx}=\frac {B\left ( t\right ) -A\left ( t\right ) }{L}\), hence the above reduces to\begin {equation} v_{t}=kv_{xx}+Q\left ( x,t\right ) -k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}+r_{t} \tag {3} \end {equation} Let \[ \tilde {Q}\left ( x,t\right ) =Q\left ( x,t\right ) +r_{t}-k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}\] Then (3) becomes\begin {align} v_{t} & =kv_{xx}+\tilde {Q}\left ( x,t\right ) \tag {4}\\ v_{t}\left ( 0,t\right ) & =0\nonumber \\ v_{t}\left ( L,t\right ) & =0\nonumber \end {align}
And initial condition is \begin {align*} v\left ( x,0\right ) & =F\left ( x\right ) \\ & =u\left ( x,0\right ) -r\left ( x,0\right ) \\ & =f\left ( x\right ) -\left ( A\left ( 0\right ) x+\left ( \frac {B\left ( 0\right ) -A\left ( 0\right ) }{2L}\right ) x^{2}\right ) \end {align*}
PDE (4) with its homogenous boundary conditions is standard one, its corresponding eigenvalue boundary value ODE \(X^{\prime \prime }+\lambda X=0\) has \(\lambda =0\) as eigenvalue with corresponding eigenfunction \(\Phi _{0}\left ( x\right ) =1\) and \(\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2}\) for \(n=1,2,3,\cdots \) with corresponding eigenfunctions \(\Phi _{n}\left ( x\right ) =\cos \left ( \sqrt {\lambda _{n}}x\right ) \). Using these, we can write the solution to (4) using eigenfunction expansion as\begin {equation} v\left ( x,t\right ) =\sum _{n=0}^{\infty }c_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \tag {4A} \end {equation} Hence \(v_{t}\left ( x,t\right ) =\sum _{n=0}^{\infty }c_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) \) and \(v_{xx}\left ( x,t\right ) =\sum _{n=0}^{\infty }c_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) \). Substituting these into (4) gives\[ \sum _{n=0}^{\infty }c_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) =\sum _{n=0}^{\infty }c_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +\tilde {Q}\left ( x,t\right ) \] Expanding \(\tilde {Q}\left ( x,t\right ) \) using same eigenfunctions since they are complete, the above becomes\[ \sum _{n=0}^{\infty }c_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) =\sum _{n=0}^{\infty }c_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +\sum _{n=0}^{\infty }b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \] But \(\Phi _{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\Phi _{n}\left ( x\right ) \) and the above becomes\begin {align} \sum _{n=0}^{\infty }c_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) & =-\sum _{n=0}^{\infty }c_{n}\left ( t\right ) \lambda _{n}\Phi _{n}\left ( x\right ) +\sum _{n=0}^{\infty }b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \nonumber \\ c_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) +c_{n}\left ( t\right ) \lambda _{n}\Phi _{n}\left ( x\right ) & =b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \nonumber \\ c_{n}^{\prime }\left ( t\right ) +c_{n}\left ( t\right ) \lambda _{n} & =b_{n}\left ( t\right ) \nonumber \\ c_{n}^{\prime }\left ( t\right ) +c_{n}\left ( t\right ) \frac {n^{2}\pi ^{2}}{L^{2}} & =b_{n}\left ( t\right ) \tag {5} \end {align}
To find \(b_{n}\left ( t\right ) \), since \(\tilde {Q}\left ( x,t\right ) =Q\left ( x,t\right ) +\frac {\partial r}{\partial t}-k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}\) then\[ Q\left ( x,t\right ) +\frac {\partial r}{\partial t}-k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}=\sum _{n=0}^{\infty }b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \] Multiplying both sides by \(\Phi _{m}\left ( x\right ) \) and integrating gives\begin {align*} \int _{0}^{L}\left ( Q\left ( x,t\right ) +\frac {\partial r}{\partial t}-k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}\right ) \Phi _{m}\left ( x\right ) dx & =\int _{0}^{L}\sum _{n=0}^{\infty }b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx\\ & =\sum _{n=0}^{\infty }b_{n}\left ( t\right ) \left ( \int _{0}^{L}\Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx\right ) \end {align*}
