MHB Solving the Wave Equation: Finding $u(x,t)$

evinda
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Hello! (Wave)

Let $$u_{tt}-c^2 u_{xx}=0, x \in \mathbb{R}, t>0 \\ u(x,0)=0, u_t(x,0)=g(x)$$
where $g \in C^1(\mathbb{R})$ with $g(x)>0$ for $x \in (0,1)$, $g(x)=0$ for $x \geq 1$ and $g(x)=-g(-x)$ for $x \leq 0$. I want to find the sets of $\{ (x,t): x \in \mathbb{R}, t \geq 0 \}$ where $u=0, u<0$ and $u>0$.

I have thought the following.

We have that $u(x,t)=\frac{\phi(x+ct)+ \phi(x-ct)}{2}+ \frac{1}{2c} \int_{x-ct}^{x+ct} \psi(\zeta) d{\zeta}$

where $u(x,0)=\phi(x)$ and $u_t(x,0)=\psi(x)$.

So in our case, $\phi(x)=0$ and $\psi(x)=g(x)$.

So we have that $u(x,t)= \frac{1}{2c} \int_{x-ct}^{x+ct} \psi(\zeta) d{\zeta}$.

We have that $u>0$ when $x-ct \in (0,1)$ and $x+ ct \in (0,+\infty)$, $u=0$ when ( $x+ct \geq 1$ and $x-ct \geq 1)$ or $(x+ct \leq -1$ and $x-ct \leq -1)$ , and $u<0$ when $x+ct \in (-1,0)$ and $x-ct \in (-\infty,0)$ .Am I right? (Thinking)
 
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Hey evinda! (Smile)

It looks correct to me.
 
I like Serena said:
Hey evinda! (Smile)

It looks correct to me.

Great... Thank you! (Smirk)
 
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