MHB Compact Support of Wave Equation IVP Solutions

[evinda]

Hello! (Wave)

I want to prove that if the initial data of the initial value problem for the wave equation have compact support, then at each time the solution of the equation has also compact support.

Doesn't the fact that a function has compact support mean that the function is zero outside some bounded set? (Thinking)

The initial value problem for the wave equation is the following, right?

$$u_{tt}=c^2 u_{xx} \text{ in } \mathbb{R} \times (0,\infty) \\ u(\cdot, 0)=\phi \text{ in } \mathbb{R} \\ u_t(\cdot,0)=\psi \text{ in } \mathbb{R}$$

Is it meant with "the initial data of the initial value problem for the wave equation have compact support" that $\phi$ and $\psi$ are $0$ ?

If so, it follows directly that the solution is $0$, since it is equal to $u(x,t)=\frac{1}{2} \left[ \phi(x-ct)+\phi(x+ct)\right]+\frac{1}{2c} \int_{x-ct}^{x+ct} \psi(\tau) d{\tau}$, right?But if so, why is it stated as compact support? (Thinking)

 

[I like Serena]

Hey evinda! (Smile)

Doesn't compact support mean that $\phi$ and $\psi$ are zero outside of some bounded set, as you already stated?
Or in other words that they are non-zero on some bounded set? (Wondering)

 

[HOI]

Hello! (Wave)

I want to prove that if the initial data of the initial value problem for the wave equation have compact support, then at each time the solution of the equation has also compact support.

Doesn't the fact that a function has compact support mean that the function is zero outside some bounded set? (Thinking)

Well, it means that the function is 0 outside of some compact set- so both bounded and closed.

The initial value problem for the wave equation is the following, right?

$$u_{tt}=c^2 u_{xx} \text{ in } \mathbb{R} \times (0,\infty) \\ u(\cdot, 0)=\phi \text{ in } \mathbb{R} \\ u_t(\cdot,0)=\psi \text{ in } \mathbb{R}$$

Is it meant with "the initial data of the initial value problem for the wave equation have compact support" that $\phi$ and $\psi$ are $0$ ?

Outside of some closed and bounded set.

If so, it follows directly that the solution is $0$, since it is equal to $u(x,t)=\frac{1}{2} \left[ \phi(x-ct)+\phi(x+ct)\right]+\frac{1}{2c} \int_{x-ct}^{x+ct} \psi(\tau) d{\tau}$, right?But if so, why is it stated as compact support? (Thinking)

Because your interpretation that the initial values are 0 for all R is incorrect! Also, in order to solve that equation you will need to have some initial values. They will determine whether the solution is identically 0.

 

[S.G. Janssens]

Specifically, by the most common definition, for a function $f: \mathbb{R} \to \mathbb{R}$ its support is
\[
\operatorname{supp}{f} := \overline{\{x \in \mathbb{R}\,:\, f(x) \neq 0\}},
\]
where the horizontal bar denotes the closure in $\mathbb{R}$. We say that $f$ has compact support (or: is compactly supported) if this set is compact.

You have already correctly specified the initial conditions $\phi$ and $\psi$, but as mentioned in the previous posts, you have misinterpreted the meaning of "compact support".

What you need to show is: If $\operatorname{supp}{\phi}$ and $\operatorname{supp}{\psi}$ are compact, then $\operatorname{supp}{u(\cdot,t)}$ is compact for all $t > 0$. For this you can use the representation formula for the solution and the fact that subsets of $\mathbb{R}$ are compact if, and only if, they are closed and bounded (Heine-Borel, remarked already in post #3).

I would start like this: Since $\phi$ and $\psi$ are compactly supported, their supports are bounded. Use the integral formula to show that this implies that for any $t > 0$ the function $u(\cdot,t)$ vanishes outside an interval of the form $[-L, L]$ for some $L > 0$ that may depend, of course, on $t$. This implies that
\[
\{x \in \mathbb{R}\,:\, u(x,t) \neq 0\} \subseteq [-L,L].
\]
So, what does this in turn imply for $\operatorname{supp}{u(\cdot,t)}$?

P.S. If you interpret the wave equation as generating a dynamical system on some (unspecified, shame on me) function space, then you are proving a statement about the time evolution $t \mapsto u(\cdot,t; \phi, \psi)$ of the initial conditions $\phi$ and $\psi$ that live in that function space. (Here I have appended $\phi$ and $\psi$ to $u$ in order to stress that $u$ depends on these initial conditions.)

 

[evinda]

Hey evinda! (Smile)

Doesn't compact support mean that $\phi$ and $\psi$ are zero outside of some bounded set, as you already stated?
Or in other words that they are non-zero on some bounded set? (Wondering)

I am confused now. Since the initial data of the initial value problem for the wave equation have compact support, we have that there is some bounded interval $[-L,L]$ such that $\phi=\psi =0$ in $\mathbb{R} \setminus [-L,L]$, right?

