From David Borthwick - Introduction to Partial Differential Equations, Exercise 6.3b (paraphrased),

Let $\Omega \subset \Bbb R^n$ be a bounded domain with piecewise $C^1$ boundary. Suppose $u(t,x)$ satisfies the heat equation $$u_t - \Delta u = 0, \\ u(0,x) > 0 \quad\text{for } x\in \Omega, \\ u(t,x) = 0 \quad\text{for } x\in \partial\Omega.$$ Define the total thermal energy at time $t$ by $$U(t) = \int_{\Omega}u(t,x)\,\Bbb dx.$$ Show that $U(t)$ is decreasing.

My attempts so far:

  1. Using the divergence theorem, I can start like $$U'(t) = \int_{\Omega} u_t(t, x) \,\Bbb dx = \int_{\Omega} \Delta u(t, x) \,\Bbb dx = \int_{\partial\Omega} \nabla u(t, x) \cdot \mathbf{n} \,\Bbb dx = \ldots ?$$
  2. By the maximum principle, it should end like $$\ldots = \int_{?} -u(t,x) \,(?)\,\Bbb dx \leq \int_{?} -\min_{[0,t]\times \bar{\Omega}} u(t,x) \,(?)\,\Bbb dx \leq 0$$ since $u(t,x)\geq 0$ for $(t,x) \in (\{0\}\times\Omega) \cup ([0,t]\times \partial \Omega)$.
  3. For the missing steps in the middle, I get stuck because I can no longer “integrate by parts” to bring $\nabla u$ “up” to $u$. Also, the negative sign seems to be crucial to prove $U$ is decreasing, but I have no idea where it could come from.
  4. Alternatively observe that $u_t$ also satisfies a heat equation. Maybe we can use the maximum principle on $u_t$ directly, but we don’t know $u_t(0,x)$ for $x\in \Omega$.

Are there more things to observe or are these the right tracks? Preferably only elementary results related to heat equation would be used. Any help is appreciated.


1 Answer 1


Have you tried the Hopf's lemma? (link: https://en.wikipedia.org/wiki/Hopf_lemma)

  • $\begingroup$ Thank you! The lemma seems powerful. Since we haven't learnt about harmonic functions / Laplace equation yet (I see the proof of the lemma needs Harnack's inequality which in turn needs Poisoon's formula), I wonder if there is a proof to the question not relying on such? Sorry that I cannot accept the answer although I appreciate it. I have edited the question to make it more specific. $\endgroup$
    – Jonat789
    Mar 22, 2021 at 0:08
  • $\begingroup$ no worries. I appreciate your feedback $\endgroup$
    – Fei Cao
    Mar 22, 2021 at 0:28

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