Solve the following PDE.
$u_t + (u_x)^2 = -u$ with $u(x,0)=x$
My attempt:
After substituting $u_{x}=v$ and solving new PDE $v_t + v v_x = -v$ with $v(x,0)=1$ , I got $v(x,t)=e^{-t}$.
so $u_{x}=e^{-t}$. I know I have to integrate it but I am getting a solution which doesn't satisfy original PDE.
Any help will be appreciated.
$u_t+(u_x)^2=-u$
$u_{xt}+2u_xu_{xx}=-u_x$
Let $v=u_x$ ,
Then $v_t+2vv_x=-v$ with $v(x,0)=1$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dv}{ds}=-v$ , letting $v(0)=v_0$ , we have $v=v_0e^{-s}=v_0e^{-t}$
$\dfrac{dx}{ds}=2v=2v_0e^{-s}$ , letting $x(0)=f(v_0)$ , we have $x=f(v_0)+2v_0(1-e^{-s})=f(ve^t)+2v(e^t-1)$ , i.e. $v=e^{-t}F(x+2v(1-e^t))$
$v(x,0)=1$ :
$F(x)=1$
$\therefore v(x,t)=e^{-t}$
$u_x(x,t)=e^{-t}$
$u(x,t)=xe^{-t}+g(t)$
$u_t(x,t)=-xe^{-t}+g_t(t)$
$\therefore-xe^{-t}+g_t(t)+e^{-2t}=-xe^{-t}-g(t)$
$g_t(t)+g(t)=-e^{-2t}$
$(e^tg(t))_t=-e^{-t}$
$e^tg(t)=e^{-t}+C$
$g(t)=e^{-2t}+Ce^{-t}$
$\therefore u(x,t)=xe^{-t}+e^{-2t}+Ce^{-t}$
$u(x,0)=x$ :
$x+1+C=x$
$C=-1$
$\therefore u(x,t)=(x-1)e^{-t}+e^{-2t}$
