MHB Is this the desired bounded set of the wave equation?

[evinda]

Hello! (Wave)

I want to show for the initial value problem of the wave equation

$$u_{tt}=u_{xx}+f(x,t), x \in \mathbb{R}, 0<t<\infty$$

that if the data (i.e. the initial data and the non-homogeneous term $f$) have compact support, then, at each time, the solution has also compact support.

I have thought the following:

The initial data are these, right?

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

The functions $f, \phi, \psi$ have compact support and so they are zero outside a bounded set $[-L,L]$.

The solution of the initial value problem is

$$u(x,t)=\frac{1}{2}[\phi(x+t)+\phi(x-t)]+\frac{1}{2} \int_{x-t}^{x+t} \psi(y) dy+\frac{1}{2} \int_0^t \int_{x-(t-s)}^{x+(t-s)} f(y,s) dy ds.$$

Let $t=T$ arbitrary.

Then

$$u(x,T)=\frac{1}{2}[\phi(x+T)+\phi(x-T)]+\frac{1}{2} \int_{x-T}^{x+T} \psi(y) dy+\frac{1}{2} \int_0^T \int_{x-(T-s)}^{x+(T-s)} f(y,s) dy ds.$$

We check when $u(x,T)=0$.

We have $u(x,T)=0$ if

1. $x+T, x-T \in \mathbb{R} \setminus{[-L,L]} \Rightarrow ((x+T<-L \text{ or } x+T>L)) \text{ and } (x-T<-L \text{ or } x-T>L)$
   
2. $(x-T<-L \text{ and } x+T<-L) \text{ or } (x-T>L \text{ and } x+T>L)$
   
3. $(x-T+s<-L \text{ and } x+T-s<-L) \text{ or } (x-T+s>L \text{ and } x+T-s>L) \text{ for } 0 \leq s \leq T$.

From $2$ we get that $x<-L-T$ or $x>L+T$.

From $3$ we get that $x<-L+s-T$ or $x>L+T-s$ and thus $x<-L$ or $x>L$.Thus the bounded set outside of which $u$ is zero is $[-L-T,L]$, right?

Or have I done something wrong? (Thinking)

 

[I like Serena]

From $3$ we get that $x<-L+s-T$ or $x>L+T-s$ and thus $x<-L$ or $x>L$.

Hey evinda!

Since $0 \le s \le T$, shouldn't that be $x<-L-T$ or $x>L+T$ ? (Wondering)

Thus the bounded set outside of which $u$ is zero is $[-L-T,L]$, right?

Shouldn't that be $[-L-T,L+T]$ ? (Wondering)

 

[evinda]

Hey evinda!

Since $0 \le s \le T$, shouldn't that be $x<-L-T$ or $x>L+T$ ? (Wondering)
Shouldn't that be $[-L-T,L+T]$ ? (Wondering)

Oh yes, right... (Nod)

And from $1$ we get that $(x<-L-T \text{ or } x>L-T) \text{ and } (x<T-L \text{ or } x>L+T)$, right? (Thinking)

Do the fact that $u$ is zero when $x>L-T$ and $x<T-L$ change something?

 

[I like Serena]

Oh yes, right...

And from $1$ we get that $(x<-L-T \text{ or } x>L-T) \text{ and } (x<T-L \text{ or } x>L+T)$, right?

Do the fact that $u$ is zero when $x>L-T$ and $x<T-L$ change something?

Yes.
What should it change? And where did you get that?
The point was that we would find that $u$ has compact support, wasn't it? (Wondering)

 

[evinda]

Yes.
What should it change? And where did you get that? (Wondering)

We get that $u$ is zero outside the boundet set $[T-L,L-T]$, right?

If so, then this holds only if $T<L$. But isn't this a contradiction since $t=T$ was arbitrary? (Worried)

 

[I like Serena]

Thus the bounded set outside of which $u$ is zero is $[-L-T,L+T]$, right?

We get that $u$ is zero outside the boundet set $[T-L,L-T]$, right?

If so, then this holds only if $T<L$. But isn't this a contradiction since $t=T$ was arbitrary?

Didn't we find that $u$ was zero outside $[-L-T,L+T]$?
That holds for any $L\ge 0$ and $T \ge 0$ doesn't it? (Wondering)

 

[evinda]

Didn't we find that $u$ was zero outside $[-L-T,L+T]$?
That holds for any $L\ge 0$ and $T \ge 0$ doesn't it? (Wondering)

Ah, from $1$ we get that $u$ is zero outside $[T-L,L-T]$ but the latter is a subset of $[-L-T,L+T]$, right? (Thinking)

 

[I like Serena]

Ah, from $1$ we get that $u$ is zero outside $[T-L,L-T]$ but the latter is a subset of $[-L-T,L+T]$, right?

(Nod)

 

[evinda]

(Nod)

Great... Thank you very much! (Happy)