5.1.2 Semi-infinite domain

5.1.2.1 [364] Left end fixed, (general case)

problem number 364

Added July 12, 2019 Solve for \(u(x,t)\) with \(t>0\) and \(x>0\) \[ u_{tt} = c^2 u_{xx} \] With boundary conditions \begin {align*} u(0,t) &= 0 \end {align*}

With initial conditions \begin {align*} u_t(x,0) &=0 \\ u(x,0) &= f(x) \end {align*}

Figure 5.47: PDE specification

Mathematica ✓

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

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {1}{2} (f(x-c t)+f(c t+x)) & x>c t \\ \frac {1}{2} (f(c t+x)-f(c t-x)) & x\leq c t \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✗

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

sol=()

Hand solution

Solving on semi-infinite domain for \(u(x,t)\)\begin {equation} \frac {\partial ^{2}u}{\partial t^{2}}=c^{2}\frac {\partial ^{2}u}{\partial x^{2}}\qquad 0<x<\infty ,t>0\tag {1} \end {equation} With BC \begin {align*} u\left ( 0,t\right ) & =0\\ u\left ( \infty ,t\right ) & <\infty \end {align*}

And initial conditions\begin {align*} u\left ( x,0\right ) & =f\left ( x\right ) \\ u_{t}\left ( x,0\right ) & =0 \end {align*}

Seperation of variables method

Let \(u\left ( x,t\right ) =X\left ( x\right ) T\left ( t\right ) \). The PDE in (1) becomes\[ \frac {T^{\prime \prime }}{c^{2}T}=\frac {X^{\prime \prime }}{X}=-\lambda \] Hence \begin {align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( \infty \right ) & <\infty \end {align*}

It is clear that \(\lambda \) can not be negative because it gives a solution that blows up. For \(\lambda =0\), the solution is \(X\left ( x\right ) =Ax+B\) and because \(X\left ( 0\right ) =0\) this implies \(B=0\). Hence solution is \(X\left ( x\right ) =Ax\). And this blows up as \(x\) increases unless \(A=0\). Hence \(\lambda =0\) is not valid eigenvalue. Therefore \(\lambda >0\). Let \(\lambda =\alpha ^{2},\alpha >0\) and the solution becomes\[ X\left ( x\right ) =A_{\alpha }\cos \left ( \alpha x\right ) +B_{\alpha }\sin \left ( \alpha x\right ) \] At \(x=0\) the above gives\[ 0=A_{\alpha }\] Therefore the solution becomes\begin {equation} X_{\alpha }\left ( x\right ) =B_{\alpha }\sin \left ( \alpha x\right ) \qquad \alpha >0\tag {1} \end {equation} The time domain ODE becomes\begin {align*} T^{\prime \prime }+c^{2}\alpha ^{2}T & =0\\ T & =C_{\alpha }\cos \left ( \alpha ct\right ) +D_{\alpha }\sin \left ( \alpha ct\right ) \\ T^{\prime }\left ( t\right ) & =-c\alpha C_{\alpha }\sin \left ( \alpha ct\right ) +c\alpha D_{\alpha }\cos \left ( \alpha ct\right ) \end {align*}

And at \(t=0,T^{\prime }\left ( 0\right ) =0\), hence the above becomes\[ 0=c\alpha D_{\alpha }\] Which means \(D_{\alpha }=0\). Therefore\begin {equation} T=C_{\alpha }\cos \left ( \alpha ct\right ) \qquad \alpha >0\tag {2} \end {equation} From (1,2) the complete solution is therefore\begin {equation} u\left ( x,t\right ) =\int _{0}^{\infty }A_{\alpha }\cos \left ( \alpha ct\right ) \sin \left ( \alpha x\right ) d\alpha \tag {3} \end {equation} Where \(A_{\alpha },C_{\alpha }\) are merged into one constant. Now the last initial condition is applied, which is \(u\left ( x,0\right ) =f\left ( x\right ) \) to the above which gives\[ f\left ( x\right ) =\int _{0}^{\infty }A_{\alpha }\sin \left ( \alpha x\right ) d\alpha \] Hence\[ A_{\alpha }=\frac {2}{\pi }\int _{0}^{\infty }f\left ( x\right ) \sin \left ( \alpha x\right ) dx \] Using the above in (3) gives the final solution as\[ u\left ( x,t\right ) =\frac {2}{\pi }\int _{0}^{\infty }\left ( \int _{0}^{\infty }f\left ( s\right ) \sin \left ( \alpha s\right ) ds\right ) \cos \left ( \alpha ct\right ) \sin \left ( \alpha x\right ) d\alpha \] D’Alambert’s formula method

