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Wikipedia claims that: In one variable, the Green's function is a solution of the initial value problem

\begin{cases}u_{t}(x,t)-ku_{xx}(x,t)=0&(x,t)\in \mathbf {R} \times (0,\infty )\\u(x,0)=\delta (x)&\end{cases} where δ is the Dirac delta function. The solution to this problem is the fundamental solution

$$\Phi (x,t)={\frac {1}{\sqrt {4\pi kt}}}\exp \left(-{\frac {x^{2}}{4kt}}\right).$$ And this article I found(https://www.rose-hulman.edu/~bryan/lottamath/heatkern.pdf) seems to agree. In short this function satisfies the initial conditions.

However, how do we even prove this claim? Surely we can't send t to 0+ and claim both sides are the same because they're both $\infty$. Dirac delta isn't the only type of impulse, as far as I know.

Are we...uh...supposed to say "This is effectively a dirac delta because convolving this with another arbitrary function yields the below:"

$$\lim_{t\to 0+} \int_{-\infty}^{\infty} \Phi(x-y,t)f(y) dy = f(x)$$

But we aren't convolving with $\Phi(x,0)$! We convolved with $\Phi$ and took a limit instead, but that can be a whole different thing from convolving with $\Phi(x-y,0)$ - at least I believe so. Especially when $\Phi(x,0)$ isn't even defined.

My question boils down to this: can anyone please convince me (with rigorous arguments) why $\Phi$ can be considered as satisfying the initial conditions.

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  • $\begingroup$ The initial condition should rather be written $\lim_{t\to 0} u(x,t) = \delta(x).$ $\endgroup$ – md2perpe Apr 27 at 9:47
  • $\begingroup$ Ok, but still, 1. Taking a limit of some function and then convolving and 2. Convolving and then taking a limit look like two different things to me in this case. The function isn't continuous at t=0. $\endgroup$ – Jayeon Yi Apr 27 at 9:54
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    $\begingroup$ I think that in the space of distributions (in which both $\Phi$ and $\delta$ belong) limit and convolution commute. $\endgroup$ – md2perpe Apr 27 at 10:00
  • $\begingroup$ @md2perpe Thank you very much for the answer, and for supplying a keyword to google with! I am truly grateful. $\endgroup$ – Jayeon Yi Apr 27 at 12:02

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