But how does this help? (Worried)

 

[I like Serena]

I am confused now. Since the initial data of the initial value problem for the wave equation have compact support, we have that there is some bounded interval $[-L,L]$ such that $\phi=\psi =0$ in $\mathbb{R} \setminus [-L,L]$, right?

But how does this help? (Worried)

Indeed. (Nod)

Now let's try to use that in your solution as Krylov suggested:
$$u(x,t)=\frac{1}{2} \left[ \phi(x-ct)+\phi(x+ct)\right]+\frac{1}{2c} \int_{x-ct}^{x+ct} \psi(\tau) d{\tau}$$

Given that $\phi(x)=\psi(x) =0$ for $x$ in $\mathbb{R} \setminus [-L,L]$.
And given a specific $t=T$.
For which $x$ will $u(x,T)=0$? (Wondering)

 

[evinda]

Indeed. (Nod)

Now let's try to use that in your solution as Krylov suggested:
$$u(x,t)=\frac{1}{2} \left[ \phi(x-ct)+\phi(x+ct)\right]+\frac{1}{2c} \int_{x-ct}^{x+ct} \psi(\tau) d{\tau}$$

Given that $\phi(x)=\psi(x) =0$ for $x$ in $\mathbb{R} \setminus [-L,L]$.
And given a specific $t=T$.
For which $x$ will $u(x,T)=0$? (Wondering)

It holds that $u(x,T)=0$ when $x-cT, x+cT \in \mathbb{R} \setminus [-L,L]$. Right?

 

[evinda]

But if so, how can we find a closed and bounded interval so that $u$ has a compact support? (Thinking)

 

[I like Serena]

It holds that $u(x,T)=0$ when $x-cT, x+cT \in \mathbb{R} \setminus [-L,L]$. Right?

Suppose $x-cT<-L$ and $x+cT>L$, don't we have that $\int_{x-cT}^{x+cT}\psi(\tau)d\tau$ and therefore $u(x,T)$ might be non-zero? (Worried)

But if so, how can we find a closed and bounded interval so that $u$ has a compact support? (Thinking)

Suppose we can find an $M$ such that $u(x,T)=0$ if either $x<-M$ or $x>M$.
Wouldn't that mean that $u(\cdot,T)$ has compact support? (Wondering)

 

[evinda]

Suppose $x-cT<-L$ and $x+cT>L$, don't we have that $\int_{x-cT}^{x+cT}\psi(\tau)d\tau$ and therefore $u(x,T)$ might be non-zero? (Worried)

Ah I see... In order $\int_{x-cT}^{x+cT}\psi(\tau)d\tau$ to be zero, there are two possible cases:

either $x-cT<-L \Rightarrow x<cT-L$ and $x+cT<-L \Rightarrow x<-cT-L$

or $x-cT>L \Rightarrow x>L+cT$ and $x+cT>L \Rightarrow x>L-cT$.

The first case holds for $x<-cT-L$ and the second for $x>L+cT$, if we suppose that $c$ is positive, right?

So it follows that $u(x,T)=0$ outside the bounded set $[-cT-L,cT+L]$, i.e. $u(x,T)$ has compact support.

So since $T$ is arbitrary, we see that at each time the solution of the equation has compact support.

Is the above right? (Thinking)

 

[I like Serena]

So it follows that $u(x,T)=0$ outside the bounded set $[-cT-L,cT+L]$, i.e. $u(x,T)$ has compact support.

So since $T$ is arbitrary, we see that at each time the solution of the equation has compact support.

Is the above right?

All correct. (Happy)

 

[evinda]

Indeed. (Nod)

I am thinking about it again now. Couldn't it be that the bounded set where $\phi$ is non-zero and the bounded set where $\psi$ is non-zero are disjoint? (Thinking)

 

[I like Serena]

I am thinking about it again now. Couldn't it be that the bounded set where $\phi$ is non-zero and the bounded set where $\psi$ is non-zero are disjoint?

Can't we take the union of those 2 bounded sets?
That is, pick $L$ such that it's bigger than any $x$-value for which either $\phi(|x|)$ or $\psi(|x|)$ is non-zero? (Wondering)

 

[evinda]

Can't we take the union of those 2 bounded sets?
That is, pick $L$ such that it's bigger than any $x$-value for which either $\phi(|x|)$ or $\psi(|x|)$ is non-zero? (Wondering)

The closed and bounded sets where the functions are non-zero don't have to be of the form $[-L,L]$, do they?
But are we sure that the sets are in the form $[a,b]$ for $a,b \in \mathbb{R}$ ? Or could we also have intervals in some other form? If not, then is it as follows?$\phi=0$ for $x \in \mathbb{R} \setminus [\alpha, \beta]$
and
$\psi=0$ for $x \in \mathbb{R} \setminus [\gamma, \delta]$, for $\alpha, \beta,\gamma, \delta \in \mathbb{R}$.

