MHB Find Solution to Cauchy Problem & Determine Space in $\mathbb{R}^2$

[evinda]

Hello! (Wave)

I want to find the solution of the following Cauchy problem and determine the space in $\mathbb{R}^2$ where the initial condition defines the solution.

$$u_t+ u_x=4, u|_{t=0}=\sin{x} \text{ for } |x|<1$$

I found that the solution of the above initial value problem is $u(t,x)=4t+\sin{(x-t)}$.

What is meant with the space where the initial condition defines the solution? (Thinking)

 

[I like Serena]

Hey evinda! (Smile)

It seems a bit like an odd question...

Best I can think of, is that
$$\{(t,x)\in \mathbb R^2\mid |x|<1\} = \mathbb R \times (-1,1)$$
is intended.
A rectangle of infinite length in both directions excluding the boundary. (Thinking)

 

[evinda]

Hey evinda! (Smile)

It seems a bit like an odd question...

Best I can think of, is that
$$\{(t,x)\in \mathbb R^2\mid |x|<1\} = \mathbb R \times (-1,1)$$
is intended.
A rectangle of infinite length in both directions excluding the boundary. (Thinking)

A ok... (Thinking)
Since $t$ represents the time, will we then have that the space in which the solution is defined is this one:

$$\{(t,x)\in \mathbb{R}^2 \mid |x|<1, t \geq 0\} = [0,+\infty)\times (-1,1)$$

? If so, will it represent a rectangle of infinite length in the upward direction excluding the boundary?

 

[I like Serena]

A ok... (Thinking)
Since $t$ represents the time, will we then have that the space in which the solution is defined is this one:

$$\{(t,x)\in \mathbb{R}^2 \mid |x|<1, t \geq 0\} = [0,+\infty)\times (-1,1)$$

? If so, will it represent a rectangle of infinite length in the upward direction excluding the boundary?

Could be.
It's fairly common that these problems start at time $t=0$, at which time something special happens, such as heat starting to be applied to a body.
But then again, that does not seem to be given, and the solution also holds for negative $t$.
That is, $t=0$ is ultimately only an arbritary reference point in time. (Thinking)

 

[evinda]

Could be.
It's fairly common that these problems start at time $t=0$, at which time something special happens, such as heat starting to be applied to a body.
But then again, that does not seem to be given, and the solution also holds for negative $t$.
That is, $t=0$ is ultimately only an arbritary reference point in time. (Thinking)

So with $t=0$ we mean the time at which something happens at the solution and we consider that before this time the same happens as after it? (Thinking)

 

[I like Serena]

So with $t=0$ we mean the time at which something happens at the solution and we consider that before this time the same happens as after it? (Thinking)

Let's pick a couple of examples... (Thinking)

Suppose a vertical metal cylinder is originally at room temperature.
And at $t=0$ we start heating it at the bottom, heating it up from the bottom to the top.
Then we are not interested in what happens before $t=0$. We already know.
And it doesn't fit into the differential equation either, since the solution would then not be differentiable at $t=0$.

Another example, suppose rays of light are passing through a glass fiber.
At $t=0$ they pass through the fiber where we are sitting, and it's a sine wave.
Then we can see what happens afterwards ($t > 0$), and we can also trace it back to where it came from ($t < 0$).

Long story short, it should be specified in the problem statement whether $t<0$ is included or not.

 

[evinda]

Let's pick a couple of examples... (Thinking)

Suppose a vertical metal cylinder is originally at room temperature.
And at $t=0$ we start heating it at the bottom, heating it up from the bottom to the top.
Then we are not interested in what happens before $t=0$. We already know.
And it doesn't fit into the differential equation either, since the solution would then not be differentiable at $t=0$.

Why woulnd't the solution be differentiable at $t=0$ ?

Another example, suppose rays of light are passing through a glass fiber.
At $t=0$ they pass through the fiber where we are sitting, and it's a sine wave.
Then we can see what happens afterwards ($t > 0$), and we can also trace it back to where it came from ($t < 0$).

So we have a sine wave when rays of light pass through the glass fiber? And this happens for $t=0$ ? And since rays of light don't change something at the glass fiber, the solution will be the same before and after $t=0$?

Long story short, it should be specified in the problem statement whether $t<0$ is included or not.

I see... (Nod)

 

[I like Serena]

Why woulnd't the solution be differentiable at $t=0$ ?

For $x=0$ he solution would be something like:
$$y(t, 0) = \begin{cases}0 &\text{ if }t<0 \\ 1-e^{-t} &\text{ if }t\ge 0\end{cases}$$
And:
$$\lim_{t\to 0^-} y_t(t,0) = 0$$
while
$$\lim_{t\to 0^+} y_t(t,0) = 1$$
Ergo, not differentiable in $t=0$.

So we have a sine wave when rays of light pass through the glass fiber? And this happens for $t=0$ ? And since rays of light don't change something at the glass fiber, the solution will be the same before and after $t=0$?

Since at $t=0$ nothing special happens, the solution doesn't change either around $t=0$. (Thinking)

 

[evinda]

For $x=0$ he solution would be something like:
$$y(t, 0) = \begin{cases}0 &\text{ if }t<0 \\ 1-e^{-t} &\text{ if }t\ge 0\end{cases}$$
And:
$$\lim_{t\to 0^-} y_t(t,0) = 0$$
while
$$\lim_{t\to 0^+} y_t(t,0) = 1$$
Ergo, not differentiable in $t=0$.

So this problem isn't somehow related with the problem of the #post 1, is it?

How would the differential equation look like? (Thinking)

Since at $t=0$ nothing special happens, the solution doesn't change either around $t=0$. (Thinking)

So what would happen at the solution when $t \neq 0$ so that we have a sine wave?
For $t=0$ when the rays of light are passing through the glass fiber the solution will be 0? Or have I understood it wrong? (Worried)

 

[I like Serena]

So this problem isn't somehow related with the problem of the #post 1, is it?

How would the differential equation look like? (Thinking)

No, neither of the examples is for the problem in #1.
They are just examples of IVP's.

It's an example of the heat equation:
$$u_t - \alpha u_{xx} = 0$$

The other example is from the wave equation:
$$u_{tt} = c^2 u_{xx}$$

So what would happen at the solution when $t \neq 0$ so that we have a sine wave?
For $t=0$ when the rays of light are passing through the glass fiber the solution will be 0? Or have I understood it wrong? (Worried)

Not for the problem in post #1, but for the wave equation $\sin(x-ct)$ is a solution everywhere.