A problem on infinite domain diffusion equation

Question

Consider the following problem $$u_t-u_{xx}=p(x,t), -\infty<x<\infty,t>0$$ $$u(x,0)=0$$ $$u\rightarrow0 \text{ as } x\rightarrow \pm \infty$$ This can be solved using many sub problems as follows.

Let $\zeta>0, \tau >0$. Consider the neighborhood $\zeta\leq x \leq \Delta \zeta+\zeta,\tau\leq t \leq \Delta \tau+\tau$ and $p(\zeta,\tau) $ is constant in this neighborhood. Now we build a new problem as follows.

$$u_{1t}-u_{1xx}=p(\zeta,\tau) \delta(x-\zeta)\delta(t-\tau)d\zeta d\tau, -\infty<x<\infty,t>0$$ $$u_1(x,0)=0$$ $$u_1\rightarrow0 \text{ as } x\rightarrow \pm \infty$$ Then the solution is

$$u_1(x,t)=\frac{p(\zeta,\tau) d\zeta d\tau}{\sqrt{4\pi(t-\tau)}}exp(-\frac{(x-\zeta)^2}{4(t-\tau)})$$

So the solution to main problem is $$u(x,t)=\int_0^t \int_{-\infty}^{\infty}\frac{p(\zeta,\tau)}{\sqrt{4\pi(t-\tau)}}exp(-\frac{(x-\zeta)^2}{4(t-\tau)}) d\zeta d\tau$$ I am not sure whether everything I have done here is right. But a similar way of solving is given below. $$u_{1t}-u_{1xx}=p(\zeta,\tau) \delta(x-\zeta)\delta(t-\tau), -\infty<x<\infty,t>0$$ $$u_1(x,0)=0$$ $$u_1\rightarrow0 \text{ as } x\rightarrow \pm \infty$$ Then the solution is

$$u_1(x,t)=\frac{p(\zeta,\tau) }{\sqrt{4\pi(t-\tau)}}exp(-\frac{(x-\zeta)^2}{4(t-\tau)})$$ $$u(x,t)=\int_0^t \int_{-\infty}^{\infty}\frac{p(\zeta,\tau)}{\sqrt{4\pi(t-\tau)}}exp(-\frac{(x-\zeta)^2}{4(t-\tau)}) d\zeta d\tau$$ Are both of these methods correct or one of them is? Any help will me much appreciated as this has confused me for many days! thanks!

Answer 1

Let $p'$ be the incremental contribution to $p$ from the given domain $D$. $$p'=p(\zeta,\tau)[H(x-\zeta)-H(x-\zeta -\zeta \tau][H(t-\zeta)-H(t-\tau -\Delta \tau]$$ Send $\Delta \tau$ and $\Delta \zeta$ to zero get the delta functions and define the new problem using this $p'$

So $$u_{1t}-u_{1xx}=p(\zeta,\tau) \delta(x-\zeta)\delta(t-\tau)d\zeta d\tau, -\infty<x<\infty,t>0$$ $$u_1(x,0)=0$$ $$u_1\rightarrow0 \text{ as } x\rightarrow \pm \infty$$ Then the solution is

$$u_1(x,t)=\frac{p(\zeta,\tau) d\zeta d\tau}{\sqrt{4\pi(t-\tau)}}exp(-\frac{(x-\zeta)^2}{4(t-\tau)})$$ is correct

Answer 2

The original equation can be rewritten as

$$u_{t}-u_{xx}=\int_{0}^{t}\int_{-\infty }^{\infty }p(\zeta,\tau) \delta(x-\zeta)\delta(t-\tau)d\zeta d\tau, -\infty<x<\infty,t>0$$

and then for the first procedure the correct expression is

$$u_{1t}-u_{1xx}=p(\zeta,\tau) \delta(x-\zeta)\delta(t-\tau), -\infty<x<\infty,t>0$$

and identical expression for the second procedure. Then the two procedures are exactly the same.

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\,{\rm u}_{t}\pars{x,t} - \,{\rm u}_{xx}\pars{x,t} =\,{\rm p}\pars{x,t}\,,\ x \in {\mathbb R}\,,\ t > 0\ \mbox{with}\ \,{\rm u}\pars{x,0}=0\ \mbox{and}\ \lim_{x\ \to\ \pm\infty}\,{\rm u}\pars{x,t} = 0.}$

Multiply both members of the equation by $\expo{-st}$ and integrate over $\ds{t \in \pars{0,\infty}}$. Well get

\begin{align} \,\tilde{{\rm u}}_{xx}\pars{x,s} - s\,\tilde{{\rm u}}\pars{x,s} =-\,\tilde{{\rm p}}\pars{x,s}\quad\mbox{where}\quad \left\{\begin{array}{rcl} \tilde{\fermi}\pars{x,s} & \equiv & \int_{0}^{\infty}\fermi\pars{x,t}\expo{-st}\,\dd t \\[1mm] \fermi\pars{x,t} & \equiv & \int_{\gamma - \infty\ic}^{\gamma + \infty\ic} \,\tilde{\fermi}\pars{x,s}\expo{st}\,{\dd s \over 2\pi\ic} \end{array}\right. \end{align} are usual Laplace transforms.