By orthogonality\[ \int _{0}^{L}\left ( Q\left ( x,t\right ) +r_{t}-k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}\right ) \Phi _{m}\left ( x\right ) dx=b_{m}\left ( t\right ) \int _{0}^{L}\Phi _{m}^{2}\left ( x\right ) dx \] When \(m=0,\Phi _{0}\left ( x\right ) =1\) and the above gives\begin {align*} \int _{0}^{L}Q\left ( x,t\right ) +r_{t}-k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}dx & =b_{0}\left ( t\right ) \int _{0}^{L}dx\\ b_{0}\left ( t\right ) & =\frac {1}{L}\int _{0}^{L}Q\left ( x,t\right ) +r_{t}-k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}dx \end {align*}
When \(m=1,2,3,\cdots \)\begin {align*} \int _{0}^{L}\left ( Q\left ( x,t\right ) +r_{t}-k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}\right ) \cos \left ( \frac {m\pi }{L}x\right ) dx & =b_{m}\left ( t\right ) \int _{0}^{L}\cos ^{2}\left ( \frac {m\pi }{L}x\right ) dx\\ \int _{0}^{L}\left ( Q\left ( x,t\right ) +r_{t}-k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}\right ) \cos \left ( \frac {m\pi }{L}x\right ) dx & =b_{m}\left ( t\right ) \frac {L}{2}\\ b_{m}\left ( t\right ) & =\frac {2}{L}\int _{0}^{L}\left ( Q\left ( x,t\right ) +r_{t}-k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}\right ) \cos \left ( \frac {m\pi }{L}x\right ) dx \end {align*}
Therefore (5) is now solved. When \(n=0\) (5) becomes\begin {align*} c_{0}^{\prime }\left ( t\right ) +c_{0}\left ( t\right ) \frac {n^{2}\pi ^{2}}{L^{2}} & =b_{0}\left ( t\right ) \\ c_{0}^{\prime }\left ( t\right ) & =b_{0}\left ( t\right ) \\ c_{0}^{\prime }\left ( t\right ) & =\frac {1}{L}\int _{0}^{L}Q\left ( x,t\right ) +r_{t}-k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}dx \end {align*}
Hence\[ c_{0}\left ( t\right ) =\int _{0}^{t}\left ( \frac {1}{L}\int _{0}^{L}Q\left ( x,\tau \right ) +r_{\tau }-k\frac {B\left ( \tau \right ) -A\left ( \tau \right ) }{L}dx\right ) dt+C_{0}\] For \(n=1,2,3,\cdots \) (5) becomes\begin {align*} c_{n}^{\prime }\left ( t\right ) +c_{n}\left ( t\right ) \frac {n^{2}\pi ^{2}}{L^{2}} & =b_{n}\left ( t\right ) \\ & =\frac {2}{L}\int _{0}^{L}\left ( Q\left ( x,\tau \right ) +r_{\tau }-k\frac {B\left ( \tau \right ) -A\left ( \tau \right ) }{L}\right ) \cos \left ( \frac {n\pi }{L}x\right ) dx \end {align*}
Integrating factor is \(I=e^{\int \frac {n^{2}\pi ^{2}}{L^{2}}dt}=e^{\frac {n^{2}\pi ^{2}}{L^{2}}t}\) and the solution to the above becomes
\begin {align*} \frac {d}{dt}\left ( c_{n}\left ( t\right ) e^{\frac {n^{2}\pi ^{2}}{L^{2}}t}\right ) & =\frac {2e^{\frac {n^{2}\pi ^{2}}{L^{2}}t}}{L}\int _{0}^{L}\left ( Q\left ( x,t\right ) +r_{t}-k\frac {B\left ( t\right ) -A\left ( t\right ) }{L}\right ) \cos \left ( \frac {n\pi }{L}x\right ) dx\\ c_{n}\left ( t\right ) e^{\frac {n^{2}\pi ^{2}}{L^{2}}t} & =\int _{0}^{t}\left ( \frac {2e^{\frac {n^{2}\pi ^{2}}{L^{2}}\tau }}{L}\int _{0}^{L}\left ( Q\left ( x,\tau \right ) +r_{\tau }-k\frac {B\left ( \tau \right ) -A\left ( \tau \right ) }{L}\right ) \cos \left ( \frac {n\pi }{L}x\right ) dx\right ) dt+C_{n}\\ c_{n}\left ( t\right ) & =e^{-\frac {n^{2}\pi ^{2}}{L^{2}}t}\int _{0}^{t}\left ( \frac {2e^{\frac {n^{2}\pi ^{2}}{L^{2}}\tau }}{L}\int _{0}^{L}\left ( Q\left ( x,\tau \right ) +r_{\tau }-k\frac {B\left ( \tau \right ) -A\left ( \tau \right ) }{L}\right ) \cos \left ( \frac {n\pi }{L}x\right ) dx\right ) dt+C_{n}e^{-\frac {n^{2}\pi ^{2}}{L^{2}}t} \end {align*}
Now that we found \(c_{n}\left ( t\right ) \) for \(n=0,1,2,3,\cdots \) the solution for \(v\left ( x,t\right ) \) is found from 4A.
