A problem on infinite domain diffusion equation 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!
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\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,
  $$
\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{\expo{-\root{s}\verts{x - x'}} \over 2\root{s}}\,\expo{s\pars{t - t'}}
\,{\dd s \over 2\pi\ic}
=\Theta\pars{t - t'}\,{\exp\pars{-\pars{x - x'}^{2}/\bracks{4\pars{t - t'}}}
\over 2\root{\pi\pars{t - t'}}}
$$

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