4.1.6 Semi-infinite domain

4.1.6.1 [235] left end constant (general case)

problem number 235

Added July 6, 2019 Solve the heat equation for $x>0,t>0$ $$u_t=k{u}_{xx}$$ The boundary conditions are $u(0,t)=A$ and initial conditions $u(x,0)=0$

Figure 4.116: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == k*D[u[x, t], {x, 2}]; 
bc  = u[0, t] == A; 
ic  = u[x, 0] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}, Assumptions -> {k>0,x > 0,t>0}], 60*10]];
 

$$\left\{\{u(x,t)\to \frac{Ax{E}_{\frac{1}{2}}\left(\frac{x^2}{4kt}\right)}{2\sqrt{\pi }\sqrt{kt}}\}\right\}$$

Maple ✓

restart; 
interface(showassumed=0); 
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2); 
ic  := u(x,0)=0; 
bc  := u(0,t)=A; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t)) assuming k>0,x>0,t>0),output='realtime'));
 

$$u(x,t)=-A\mathrm{erf}\left(\frac{x}{2\sqrt{k}\sqrt{t}}\right)+A$$

Hand solution

Solving $$\begin{array}{rl}\text{(1)}& u_t& =k{u}_{xx}t>0,x>0u(0,t)& =A\\ u(x,0)& =0\end{array}$$

And $u(x,t)<\mathrm{\infty }$ as $x\to \mathrm{\infty }$. This means $u(x,t)$ is bounded. This conditions is always needed to solve these problems.

Let $U(x,s)$ be the Laplace transform of $u(x,t)$. Defined as $$\mathcal{L}(u,t)={\int }_0^{\mathrm{\infty }}{e}^{-st}u(x,t)dt$$ Applying Laplace transform to the original PDE (1) gives$$sU(x,s)-u(x,0)=k{U}_{xx}(x,s)$$ But $u(x,0)=0$, therefore the above becomes$$U_{xx}-\frac{s}{k}U=0$$ The solution to this differential equation is$$U(x,s)=c_1{e}^{\sqrt{\frac{s}{k}}x}+c_2{e}^{-\sqrt{\frac{s}{k}}x}$$ Since $u(x,t)$ is bounded in the limit as $x\to \mathrm{\infty }$ and $k>0$, therefore it must be that $c_1=0$ to keep the solution bounded. The above simplifies to$$\begin{array}{}\text{(2)}& U(x,s)=c_2{e}^{-\sqrt{\frac{s}{k}}x}\end{array}$$ At $x=0,u(0,t)=A$. Therefore $U(0,s)=\mathcal{L}\left(u(0,t)\right)=\mathcal{L}\left(A\right)=\frac{1}{s}A$. Hence at $x=0$ the above gives$$\frac{1}{s}A=c_2$$ Therefore (2) becomes$$\begin{array}{}\text{(3)}& U(x,s)=\frac{A}{s}{e}^{-\sqrt{\frac{s}{k}}x}\end{array}$$ From tables, the inverse Laplace transform of the above is (since $x>0,k>0$)$$\begin{array}{rl}u(x,t)& =A\mathrm{erfc}\left(\frac{x}{2\sqrt{kt}}\right)\\ & =A(1-\mathrm{erf}\left(\frac{x}{2\sqrt{kt}}\right))\end{array}$$

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4.1.6.2 [236] left end constant (special case)

problem number 236

Added July 6, 2019 Solve the heat equation for $x>0,t>0$ $$u_t=k{u}_{xx}$$ The boundary conditions are $u(0,t)=A$ and initial conditions $u(x,0)=0$, using $$\begin{array}{rl}A& =60\\ k& =\frac{1}{10}\end{array}$$

Figure 4.117: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
k=1/10; A=60; 
pde =  D[u[x, t], t] == k*D[u[x, t], {x, 2}]; 
bc  = u[0, t] == A; 
ic  = u[x, 0] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
 

$$\{\{u(x,t)\to \boxed{\text{Indeterminate}\text{ if }x\le 0}\}\}$$ It fail if assumption $x>0$ is given. A bug

