Consider that
$$E(t,x)=\dfrac{H(t)}{2\sqrt{\pi t}}e^{-|x|^2/4t}.$$
I want to show that $E_t - E_{xx} = \delta(t)\delta(x)$. This means that we need to show that if $\phi\in \mathcal{D}(\mathbb{R}^2)$ is a test-function, then
$$(E_t-E_{xx},\phi)=(\delta(t)\delta(x),\phi)=\phi(0,0).$$
Now, what I know is that
$$(E_t,\phi)=-(E(t,x),\phi_t)=-\int_{-\infty}^{\infty}\int_{0}^{\infty}\dfrac{1}{2\sqrt{\pi t}}e^{-|x|^2/4t}\dfrac{\partial \phi}{\partial t}(t,x) dtdx.$$
While we have that
$$(E_{xx},\phi)=(E,\phi_{xx})=\int_{-\infty}^{\infty}\int_{0}^{\infty}\dfrac{1}{2\sqrt{\pi t}}e^{-|x|^2/4t}\dfrac{\partial^2\phi}{\partial x^2}(t,x)dtdx.$$
Now I have no idea on how to procede with this. I mean, we have the derivatives of $\phi$ inside which might depend on the two variables. And we have no information about $\phi$, we just know it is a test function.
I've tried integration by parts but I didn't get very far. How can these integrals be solved in order to show that the result is just $\phi(0,0)$?
I believe there's some easy way, but I can't find it. The integration by parts just got me that
$$E_t-E_{xx} = \dfrac{H(t)e^{-x^2/4t}}{16\pi t^3}(x^2(2+\sqrt{\pi t} - 4t(1+\sqrt{\pi t})),$$
but this is not of much help. It is certainly not obvious that when we integrate $\phi$ against this we have $\phi(0,0)$. I already tried everything but it didn't work