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Let's consider two problems for diffusion equation.

The first one: $$ u_t = a^2u_{xx},\qquad 0<x<l,\quad 0<t\leq T $$ $$ u(x,0) = \phi(x), \qquad 0 \leq x \leq l $$ \begin{equation} u(0,t)=0,\quad u(l,t)=0, \quad 0 \leq t \leq T \end{equation} and the second one: $$ u_t = a^2u_{xx},\qquad -\infty <x<+\infty,\quad t>0 $$ $$ u(x,0) = \phi(x), \qquad -\infty < x < +\infty $$ For both these cases there are well known expressions for Green's function: $$ G_1(x,\xi,t) = \frac{2}{l}\sum_{n=1}^{\infty} \exp\left({-\left(\frac{\pi n}{l}\right)^2}a^2t\right)\sin\frac{\pi nx}{l}\sin\frac{\pi n\xi}{l} $$ $$ G_2(x,\xi,t) = \frac{1}{\sqrt{4\pi a^2 t}}\exp{\left(-\cfrac{(x-\xi)^2}{4a^2t}\right)} $$ Thus we obtain solutions for first (finite) problem: \begin{equation} u(x,t)=\int\limits_0^lG_1(x,\xi,t)\phi(\xi)\,d\xi \end{equation} and for the second (infinite): \begin{equation} u(x,t)=\int\limits_{-\infty}^{+\infty}G_2(x,\xi,t)\phi(\xi)\,d\xi \end{equation}

Is there any way to obtain Green's function for infinite case from Green's function for range $[0,l]$ (for example by stating $\quad l\rightarrow+\infty$, but I have not achieved any success with this idea)? Or may be second solution from first?

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up vote 2 down vote accepted

Yes, $G_2$ can be obtained as a limit of $G_1$ but I will do a slightly easier limit first.

Write $${G}_1(x,\xi,t)=\frac{2}{l}\sum_{q} \exp\left(-q^2a^2t\right)\sin q x\sin q\xi,$$ where $q=\frac{\pi n}{l}$, $n=1,2,\ldots$ Now since $q_{j+1}-q_j=\frac{\pi}{l}$, for $l\rightarrow\infty$ we have $$\frac{\pi}{l}\sum_{q}\rightarrow \int_0^{\infty}dq,$$ and therefore \begin{align}\lim_{l\rightarrow\infty}{G}_1(x,\xi,t)&=\frac{2}{\pi}\int_0^{\infty}\exp\left(-q^2a^2t\right)\sin q x\sin q\xi\;dq=\\ &=\frac{1}{\sqrt{4\pi a^2t}}\left[\exp{\left(-\cfrac{(x-\xi)^2}{4a^2t}\right)}- \exp{\left(-\cfrac{(x+\xi)^2}{4a^2t}\right)}\right]. \end{align} As you may guess, this is the Green function for the heat equation on the half-line. Now the modifications for the whole line will be as follows:

  • Instead of $G_1(x,\xi,t)$ one should start with translated Green function $$\tilde{G}_1(x,\xi,t)=G_1(x-l/2,\xi-l/2,t).$$ It corresponds to the solution of the heat equation on $[-l/2,l/2]$.

  • Similarly transform the sum over $n$ into an integral over $q$ (you will need to consider odd and even values of $n$ separately). Evaluating two resulting gaussian integrals, you will recover $G_2(x,\xi,t)$.

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Wow! It is terrific idea! But one question appears: – mechanician May 30 '13 at 9:47
when we get integral from sum we do the substitution q=pi*n/l. But following the definition of Riemann integral we do the summation of values of subintegratal function in some point of interval of separation. So in this case we choose in every interval left point, don't we? And therefore we obtain integral limits as 0 and infinity, but not because of boundaries in beginning equation (0 and l). Am I right? – mechanician May 30 '13 at 9:58
@mechanician Not sure to understand the question... The limits $0$ and $\infty$ are related to values of $q$. When you will do the 2nd problem (on $[-l/2,l/2]$) they still will be $0$ and $\infty$. – Start wearing purple May 30 '13 at 12:41
Yes. I have already done it and both integrals had limits $0$ and $\infty$. Firstly I thought, that limits are related to bounds of interval and I was a little bit confused... But now I get it! Thanks a lot! – mechanician May 30 '13 at 14:08
@O.L., please look this… – alekhine Feb 5 '15 at 19:15

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