Maple ✓

restart; 
k:=1/10; 
A:=60; 
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2); 
ic  := u(x,0)=0; 
bc  := u(0,t)=A; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t)) assuming x>0),output='realtime'));
 

$$u(x,t)=-60\mathrm{erf}\left(\frac{\sqrt{10}x}{2\sqrt{t}}\right)+60$$

Hand solution

Solving on semi-infinite domain $$\begin{array}{rl}{u}_t& =k{u}_{xx}t>0,x>0\\ u(0,t)& =A\\ u(x,0)& =0\end{array}$$

With $A=60,k=\frac{1}{10}$

The general problem above was solved in 4.1.6.1 on page 791 and the solution is$$u(x,t)=A(1-\mathrm{erf}\left(\frac{x}{2\sqrt{kt}}\right))$$ Substituting the specific values given above into this solution gives$$u(x,t)=60(1-\mathrm{erf}\left(\frac{x}{2\sqrt{\frac{t}{10}}}\right))$$ Animation is below

Source code used for the above

Figure 4.118: Source code

Figure 4.119: Code for animation

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4.1.6.3 [237] Logan p. 76. Left end general function of time (general case)

problem number 237

This is problem at page 76 from David J Logan text book.

Solve the heat equation for $x>0,t>0$ $$u_t=k{u}_{xx}$$ The boundary conditions are $u(0,t)=f(t)$ and initial conditions $u(x,0)=0$

Figure 4.120: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == k*D[u[x, t], {x, 2}]; 
bc  = u[0, t] == f[t]; 
ic  = u[x, 0] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}, Assumptions -> {t > 0, x > 0,k>0}], 60*10]]; 
sol =  sol /. {K[2] -> z}
 

$$\left\{\{u(x,t)\to \frac{x\text{Integrate}[\frac{f(z)e^{-\frac{x^2}{4kt-4kz}}}{(t-z{)}^{3/2}},\{z,0,t\},\text{Assumptions}\to \text{True}]}{2\sqrt{\pi }\sqrt{k}}\}\right\}$$

Maple ✓

restart; 
interface(showassumed=0); 
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2); 
ic  := u(x,0)=0; 
bc  := u(0,t)=f(t); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t)) assuming t>0,x>0,k>0),output='realtime'));
 

$$u(x,t)=\frac{x\left({\int }_0^t\frac{{\mathrm{e}}^{\frac{x^2}{4(-t+\zeta )k}}f\left(\zeta \right)}{{(t-\zeta )}^{\frac{3}{2}}}d\zeta \right)}{2\sqrt{\pi }\sqrt{k}}$$

Hand solution

Solving on semi-infinite domain $$\begin{array}{rl}\text{(1)}& u_t& =k{u}_{xx}t>0,x>0u(0,t)& =f\left(t\right)\\ u(x,0)& =0\end{array}$$

With $k>0$ and $u(x,t)<\mathrm{\infty }$ as $x\to \mathrm{\infty }$. This means $u(x,t)$ is bounded. This conditions is always needed to solve these problems.

