# 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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