# Compact Support of Wave Equation IVP Solutions

• MHB
• evinda
In summary: If $\phi$ and $\psi$ are zero on $[-L,L]$, then the solution must be zero too, right?The compact support of $\phi$ and $\psi$ means that they are zero outside of some closed and bounded set- so both bounded and closed. This implies that the solution is zero, since it is equal to $u(x,t)=\frac{1}{2} \left[ \phi(x-ct)+\phi(x+ct)\right]+\frac{1}{2c} \int_{x-ct}^{x+ct} \psi(\tau) d{\tau}$.
evinda
Gold Member
MHB
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)

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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)

evinda said:
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.

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.)

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I like Serena said:
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)

evinda said:
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)

I like Serena said:
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?

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

evinda said:
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)

evinda said:
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)

I like Serena said:
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)

evinda said:
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)

I like Serena said:
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)

evinda said:
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)

I like Serena said:
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?

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)

evinda said:
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?

I like Serena said:
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)

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evinda said:
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)

I like Serena said:
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)

evinda said:
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 $x \ge -\max(|\alpha|,|\beta|)$.

## 1. What is the wave equation and what does it represent?

The wave equation is a mathematical model used to describe the behavior of waves, such as sound waves or electromagnetic waves. It represents the relationship between the wave's properties, such as amplitude and frequency, and its propagation through a medium.

## 2. What is the initial value problem (IVP) for the wave equation?

The IVP for the wave equation refers to the set of conditions that need to be specified in order to uniquely determine a solution to the equation. These conditions typically include the initial displacement and velocity of the wave at a given point in time and space.

## 3. What is the compact support property of wave equation IVP solutions?

The compact support property states that the solution to the wave equation will only be non-zero within a finite region of space and time. In other words, the wave will only exist in a limited area and for a limited amount of time.

## 4. Why is the compact support property important in wave equation analysis?

The compact support property is important because it allows us to make simplifications and approximations in the analysis of wave behavior. It also ensures that the solution to the wave equation is physically realistic, as waves in the real world are typically localized in space and time.

## 5. How is the compact support property applied in practical situations?

In practical situations, the compact support property can be used to determine the behavior of waves in specific scenarios, such as in the design of acoustic or electromagnetic devices. It can also be used to study the effects of boundary conditions on wave propagation and to analyze the stability of wave solutions.

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