Let $U(x,s)$ be the Laplace transform of $u(x,t)$. Defined as $$\mathcal{L}(u,t)={\int }_0^{\mathrm{\infty }}{e}^{-st}u(x,t)dt$$ Applying Laplace transform to the original PDE (1) gives$$sU(x,s)-u(x,0)=k{U}_{xx}(x,s)$$ But $u(x,0)=0$, therefore the above becomes$$U_{xx}-\frac{s}{k}U=0$$ The solution to this differential equation is$$U(x,s)=c_1{e}^{\sqrt{\frac{s}{k}}x}+c_2{e}^{-\sqrt{\frac{s}{k}}x}$$ Since $u(x,t)$ is bounded in the limit as $x\to \mathrm{\infty }$ and $k>0$, therefore it must be that $c_1=0$ to keep the solution bounded. The above simplifies to$$\begin{array}{}\text{(2)}& U(x,s)=c_2{e}^{-\sqrt{\frac{s}{k}}x}\end{array}$$ At $x=0,u(0,t)=f\left(t\right)$. Therefore $U(0,s)=\mathcal{L}\left(f\left(t\right)\right)=F\left(s\right)$. Hence at $x=0$ the above gives$$F\left(s\right)=c_2$$ Therefore (2) becomes$$\begin{array}{}\text{(3)}& U(x,s)=F\left(s\right)e^{-\sqrt{\frac{s}{k}}x}\end{array}$$ By convolution, the above becomes$$\begin{array}{}\text{(4)}& u(x,t)=f\left(t\right)⊛G(x,t)\end{array}$$ Where $G(x,t)$ is the inverse transform of $e^{-\sqrt{\frac{s}{k}}x}$ which is $\frac{x{e}^{\frac{-x^2}{4kt}}}{2\sqrt{k\pi }{t}^{\frac{3}{2}}}$. Hence (4) becomes$$\begin{array}{rl}u(x,t)& =f\left(t\right)⊛\frac{x{e}^{\frac{-x^2}{4kt}}}{2\sqrt{k\pi }{t}^{\frac{3}{2}}}\\ & =\frac{x}{2\sqrt{k\pi }}{\int }_0^t\frac{f\left(\tau \right)}{{(t-\tau )}^{\frac{3}{2}}}{e}^{\frac{-x^2}{4k(t-\tau )}}d\tau \end{array}$$

For $k=1$$$u(x,t)=\frac{x}{2\sqrt{\pi }}{\int }_0^t\frac{f\left(\tau \right)}{{(t-\tau )}^{\frac{3}{2}}}{e}^{\frac{-x^2}{4(t-\tau )}}d\tau$$

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4.1.6.4 [238] Left end function of time (special case)

problem number 238

Added July 7, 2019 Solve the heat equation for $x>0,t>0$ $$u_t=k{u}_{xx}$$ The boundary conditions are $u(0,t)=\sin (t)$ and initial conditions $u(x,0)=0$ using $k=\frac{1}{10}$

Figure 4.121: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
k=1/10; 
pde =  D[u[x, t], t] == k*D[u[x, t], {x, 2}]; 
bc  = u[0, t] == Sin[t]; 
ic  = u[x, 0] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}, Assumptions -> {t > 0, x > 0}], 60*10]];
 

$$\left\{\{u(x,t)\to \sqrt{\frac{5}{2\pi }}x\text{Integrate}[\frac{\sin (K[2])e^{-\frac{5x^2}{2(t-K[2])}}}{(t-K[2]{)}^{3/2}},\{K[2],0,t\},\text{Assumptions}\to \text{True}]\}\right\}$$

Maple ✓

restart; 
interface(showassumed=0); 
k:=1/10; 
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2); 
ic  := u(x,0)=0; 
bc  := u(0,t)=sin(t); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t)) assuming t>0,x>0),output='realtime'));
 

$$u(x,t)=-\frac{\sqrt{10}x({\int }_0^t-\frac{{\mathrm{e}}^{-\frac{5x^2}{2\zeta }}\sin (t-\zeta )}{{\zeta }^{\frac{3}{2}}}d\zeta )}{2\sqrt{\pi }}$$

Hand solution

Solving

$$\begin{array}{rl}\text{(1)}& u_t& =k{u}_{xx}t>0,x>0u(0,t)& =f\left(t\right)\\ u(x,0)& =0\end{array}$$

Using $k=\frac{1}{10}$ and $f\left(t\right)=\sin \left(t\right)$.