For the half line, the D’Alambert’s is given by, using \(v_{0}\left ( x\right ) =u_{t}\left ( x,0\right ) \) as the initial velocity\[ u\left ( x,t\right ) =\left \{ \begin {array} [c]{ccc}\frac {1}{2}\left ( f\left ( x+ct\right ) +f\left ( x-ct\right ) \right ) +\frac {1}{2c}\int _{x-ct}^{x+ct}v_{0}\left ( s\right ) ds & & x>ct\geq 0\\ & & \\ \frac {1}{2}\left ( f\left ( x+ct\right ) -f\left ( ct-x\right ) \right ) +\frac {1}{2c}\int _{ct-x}^{x+ct}v_{0}\left ( s\right ) ds & & x\leq ct \end {array} \right . \] Hence, since \(u_{t}\left ( x,0\right ) =0\) in this problem\[ u\left ( x,t\right ) =\left \{ \begin {array} [c]{ccc}\frac {1}{2}\left ( f\left ( x+ct\right ) +f\left ( x-ct\right ) \right ) & & x>ct\geq 0\\ & & \\ \frac {1}{2}\left ( f\left ( x+ct\right ) -f\left ( ct-x\right ) \right ) & & x\leq ct \end {array} \right . \]

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5.1.2.2 [365] Left end fixed with specific initial position

problem number 365

Taken from Mathematica DSolve help pages.

Solve for \(u(x,t)\) initial value wave PDE on infinite domain with \(t>0\) and \(x>0\). \[ u_{tt} = c^2 u_{xx} \] With initial conditions \begin {align*} u(x,0) &= \sin ^2(x) \hspace {20pt} \pi <x< 2\pi \\ \frac {\partial u}{\partial t}(x,0) &= 0 \end {align*}

And boundary conditions \(u(0,t)=0\)

Figure 5.48: PDE specification

Mathematica ✓

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

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {1}{2} \left (\left (\begin {array}{cc} \{ & \begin {array}{cc} \sin ^2(c t-x) & \pi <x-c t<2 \pi \\ 0 & \text {True} \\\end {array} \\\end {array}\right )+\left (\begin {array}{cc} \{ & \begin {array}{cc} \sin ^2(c t+x) & \pi <c t+x<2 \pi \\ 0 & \text {True} \\\end {array} \\\end {array}\right )\right ) & x>c t\geq 0 \\ \frac {1}{2} \left (\left (\begin {array}{cc} \{ & \begin {array}{cc} \sin ^2(c t+x) & \pi <c t+x<2 \pi \\ 0 & \text {True} \\\end {array} \\\end {array}\right )-\left (\begin {array}{cc} \{ & \begin {array}{cc} \sin ^2(c t-x) & \pi <c t-x<2 \pi \\ 0 & \text {True} \\\end {array} \\\end {array}\right )\right ) & 0\leq x\leq c t \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

restart; 
pde := diff(u(x, t), t$2) = c^2 * diff(u(x, t), x$2); 
ic  := u(x,0)= piecewise(Pi<x and x<2*Pi,sin(x)^2),(D[2](u))(x,0)=0; 
bc  := 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'));
 

\[u \left (x , t\right ) = \frac {\left (\left \{\begin {array}{cc}-\left (\left \{\begin {array}{cc}0 & c t -x \le \pi \\\sin ^{2}\left (c t -x \right ) & c t -x <2 \pi \\0 & 2 \pi \le c t -x \end {array}\right .\right )+\left (\left \{\begin {array}{cc}0 & c t +x \le \pi \\\sin ^{2}\left (c t +x \right ) & c t +x <2 \pi \\0 & 2 \pi \le c t +x \end {array}\right .\right ) & x <c t \\\left (\left \{\begin {array}{cc}0 & -c t +x \le \pi \\\sin ^{2}\left (c t -x \right ) & -c t +x <2 \pi \\0 & 2 \pi \le -c t +x \end {array}\right .\right )+\left (\left \{\begin {array}{cc}0 & c t +x \le \pi \\\sin ^{2}\left (c t +x \right ) & c t +x <2 \pi \\0 & 2 \pi \le c t +x \end {array}\right .\right ) & c t <x \end {array}\right .\right )}{2}\]

