HW5 Q2 Solution (MTH847 Fall 2017)

Question

Consider the nonlinear wave equation on \(\mathbb{R}\times\mathbb{R}\): \begin{gather} \Box u = \lambda u^p \newline u(0,x) = g(x)\newline \partial_t u(0,x) = h(x) \end{gather} where \(\lambda\in \mathbb{R}\) and \(p\geq 2\) is an integer. We can rewrite this using Duhamel's formula applied to D'Alembert's solution as the integral equation \[ u(t,x) = \frac12 [ g(x+t) + g(x-t) + \int_{x-t}^{x+t} h(y) ~\mathrm{d}y ] - \frac12 \int_0^t \int_{x - (t-s)}^{x + (t-s)} \lambda u(s,y)^p ~\mathrm{d}y ~\mathrm{d}s.\] Prove that for every \(g, h\in C^0_c(\mathbb{R})\), there exists a time \(T\) and a function \( u\in C^0([0,T]\times \mathbb{R})\) such that \(u\) verifies the integral equation given above.
(Hint: emulate our proof of local existence for the nonlinear heat equation. Construct a sequence of functions by setting \[ u_{i+1}(t,x) = \frac12 [ g(x+t) + g(x-t) + \int_{x-t}^{x+t} h(y) ~\mathrm{d}y ] - \frac12 \int_0^t \int_{x - (t-s)}^{x + (t-s)} \lambda u_i(s,y)^p ~\mathrm{d}y ~\mathrm{d}s \] and showing that the sequence converges uniformly on \([0,T]\times\mathbb{R}\) if \(T\) is chosen small enough.)

Solution

Let \(u_0 = \frac12 [g(x+t) + g(x-t) + \int_{x-t}^{x+t} h(y) ~\mathrm{d}y\) and define \(u_i\) for \(i \geq 1\) inductively as above. To reach the desired conclusion we will show that

  1. The sequence \( (u_i)\) is uniformly bounded on \( [0,T_0] \times \mathbb{R}\) for some \(T_0 > 0\).
  2. The telescoping series \( \sum u{i+1} - u{i} \) is absolutely convergent on \( [0,T] \times \mathbb{R}\) for some \(T \in (0, T_0]\), this will be achieved by uniform comparison with a geometric series.

First, however, note that if \(g,h\in C^0_c(\mathbb{R})\) are as hypothesized, then there exists some large number \(M\) such that \[ |g(x)| < M, \quad |h(x)| < M, \quad \int |h(y)| ~\mathrm{d}y < M.\] Thus by construction we must have \[ |u_0(t,x)| < \frac32 M \] for every \(t,x\).

Claim 1: There exists \(T_0 > 0\) such that for every \(t\in [0,T_0]\) and \(x\in \mathbb{R}\), \( |u_i(t,x)| < 2M\).

By induction, it suffices to show that there exists \(T_0 > 0\) such that the statement \[ \sup_{(t,x)\in [0,T_0]\times\mathbb{R}} |u_{i}(t,x)| < 2M \implies \sup_{(t,x)\in [0,T_0]\times\mathbb{R}} |u_{i+1}(t,x)| < 2M \] is true. Observe that by our formula \[ |u_{i+1}(t,x)| < \frac32 M + \frac{|\lambda|}{2} \int_0^t \int_{x - (t-s)}^{x+(t-s)} |u_i(s,y)|^p ~\mathrm{d}y ~\mathrm{d}s. \] By induction hypothesis we have then \[ |u_{i+1}(t,x)| < \frac32 M + 2^{p-1}|\lambda| M^p \int_0^t \int_{x - (t-s)}^{x+(t-s)} 1 ~\mathrm{d}y ~\mathrm{d}s. \] The integral \[ \int_0^t \int_{x - (t-s)}^{x+(t-s)} 1 ~\mathrm{d}y ~\mathrm{d}s = t^2 \] since it is just the area of a triangle. So provided we choose \[ t \leq T_0 = \frac{1}{(2M)^{(p-1)/2} |2\lambda|^{1/2}} \] we guarantee that \[ | u_{i+1}(t,x)| < 2 M \] as desired.

Claim 2: There exists \(T \in (0,T_0]\) such that \( \sup |u_{i+1} - u_i| \leq \frac12 \sup |u_{i} - u_{i-1}|\) on $(0,T]\times \mathbb{R}$.

Observe that \[ u_{i+1}(t,x) - u_i(t,x) = - \frac{\lambda}{2} \int_0^t \int_{x - (t-s)}^{x+(t-s)} u_i^p(s,y) - u_{i-1}^p(s,y) ~\mathrm{d}y ~\mathrm{d}s.\] Notice that \[ u_i^p - u_{i-1}^p = (u_i - u_{i-1}) (u_i^{p-1} + u_i^{p-2} u_{i-1} + \cdots + u_{i-1}^{p-1}) \] so that \ \sup |u_i^p - u_{i-1}^p | \leq p (2M)^{p-1} \sup |u_i - u_{i-1}|.\ Arguing as before we then have, for any \(T \in (0,T_0]\), \[ \sup_{[0,T] \times \mathbb{R}} | u_{i+1} - u_{i} | \leq \frac{|\lambda|}{2} p (2M)^{p-1} \sup_{[0,T] \times \mathbb{R}} | u_i - u_{i-1} | T^2.\] So provided we choose \[ T = \min( \frac{1}{\sqrt{|\lambda|p} (2M)^{(p-1)/2}}, T_0) \] we prove the claim.

Claim 1 and Claim 2 together implies that the sequence of functions \( (u_i)\) converges uniformly on the domain \( [0,T]\times \mathbb{R}\), and hence the integral equation has a solution.

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