The general solution was solved in problem 4.1.6.3 on page 803 and the solution was found to be$$u(x,t)=\frac{x}{2\sqrt{k\pi }}{\int }_0^t\frac{f\left(\tau \right)}{{(t-\tau )}^{\frac{3}{2}}}{e}^{\frac{-x^2}{4k(t-\tau )}}d\tau$$ Replacing the given values above, the solution becomes$$u(x,t)=\sqrt{\frac{5}{2}}\frac{x}{\sqrt{\pi }}{\int }_0^t\frac{\sin \left(\tau \right)}{{(t-\tau )}^{\frac{3}{2}}}{e}^{\frac{-5x^2}{2(t-\tau )}}d\tau$$ We could also use the following form of the solution$$u(x,t)=\sqrt{\frac{5}{2}}\frac{x}{\sqrt{\pi }}{\int }_0^t\frac{\sin (t-\tau )}{{\tau }^{\frac{3}{2}}}{e}^{\frac{-5x^2}{2\tau }}d\tau$$ Animation is below

Source code used for the above

Figure 4.122: Source code

Figure 4.123: Code for animation

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4.1.6.5 [239] nonhomogeneous BC

problem number 239

Solve the heat equation $$\frac{\partial u}{\partial t}=k\frac{{\partial }^{2}u}{\partial x^2}$$ For $x>0$ and $t>0$. The boundary conditions is $u(0,t)=1$ and And initial condition $u(x,0)=0$

Figure 4.124: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == k*D[u[x, t], {x, 2}]; 
bc  = u[0, t] == 1; 
ic  = u[x, 0] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}, Assumptions -> {t > 0, k > 0, x > 0}], 60*10]];
 

$$\left\{\{u(x,t)\to \frac{x{E}_{\frac{1}{2}}\left(\frac{x^2}{4kt}\right)}{2\sqrt{\pi }\sqrt{kt}}\}\right\}$$

Maple ✓

restart; 
interface(showassumed=0); 
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2); 
ic  := u(x,0)=0: 
bc  := u(0,t)=1; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t),HINT = boundedseries) assuming t>0,x>0,k>0),output='realtime'));
 

$$u(x,t)=\mathrm{erfc}\left(\frac{x}{2\sqrt{kt}}\right)$$

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4.1.6.6 [240] I.C. not zero

problem number 240

Added December 20, 2018.

From https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Solve the heat equation for $u(x,t)$ $$\frac{\partial u}{\partial t}=\frac{1}{4}\frac{{\partial }^{2}u}{\partial x^2}$$ With initial condition $$u(x,t_0)=10;$$ And boundary conditions $$u(-x_0,t)=0$$ For $x>|x_0|$ and $t>|t_0|$.

Figure 4.125: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == (1/4)*D[u[x, t], {x, 2}]; 
bc  = u[-x0, t] == 0; 
ic  = u[x, t0] == 10; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], x, t, Assumptions -> {t > Abs[t0], x > Abs[x0]}], 60*10]];
 

$$\left\{\{u(x,t)\to \begin{array}{cc}\{& \begin{array}{cc}10\text{erf}\left(\frac{x+\text{x0}}{\sqrt{t-\text{t0}}}\right)& x+\text{x0}>0\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$ due to IC/BC not zero

Maple ✓

restart; 
pde := diff(u(x, t), t) = (1/4)*(diff(u(x, t), x$2)); 
bc  := u(-x0, t) = 0; 
ic  := u(x, t0) = 10; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde, bc,ic],u(x,t)) assuming x>abs(x0), t>abs(t0)),output='realtime'));
 

$$u(x,t)=10\mathrm{erf}\left(\frac{x+\mathit{x}\mathit{0}}{\sqrt{t-\mathit{t}\mathit{0}}}\right)$$

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4.1.6.7 [241] nonhomogeneous BC

problem number 241

Solve the heat equation

$$\frac{\partial u}{\partial t}=k\frac{{\partial }^{2}u}{\partial x^2}$$ For $x>0$ and $t>0$. The boundary conditions is $u(0,t)=\mu$ and And initial condition $u(x,0)=\lambda$

