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I want to find the functional form of the Green function G(x,t) for a parabolic differential equation:

$$ \frac{\partial{}G(x,t)}{\partial{}t}=a\frac{\partial{}^2G(x,t)}{\partial{}x^2}+\delta(t)\delta(x)$$

Then, I'd like to write the general solution of the heat equation:

$$\frac{\partial{}T(x,t)}{\partial{}t}=a\frac{\partial{}^2T(x,t)}{\partial{}x^2}+f(x,t) $$

where f is a known source function.

I think I need to use a transform like that of Laplace or Fourier. How can I handle the Dirac delta function in this equation?

Jackson Hart
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2 Answers2

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You can separate the dimensions. You start by trying to find a function $G(x,t)$ such that

  • $G(x,t)$ satisfies the homogeneous heat equation in all of space and time, except formally at $x=t=0$;
  • At $t=0$, $G(x,t)$ is zero forall non-zero values of $x$;
  • For any nonzero real $(a,b)$, $\int_{-|a|}^{|b} G(x,0) dx = 1$.

I'm not saying yet that this function will be the Green's function you are looking for, but say you find such a function. An example function would be something like $$ G(x,t) = \frac{1}{4\pi\alpha} \frac{1}{\sqrt{t}} e^{-\frac{x^2}{4\alpha t}} $$ where the powers of 2 and $\pi$ are chosen carefully such that for all non-zero $t = t_0$, $$\int_{-\infty}^\infty G(x,t_0) dx = 1 $$ thus ensuring (at least plausibly) that when $t = 0$ the integral remains $1$. And when $t=0$, since outside of $x=0$ the function is zero, yet it's integral is $1$, that means it is a $\delta$-function in one dimension.

his would work if all your heat sources were instantaneous at $t=0$. But notice that the heat equation is time invariant, in the sense that you can make a substitution of $t' = t+\tau$ and the form remains the same.

So in terms of that $G(x,t)$ your general solution will be: $$ h(x,y) = \int_{\tau = -\infty }^\infty \int_{x'=-\infty}^\infty f(x,\tau) G(x-x',t-\tau) dx' \, d\tau $$

There will be issues of solution stability as the problem is ill-posed if you try to find initial conditions that lead to arbitrary future heat distributions, of course.

Mark Fischler
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  • Im confused. How does this address the dirac delta part? – Jackson Hart Aug 12 '15 at 01:28
  • The delta part is addressed by fining a function (that solves the homgeneous equation) which at time zero is zero for all non-zero values of $x$, yet which integrates to 1. That embodies the relevant features of the delta function. – Mark Fischler Aug 19 '15 at 16:54
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you start with a homogeneous equation $$ \frac{\partial{}G(x,t)}{\partial{}t} - a\frac{\partial{}^2G(x,t)}{\partial{}x^2} =0 $$ the solution is $$ G(x,t) = \frac{\mathrm{exp}\left( -\frac{{{x}^{2}}}{ 4 \, a \, t}\right) }{\sqrt{t \; 2 \; \pi }}$$ The area below the function graph is $$ A_t = \frac{\int_{-\infty }^{\infty }{{e}^{-\frac{{{x}^{2}}}{4\cdot a\cdot t}}}dx}{\sqrt{2 \pi t}} = \sqrt{2 \pi a} $$ Therefore, $$ \lim_{t \rightarrow 0^+} \frac{G(x,t)}{A_t} =\delta(t) $$

Notice that there is no product of delta's.

Then you look for a solution in a convolution form

$$ T(x,t) = \int_{ -\infty }^\infty \int_{-\infty}^\infty f(x' -x,\tau'-t) G(x',t') dx' \, dt' $$

and you require that both satisfy the same equation.

That is

$$ \frac{\int_{-\infty }^{\infty }\frac{\left( \frac{d}{d\,t} \mathrm{f}\left( x-z,t-u\right) \right) {{e}^{-\frac{{{z}^{2}}}{4\cdot a\cdot u}}}}{\sqrt{u}}du}{\sqrt{2 \, \pi }}-\frac{a \int_{-\infty }^{\infty }\frac{\left( \frac{{{d}^{2}}}{d\,{{x}^{2}}} \mathrm{f}\left( x-z,t-u\right) \right) \cdot {{e}^{-\frac{{{z}^{2}}}{4\cdot a\cdot u}}}}{\sqrt{u}}du}{\sqrt{2\,\pi }}= \sqrt{2 \pi a} \delta(x- z) $$

user48672
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