- Thread starter
- #1

- Apr 13, 2013

- 3,836

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?

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: