3.1.7 Infinite domain

3.1.7.1 [247] Inverse exponential I.C.

problem number 247

From Mathematica DSolve help pages. Solve the heat equation for \(u(x,t)\) on real line with \(t>0\)

\[ u_t= u_{xx} \]

With initial condition

\[ u(x,0)= e^{-x^2} \]

Figure 3.133: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == D[u[x, t], {x, 2}]; 
ic  = u[x, 0] == E^(-x^2); 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, u[x, t], {x, t}], 60*10]];
 

\[\left \{\left \{u(x,t)\to \fbox {$\frac {e^{-\frac {x^2}{4 t+1}}}{\sqrt {4+\frac {1}{t}} \sqrt {t}}\text { if }\Re \left (\frac {1}{t}\right )\geq -4$}\right \}\right \}\]

Maple ✓

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

\[u \left (x , t\right ) = \frac {{\mathrm e}^{-\frac {x^{2}}{1+4 t}}}{\sqrt {1+4 t}}\]

Hand solution

Solve

\[ \frac {\partial u}{\partial t}=\frac {\partial ^{2}u}{\partial x^{2}}\]

On \(-\infty <x<\infty ,t>0\) with \(u\left ( x,0\right ) =f\left ( x\right ) =e^{-x^{2}}\). The first step is to find Green function for the above PDE. Taking Fourier transform of both sides w.r.t. \(x\), using \(\hat {u}\left ( k,t\right ) \) as the Fourier transform of \(u\left ( x,t\right ) \) gives

\begin{align*} \frac {d}{dt}\hat {u}\left ( k,t\right ) & =\left ( ik\right ) ^{2}\hat {u}\left ( k,t\right ) \\ & =-k^{2}\hat {u}\left ( k,t\right ) \\ \frac {d}{dt}\hat {u}\left ( k,t\right ) +k^{2}\hat {u}\left ( k,t\right ) & =0 \end{align*}

The solution to the above is

\begin{equation} \hat {u}\left ( k,t\right ) =Ce^{-k^{2}t}\tag {1}\end{equation}

At \(t=0,\)

\[ \hat {u}\left ( k,0\right ) =\mathcal {F}\left ( h\left ( x\right ) \right ) \]

Therefore

\[ C=\mathcal {F}\left ( h\left ( x\right ) \right ) \]

And (1) becomes

\[ \hat {u}\left ( k,t\right ) =\mathcal {F}\left ( h\left ( x\right ) \right ) e^{-k^{2}t}\]

To find Green function, we replace \(h\left ( x\right ) \) by \(\delta \left ( x-\xi \right ) \) where \(\xi \) is the location of the pulse. But \(\mathcal {F}\left ( \delta \left ( x-\xi \right ) ;k\right ) =\frac {1}{\sqrt {2\pi }}\int _{-\infty }^{\infty }\delta \left ( x-\xi \right ) e^{-ikx}dx=\frac {1}{\sqrt {2\pi }}e^{-i\xi k}\). Therefore the above becomes

\[ \hat {G}\left ( k,t\right ) =\frac {1}{\sqrt {2\pi }}e^{-i\xi k}e^{-k^{2}t}\]

The above is the Fourier transform of the Green function. Now we invert it

\begin{align} G\left ( x,t\right ) & =\frac {1}{\sqrt {2\pi }}\int _{-\infty }^{\infty }\left ( \frac {1}{\sqrt {2\pi }}e^{-i\xi k}e^{-k^{2}t}\right ) e^{ikx}dk\nonumber \\ & =\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{-i\xi k-k^{2}t+ikx}dk\tag {2}\end{align}

We would like to use Gaussian as the integrand, hence we want to change \(-i\xi k-k^{2}t+ikx\) to \(-\left ( k\sqrt {t}-A\right ) ^{2}\). We do this by completing the square.

\begin{align*} -i\xi k-k^{2}t+ikx & =-\left ( k\sqrt {t}-A\right ) ^{2}\\ & =-\left ( k^{2}t+A^{2}-2Ak\sqrt {t}\right ) \\ & =-k^{2}t-A^{2}+2Ak\sqrt {t}\end{align*}

Comparing sides then \(2Ak\sqrt {t}=k\left ( -i\xi +ix\right ) \) or \(A=\frac {-i\xi +ix}{2\sqrt {t}}\). Therefore

\begin{align*} -i\xi k-k^{2}t+ikx & =-\left ( k\sqrt {t}-\frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}+A^{2}\\ & =-\left ( k\sqrt {t}-\frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}+\left ( \frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}\end{align*}

Hence

\begin{align*} e^{-i\xi k-k^{2}t+ikx} & =e^{-\left ( k\sqrt {t}-\frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}+\left ( \frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}\\ & =e^{-\left ( k\sqrt {t}-\frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}e^{\left ( \frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}\end{align*}

Substituting the above into (2) gives

\begin{align*} G\left ( x,t\right ) & =\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{-\left ( k\sqrt {t}-\frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}e^{\left ( \frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}dk\\ & =\frac {1}{2\pi }e^{\left ( \frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}\int _{-\infty }^{\infty }e^{-\left ( k\sqrt {t}-\frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}dk \end{align*}

To evaluate \(\int _{-\infty }^{\infty }e^{-\left ( k\sqrt {t}-\frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}dk\), let \(u=k\sqrt {t}\), then \(du=\sqrt {t}dk\). The above becomes

\begin{align*} G\left ( x,t\right ) & =\frac {1}{2\pi }e^{\left ( \frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}\int _{-\infty }^{\infty }e^{-\left ( u-\frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}\frac {du}{\sqrt {t}}\\ & =\frac {1}{2\pi \sqrt {t}}e^{\left ( \frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}\int _{-\infty }^{\infty }e^{-\left ( u-\frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}du \end{align*}

Now the integral is Gaussian. \(\int _{-\infty }^{\infty }e^{-\left ( u-\frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}du=\sqrt {\pi }\) and the above becomes

\begin{align*} G\left ( x,t\right ) & =\frac {\sqrt {\pi }}{2\pi \sqrt {t}}e^{\left ( \frac {-i\xi +ix}{2\sqrt {t}}\right ) ^{2}}\\ & =\frac {1}{2\sqrt {\pi t}}e^{\left ( i\left ( \frac {-\xi +x}{2\sqrt {t}}\right ) \right ) ^{2}}\\ & =\frac {1}{2\sqrt {\pi t}}e^{-\frac {\left ( x-\xi \right ) ^{2}}{4t}}\end{align*}

Now that we found the Green function for the PDE, we can find the solution as

\begin{align*} u\left ( x,t\right ) & =\int _{-\infty }^{\infty }G\left ( \xi ,t\right ) h\left ( \xi \right ) d\xi \\ & =\int _{-\infty }^{\infty }\frac {1}{2\sqrt {\pi t}}e^{-\frac {\left ( x-\xi \right ) ^{2}}{4t}}h\left ( \xi \right ) d\xi \\ & =\frac {1}{\sqrt {4\pi t}}\int _{-\infty }^{\infty }e^{-\frac {\left ( x-\xi \right ) ^{2}}{4t}}e^{-\xi ^{2}}d\xi \end{align*}

But \(\int _{-\infty }^{\infty }e^{-\frac {\left ( x-\xi \right ) ^{2}}{4t}}e^{-\xi ^{2}}d\xi =\frac {2e^{-\frac {x^{2}}{1+4t}\sqrt {\pi }}}{\sqrt {\frac {1+4t}{t}}}\), hence the above becomes

\begin{align*} u\left ( x,t\right ) & =\frac {1}{\sqrt {4\pi t}}\frac {2e^{-\frac {x^{2}}{1+4t}\sqrt {\pi }}}{\sqrt {\frac {1+4t}{t}}}\\ & =\frac {e^{-\frac {x^{2}}{1+4t}}}{\sqrt {1+4t}}\end{align*}

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3.1.7.2 [248] Advection term

problem number 248

From Mathematica DSolve help pages. Solve the heat equation for \(u(x,t)\) on real line with \(t>0\)

\[ u_t = 12 u_{xx} + u_x \sin t \]

With initial condition

\[ u(x,0)= x \]

Figure 3.134: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == 12*D[u[x, t], {x, 2}] + Sin[t]*D[u[x, t], x]; 
ic  = u[x, 0] == x; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, u[x, t], {x, t}], 60*10]];
 

\[\{\{u(x,t)\to -\cos (t)+x+1\}\}\]

Maple ✓

restart; 
pde := diff(u(x,t),t)= 12* diff(u(x,t),x$2)+sin(t)*diff(u(x,t),x); 
ic  := u(x,0)=x; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic],u(x,t))),output='realtime'));
 

\[u \left (x , t\right ) = -\cos \left (t \right )+x +1\]

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3.1.7.3 [249] UnitBox I.C.

problem number 249

From Mathematica DSolve help pages. Solve the heat equation for \(u(x,t)\) on real line with \(t>0\)

\[ u_t= u_{xx} \]

With initial condition

\[ u(x,0)= \text {UnitBox[x]} \]

Where UnitBox is equal to 1 if \(|x| \leq \frac {1}{2}\) and zero otherwise.