\begin {align*} v\left ( x,t\right ) & =\sum _{n=0}^{\infty }c_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \\ & =\int _{0}^{t}\left ( \frac {1}{L}\int _{0}^{L}Q\left ( x,\tau \right ) +r_{\tau }-k\frac {B\left ( \tau \right ) -A\left ( \tau \right ) }{L}dx\right ) dt+C_{0}+\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \\ & \\ & =\int _{0}^{t}\left ( \frac {1}{L}\int _{0}^{L}Q\left ( x,\tau \right ) +r_{\tau }-k\frac {B\left ( \tau \right ) -A\left ( \tau \right ) }{L}dx\right ) dt+C_{0}\\ & +\sum _{n=1}^{\infty }\left ( e^{-\frac {n^{2}\pi ^{2}}{L^{2}}t}\int _{0}^{t}\left ( \frac {2e^{\frac {n^{2}\pi ^{2}}{L^{2}}\tau }}{L}\int _{0}^{L}\left ( Q\left ( x,\tau \right ) +r_{\tau }-k\frac {B\left ( \tau \right ) -A\left ( \tau \right ) }{L}\right ) \cos \left ( \frac {n\pi }{L}x\right ) dx\right ) dt+C_{n}e^{-\frac {n^{2}\pi ^{2}}{L^{2}}t}\right ) \cos \left ( \frac {n\pi }{L}x\right ) \end {align*}
But \[ u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x,t\right ) \] Hence
\begin {align*} u\left ( x,t\right ) & =A\left ( t\right ) x+\left ( \frac {B\left ( t\right ) -A\left ( t\right ) }{2L}\right ) x^{2}+\int _{0}^{t}\left ( \frac {1}{L}\int _{0}^{L}Q\left ( x,\tau \right ) +r_{\tau }-k\frac {B\left ( \tau \right ) -A\left ( \tau \right ) }{L}dx\right ) dt+C_{0}\\ & +\sum _{n=1}^{\infty }\left ( e^{-\frac {n^{2}\pi ^{2}}{L^{2}}t}\int _{0}^{t}\left ( \frac {2e^{\frac {n^{2}\pi ^{2}}{L^{2}}\tau }}{L}\int _{0}^{L}\left ( Q\left ( x,\tau \right ) +r_{\tau }-k\frac {B\left ( \tau \right ) -A\left ( \tau \right ) }{L}\right ) \cos \left ( \frac {n\pi }{L}x\right ) dx\right ) dt+C_{n}e^{-\frac {n^{2}\pi ^{2}}{L^{2}}t}\right ) \cos \left ( \frac {n\pi }{L}x\right ) \end {align*}
But \begin {align*} r_{\tau } & =\frac {d}{dt}\left ( A\left ( \tau \right ) x+\left ( \frac {B\left ( \tau \right ) -A\left ( \tau \right ) }{2L}\right ) x^{2}\right ) \\ & =\frac {2LA^{\prime }\left ( \tau \right ) x+\left ( B^{\prime }\left ( \tau \right ) -A^{\prime }\left ( \tau \right ) \right ) x^{2}}{2L} \end {align*}
Hence
\begin {gather*} u\left ( x,t\right ) =C_{0}+A\left ( t\right ) x+\left ( \frac {B\left ( t\right ) -A\left ( t\right ) }{2L}\right ) x^{2}\\ +\frac {1}{2L^{2}}\int _{0}^{t}\left ( \int _{0}^{L}2LQ\left ( x,\tau \right ) +2LA^{\prime }\left ( \tau \right ) x+\left ( B^{\prime }\left ( \tau \right ) -A^{\prime }\left ( \tau \right ) \right ) x^{2}-2k\left ( B\left ( \tau \right ) -A\left ( \tau \right ) \right ) dx\right ) dt+\\ \sum _{n=1}^{\infty }\cos \frac {n\pi }{L}x\left ( e^{-\frac {n^{2}\pi ^{2}}{L^{2}}t}\int _{0}^{t}\left ( \frac {e^{\frac {n^{2}\pi ^{2}}{L^{2}}\tau }}{L^{2}}\int _{0}^{L}\left ( 2LQ\left ( x,\tau \right ) +2LA^{\prime }\left ( \tau \right ) x+\left ( B^{\prime }\left ( \tau \right ) -A^{\prime }\left ( \tau \right ) \right ) x^{2}-2k\left ( B\left ( \tau \right ) -A\left ( \tau \right ) \right ) \right ) \cos \left ( \frac {n\pi }{L}x\right ) dx\right ) dt+C_{n}e^{-\frac {n^{2}\pi ^{2}}{L^{2}}t}\right ) \end {gather*}
The constants \(C_{0},C_{n}\) are found from initial conditions \(u\left ( x,0\right ) =f\left ( x\right ) \).
