Heat Kernel Property Let $\phi$ be the Heat Kernel in $\mathbb{R}^n$, i. e.
$$\phi (x,t)={(4\pi t)}^{-n/2}\exp\left( - \frac{\mid x \mid ^2}{4t}\right)$$
and let $u$ satisfy Heat equation. Show that:
$$\frac{d}{dt}\int\limits_{\mathbb{R}^n}\phi |Du|^2 dx=-2\int\limits_{\mathbb{R}^n} \phi (\Delta u)^2 dx$$
What I have tried:
We know $\phi_{t}=\Delta\phi$. We get:
$$\frac{d}{dt}\int\limits_{\mathbb{R}^n}\phi |Du|^2dx=\int\limits_{\mathbb{R}^n}\phi_{t}\langle Du,Du \rangle dx+\int\limits_{\mathbb{R}^n}2\phi\langle Du_{t},Du \rangle dx$$
Then you can put $u_{t}=\Delta u$ and apply integration by parts for the second term. But I do not know how to get rid of $\phi$.
 A: Unless I overlook a property of $\phi$, I think what you are stating is not true.
Consider the 1D case, $n = 1$.
Then, starting from what you wrote down
$$\partial_t\int_\mathbb{R} \phi (u_x)^2 dx = \int_\mathbb{R} \phi_t (u_x)^2  + 2 \phi u_x u_{xt} dx$$
Focus on the first summand in the integral: $\phi_t = \Delta \phi \overset{1D}{=} \phi''$ and define $f = (u_x)^2$
$$\int_\mathbb{R} \phi_{xx} f dx \overset{\text{Int. by parts}}{=} \int_\mathbb{R} (\phi_x f)_x dx - \int_\mathbb{R} \phi_x f_x dx = \phi_x f \Big \vert_\mathbb{R} - \int_\mathbb{R} \phi_x f_x dx $$
For $\phi_x = \frac{1}{\sqrt{4 \pi t}} \frac{-x}{2t} \exp \Big( \frac{-x^2}{4t} \Big)$ and $f= (u')^2$ bounded, $ \phi_x f \Big \vert_\mathbb{R} = 0$ and thus (again integration by parts)
$$\int_\mathbb{R} \phi_{xx} f dx = - \int_\mathbb{R} \phi_x f_x dx \overset{\text{Int. by parts}}{=}  - \int_\mathbb{R} (\phi f_x)_x dx +  \int_\mathbb{R} \phi f_{xx} dx$$
Since $- \int_\mathbb{R} (\phi f_x)_x dx = -\phi f_x \Big \vert_\mathbb{R} = 0$  (for bounded $f_x = (u_x)^2_x = 2 u_x  u_{xx})$
So you have
$$\partial_t\int_\mathbb{R} \phi (u_x)^2 dx =\int_\mathbb{R} \phi \big( f_{xx} + 2 u_xu_{xxx} \big) dx$$
Since $f_{xx} = \partial_x \partial_x (u_x)^2 = \partial_x 2(u_x \partial_x u_x )= 2\partial_x (u_x u_{xx})$
and $2u_xu_{xxx} = 2\partial_x(u_x u_{xx}) - 2 u_{xx}u_{xx}$
$$\partial_t \int_\mathbb{R} \phi (u_x)^2 dx =\int_\mathbb{R} \phi \big( 2\partial_x (u_x u_{xx}) + 2\partial_x(u_x u_{xx}) - 2 u_{xx}u_{xx} \big) dx \\
= 4\int_\mathbb{R} \phi \partial_x (u_x u_{xx}) dx - 2\int_\mathbb{R} \phi u_{xx}u_{xx} dx$$
Now consider $u(x, t) = \sin(x) e^{-t}$. Clearly, $u_t = u_{xx} = -\sin(x)e^{-t}.$
Compute $u_x = \cos(x)e^{-t}, u_{xx} = -\sin(x) e^{-t}$ and
$u_x u_{xx} = -\sin(x)\cos(x) e^{-2t}$.
Evaluate $\partial_x (u_x u_{xx}) =  e^{-2t} \Big( - \cos(x) \cos(x) + \sin(x) \sin(x) \Big) = - \cos(2x) e^{-2t} $.
Plugging this into the "remainder"
$$4\int_\mathbb{R} \phi \partial_x (u_x u_{xx}) dx = -\frac{4 e^{-2t}}{\sqrt{4\pi t}} \int_\mathbb{R} \cos(2x) \exp \Big( \frac{-x^2}{4t} \Big) dx $$
Setting for instance $t=0.25$ and asking Wolfram Alpha for help shows that the remainder does not vanish.