Hand solution

Solving on semi-infinite domain \begin {align} u_{tt} & =c^{2}u_{xx}\qquad t>0,x>0\tag {1}\\ u\left ( 0,t\right ) & =0\nonumber \\ u\left ( x,0\right ) & =f\left ( x\right ) =\sin ^{2}\left ( x\right ) \qquad \pi <x<2\pi \nonumber \end {align}

With \(k>0\) and \(u\left ( x,t\right ) <\infty \) as \(x\rightarrow \infty \). This means \(u\left ( x,t\right ) \) is bounded. The general solution to the above PDE was given in problem 5.1.2.1 on page 1368 as (using the D’Alambert’s solution and not the Fourier integral solution)\[ u\left ( x,t\right ) =\left \{ \begin {array} [c]{ccc}\frac {1}{2}\left ( f\left ( x+ct\right ) +f\left ( x-ct\right ) \right ) & & x>ct\geq 0\\ & & \\ \frac {1}{2}\left ( f\left ( x+ct\right ) -f\left ( ct-x\right ) \right ) & & x\leq ct \end {array} \right . \] But \(f\left ( x\right ) =\sin ^{2}\left ( x\right ) \) and the above becomes\[ u\left ( x,t\right ) =\left \{ \begin {array} [c]{ccc}\frac {1}{2}\left ( \sin ^{2}\left ( x+ct\right ) +\sin ^{2}\left ( x-ct\right ) \right ) & & x>ct\geq 0\\ & & \\ \frac {1}{2}\left ( \sin ^{2}\left ( x+ct\right ) -\sin ^{2}\left ( ct-x\right ) \right ) & & x\leq ct \end {array} \right . \] But here \(f\left ( x\right ) \) is restricted to \(\pi <x<2\pi \). Hence the above solution is modified as follows\[ u\left ( x,t\right ) =\left \{ \begin {array} [c]{ccc}\left \{ \begin {array} [c]{ccc}\frac {1}{2}\sin ^{2}\left ( x+ct\right ) & & \pi <x+ct<2\pi \\ 0 & & \text {otherwise}\end {array} \right . +\left \{ \begin {array} [c]{ccc}\frac {1}{2}\sin ^{2}\left ( x-ct\right ) & & \pi <x-ct<2\pi \\ 0 & & \text {otherwise}\end {array} \right . & & x>ct\geq 0\\ & & \\ \left \{ \begin {array} [c]{ccc}\frac {1}{2}\sin ^{2}\left ( x+ct\right ) & & \pi <x+ct<2\pi \\ 0 & & \text {otherwise}\end {array} \right . -\left \{ \begin {array} [c]{ccc}\frac {1}{2}\sin ^{2}\left ( ct-x\right ) & & \pi <ct-x<2\pi \\ 0 & & \text {otherwise}\end {array} \right . & & 0<x\leq ct \end {array} \right . \]

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5.1.2.3 [366] Logan page 115, left end fixed with source

problem number 366

This is problem at page 115, David J Logan textbook, applied PDE textbook.

Falling cable lying on a table that is suddenly removed. \[ u_{tt} = c^2 u_{xx} - g \] With boundary condition \begin {align*} u(0,t) &= 0 \end {align*}

And initial conditions \begin {align*} u(x,0) &= 0 \\ \frac {\partial u}{\partial t}(x,0) &= 0 \end {align*}

Figure 5.49: PDE specification

Mathematica ✓

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

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} -\frac {g t^2}{2} & c t\leq x \\ \frac {g x (x-2 c t)}{2 c^2} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

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

\[u \left (x , t\right ) = \frac {\left (-c^{2} t^{2}+\left (c t -x \right )^{2} \theta \left (t -\frac {x}{c}\right )\right ) g}{2 c^{2}}\]

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5.1.2.4 [367] Left moving boundary condition

problem number 367

Solve for \(u(x,t)\) with \(t>0\) and \(x>0\) \[ \frac {\partial ^2 u}{\partial t^2} = c^2 \frac {\partial ^2 u}{\partial x^2} \] With boundary conditions \begin {align*} u(0,t) &= g(t) \end {align*}

With initial conditions \begin {align*} \frac {\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= 0 \end {align*}

Figure 5.50: PDE specification

Mathematica ✓

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

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} 0 & x>c t \\ g\left (t-\frac {x}{c}\right ) & x\leq c t \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