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Added December 20, 2018.
Example 8.4.3 from Partial differential equations and boundary value problems with Maple by George A. Articolo, 2nd ed.
Solve the heat equation for \(u(x,t)\)
\[ u_t = k u_{xx} + t \]
For \(0<x<1\) and \(t>0\). The boundary conditions are \begin {align*} u(0,t)&=5 \\ u(1,t)+ \frac {\partial u}{\partial x}(1,t) &= 10\\ \end {align*}
Initial condition is \(u(x,0)=\frac {- 40 x^2}{3}+ \frac {45 x}{2}+5\) and \(k=\frac {1}{20}\)
Mathematica ✓
ClearAll["Global`*"]; k = 1/20; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}] + t; bc = {u[0, t] == 5, u[1, t] + Derivative[1, 0][u][1, t] == 10}; ic = u[x, 0] == (-40*x^2)/3 + (45*x)/2 + 5; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], x, t], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {5 x}{2}+\underset {K[1]=1}{\overset {\infty }{\sum }}-\frac {\sqrt {2} \left (\frac {40 e^{-\frac {1}{20} t K[2,K[1]]} \left (\cos \left (\sqrt {K[2,K[1]]}\right )-1\right ) \left (e^{\frac {1}{20} t K[2,K[1]]} (t K[2,K[1]]-20)+20\right )}{\sqrt {\cos \left (2 \sqrt {K[2,K[1]]}\right )+3} K[2,K[1]]^{5/2}}+\frac {40 e^{-\frac {1}{20} t K[2,K[1]]} \left (\cos \left (\sqrt {K[2,K[1]]}\right ) (K[2,K[1]]+4)+\sqrt {K[2,K[1]]} \sin \left (\sqrt {K[2,K[1]]}\right )-4\right )}{3 \sqrt {\cos \left (2 \sqrt {K[2,K[1]]}\right )+3} K[2,K[1]]^{3/2}}\right ) \sin \left (x \sqrt {K[2,K[1]]}\right )}{\sqrt {\cos ^2\left (\sqrt {K[2,K[1]]}\right )+1}}+5 & \tan \left (\sqrt {K[2,K[1]]}\right )+\sqrt {K[2,K[1]]}=0\land K[1]\in \mathbb {Z}\land K[1]\geq 1\land K[2,K[1]]>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x, t), t) = (1/20)*(diff(u(x, t), x$2))+t; bc := u(0, t) = 5, (u(1, t)+ eval( diff(u(x,t),x),x=1)) = 10; ic := u(x, 0) = -40*x^2/3+45*x/2+5; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde, bc,ic], u(x, t))),output='realtime'));
\[u \left (x , t\right ) = \frac {5 x}{2}+\int _{0}^{t}\left (\sum _{\mathit {n1} =0}^{\infty }\left \{\begin {array}{cc}0 & \lambda _{n}=0 \\\frac {4 \left (\cos \left (\lambda _{n}\right )-1\right ) \tau \,{\mathrm e}^{-\frac {\left (t -\tau \right ) \lambda _{n}^{2}}{20}} \sin \left (x \lambda _{n}\right )}{-2 \lambda _{n}+\sin \left (2 \lambda _{n}\right )} & \mathit {otherwise} \end {array}\right .\right )d \tau +\left (\sum _{n=0}^{\infty }\left \{\begin {array}{cc}0 & \lambda _{n}=0 \\\frac {80 \left (\lambda _{n}^{2} \cos \left (\lambda _{n}\right )+\lambda _{n} \sin \left (\lambda _{n}\right )+4 \cos \left (\lambda _{n}\right )-4\right ) {\mathrm e}^{-\frac {t \lambda _{n}^{2}}{20}} \sin \left (x \lambda _{n}\right )}{3 \left (-2 \lambda _{n}+\sin \left (2 \lambda _{n}\right )\right ) \lambda _{n}^{2}} & \mathit {otherwise} \end {array}\right .\right )+5\boldsymbol {\mathrm {where}}\left \{\lambda _{n}+\tan \left (\lambda _{n}\right )=0\wedge 0<\lambda _{n}\right \}\]
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