Figure 4.126: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == k*D[u[x, t], {x, 2}]; 
bc  = u[0, t] == lambda; 
ic  = u[x, 0] == mu; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}, Assumptions -> {t > 0, k > 0, x > 0}], 60*10]];
 

$$\left\{\{u(x,t)\to \mu \text{erf}\left(\frac{x}{2\sqrt{kt}}\right)+\frac{\lambda x{E}_{\frac{1}{2}}\left(\frac{x^2}{4kt}\right)}{2\sqrt{\pi }\sqrt{kt}}\}\right\}$$

Maple ✓

restart; 
interface(showassumed=0); 
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2); 
ic  := u(x,0)=mu: 
bc  := u(0,t)=lambda; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t),HINT = boundedseries) assuming t>0,x>0,k>0),output='realtime'));
 

$$u(x,t)=\mu +(\lambda -\mu )\mathrm{erfc}\left(\frac{x}{2\sqrt{kt}}\right)$$

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4.1.6.8 [242] nonhomogeneous BC

problem number 242

From Mathematica DSolve help pages. Solve the heat equation for $u(x,t)$ on half the line $x>0$ and $t>0$ $$\frac{\partial u}{\partial t}=\frac{{\partial }^{2}u}{\partial x^2}$$ With initial condition $$u(x,0)=\cos x$$ And boundary conditions $$u(0,t)=1$$

Figure 4.127: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == D[u[x, t], {x, 2}]; 
ic  = u[x, 0] == Cos[x]; 
bc  = u[0, t] == 1; 
sol =  AbsoluteTiming[TimeConstrained[FullSimplify[DSolve[{pde, ic, bc}, u[x, t], {x, t}]], 60*10]];
 

$$\left\{\{u(x,t)\to \begin{array}{cc}\{& \begin{array}{cc}\frac{\text{Integrate}[(e^{-\frac{(x-K[1]{)}^2}{4t}}-e^{-\frac{(x+K[1]{)}^2}{4t}})\cos (K[1]),\{K[1],0,\mathrm{\infty }\},\text{Assumptions}\to \text{True}]+\sqrt{t}x\text{Integrate}[\frac{e^{-\frac{x^2}{4(t-K[2])}}}{(t-K[2]{)}^{3/2}},\{K[2],0,t\},\text{Assumptions}\to \text{True}]}{2\sqrt{\pi }\sqrt{t}}& x>0\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✓

restart; 
pde := diff(u(x, t), t)=diff(u(x, t), x$2); 
ic  := u(x,0)=cos(x); 
bc  := u(0,t)=1; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t)) assuming t>0 and x>0),output='realtime'));
 

$$u(x,t)=-\frac{\mathrm{erf}\left(\frac{2it-x}{2\sqrt{t}}\right){\mathrm{e}}^{-ix-t}}{2}+\frac{\mathrm{erf}\left(\frac{2it+x}{2\sqrt{t}}\right){\mathrm{e}}^{ix-t}}{2}-\mathrm{erf}\left(\frac{x}{2\sqrt{t}}\right)+1$$

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4.1.6.9 [243] nonhomogeneous B.C.

problem number 243

Solve the heat equation for $u(x,t)$ on half the line $x>0$ and $t>0$ $$u_t=k{u}_{xx}$$ With initial condition $$u(x,0)=0$$ And boundary conditions $u(0,t)=t$. Solution is bounded at infinity.

Figure 4.128: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == k*D[u[x, t], {x, 2}]; 
ic  = u[x, 0] == 0; 
bc  = u[0, t] == t; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}, Assumptions -> {k > 0, x > 0, t > 0}], 60*10]];
 

$$\left\{\{u(x,t)\to \frac{(2kt+x^2)\mathrm{Erfc}\left(\frac{x}{2\sqrt{kt}}\right)-\frac{2x\sqrt{kt}{e}^{-\frac{x^2}{4kt}}}{\sqrt{\pi }}}{2k}\}\right\}$$

