$L^2$-norm of a solution of the heat equation

Question

Let $u\in\mathcal{C}^{2,1}(\mathbb{R}^n\times (0,\infty))$ be a solution of the heat equation $$\left[\begin{array}{ll}u_t-\Delta u=0& \mathrm{in}\ \mathbb{R}^n\times(0,\infty)\\ u(x,0)=u_0(x),&x\in\mathbb{R}^n\end{array}\right.$$ where $u_0\in\mathcal{C}^0_c(\mathbb{R}^n)$. The boundary conditions above are meant in the following sense: $\lim_{t\to 0, x\to x_0}u(x,t)=u_0(x_0)$ for all $x_0\in\mathbb{R}^n$. If we assume in advance that $|u|\to 0$ as $|x|\to\infty$ then the following relation holds for all $t>0$: $$\|u(\cdot,t)\|_{L^2(\mathbb{R}^n)}\leq\|u_0\|_{L^2(\mathbb{R}^n)}.$$ How to prove this result? I am required to use an energy method whence I started with $E(t):=\int_{\mathbb{R}^n}u(x,t)^2dx$ and tried to prove that this function is differentiable for $t>0$ with non-positive derivative. Therefore I introduced $\Omega_n:=B_n(0)$ and considered $E_n(t):=\int_{\Omega_n}u(x,t)^2dx$ (although it is not clear to me why the derivative of $E_n$ converges to the one of $E$). Differentiating $E_n$ and using the PDE as well as integratio by parts I arrive at: $$E_n'(t)=\int_{\Omega_n}2uu_tdx=2\int_{\Omega_n}2u\Delta udx=-2\int_{\Omega_n}|\nabla u|^2dx+2\oint_{\partial\Omega_n}u\frac{\partial u}{\partial\nu}dS.$$ How to proceed? Although I know that the modulus of $u$ converges to 0, I do not know how to conclude something similar for the normal derivative. Is this ansatz reasonable after all? Thank you very in advance!

Answer 1

Since $u_0$ has compact support, it follows that $u$ and its derivatives decay at infinity. This justifies your argument.

You can arrive to the same conclussion using that $$ u=G_t\ast u_0, $$ where $$ G_t(x)=(4\,\pi\,t)^{-n/2}\,e^{-|x|^2|/4t} $$ is the heat kernel: $$ \|u\|_2=\|G_t\ast u_0\|_2\le\|G_t\|_1\,\|u_0\|_2=\|u_0\|_2. $$