# [SOLVED]Determine space of solution

#### evinda

##### Well-known member
MHB Site Helper
Hello!!!

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?

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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?
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?
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!!

#### evinda

##### Well-known member
MHB Site Helper
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?
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!!
So the solution is defined for $(x,t) \in (0,e^t) \cup (2e^t, 3e^t) \times \mathbb{R}$, right?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I used the method of characteristics.
I see.
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.
$(0,e^t) \cup (2e^t, 3e^t) \times \mathbb{R}$ is only a set for a specific $t$.
And if we pick a specific $t$, we get a rectangular set, don't we?

#### evinda

##### Well-known member
MHB Site Helper
I'm afraid that is bad notation.
$(0,e^t) \cup (2e^t, 3e^t) \times \mathbb{R}$ is only a set for a specific $t$.
And if we pick a specific $t$, we get a rectangular set, don't we?
Ok... But how else could we write down the space ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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:
$$\{(x,t)\in \mathbb R^2 \mid 0 \le x \le e^t \ \lor\ 2e^t \le x \le 3e^t\}$$

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

#### evinda

##### Well-known member
MHB Site Helper
How about using set builder notation:
$$\{(x,t)\in \mathbb R^2 \mid x\in [0,e^t]\cup [2e^t,3e^t]\}$$
or:
$$\{(x,t)\in \mathbb R^2 \mid 0 \le x \le e^t \ \lor\ 2e^t \le x \le 3e^t\}$$
Ah, I see...

Btw, those intervals should include the boundaries, shouldn't they?
You mean because I wrote $x\in (0,e^t)\cup (2e^t,3e^t)$ ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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.

#### evinda

##### Well-known member
MHB Site Helper
Yes, that notation explicitly excludes the boundaries.
But in the OP the boundaries were included.
Yes, that's right... Thanks a lot!!!