Use Duhamel's principle in PDE

Question

Let $u(t,x)$ be solution for:

$$ u_{tt}-u_{xx}=q(t,x)u$$

with zero initial conditions. If $q$ is bounded in $\mathbb{R}^2$, show that $u \equiv 0$ using Duhamel's principle

Answer 1

On Duhamel's principle, we must:

$$ u_{tt}-u_{xx}=q(t,x)u \Rightarrow u(t,x) = \iint\limits_{\triangle (t,x)} {q(s,y)} {u(s,y)} ds dy $$

Let $M = \sup\limits_{\triangle (t,x)}|{u(t,x)}|$ be note that the sup is assumed in the triangle. Consider $(t_1,x_1) \in \mathbb{R}^2$ such that $M=|u(t_1,x_1)|$ then

$$M=|u(t_1,x_1)|=|\iint\limits_{\triangle} {q(s,y)} {u(s,y)} ds dy| \leq \iint\limits_{\triangle} |{q(s,y)}| |{u(s,y)}| ds dy \leq M\iint\limits_{\triangle} |{q(s,y)}| ds dy$$

as the function q is limited $\exists K \in \mathbb{R}$ such that$\iint\limits_{\triangle} |{q(s,y)}| ds dy \leq K \iint\limits_{\triangle} ds dy$. It is easy to see that there is a T such that the area of ​​the triangle is less than $1 \over2$.

$$M \leq \iint\limits_{\triangle } M |{q(s,y)}| ds dy \leq M K \iint\limits_{\triangle} ds dy = MK A(\triangle) < \frac{1}{2}M $$

So we conclude that $M = 0$ and therefore $u(t,x) \equiv 0$.