Solve initial value problem (C.S.I.R)?

Question

The initial value problem is

$$ \frac{\partial u}{\partial t} +x\frac{\partial u}{\partial x} = x, \ \ 0 \leq x \leq 1, \ \ t > 0 \ \ and$$

$$ u(x,0) = 2x \ \ $$ has

1. a unique solution $u(x,t) \ \ $ which $\rightarrow \infty \ \ as \ \ t \ \ \rightarrow \infty$
2. more than solution.
3. a solution which remains bounded as $ t \rightarrow \infty$.
4. no solution.

I have solved $\frac{dt}{1} = \frac{dx}{x} = \frac{du}{x}$, we obtain

$u -x = c_1$ and $ x = c_2 e^t$, we get $ u(x,t) = c_1 + c_2 e^t$ and use $u(x,0) = 2x$, we get $c_1 = x$, we obtain $u(x,t) = x + c_2 e^t$.

I think (2) is right answer.

Please check my answer .

Thank you

Answer 1

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=1$ , we have $t=s$

$\dfrac{dx}{ds}=x$ , letting $x(0)=x_0$ , we have $x=x_0e^s=x_0e^t$

$\dfrac{du}{ds}=x=x_0e^s$ , we have $u=x_0e^s+f(x_0)=x+f(xe^{-t})$

$u(x,0)=2x$ :

$x+f(x)=2x$

$f(x)=x$

$\therefore u(x,t)=x+xe^{-t}=x(e^{-t}+1)$ , which belongs to (3).