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Assume $u$ solves the initial-value problem $$\begin{cases}u_{tt}-\Delta u = 0 & \text{in } \mathbb{R}^n \times (0,\infty) \\ u = 0, u_t = h & \text{on }\mathbb{R}^n \times \{t=0\}. \end{cases}$$ Show that $v := u_t$ solves $$\begin{cases}v_{tt}-\Delta v = 0 & \text{in } \mathbb{R}^n \times (0,\infty) \\ v=h,v_t=0 & \text{on }\mathbb{R}^n \times \{t=0\}. \end{cases}$$ This is Stokes' rule.

This is Chapter 2, Exercise 18 of PDE Evans, 2nd edition. Here is the work I did so far:

We have $v_{tt}=(u_t)_{tt}=(u_{tt})_t$ and $\Delta v = \sum_{i=1}^n v_{x_i x_i} = \sum_{i=1}^n (u_t)_{x_i x_i} =\left(\sum_{i=1}^n u_{x_i x_i} \right)_t = (\Delta u)_t.$

Therefore, $$v_{tt}-\Delta v = (u_{tt})_t-(\Delta u)_t=(u_{tt}-\Delta u)_t=0_t=0.$$ Also, $v(x,0)=u_t(x,0)=h$. But how do I show $v_t(x,0)=0$? I can't find anything from $v_t(x,0)=u_{tt}(x,0)$.

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2 Answers2

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I think I just figured out the answer to my question: $$v_t(x,0)=u_{tt}(x,0)=[u(x,0)]_{tt}=0_{tt}=0.$$

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    $v_t(x,0) = u_{tt}(x,0) = \Delta u(x,0)$. Your initial conditions in the original equation tell you that the wave is 'flat' everywhere at time $t=0$, i.e. its spatial derivatives and therefore Laplacian are zero. – Ken Wei Feb 25 '16 at 06:06
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I am working on this problem now and I am getting bogged down with this question: How do you know that $u_{tx_ix_i}=u_{x_ix_it}$? We only know that $u\in C^2(\mathbb{R}^n \times [0,\infty))$ by theorems 2 and 3 in section 2.4 of Evans, so the hypotheses of Clairaut's theorem are not satisfied if we want to commute third partial derivatives, correct?

If we knew that the $u_{x_i}$ were twice continuously differentiable, then I would agree that $u_{tx_ix_i}=u_{x_ix_it}$.

Here is an answer to your original question, Cookie.

Luke
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