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

\[u \left (x , t\right ) = \]

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5.1.2.5 [368] moving Left end

problem number 368

Taken from Mathematica DSolve help pages. Initial value problem with a Neumann condition on the half-line. \[ u_{tt} = c^2 u_{xx} \] With initial conditions \begin {align*} u(x,0) &= \sin ^3(x) \\ \frac {\partial u}{\partial t}(x,0) &= 1-e^{- \frac {x}{10}} \end {align*}

And boundary conditions \(\frac {\partial u}{\partial x}(0,t)=1\)

Figure 5.51: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; 
ic  = {u[x, 0] == Sin[x]^3, Derivative[0, 1][u][x, 0] == 1 - E^(-(x/10))}; 
bc  = Derivative[1, 0][u][0, t] == 1; 
sol =  AbsoluteTiming[TimeConstrained[DSolveValue[{pde, ic, bc}, u[x, t], {x, t},Assumptions->c>0], 60*10]];
 

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} 0 & x>c t \\ g\left (t-\frac {x}{c}\right ) & x\leq c t \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

restart; 
pde := diff(u(x, t), t$2) = c^2 * diff(u(x, t), x$2); 
ic  := u(x,0)= sin(x)^3,(D[2](u))(x,0)=1-exp(-x/10); 
bc:=(D[1](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'));
 

\[u \left (x , t\right ) = \frac {\left \{\begin {array}{cc}-c \left (\sin ^{3}\left (c t -x \right )\right )+c \left (\sin ^{3}\left (c t +x \right )\right )+2 c t +10 \,{\mathrm e}^{-\frac {c t}{10}-\frac {x}{10}}-10 \,{\mathrm e}^{\frac {c t}{10}-\frac {x}{10}} & c t <x \\c \left (\sin ^{3}\left (c t -x \right )\right )+c \left (\sin ^{3}\left (c t +x \right )\right )-2 c^{2} t +\left (2 t +2 x \right ) c +10 \,{\mathrm e}^{-\frac {c t}{10}-\frac {x}{10}}+10 \,{\mathrm e}^{-\frac {c t}{10}+\frac {x}{10}}-20 & x <c t \end {array}\right .}{2 c}\]

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5.1.2.6 [369] I.C. at \(t=1\)

problem number 369

Added December 20, 2018.

Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018 Solve \[ u_{tt} = u_{xx} \] With initial conditions \begin {align*} u(x,1) &= e^{-(x-6)^2}+e^{-(x+6)^2} \\ \frac {\partial u}{\partial t}(x,1) &= \frac {1}{2} \end {align*}

And boundary conditions \(\frac {\partial u}{\partial x}(0,t)=1\)

Figure 5.52: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; 
ic  = {u[x, 1] == Sin[x]^3, Derivative[0, 1][u][x, 1] == 1 - E^(-(x/10))}; 
bc  = Derivative[1, 0][u][0, t] == 1; 
sol =  AbsoluteTiming[TimeConstrained[DSolveValue[{pde, ic, bc}, u[x, t], {x, t}, Assumptions -> t > 0], 60*10]];
 

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} 0 & x>c t \\ g\left (t-\frac {x}{c}\right ) & x\leq c t \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

restart; 
pde := diff(u(x, t), t$2) = diff(u(x, t), x$2); 
ic  := u(x, 1) = exp(-(x-6)^2)+exp(-(x+6)^2), eval(diff(u(x,t),t),t=1)= 1/2; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic],u(x,t)) assuming t>0),output='realtime'));
 

\[u \left (x , t\right ) = \frac {t}{2}+\frac {{\mathrm e}^{-\left (t +x -7\right )^{2}}}{2}+\frac {{\mathrm e}^{-\left (t -x +5\right )^{2}}}{2}+\frac {{\mathrm e}^{-\left (t -x -7\right )^{2}}}{2}+\frac {{\mathrm e}^{-\left (t +x +5\right )^{2}}}{2}-\frac {1}{2}\]

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5.1.2.7 [370] B.C. at \(x=1\)

problem number 370

Added December 20, 2018.

