MHB Determining Solution Space for Cauchy Problem ($u_t+xu_x=(x+t)u$)

[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.$$u_t+xu_x=(x+t)u, u|_{t=0}=\phi(x), x \in [0,1] \cup [2,3].$$

($\phi(x)$ arbitrary smooth function)

I have tried the following:

We need to solve the following two problems.

$$\frac{dx}{dt}=x, x|_{t=0}=x_0 \\ \frac{du}{dt}=(x+t)u, u|_{t=0}=\phi(x)$$

We get that $x(t)=x_0 e^t$ and $u(t,x)=\phi(xe^{-t}) e^{x(1-e^{-t})+\frac{t^2}{2}}$.

Am I right?

But how can we determine the space in $\mathbb{R}^2$ where the initial condition defines the solution?The initial condition $x \in [0,1] \cup [2,3]$ given holds for the case when $t=0$, or not?

If so, do we use this condition to find the desired space?

From $0 \leq x_0 \leq 1$ we get that $0 \leq x \leq e^t$ and from $2 \leq x_0 \leq 3$ we get that $2e^t \leq x \leq 3e^t$, right? Does this help? (Thinking)

 

[I like Serena]

We need to solve the following two problems.

$$\frac{dx}{dt}=x, x|_{t=0}=x_0 \\ \frac{du}{dt}=(x+t)u, u|_{t=0}=\phi(x)$$

We get that $x(t)=x_0 e^t$ and $u(t,x)=\phi(xe^{-t}) e^{x(1-e^{-t})+\frac{t^2}{2}}$.

Am I right?

Hey evinda!

It looks correct to me.
That is, substituting it into the original equation shows that it's a solution.

Btw, how did you get to that solution? (Wondering)
I do not recognize the method.

But how can we determine the space in $\mathbb{R}^2$ where the initial condition defines the solution?

The initial condition $x \in [0,1] \cup [2,3]$ given holds for the case when $t=0$, or not?

If so, do we use this condition to find the desired space?

From $0 \leq x_0 \leq 1$ we get that $0 \leq x \leq e^t$ and from $2 \leq x_0 \leq 3$ we get that $2e^t \leq x \leq 3e^t$, right? Does this help? (Thinking)

The solution for $u(t,x)$ that you have is well-defined iff $\phi(xe^{-t})$ is well-defined.
And you have already found where that is.

Graphing it shows:
[DESMOS]advanced: {"version":5,"graph":{"squareAxes":false,"viewport":{"xmin":-0.25,"ymin":-2.2,"xmax":4.5,"ymax":2.2}},"expressions":{"list":[{"id":"graph1","type":"expression","latex":"0\\le x\\le e^{y}","color":"#2d70b3"},{"id":"2","type":"expression","latex":"2e^{y}\\le x\\le3e^{y}","color":"#2d70b3"}]}}[/DESMOS]

So I believe you already have the solution! (Happy)

 

[evinda]

Hey evinda!

It looks correct to me.
That is, substituting it into the original equation shows that it's a solution.

Btw, how did you get to that solution? (Wondering)
I do not recognize the method.

I used the method of characteristics.

The solution for $u(t,x)$ that you have is well-defined iff $\phi(xe^{-t})$ is well-defined.
And you have already found where that is.

Graphing it shows:So I believe you already have the solution! (Happy)

So the solution is defined for $(x,t) \in (0,e^t) \cup (2e^t, 3e^t) \times \mathbb{R}$, right? (Thinking)

 

[I like Serena]

I used the method of characteristics.

I see. (Thinking)
Yep. I get the same result.

So the solution is defined for $(x,t) \in (0,e^t) \cup (2e^t, 3e^t) \times \mathbb{R}$, right?

I'm afraid that is bad notation. (Worried)
$(0,e^t) \cup (2e^t, 3e^t) \times \mathbb{R}$ is only a set for a specific $t$.
And if we pick a specific $t$, we get a rectangular set, don't we? (Wondering)

 

[evinda]

I'm afraid that is bad notation. (Worried)
$(0,e^t) \cup (2e^t, 3e^t) \times \mathbb{R}$ is only a set for a specific $t$.
And if we pick a specific $t$, we get a rectangular set, don't we? (Wondering)

Ok... But how else could we write down the space ? (Thinking)

 

[I like Serena]

Ok... But how else could we write down the space ?

How about using set builder notation:
$$\{(x,t)\in \mathbb R^2 \mid x\in [0,e^t]\cup [2e^t,3e^t]\}$$
or:
$$\{(x,t)\in \mathbb R^2 \mid 0 \le x \le e^t \ \lor\ 2e^t \le x \le 3e^t\}$$
(Wondering)

Btw, those intervals should include the boundaries, shouldn't they? (Wondering)

 

[evinda]

How about using set builder notation:
$$\{(x,t)\in \mathbb R^2 \mid x\in [0,e^t]\cup [2e^t,3e^t]\}$$
or:
$$\{(x,t)\in \mathbb R^2 \mid 0 \le x \le e^t \ \lor\ 2e^t \le x \le 3e^t\}$$
(Wondering)

Ah, I see... (Nod)

Btw, those intervals should include the boundaries, shouldn't they? (Wondering)

You mean because I wrote $x\in (0,e^t)\cup (2e^t,3e^t)$ ? (Thinking)

 

[I like Serena]

You mean because I wrote $x\in (0,e^t)\cup (2e^t,3e^t)$ ?

Yes, that notation explicitly excludes the boundaries.
But in the OP the boundaries were included. (Thinking)

 

[evinda]

Yes, that notation explicitly excludes the boundaries.
But in the OP the boundaries were included. (Thinking)

Yes, that's right... Thanks a lot! (Smirk)