Figure 3.135: PDE specification

Mathematica ✓

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

\[\left \{\left \{u(x,t)\to \frac {1}{2} \left (\text {erf}\left (\frac {2 x+1}{4 \sqrt {t}}\right )-\text {erf}\left (\frac {2 x-1}{4 \sqrt {t}}\right )\right )\right \}\right \}\]

Maple ✓

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

\[u \left (x , t\right ) = -\frac {\operatorname {erf}\left (\frac {2 x -1}{4 \sqrt {t}}\right )}{2}+\frac {\operatorname {erf}\left (\frac {1+2 x}{4 \sqrt {t}}\right )}{2}\]

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3.1.7.4 [250] No source

problem number 250

Solve the heat equation

\[ u_t = k u_{xx} \]

For \(-\infty <x<\infty \) and \(t>0\), and initial condition is \(u(x,0)=f(x)\)

Figure 3.136: PDE specification

Mathematica ✓

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

\[\left \{\left \{u(x,t)\to \frac {\int _{-\infty }^{\infty }e^{s (i x-k s t)} \int _{-\infty }^{\infty }e^{-i s x} f(x)dxds}{2 \pi }\right \}\right \}\]

Maple ✓

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

\[u \left (x , t\right ) = \frac {\int _{-\infty }^{\infty }f \left (-\zeta \right ) {\mathrm e}^{-\frac {\left (x +\zeta \right )^{2}}{4 k t}}d \zeta }{2 \sqrt {\pi }\, \sqrt {k}\, \sqrt {t}}\]

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3.1.7.5 [251] constant as source

problem number 251

Solve the heat equation

\[ u_t = k u_{xx} + m \]

For \(-\infty <x<\infty \) and \(t>0\). Initial condition is \(u(x,0)=\sin (x)\)

Figure 3.137: PDE specification

Mathematica ✓

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

\[\left \{\left \{u(x,t)\to e^{-k t} \sin (x)+m t\right \}\right \}\]

Maple ✓

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

\[u \left (x , t\right ) = \sin \left (x \right ) {\mathrm e}^{-t k}+t m\]

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3.1.7.6 [252] No intial conditions

problem number 252

Solve the heat equation for \(u(x,t)\)

\[ u_t= u_xx \]

Figure 3.138: PDE specification

Mathematica ✓

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

\[\{\{u(x,t)\to \cosh (c_2 (x+c_2 t)+c_1)+\sinh (c_2 (x+c_2 t)+c_1)+1\}\}\]

Maple ✓

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

\[u \left (x , t\right ) = c_{3} {\mathrm e}^{\textit {\_c}_{1} t -\sqrt {\textit {\_c}_{1}}\, x} \left ({\mathrm e}^{2 \sqrt {\textit {\_c}_{1}}\, x} c_{1} +c_{2} \right )\]

Hand solution

Solve

\[ \frac {\partial u}{\partial t}=\frac {\partial ^{2}u}{\partial x^{2}}\]

for \(t>0,-\infty <x<\infty \). Let \(u=X\left ( x\right ) T\left ( t\right ) \) then we obtain

\[ T^{\prime }X=X^{\prime \prime }T \]

Dividing by \(XT\neq 0\)

\[ \frac {T^{\prime }}{T}=\frac {X^{\prime \prime }}{X}=-\lambda \]

(Only positive eigenvalues are possible). The two ODE’s are

\begin{align} T^{\prime }+\lambda T & =0\tag {1}\\ X^{\prime \prime }+\lambda X & =0\tag {2}\end{align}

Solution for (2) is \(X\left ( x\right ) =C_{1}e^{\sqrt {\lambda }x}+C_{2}e^{-\sqrt {\lambda }x}\) and solution for (1) is \(T\left ( t\right ) =C_{3}e^{-\lambda t}\). Hence

\begin{align*} u\left ( x,t\right ) & =C_{3}e^{-\lambda t}\left ( C_{1}e^{\sqrt {\lambda }x}+C_{2}e^{-\sqrt {\lambda }x}\right ) \\ & =C_{3}e^{-\lambda t}C_{1}e^{\sqrt {\lambda }x}+C_{3}e^{-\lambda t}C_{2}e^{-\sqrt {\lambda }x}\\ & =C_{3}e^{-\lambda t}C_{1}e^{\sqrt {\lambda }x}+\frac {C_{3}e^{-\lambda t}C_{2}}{e^{\sqrt {\lambda }x}}\end{align*}