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

Solve \[ \frac {\partial ^2 u}{\partial t^2} = \frac {1}{4} \frac {\partial ^2 u}{\partial x^2} \] With initial conditions \begin {align*} u(x,0) &= e^{-x^2}\\ \frac {\partial u}{\partial t}(x,0) &= 0 \end {align*}

And Boundary conditions \(\frac {\partial u}{\partial x}(1,t)= 0\)

Figure 5.53: PDE specification

Mathematica ✓

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

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^{-x^2} & x-1\geq 0\land t=0 \\ \frac {2 \int _0^{\infty }\frac {1}{4} e^{-\frac {1}{4} K[1] (K[1]+4 i)} \sqrt {\pi } \cos \left (\frac {1}{2} t K[1]\right ) \cos ((x-1) K[1]) \left (\operatorname {Erfc}\left (1-\frac {1}{2} i K[1]\right )+e^{2 i K[1]} \operatorname {Erfc}\left (\frac {1}{2} i K[1]+1\right )\right )dK[1]}{\pi } & x-1\geq 0\land t>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

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

\[u \left (x , t\right ) = \frac {\left (\left \{\begin {array}{cc}{\mathrm e}^{-\frac {\left (t -2 x \right )^{2}}{4}}+{\mathrm e}^{-\frac {\left (t +2 x \right )^{2}}{4}} & \frac {t}{2}<x -1 \\{\mathrm e}^{-\frac {\left (t +2 x \right )^{2}}{4}}+{\mathrm e}^{-\frac {\left (t -2 x +4\right )^{2}}{4}} & x -1<\frac {t}{2} \end {array}\right .\right )}{2}\]

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5.1.2.8 [371] Left end free. zero initial velocity (general solution)

problem number 371

Added July 13, 2019.

Solve for \(u(x,t)\) with \(x>0,t>0\) \[ u_{tt} = c^2 u_{xx} \] With initial conditions \begin {align*} u(x,0) &= f(x)\\ \frac {\partial u}{\partial t}(x,0) &= 0 \end {align*}

And boundary condition \(\frac {\partial u}{\partial x}(0,t) = 0\).

Figure 5.54: PDE specification

Mathematica ✓

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

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} f(x) & x\geq 0\land t=0 \\ \frac {2 \int _0^{\infty }\cos (c t K[1]) \cos (x K[1]) \int _0^{\infty }\cos (x K[1]) f(x)dxdK[1]}{\pi } & x\geq 0\land t>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

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

\[u \left (x , t\right ) = \frac {\left (\left \{\begin {array}{cc}f \left (-c t +x \right )+f \left (c t +x \right ) & c t <x \\f \left (c t -x \right )+f \left (c t +x \right ) & x <c t \end {array}\right .\right )}{2}\]

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5.1.2.9 [372] Left end free. zero initial velocity (Special solution)

problem number 372

Added July 14, 2019.

Solve for \(u(x,t)\) with \(x>0,t>0\) \[ u_{tt} = c^2 u_{xx} \] With initial conditions \begin {align*} u(x,0) &= f(x)\\ \frac {\partial u}{\partial t}(x,0) &= 0 \end {align*}

And boundary condition \(\frac {\partial u}{\partial x}(0,t) = 0\) using \begin {align*} c &= 3\\ f(x) &=\left \{ \begin {array} [c]{cc}1 & 4\leq x\leq 5\\ 0 & \text {otherwise}\end {array} \right . \end {align*}

Figure 5.55: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
c=3; 
f[x_]:=Piecewise[{{1,4<x<5},{0,True}}]; 
pde =  D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; 
ic  = {u[x, 0] == f[x], Derivative[0, 1][u][x, 0] == 0}; 
bc  = Derivative[1, 0][u][0, t] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}, Assumptions -> {t > 0, x > 0}], 60*10]];
 

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {1}{2} \left (\left (\begin {array}{cc} \{ & \begin {array}{cc} 1 & 4<x-3 t<5 \\ 0 & \text {True} \\\end {array} \\\end {array}\right )+\left (\begin {array}{cc} \{ & \begin {array}{cc} 1 & 4<3 t+x<5 \\ 0 & \text {True} \\\end {array} \\\end {array}\right )\right ) & x>3 t \\ \frac {1}{2} \left (\left (\begin {array}{cc} \{ & \begin {array}{cc} 1 & 4<3 t-x<5 \\ 0 & \text {True} \\\end {array} \\\end {array}\right )+\left (\begin {array}{cc} \{ & \begin {array}{cc} 1 & 4<3 t+x<5 \\ 0 & \text {True} \\\end {array} \\\end {array}\right )\right ) & x\leq 3 t \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