We'll write $\ds{\,\tilde{\rm u}\pars{x,s}}$ as:

\begin{align} \,\tilde{\rm u}\pars{x,s} &=-\int_{-\infty}^{\infty}\,{\rm G}\pars{x,s,x'}\,\tilde{\rm p}\pars{x',s}\,\dd x' \end{align}

such that $\ds{\pars{\partiald[2]{}{x} - s}\,{\rm G}\pars{x,s,x'} = \delta\pars{x - x'}}$ and $\ds{\lim_{x\ \to\ \pm\infty}\,{\rm G}\pars{x,s,x'} = 0}$. The differential equation is equivalent to

$$ \pars{\partiald[2]{}{x} - s}\,{\rm G}\pars{x,x'} = 0\,,\quad x \not= x'\,;\qquad \left\{\begin{array}{lcl} \left.\lim_{\epsilon\ \to\ 0^{+}}\,{\rm G}\pars{x,s,x'} \right\vert_{x\ =\ x' - \epsilon}^{x\ =\ x' + \epsilon} & = & 0 \\[5mm] \left.\lim_{\epsilon\ \to\ 0^{+}}\partiald{\,{\rm G}\pars{x,s,x'}}{x} \right\vert_{x\ =\ x' - \epsilon}^{x\ =\ x' + \epsilon} & = & 1 \end{array}\right. $$

Then,

\begin{align} \,{\rm G}\pars{x,x'} &=\left\{\begin{array}{lclrcl} \,{\rm A}\pars{x',s}\expo{\root{s}x} & \mbox{if} & x & < & x' \\ \,{\rm B}\pars{x',s}\expo{-\root{s}x} & \mbox{if} & x & > & x' \end{array}\right. \end{align}

We'll get a couple of equations which determine $\ds{\,{\rm A}\pars{x',s}}$ and $\ds{\,{\rm B}\pars{x',s}}$:

\begin{align}&\left\{\begin{array}{rcrcl} \expo{\root{s}x'}\,{\rm A}\pars{x',s} & - & \expo{-\root{s}x'}\,{\rm B}\pars{x',s} & = & 0 \\[2mm] -\root{s}\expo{\root{s}x'}\,{\rm A}\pars{x',s} & - & \root{s}\expo{-\root{s}x'}\,{\rm B}\pars{x',s} & = & 1 \end{array}\right. \\[5mm] &\,{\rm A}\pars{x',s}=-\,{\expo{-\root{s}x'} \over 2\root{s}}\,,\qquad \,{\rm B}\pars{x',s}=-\,{\expo{\root{s}x'} \over 2\root{s}} \\[5mm]&\imp\ \,{\rm G}\pars{x,s,x'}= -\,{\expo{-\root{s}\verts{x - x'}} \over 2\root{s}} \end{align}

$\ds{\,\tilde{\rm u}\pars{x,s}}$ is given by:

\begin{align} \,{\rm u}\pars{x,s} &=\half\int_{-\infty}^{\infty}{\expo{-\root{s}\verts{x - x'}} \over \root{s}}\, \,\tilde{\rm p}\pars{x',s}\,\dd x' \\[5mm]&=\half\int_{-\infty}^{\infty} {\expo{-\root{s}\verts{x - x'}} \over \root{s}}\, \int_{0}^{\infty}\,{\rm p}\pars{x',t'}\expo{-st'}\,\dd t'\,\dd x' \\[5mm]&=\half\int_{0}^{\infty}\int_{-\infty}^{\infty} {\expo{-st' -\root{s}\verts{x - x'}} \over \root{s}}\, \,{\rm p}\pars{x',t'}\,\dd x'\,\dd t' \\[1cm]& \,{\rm u}\pars{x,t} =\int_{0}^{\infty}\!\!\!\!\!\int_{-\infty}^{\infty}\bracks{% \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{-\root{s}\verts{x - x'}} \over 2\root{s}}\,\expo{s\pars{t - t'}} \,{\dd s \over 2\pi\ic}} \,{\rm p}\pars{x',t'}\,\dd x'\,\dd t' \end{align}

However, $$ \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{-\root{s}\verts{x - x'}} \over 2\root{s}}\,\expo{s\pars{t - t'}} \,{\dd s \over 2\pi\ic} =\Theta\pars{t - t'}\,{\exp\pars{-\pars{x - x'}^{2}/\bracks{4\pars{t - t'}}} \over 2\root{\pi\pars{t - t'}}} $$

$$\color{#66f}{\large\,{\rm u}\pars{x,t}} =\color{#66f}{\large% \int_{0}^{t}\int_{-\infty}^{\infty} \exp\pars{-\,{\bracks{x - x'}^{2} \over 4\bracks{t - t'}}}\,{\rm p}\pars{x',t'}\, {\dd x'\,\dd t' \over 2\root{\pi\pars{t - t'}}}} $$