Maple ✓

restart; 
interface(showassumed=0); 
pde := diff(u(x, t), t)=k*diff(u(x, t), x$2); 
ic  := u(x,0)=0; 
bc  := u(0,t)=t; 
assume(x>0); 
assume(t>0); 
assume(k>0); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t))),output='realtime'));
 

$$u(x,t)=-\frac{2\sqrt{k}tx{\mathrm{e}}^{-\frac{x^2}{4kt}}+\sqrt{\pi }(\mathrm{erf}\left(\frac{x}{2\sqrt{k}\sqrt{t}}\right)-1)(2k{t}^{\frac{3}{2}}+\sqrt{t}{x}^2)}{2\sqrt{\pi }k\sqrt{t}}$$

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4.1.6.10 [244] Unit triangle I.C.

problem number 244

From Mathematica DSolve help pages. Solve the heat equation for $u(x,t)$ on half the line $x>0$ and $t>0$ $$u_t=u_{xx}$$ With initial condition $$u(x,0)=\text{UnitTriagle[x-3]}$$ And boundary conditions $$\frac{\partial u}{\partial x}(0,t)=0$$

Figure 4.129: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == D[u[x, t], {x, 2}]; 
ic  = u[x, 0] == UnitTriangle[x - 3]; 
bc  = Derivative[1, 0][u][0, t] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]];
 

$$\left\{\{u(x,t)\to \begin{array}{cc}\{& \begin{array}{cc}\frac{1}{2}(\frac{\text{erf}\left(\frac{|x-4|}{2\sqrt{t}}\right)(x-4{)}^2}{|4-x|}+(x+2)\text{erf}\left(\frac{x+2}{2\sqrt{t}}\right)-2(x+3)\text{erf}\left(\frac{x+3}{2\sqrt{t}}\right)+(x+4)\text{erf}\left(\frac{x+4}{2\sqrt{t}}\right)-\frac{2(x-3{)}^2\text{erf}\left(\frac{|x-3|}{2\sqrt{t}}\right)}{|3-x|}+\frac{(x-2{)}^2\text{erf}\left(\frac{|x-2|}{2\sqrt{t}}\right)}{|2-x|}+\frac{2(e^{-\frac{(x-4{)}^2}{4t}}-2e^{-\frac{(x-3{)}^2}{4t}}+e^{-\frac{(x-2{)}^2}{4t}}+e^{-\frac{(x+2{)}^2}{4t}}-2e^{-\frac{(x+3{)}^2}{4t}}+e^{-\frac{(x+4{)}^2}{4t}})\sqrt{t}}{\sqrt{\pi }})& x>0\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✓

restart; 
pde := diff(u(x, t), t)=diff(u(x, t), x$2); 
ic  := u(x,0)=piecewise( x>2 and x<3,-2+x, x>3 and x<4, 4-x, 0); 
bc:=(D[1](u))(0,t)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t)) assuming t>0 and x>0),output='realtime'));
 

$$u(x,t)=\frac{t{\mathrm{e}}^{-\frac{{(x-4)}^2}{4t}}-2t{\mathrm{e}}^{-\frac{{(x-3)}^2}{4t}}+t{\mathrm{e}}^{-\frac{{(x-2)}^2}{4t}}+t{\mathrm{e}}^{-\frac{{(x+2)}^2}{4t}}-2t{\mathrm{e}}^{-\frac{{(x+3)}^2}{4t}}+t{\mathrm{e}}^{-\frac{{(x+4)}^2}{4t}}+\frac{((x-4)\mathrm{erf}\left(\frac{x-4}{2\sqrt{t}}\right)+(-2x+6)\mathrm{erf}\left(\frac{x-3}{2\sqrt{t}}\right)+(x-2)\mathrm{erf}\left(\frac{x-2}{2\sqrt{t}}\right)+(x+2)\mathrm{erf}\left(\frac{x+2}{2\sqrt{t}}\right)+(-2x-6)\mathrm{erf}\left(\frac{x+3}{2\sqrt{t}}\right)+(x+4)\mathrm{erf}\left(\frac{x+4}{2\sqrt{t}}\right))\sqrt{\pi }\sqrt{t}}{2}}{\sqrt{\pi }\sqrt{t}}$$

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4.1.6.11 [245] I.C. not at $t=0$

problem number 245

Added December 20, 2018.

Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Solve for $u(x,t)$ for $t>0,x>0$ $$u_t=\frac{1}{4}{u}_{xx}$$ With initial condition $$u(x,t_0)=10e^{-x^2}$$ And boundary conditions $$\frac{\partial u}{\partial x}(x_0,t)=0$$

Figure 4.130: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == (1*D[u[x, t], {x, 2}])/4; 
ic  = u[x, t0] == 10*Exp[-x^2]; 
bc  = Derivative[1, 0][u][x0, t] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}, Assumptions -> {x > 0, t > 0}], 60*10]];
 

$$\left\{\{u(x,t)\to \begin{array}{cc}\{& \begin{array}{cc}\frac{5(e^{-\frac{x^2}{t-\text{t0}+1}}(\text{erf}\left(\frac{x+(-t+\text{t0}-1)\text{x0}}{\sqrt{t^2-2\text{t0}t+t+(\text{t0}-1)\text{t0}}}\right)+1)+e^{-\frac{(x-2\text{x0}{)}^2}{t-\text{t0}+1}}\mathrm{Erfc}\left(\frac{x+(t-\text{t0}-1)\text{x0}}{\sqrt{(t-\text{t0})(t-\text{t0}+1)}}\right))}{\sqrt{t-\text{t0}+1}}& x>\text{x0}\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✓

restart; 
pde := diff(u(x, t), t) = 1/4*(diff(u(x, t), x$2)); 
bc  := eval( diff(u(x,t),x),x=x0)=0; 
ic  := u(x,t0)=10*exp(-x^2); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc,ic],u(x,t)) assuming x>0,t>0),output='realtime'));
 

$$u(x,t)=-\frac{5(-\mathrm{erf}\left(\frac{x+(-t+\mathit{t}\mathit{0}-1)\mathit{x}\mathit{0}}{\sqrt{t-\mathit{t}\mathit{0}}\sqrt{t-\mathit{t}\mathit{0}+1}}\right){\mathrm{e}}^{\frac{x^2}{-t+\mathit{t}\mathit{0}-1}}-{\mathrm{e}}^{\frac{x^2}{-t+\mathit{t}\mathit{0}-1}}+(\mathrm{erf}\left(\frac{-\mathit{t}\mathit{0}\mathit{x}\mathit{0}+x+(t-1)\mathit{x}\mathit{0}}{\sqrt{t-\mathit{t}\mathit{0}}\sqrt{t-\mathit{t}\mathit{0}+1}}\right)-1){\mathrm{e}}^{-\frac{{(x-2\mathit{x}\mathit{0})}^2}{t-\mathit{t}\mathit{0}+1}})}{\sqrt{t-\mathit{t}\mathit{0}+1}}$$

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4.1.6.12 [246] Diffusion with advection

problem number 246

Added April 5, 2019.

Solve for $u(x,t)$ in $$u_t=u_{xx}-u_x$$

For $t>0,x>0$. With boundary conditions $u(0,t)=0$ and intitial conditions $u(x,0)=f(x)$

Figure 4.131: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde = D[u[x, t], t] == D[u[x, t], {x, 2}] - D[u[x, t], x]; 
ic  = u[x, 0] == f[x]; 
bc  = u[0, t] == 0; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}, Assumptions -> {x > 0, t > 0}], 60*10]];
 

$$\left\{\{u(x,t)\to e^{\frac{x}{2}-\frac{t}{4}}\left(\begin{array}{cc}\{& \begin{array}{cc}\frac{\text{Integrate}[e^{-\frac{K[1]}{2}}(e^{-\frac{(x-K[1]{)}^2}{4t}}-e^{-\frac{(x+K[1]{)}^2}{4t}})f(K[1]),\{K[1],0,\mathrm{\infty }\},\text{Assumptions}\to \text{True}]}{2\sqrt{\pi }\sqrt{t}}& x>0\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\right)\}\right\}$$