restart; 
c:=3; 
f:=piecewise(4<x and x<5,1,true,0); 
pde := diff(u(x, t), t$2) = c^2*(diff(u(x, t), x$2)); 
bc  := eval( diff(u(x,t),x),x=0)=0; 
ic  := u(x,0)=f(x),eval(diff(u(x,t),t),t=0)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc,ic],u(x,t)) assuming x>0,t>0),output='realtime'));
 

\[u \left ( x,t \right ) ={\frac { \begin {cases} \left ( \begin {cases}1 & 4<6\,t+x \land 6\,t+x<5\\0 & \text {otherwise}\end {cases} \right ) \left ( 3\,t+x \right ) + \left ( \begin {cases}1 & 4<x-6\,t \land x-6\,t<5\\0 & \text {otherwise}\end {cases} \right ) \left ( x-3\,t \right ) & 3\,t<x\\ \left ( \begin {cases}1 & 4<6\,t+x \land 6\,t+x<5\\0 & \text {otherwise}\end {cases} \right ) \left ( 3\,t+x \right ) + \left ( \begin {cases}1 & 4<x \land x<5\\0 & \text {otherwise}\end {cases} \right ) \left ( 3\,t-x \right ) & x<3\,t\end {cases} }{2}}\]

Hand solution

Solve \(\frac {\partial ^{2}u}{\partial t^{2}}=c^{2}\frac {\partial ^{2}u}{\partial x^{2}}\) for \(x\geq 0,t\geq 0\) with initial conditions \(u\left ( x,0\right ) =f\left ( x\right ) =\left \{ \begin {array} [c]{cc}1 & 4\leq x\leq 5\\ 0 & \text {otherwise}\end {array} \right . \) and \(\frac {\partial u\left ( x,0\right ) }{\partial t}=0\) and boundary condition \(\frac {\partial u\left ( 0,t\right ) }{\partial x}=0\) (Free end)

The general solution by D’Almbert’s is given by\[ u\left ( x,t\right ) =\frac {1}{2}\left \{ \begin {array} [c]{ccc}f\left ( x+ct\right ) +f\left ( x-ct\right ) & & x>ct\geq 0\\ f\left ( x+ct\right ) +f\left ( ct-x\right ) & & x\leq ct \end {array} \right . \] But \(f\left ( x\right ) \) is defined for \(4\leq x\leq 5\) only, hence the solution becomes\[ u\left ( x,t\right ) =\left \{ \begin {array} [c]{ccc}\left \{ \begin {array} [c]{ccc}\frac {1}{2}f\left ( x+ct\right ) & & 4\leq x\leq 5\\ 0 & & \text {otherwise}\end {array} \right . +\left \{ \begin {array} [c]{ccc}\frac {1}{2}f\left ( x-ct\right ) & & 4\leq x\leq 5\\ 0 & & \text {otherwise}\end {array} \right . & & x>ct\geq 0\\ & & \\ \left \{ \begin {array} [c]{ccc}\frac {1}{2}f\left ( x+ct\right ) & & 4\leq x\leq 5\\ 0 & & \text {otherwise}\end {array} \right . +\left \{ \begin {array} [c]{ccc}\frac {1}{2}f\left ( ct-x\right ) & & 4\leq x\leq 5\\ 0 & & \text {otherwise}\end {array} \right . & & 0<x\leq ct \end {array} \right . \] Animation is below

Source code used for the above

Figure 5.56: Source code

Figure 5.57: Code for animation

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5.1.2.10 [373] Left end fixed. zero initial velocity (Special solution)

problem number 373

Added January 9, 2020.

Solve for \(u(x,t)\) with \(x>0,t>0\) \[ u_{tt} = c^2 u_{xx} \] With initial conditions \begin {align*} u(x,0) &= f(x)\\ \frac {\partial u}{\partial t}(x,0) &= 0 \end {align*}

And boundary condition \(u(0,t) = 0\) using \begin {align*} c &= 3\\ f(x) &=\left \{ \begin {array} [c]{cc}1 & 4\leq x\leq 5\\ 0 & \text {otherwise}\end {array} \right . \end {align*}

Figure 5.58: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
c=3; 
f[x_]:=Piecewise[{{1,4<x<5},{0,True}}]; 
pde =  D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; 
ic  = {u[x, 0] == f[x], Derivative[0, 1][u][x, 0] == 0}; 
bc  = u[0, t] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}, Assumptions -> {t > 0, x > 0}], 60*10]];
 

