# PDEMOC7

$\displaystyle xu_x + u_y = y, u(x,0)=x^2\,$

The initial curve will be parameterized by $\displaystyle \Gamma(s,0,s^2)\,$ .

The characteristics are

$\displaystyle \frac{dx}{dt}=x, \frac{dy}{dt}=1, \frac{dz}{dt}=y\,$

For $\displaystyle x\,$ ,

$\displaystyle \frac{dx}{dt}=x\,$

$\displaystyle x^{-1}dx = dt\,$

$\displaystyle \ln x = t+c_1(s)\,$

$\displaystyle x(s,t) = c_4(s)e^t\,$

$\displaystyle x(s,0) = c_4(s) = s\,$

So $\displaystyle x(s,t) = se^t\,$

For $\displaystyle y\,$ ,

$\displaystyle \frac{dy}{dt}=1$

$\displaystyle y(s,t) = t+c_2(s)\,$

$\displaystyle y(s,0) = c_2(s) = 0\,$

So $\displaystyle y(s,t) = t\,$

For $\displaystyle z\,$ ,

$\displaystyle \frac{dz}{dt}=y=t\,$

$\displaystyle z(s,t) = \frac{1}{2}t^2+c_s(3)\,$

$\displaystyle z(s,0) = c_s(0) = s^2\,$

$\displaystyle z(s,t) = \frac{1}{2}t^2+s^2\,$

The solution is

$\displaystyle z(s,t) = u(x,y) =\frac{1}{2}y^2 + (xe^{-y})^2\,$