Right?

Then we know that there is some bounded set $[M,L]$ such that the union $[\alpha, \beta] \cup [\gamma, \delta]$ is a subset of $[M,L]$, right?

So $\phi=0$ for $x \in \mathbb{R} \setminus{[M,L]}$ and $\psi=0$ for $x \in \mathbb{R} \setminus [M,L]$.

Now let $t=T$, where $T$ arbitrary. Then

$u(x,T)=\frac{1}{2} [\phi(x-cT)+\phi(x+cT)]+\frac{1}{2c} \int_{x-cT}^{x+cT} \psi(\tau) d{\tau}$.$\phi(x-cT)=0$ for $x-cT<M$ or $x-cT>L$

$\phi(x+cT)=0$ for $x+cT<M$ or $x+cT>L$

$\int_{x-cT}^{x+cT} \psi(\tau) d{\tau}=0$ if ($x-cT<M$ and $x+cT<M$) or ($x-cT>L$ and $x+cT>L$).Then $u(x,T)=0$ if

- $x<M+cT$ or $x>L+cT$ and

- $x<M-cT$ or $x>L-cT$ and

- ($x<M+cT$ and $x<M-cT$) or ($x>L+cT$ and $x>L-cT$)

Right? If so , then for $c>0$ we get that $u(x,T)=0$ for $x>L+cT$ and $x<M-cT$, i.e. $x \in \mathbb{R} \setminus{[M-cT,L+cT]}$, right?

 

[evinda]

No, I am wrong... Because if we have $\phi(x)=0$ for $x \in \mathbb{R} \setminus{[\alpha, \beta]}$, it doesn't imply that $\phi(x)=0$ for $x \in \mathbb{R} \setminus{[\alpha, \beta] \cup [\gamma, \delta]}$... (Worried)

 

[I like Serena]

The closed and bounded sets where the functions are non-zero don't have to be of the form $[-L,L]$, do they?
But are we sure that the sets are in the form $[a,b]$ for $a,b \in \mathbb{R}$ ? Or could we also have intervals in some other form? If not, then is it as follows?$\phi=0$ for $x \in \mathbb{R} \setminus [\alpha, \beta]$
and
$\psi=0$ for $x \in \mathbb{R} \setminus [\gamma, \delta]$, for $\alpha, \beta,\gamma, \delta \in \mathbb{R}$.

Right?

How about picking $L= \max(|\alpha|, |\beta|, |\gamma|, |\delta|)$?
Then $\phi=\psi=0$ on $\mathbb R\setminus [-L,L]$ isn't it? (Wondering)

Btw, we could also have e.g. that $\{x\mid \phi(x) \ne 0\} = \{x \in \mathbb R \setminus \mathbb Q \mid x \in (-1,3) \lor x \in [5,8)\}$, couldn't we?
In that case we can refer to $[-8,8]$ as a compact interval that supports $\phi$ can't we?

 

[evinda]

How about picking $L= \max(|\alpha|, |\beta|, |\gamma|, |\delta|)$?
Then $\phi=\psi=0$ on $\mathbb R\setminus [-L,L]$ isn't it? (Wondering)

I am a little confused right now.
Could you explain to me why this holds? (Worried)

 

[I like Serena]

I am a little confused right now.
Could you explain to me why this holds? (Worried)

Suppose $\phi(x)=0$ for $x\in \mathbb R \setminus [\alpha,\beta]$.

Then that means that we have that $\phi(x)=0$ for $x >\max(|\alpha|,|\beta|)$ don't we?
And also that $\phi(x)=0$ for $x < -\max(|\alpha|,|\beta|)$.

In other words: $\phi(x)=0$ for $x \in \mathbb R \setminus [-\max(|\alpha|,|\beta|), \max(|\alpha|,|\beta|)]$. (Thinking)

 

[evinda]

Then that means that we have that $\phi(x)=0$ for $x >\max(|\alpha|,|\beta|)$ don't we?
And also that $\phi(x)=0$ for $x < -\max(|\alpha|,|\beta|)$.

How does this follow? That's what I don't understand... (Worried)

 

[I like Serena]

How does this follow? That's what I don't understand... (Worried)

If $\phi(x)\ne 0$ then we must have that $x\le \beta$ yes?
And $\beta \le |\beta| \le \max(|\alpha|,|\beta|)$ isn't it?
So if $\phi(x)\ne 0$ then we must have that $x \le \max(|\alpha|,|\beta|)$ as well, don't we? (Thinking)

Similarly, if $\phi(x)\ne 0$, we must have that $x \ge -\max(|\alpha|,|\beta|)$.