From PDE Evans, 2nd edition, page 204

Example 9 (Wave equation from the heat equation). Next we employ some Laplace transform ideas to provide a new derivation of the solution for the wave equation, based--surprisingly--upon the initial-value problem.

$\quad$ Suppose $u$ is a bounded, smooth solution of the initial-value problem: \begin{cases} u_{tt}-\Delta u = 0 & \text{in }\mathbb{R}^n \times (0,\infty) \\ u=g,u_t=0 &\text{on }\mathbb{R}^n \times \{t=0\} \tag{46} \end{cases} where $n$ is odd and $g$ is smooth, with compact support. We extend $u$ to negative times by writing \begin{align} u(x,t)=u(x,-t) & \quad\text{if }x\in \mathbb{R}^n, t < 0. \tag{47} \end{align} Then $$u_{tt}-\Delta u = 0 \quad\text{in }\mathbb{R}^n \times \mathbb{R}.$$ $\quad$ Next define $$v(x,t) := \frac{1}{(4\pi t)^{1/2}} \int_{-\infty}^{\infty} e^{-s^2/4t} u(x,s) \, ds \quad (x \in \mathbb{R}^n, t > 0) \tag{48}$$ Hence $$\lim_{t \rightarrow 0} v = g \quad \text{uniformly on } \mathbb{R}^n.$$ In addition \begin{align} \Delta v(x,t) &= \frac{1}{(4\pi t)^{1/2}} \int_{-\infty}^{\infty} e^{-s^2/4t} \Delta u(x,s) \, ds \\ &= \frac{1}{(4\pi t)^{1/2}} \int_{-\infty}^{\infty} e^{-s^2/4t} u_{ss} \, ds \\ &= \frac{1}{(4\pi t)^{1/2}} \left[-\int_{-\infty}^{\infty} e^{-s^2/4t} u_{s} \left(-\frac{s}{2t} \right) \, ds + \require{cancel} \color{blue}{\cancelto{0}{e^{-\frac{s^2}{4t}} u_s \vert_{-\infty}^{\infty}}}\right] \\ &= \frac{1}{(4\pi t)^{1/2}} \left[ -\int_{-\infty}^{\infty} e^{-s^2/4t} u(x,s) \left(\frac{s^2}{4t^2}-\frac{1}{2t} \right) \, ds + \require{cancel} \color{blue}{\cancelto{0}{\frac{s}{2t}e^{-\frac{s^2}{4t}} u(x,s) \vert_{-\infty}^{\infty}}} \right] \\ &= \frac{1}{(4\pi t)^{1/2}} \int_{-\infty}^{\infty} e^{-s^2/4t} u(x,s) \left(\frac{1}{2t}-\frac{s^2}{4t^2} \right) \, ds \\ &= v_t(x,t) \end{align}

Consequently $v$ solves this initial-value problem for the heat equation: \begin{cases}v_t - \Delta v = 0 & \text{in } \mathbb{R}^n \times (0,\infty) \\ \qquad \quad v=g &\text{on } \mathbb{R}^n \times \{t=0\} \end{cases}

The proof in the textbook actually does not stop here; it continues on for a good two whole pages. I only provided the beginning part of the proof for context.

My question concerns the terms above, which I highlighted in $\color{blue}{\textbf{blue}}$, that cancel to $0$. (By the way, those lines are due to integration by parts.) I realize this cancellation to $0$ is due to the fact that the function $g$ has compact support. But how is this so exactly?

  • $\begingroup$ I didn't look that carefully but I would say that it is because 1) u is symmetric with respect to t and 2) u is assumed to be bounded $\endgroup$ – Thomas Jul 3 '14 at 21:35
  • $\begingroup$ @glace Why bring $v$ into this? You wrote yourself: "Suppose $u$ is a bounded, smooth solution..." The question is what to say about $u_s$. $\endgroup$ – user147263 Jul 3 '14 at 22:13
  • 1
    $\begingroup$ I think we should just assume that $u_s$ is also assumed to be bounded. This is a true fact (follows from having these nice initial values), but I would not want to prove it just for the sake of this illustration. This is an example, not a theorem. The author wanted to focus on what's essential here. $\endgroup$ – user147263 Jul 3 '14 at 22:37
  • $\begingroup$ I guess if $u_s$ is bounded, then for some finite $M \ge 0$, $$|e^{-s^2/4t}u_s \vert_{-\infty}^{\infty}| \le |e^{-s^2/4t}M \vert_{-\infty}^{\infty}| = M|e^{-s^2/4t} \vert_{-\infty}^{\infty}|$$ and $|e^{-s^2/4t} \vert_{-\infty}^{\infty}|=0$ whenever $t > 0$. Hence $e^{-s^2/4t}u_s \vert_{-\infty}^{\infty} = 0$? $\endgroup$ – Cookie Jul 3 '14 at 22:42
  • $\begingroup$ @glace Yes, that's what I meant. The point is, any product involving the Gaussian will vanish at infinity unless the solution does something very weird. With the heat equation, there is a mathematical possibility of such weirdness happening, but one usually assumes it away. $\endgroup$ – user147263 Jul 14 '14 at 5:52

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