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I want to show that $\frac{\partial{E}}{\partial{t}}-\Delta E=\delta(t,x) $.

So it suffices to show that $\langle \frac{\partial{E}}{\partial{t}}-\Delta E, \phi \rangle=\phi(0,0) $.

So far I have shown that $ \langle \frac{\partial{E}}{\partial{t}}-\Delta E, \phi \rangle=\lim_{\epsilon \to 0} \int_{\mathbb{R}^n} \frac{1}{(2 \sqrt{\pi \epsilon})^n} e^{-\frac{|x|^2}{4 \epsilon}} \phi(\epsilon,x) dx$

$E$ is the heat kernel, and it's a fundamental solution of the heat equation in the sense of distributions

How could we continue?

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Hint: proof that

1) $(\partial_t - \Delta)K_t(x)=0$ $\forall x \in \mathbb{R}^n$, $\forall t >0$

2) $\displaystyle \int_{\mathbb{R}^n} K_t(x)dx)=1$ $\forall x \in \mathbb{R}^n$, $\forall t >0$

3) $\forall \delta > 0$ we have $\displaystyle \lim_{t \rightarrow 0+} \int_{\mathbb{R}^n \setminus B(0,\delta)} K_t(x) dx = 0$

where $K_t(x):=1/(4\pi t)^{n/2} e^{-|x|^2/4t}$ if $(x,t) \in \mathbb{R}^n \times (0,\infty)$ and zero otherwise.

You can conclude that $K_t(x) \in L^1_{loc}(\mathbb{R}^{n+1})$ is a distribution, and the distributional derivative can be written as

(i) $\displaystyle (\partial_t K_t)(\varphi):=-\lim_{\epsilon \rightarrow 0^+} \int_{\epsilon}^{\infty} \int_{\mathbb{R}^n} \partial_t \varphi(x,t) K_t(x) dx dt$ , $\forall \varphi \in \mathcal{D}(\mathbb{R}^{n+1})$

integrates parts respect $t$ the inner integral function, place $x=y\sqrt{4 \epsilon}$, and use Lebesgue theorem to show that

$\displaystyle \int_{\mathbb{R}^n} \varphi(x,\epsilon)K_\epsilon(x) dx \rightarrow \varphi(0) \cdot 1/\pi^{n/2} \int_{\mathbb{R}^n} e^{-|y|^2} dy =\varphi(0)$

in the previous step using a well-known integral. Now, uses the property 1) the heat kernel, and with an integration by parts, you can conclude that

$\displaystyle \int_{t > \epsilon} \int_{\mathbb{R}^n} \varphi(x,t) \partial_t K_t(x) dx dt = \int _{t > \epsilon} \int_{\mathbb{R}^n} \Delta \varphi(x,t) K_t(x) dx dt$

back to (i) and you can conclude.

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  • $\begingroup$ Could you explain it further to me how you got the following? @JohnMartin $$\int_{\mathbb{R}^n} \phi(x, \epsilon) K_{\epsilon}(x) dx \to \phi(0) \cdot \frac{1}{\pi^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-|y|^2} dy=\phi(0)$$ Also with $\phi(0)$ do you mean $\phi(0, \epsilon)$ ? $\endgroup$ – Evinda Mar 15 '16 at 18:20
  • $\begingroup$ Yes, after replacing you have $1 /(\pi/2) \int \varphi (y \sqrt{4\epsilon},\epsilon) e^{-|y|^2} dy$, then pass to the limit for $\epsilon \rightarrow 0$, and $\varphi(0) \in \mathcal{D}(\mathbb{R}^{n+1})$. In final step proves that $(\partial_t K_t)(\varphi)=(\Delta K_t)(\varphi)+\varphi(0)$, that is our thesis. $\endgroup$ – user288972 Mar 15 '16 at 18:33
  • $\begingroup$ Yes, it's correct, and $\int_{\mathbb{R}^n} e^{-|y|^2} dy = \pi^{n/2}$ $\endgroup$ – user288972 Mar 15 '16 at 19:29
  • $\begingroup$ I have also a question about the proof that $supp(u \ast v) \subset supp u+ supp v$. Could I ask you my question? $\endgroup$ – Evinda Mar 15 '16 at 21:17
  • $\begingroup$ Yes, I had read the question, but now I can not find it. However, I have studied this fact in a slightly different way $\endgroup$ – user288972 Mar 15 '16 at 21:22

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