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Consider heat flow on an infinite, 1D wire. The temperature T(x,t) obeys the diffusion equation,
$$ \frac{∂T}{∂t} = D \frac{∂^2T}{∂x^2} $$ with initial condition $T(x,0) = δ(x)$.

The heat kernel is given by: $$ T[x, t] =\frac{ 1}{\sqrt{4πDt}} \exp\left(-\frac{x^2}{4Dt}\right) $$

I was asked to verify the heat kernel is a solution. It is easy to show that this satisfies the heat equation. However, to check the initial condition at $t=0$, I must take the limit as $t\to0$ (it shoots to infinity from the look of it). Could someone give me a hint on how to do this?

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  • $\begingroup$ the exponential decay should be much faster than the inverse square root tending to infinity - so I don't think your assumption of divergence is correct. And $T(x,t) = \delta (x)$ for $t \to 0$, so I don't understand your comment @lemon $\endgroup$ – Sanya Sep 16 '17 at 9:57
  • $\begingroup$ Indeed you are correct! It converges to DiracDelta[x] as t->0, even though it looks like it diverges. My question is how will I prove show that the limit converges to DiracDelta. Moretti had given a comprehensive proof below:) $\endgroup$ – Z.Yang Sep 18 '17 at 2:56
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It is easy. Observe that $$T[x,t] = \frac{f(x/s)}{s}$$ where $s = \sqrt{t}$ and $f(x) = T[x,1]$. Since $$\int_{\mathbb R} f(x) dx=1 $$ we also have $$\int_{\mathbb R} \frac{f(x/s)}{s} dx=1 $$ simply by means of a trivial change of variables, defining $z = x/s$ where $s>0$. Now take a bounded continuous function $g : \mathbb R \to \mathbb C$, with the aforementioned change of variables we have $$\int_{\mathbb R} \frac{f(x/s)}{s} g(x) dx= \int_{\mathbb R} \frac{f(x/s)}{s} g(sx/s) s dx/s = \int_{\mathbb R} f(z) g(sz) dz\:.$$ Therefore $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) dx = \lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} f(z) g(0) dz = g(0) \int_{\mathbb R} f(z) dz = g(0) 1 = g(0)\:.$$ In other words $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) = g(0)\:. \tag{1}$$ The only crucial passage is $$\lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} \lim_{s\to 0^+} f(z) g(sz) dz$$ A quite mild condition that guarantees the passage is that $g$ is bounded as already required (as consequence of Lebesgue's dominated convergence theorem).
I stress that (1) (where $g$ is a smooth compactly supported functions and we obtained the result with much weaker hypotheses) is one of the possible ways to rigorously state that $$T[x,0^+] = \delta(x)\:.$$ The heat kernel is used to construct the solution of the heat equation $g=g(x,t)$ out of the initial condition $g(x)$: $$g(x,t) = \int_{\mathbb R} T[x-y,t]g(y) dy $$ satisfies $$ \frac{∂g}{∂t} = D \frac{∂^2g}{∂x^2} $$ for $t>0$ with initial condition $$g(x,0) = g(x)\:.$$ The proof is immediate from (1).

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  • $\begingroup$ Thanks very much for your help! I will work through this first thing tomorrow morning :) $\endgroup$ – Z.Yang Sep 17 '17 at 12:37
  • $\begingroup$ So I've transformed the heat kernel to a periodic version so that it now models a circumference. I have made x -> x+n, and n is evaluated over Z. I.E. T(x,t) -> T(x+n, t) and this can be reduced to ∑ Exp(-Dt(2πn)^2) * Exp(2πinx) using Poisson’s formula (summed over Z), I shall call this new equation S(x, t). $\endgroup$ – Z.Yang Sep 20 '17 at 9:15
  • $\begingroup$ If now I impose a new initial value: g(x) = ∑dnExp(2πinx) And I want to rewrite S(x, t) using this new initial condition. I was told that this can be done using convolution/Greens Method? So I did the following: ∑ ∫dy. { [Exp(-Dt(2πn)^2) * Exp(2πin (x-y) )] * [dnExp(2πiny)] } where the integral is evaluated wrt the boundary condition, in this case it is just [-0.5, 0.5] since f(x+n) has period T=1; and the summation is over Z. But I am not sure whether I could do the convolution of what is inside the summations, and leaving the summation outside integral. $\endgroup$ – Z.Yang Sep 20 '17 at 9:15
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Another way of getting at Valter Moretti's answer using a somewhat outdated (but altogether rigorous) notion of generalized functions is simply to witness that:

$$\int_{-\infty}^\infty T(x, t)\,\mathrm{d} t=1$$

i.e. $T$ is a normalized Gaussian function of $x$ and therefore that we can think of the equivalence class of sequences prototyped by the sequence $T\left(x, 1\right) ,\,T\left(x, \frac{1}{2}\right),\,T\left(x, \frac{1}{3}\right),\,\cdots$ as the generalized function $\delta(x)$ in the way defined in M. J. Lighthill, "An Introduction to Fourier Analysis and Generalized Functions". Here, we conceive of a generalized function not foremost as a member of the algebraic dual of the Schwartz space but rather as an equivalence class of sequences of functions, where the equivalence relationship is $f=\{f_n(x)\}_{n=0}^\infty\sim g=\{g_n(x)\}_{n=0}^\infty$ iff $\lim\limits_{n\to\infty}\langle f_n,\,h\rangle=\langle g_n,\,h\rangle\,\forall h\in\mathscr{S}$ where $\mathscr{S}$ is the Schwartz space. So we have, by the Lighthill definition of $\delta$, that $T(x,\,0^+)=\delta(x)$ and the rest of Valter's answer follows.

Now, I admit that this may seem a bit of a daft answer, because essentially all I am doing is saying in fancy "it's true because that is one possible definition of Dirac delta", but it does recall one approach to the introduction to the notion of generalized functions (the Lighthill / Temple approach) that is still sometimes used in introductory expositions of the idea. When this approach is discussed, the heat kernel is often explicitly singled out as Lighthill's "prototype" for the $\delta$ equivalence class. I sometimes find it helpful to think of generalized functions in this way to see certain results. So I answered, because your question evoked fond memories of my first grasp of the rigorous notion of a generalized function through Lighthill's approach.

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  • $\begingroup$ Thanks very much for your reply! I will try to work on this:) $\endgroup$ – Z.Yang Sep 17 '17 at 12:37
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Take the integral of T with respect to x from - infinity to + infinity and show that it is equal to 1 at all times.

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