Maple ✓

restart; 
pde := diff(u(x,t),t)=diff(u(x,t),x$2)- diff(u(x,t),x); 
ic  := u(x,0)=f(x); 
bc  := u(0,t)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde, ic, bc], u(x, t))assuming t>0,x>0),output='realtime'));
 

$$u(x,t)=({\mathcal{L}}^{-1}(\frac{\int \frac{\sqrt{{\mathrm{e}}^{\sqrt{4s+1}x}}f\left(x\right)}{\sqrt{{\mathrm{e}}^x}}dx}{\sqrt{4s+1}\sqrt{{\mathrm{e}}^{\sqrt{4s+1}x}}},s,t)-{\mathcal{L}}^{-1}(\frac{(\int \frac{f\left(x\right)}{\sqrt{{\mathrm{e}}^x}\sqrt{{\mathrm{e}}^{\sqrt{4s+1}x}}}dx)\sqrt{{\mathrm{e}}^{\sqrt{4s+1}x}}}{\sqrt{4s+1}},s,t)+{\mathcal{L}}^{-1}(\frac{{\int }_{}^0\frac{f\left(\mathit{\text{\_a}}\right)}{\sqrt{{\mathrm{e}}^{\mathit{\text{\_a}}}}\sqrt{{\mathrm{e}}^{\sqrt{4s+1}\mathit{\text{\_a}}}}}d\mathit{\text{\_a}}}{\sqrt{4s+1}\sqrt{{\mathrm{e}}^{\sqrt{4s+1}x}}},s,t)-{\mathcal{L}}^{-1}(\frac{{\int }_{}^0\frac{\sqrt{{\mathrm{e}}^{\sqrt{4s+1}\mathit{\text{\_a}}}}f\left(\mathit{\text{\_a}}\right)}{\sqrt{{\mathrm{e}}^{\mathit{\text{\_a}}}}}d\mathit{\text{\_a}}}{\sqrt{4s+1}\sqrt{{\mathrm{e}}^{\sqrt{4s+1}x}}},s,t)){\mathrm{e}}^{\frac{x}{2}}$$

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4.1.6.13 [247] Practice exam problem

problem number 247

Added May 23, 2019.

From Math 5587 midterm I, Fall 2016, practice exam, problem 13.

Solve for $u(x,t)$ with IC $u(x,0)=x^2+1$ and BC $u_t(0,t)=1$ for $x>0,t>0$ $$\begin{array}{r}{u}_t=u_{xx}\end{array}$$

Figure 4.132: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde = D[u[x, t], t] == D[u[x,t],{x,2}]; 
ic  = u[x,0]==x^2+1; 
bc  = u[0,t]==1; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{pde,ic,bc}, u[x, t], {x, t}], 60*10]];
 

$$\left\{\{u(x,t)\to \begin{array}{cc}\{& \begin{array}{cc}\frac{\text{Integrate}[(e^{-\frac{(x-K[1]{)}^2}{4t}}-e^{-\frac{(x+K[1]{)}^2}{4t}})(K[1{]}^2+1),\{K[1],0,\mathrm{\infty }\},\text{Assumptions}\to \text{True}]+\sqrt{t}x\text{Integrate}[\frac{e^{-\frac{x^2}{4(t-K[2])}}}{(t-K[2]{)}^{3/2}},\{K[2],0,t\},\text{Assumptions}\to \text{True}]}{2\sqrt{\pi }\sqrt{t}}& x>0\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✓

restart; 
pde := diff(u(x,t),t)= diff(u(x,t),x$2); 
ic  := u(x,0)=x^2+1; 
bc :=u(0,t)=1; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t))),output='realtime'));
 

$$u(x,t)=x^2+2t-2{\mathcal{L}}^{-1}(\frac{{\mathrm{e}}^{-\sqrt{s}x}}{s^2},s,t)+1$$

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