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {1}{2} \left (\left (\begin {array}{cc} \{ & \begin {array}{cc} 1 & 4<x-3 t<5 \\ 0 & \text {True} \\\end {array} \\\end {array}\right )+\left (\begin {array}{cc} \{ & \begin {array}{cc} 1 & 4<3 t+x<5 \\ 0 & \text {True} \\\end {array} \\\end {array}\right )\right ) & x>3 t \\ \frac {1}{2} \left (\left (\begin {array}{cc} \{ & \begin {array}{cc} 1 & 4<3 t+x<5 \\ 0 & \text {True} \\\end {array} \\\end {array}\right )-\left (\begin {array}{cc} \{ & \begin {array}{cc} 1 & 4<3 t-x<5 \\ 0 & \text {True} \\\end {array} \\\end {array}\right )\right ) & x\leq 3 t \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

restart; 
c:=3; 
f:=piecewise(4<x and x<5,1,true,0); 
pde := diff(u(x, t), t$2) = c^2*(diff(u(x, t), x$2)); 
bc  := u(0,t)=0; 
ic  := u(x,0)=f(x),D[2](u)(x,0)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc,ic],u(x,t)) assuming x>0,t>0),output='realtime'));
 

\[u \left ( x,t \right ) ={\frac { \begin {cases} \left ( \begin {cases}1 & 4<6\,t+x \land 6\,t+x<5\\0 & \text {otherwise}\end {cases} \right ) \left ( 3\,t+x \right ) - \left ( \begin {cases}1 & 4<x \land x<5\\0 & \text {otherwise}\end {cases} \right ) \left ( 3\,t-x \right ) & x<3\,t\\ \left ( \begin {cases}1 & 4<6\,t+x \land 6\,t+x<5\\0 & \text {otherwise}\end {cases} \right ) \left ( 3\,t+x \right ) + \left ( \begin {cases}1 & 4<x-6\,t \land x-6\,t<5\\0 & \text {otherwise}\end {cases} \right ) \left ( x-3\,t \right ) & 3\,t<x\end {cases} }{2}}\]

Hand solution

Solve \(\frac {\partial ^{2}u}{\partial t^{2}}=c^{2}\frac {\partial ^{2}u}{\partial x^{2}}\) for \(x\geq 0,t\geq 0\) with initial conditions \(u\left ( x,0\right ) =f\left ( x\right ) =\left \{ \begin {array} [c]{cc}1 & 4\leq x\leq 5\\ 0 & \text {otherwise}\end {array} \right . \) and \(\frac {\partial u\left ( x,0\right ) }{\partial t}=0\) and boundary condition \(u\left ( 0,t\right ) =0\) (Fixed end)

The general solution by D’Almbert’s is given by\[ u\left ( x,t\right ) =\frac {1}{2}\left \{ \begin {array} [c]{ccc}f\left ( x+ct\right ) +f\left ( x-ct\right ) & & x>ct\geq 0\\ f\left ( x+ct\right ) -f\left ( ct-x\right ) & & x\leq ct \end {array} \right . \] But \(f\left ( x\right ) \) is defined for \(4\leq x\leq 5\) only, hence the solution becomes\[ u\left ( x,t\right ) =\left \{ \begin {array} [c]{ccc}\left \{ \begin {array} [c]{ccc}\frac {1}{2}f\left ( x+ct\right ) & & 4\leq x\leq 5\\ 0 & & \text {otherwise}\end {array} \right . +\left \{ \begin {array} [c]{ccc}\frac {1}{2}f\left ( x-ct\right ) & & 4\leq x\leq 5\\ 0 & & \text {otherwise}\end {array} \right . & & x>ct\geq 0\\ & & \\ \left \{ \begin {array} [c]{ccc}\frac {1}{2}f\left ( x+ct\right ) & & 4\leq x\leq 5\\ 0 & & \text {otherwise}\end {array} \right . +\left \{ \begin {array} [c]{ccc}-\frac {1}{2}f\left ( ct-x\right ) & & 4\leq x\leq 5\\ 0 & & \text {otherwise}\end {array} \right . & & 0<x\leq ct \end {array} \right . \] Animation is below

Source code used for the above

Figure 5.59: Source code

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5.1.2.11 [374] Left end free. zero initial position (general solution)

problem number 374

Added July 14, 2019.

