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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!

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    $\begingroup$ Would not passing to Fourier transforms help? $\endgroup$
    – Urgje
    Oct 11, 2015 at 12:05
  • $\begingroup$ @Urgje Unfortunately, the Fourier transform was not introduced in my course. I am required to use an "energy method" which brought me to my ansatz above. $\endgroup$
    – JohnSmith
    Oct 11, 2015 at 12:29
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    $\begingroup$ I guess it is quite difficult. I can't see how you can show that $E$ is finite. $\endgroup$
    – user99914
    Oct 11, 2015 at 12:51

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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: $$ \|u\|_2=\|G_t\ast u_0\|_2\le\|G_t\|_1\,\|u_0\|_2=\|u_0\|_2. $$

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  • $\begingroup$ Thank you for your answer! How can one prove the decay of the derivatives at infinity? $\endgroup$
    – JohnSmith
    Oct 11, 2015 at 18:04
  • $\begingroup$ From the formula $u=G_t\ast u_0$. If $u_0$ is not smooth, there will be a dependence on $t$. $\endgroup$ Oct 11, 2015 at 18:05
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    $\begingroup$ Ok, I know that $u=G_t\ast u_0$ solves the present problem. But, unfortunately I have no uniqueness result available. Is there a way to prove the decay of the derivatives for an a priori general solution of the problem? $\endgroup$
    – JohnSmith
    Oct 11, 2015 at 18:11
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    $\begingroup$ There is no uniqueness. But the other solutions are "wild", and they are not in $L^2$ for $t>0$. There is uniqueness in the class of functions bounded by a quadratic exponential. $\endgroup$ Oct 11, 2015 at 18:35
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    $\begingroup$ Ok, this uniqueness result is new to me. So for only having a general solution of the given problem, it appears to be rather difficult to say anything if one does not know that it actually coincides with the convolution of heat kernel and initial data. $\endgroup$
    – JohnSmith
    Oct 11, 2015 at 21:52

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