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3.1.7.7 [253] piecewise I.C.

problem number 253

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 \(u(x,t)\) on real line with \(t>0\)

\[ u_t = \mu u_{xx} - 1 \]

With initial condition

\[ u(x,1)= \begin {cases} 0 & x \geq 0 \\ 1 & x < 0 \end {cases} \]

Figure 3.139: PDE specification

Mathematica ✓

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

\[\left \{\left \{u(x,t)\to \frac {\int _{-\infty }^{\infty }\frac {i e^{K[1] (i x-\mu (t-1) K[1])}}{K[1]}dK[1]}{2 \pi }\right \}\right \}\]

due to i.c. not at zero

Maple ✓

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

\[u \left (x , t\right ) = \frac {3}{2}-\frac {\operatorname {erf}\left (\frac {x}{2 \sqrt {\mu }\, \sqrt {t -1}}\right )}{2}-t\]

Hand solution

Solve

\[ u_{t}=\mu u_{xx}-1 \]

for \(t>0,-\infty <x<\infty \) with initial conditions \(u\left ( x,0\right ) =f\left ( x\right ) =\left \{ \begin {array} [c]{cc}0 & x\geq 0\\ 1 & x<0 \end {array} \right . \)

Let \(v=u+t\). Hence \(u=v-t\) and \(u_{t}=v_{t}-1\) and \(u_{x}=v_{x}\) and \(u_{xx}=v_{xx}\). The above PDE becomes

\begin{align} v_{t}-1 & =\mu v_{xx}-1\nonumber \\ v_{t} & =\mu v_{xx}\tag {1}\end{align}

Initial conditions do not change. They are \(v\left ( x,0\right ) =u\left ( x,0\right ) =\left \{ \begin {array} [c]{cc}0 & x\geq 0\\ 1 & x<0 \end {array} \right . \). Using Green function for 1D heat PDE on the real line, (also called heat Kernel)

\[ G\left ( x,t\right ) =\frac {1}{\sqrt {4\pi \mu t}}e^{\frac {-x^{2}}{4\mu t}}\]

Then the solution to (1) is

\begin{align*} v\left ( x,t\right ) & =\int _{-\infty }^{\infty }f\left ( x^{\prime }\right ) G\left ( x-x^{\prime },t\right ) dx^{\prime }\\ & =\int _{-\infty }^{0}\frac {1}{\sqrt {4\pi \mu t}}e^{\frac {-\left ( x-x^{\prime }\right ) ^{2}}{4\mu t}}dx^{\prime }\\ v\left ( x,t\right ) & =\frac {-1}{\sqrt {4\pi \mu t}}\int _{0}^{\infty }e^{\frac {-\left ( x-x^{\prime }\right ) ^{2}}{4\mu t}}dx^{\prime }\end{align*}

But \(\int _{0}^{\infty }e^{\frac {-\left ( x-x^{\prime }\right ) ^{2}}{4\mu t}}dx^{\prime }=\sqrt {\pi \mu t}\left ( 1+\operatorname {erf}\left ( \frac {x}{2\sqrt {\mu t}}\right ) \right ) \), hence

\[ v\left ( x,t\right ) =\frac {-1}{2}\left ( 1+\operatorname {erf}\left ( \frac {x}{2\sqrt {\mu t}}\right ) \right ) \]

Since \(u=v-t\) then

\begin{align*} u\left ( x,t\right ) & =\frac {-1}{2}\left ( 1+\operatorname {erf}\left ( \frac {x}{2\sqrt {\mu t}}\right ) \right ) -t\\ & =-\frac {1}{2}-\frac {1}{2}\operatorname {erf}\left ( \frac {x}{2\sqrt {\mu t}}\right ) -t \end{align*}

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3.1.7.8 [254] Practice exam problem

problem number 254

Added May 23, 2019.

From Math 5587 midterm I, Fall 2016, practice exam, problem 14.

Solve for \(u(x,t)\) with IC \(u(x,0)=x\) for \(-\infty <x<\infty ,t>0\)

\begin{align*} u_{t} = u_{xx} \end{align*}

Figure 3.140: PDE specification

Mathematica ✓

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

\[\{\{u(x,t)\to \fbox {$x\text { if }\Re (t)\geq 0$}\}\}\]

Maple ✓

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

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