Solve for \(u(x,t)\) with \(x>0,t>0\) \[ u_{tt} = c^2 u_{xx} \] With initial conditions \begin {align*} u(x,0) &= 0\\ \frac {\partial u}{\partial t}(x,0) &= g(x) \end {align*}

And boundary condition \(\frac {\partial u}{\partial x}(0,t) = 0\).

Figure 5.60: PDE specification

Mathematica ✓

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

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} 0 & x\geq 0\land t=0 \\ \frac {2 \int _0^{\infty }\frac {\cos (x K[1]) \sin (c t K[1]) \int _0^{\infty }\cos (x K[1]) g(x)dx}{c K[1]}dK[1]}{\pi } & x\geq 0\land t>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

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

\[u \left (x , t\right ) = \frac {\left \{\begin {array}{cc}\int _{-c t +x}^{c t +x}g \left (\zeta \right )d \zeta & c t <x \\\int _{0}^{c t -x}g \left (\zeta \right )d \zeta +\int _{0}^{c t +x}g \left (\zeta \right )d \zeta & x <c t \end {array}\right .}{2 c}\]

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5.1.2.12 [375] Left end free. Non zero initial position and velocity (general solution)

problem number 375

Added July 14, 2019.

Solve for \(u(x,t)\) with \(x>0,t>0\) \[ u_{tt} = c^2 u_{xx} \] With initial conditions \begin {align*} u(x,0) &= f(x)\\ \frac {\partial u}{\partial t}(x,0) &= g(x) \end {align*}

And boundary condition \(\frac {\partial u}{\partial x}(0,t) = 0\).

Figure 5.61: PDE specification

Mathematica ✓

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

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} f(x) & x\geq 0\land t=0 \\ \frac {2 \int _0^{\infty }\frac {\sqrt {\frac {\pi }{2}} \cos (x K[1]) \left (c \sqrt {\frac {2}{\pi }} \cos (c t K[1]) K[1] \int _0^{\infty }\cos (x K[1]) f(x)dx+\sqrt {\frac {2}{\pi }} \sin (c t K[1]) \int _0^{\infty }\cos (x K[1]) g(x)dx\right )}{c K[1]}dK[1]}{\pi } & x\geq 0\land t>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

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

\[u \left (x , t\right ) = \frac {\left (\left \{\begin {array}{cc}f \left (-c t +x \right )+f \left (c t +x \right )+\frac {\int _{-c t +x}^{c t +x}g \left (\zeta \right )d \zeta }{c} & c t <x \\\frac {c f \left (c t -x \right )+c f \left (c t +x \right )+\int _{0}^{c t -x}g \left (\zeta \right )d \zeta +\int _{0}^{c t +x}g \left (\zeta \right )d \zeta }{c} & x <c t \end {array}\right .\right )}{2}\]

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5.1.2.13 [376] Left end free with source

problem number 376

Added December 20, 2018.

Example 17, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Left end free with initial position and velocity given.

Solve for \(u(x,t)\) with \(x>0,t>0\) \[ \frac {\partial ^2 u}{\partial t^2} = 9 \frac {\partial ^2 u}{\partial x^2} + f(x,t) \] With initial conditions \begin {align*} u(x,0) &= 0\\ \frac {\partial u}{\partial t}(x,0) &= x^3 \end {align*}

And boundary condition \(\frac {\partial u}{\partial x}(0,t) = 0\).

Figure 5.62: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], {t, 2}] == 9*D[u[x, t], {x, 2}]; 
ic  = {u[x, 0] == 0, Derivative[0, 1][u][x, 0] == x^3}; 
bc  = Derivative[1, 0][u][0, t] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}, Assumptions -> {t > 0, x > 0}], 60*10]];
 

\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} t x \left (9 t^2+x^2\right ) & x>3 t \\ \frac {1}{12} \left (81 t^4+54 x^2 t^2+x^4\right ) & x\leq 3 t \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

restart; 
pde := diff(u(x, t), t$2) = 9*(diff(u(x, t), x$2)); 
bc  := eval( diff(u(x,t),x),x=0)=0; 
ic  := u(x,0)=0,eval(diff(u(x,t),t),t=0)=x^3; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc,ic],u(x,t)) assuming x>0,t>0),output='realtime'));
 

\[u \left (x , t\right ) = \frac {\left (\left \{\begin {array}{cc}108 t^{3} x +12 t x^{3} & 3 t <x \\81 t^{4}+54 t^{2} x^{2}+x^{4} & x <3 t \end {array}\right .\right )}{12}\]

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