#### 3.1.7 Inﬁnite domain

3.1.7.3 [249] UnitBox I.C.
3.1.7.4 [250] No source
3.1.7.5 [251] constant as source
3.1.7.6 [252] No intial conditions
3.1.7.7 [253] piecewise I.C.
3.1.7.8 [254] Practice exam problem

##### 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 $u(x,0)= e^{-x^2}$

Mathematica

$\left \{\left \{u(x,t)\to \frac {e^{-\frac {x^2}{4 t+1}}}{\sqrt {4 t+1}}\right \}\right \}$

Maple

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

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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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 $u(x,0)= x$

Mathematica

$\{\{u(x,t)\to t \sin (t)+x\}\}$

Maple

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

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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.

Mathematica

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

Maple

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

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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 $$u(x,0)=f(x)$$

Mathematica

$\left \{\left \{u(x,t)\to \int _{-\infty }^{\infty } \frac {f(s) e^{-\frac {(s-x)^2}{4 k t}}}{2 \sqrt {\pi } \sqrt {k t}} \, ds\right \}\right \}$

Maple

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

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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 $$u(x,0)=\sin (x)$$

Mathematica

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

Maple

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

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##### 3.1.7.6 [252] No intial conditions

problem number 252

Solve the heat equation for $$u(x,t)$$ $u_t= u_xx$

Mathematica

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

Maple

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

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

Solve the heat equation for $$u(x,t)$$ on real line with $$t>0$$ $u_t = \mu u_{xx} - 1$ With initial condition $u(x,1)= \begin {cases} 0 & x \geq 0 \\ 1 & x < 0 \end {cases}$

Mathematica

$\left \{\left \{u(x,t)\to \frac {\int _{-\infty }^{\infty }\frac {i e^{K[1] (i x-\mu (t-1) K[1])}}{K[1]}dK[1]}{2 \pi }\right \}\right \}$ due to i.c. not at zero

Maple

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

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

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*}

Mathematica

$\{\{u(x,t)\to x\}\}$

Maple

$u \left (x , t\right ) = x$