Consider

$$\begin{cases} ux+uy&=2\ u(x,0)&=x^2 \end{cases} \tag{1}$$

We have the initial curve

$$\Gamma(s)=(s,0,s^2) \tag{2}$$

We parametrize the initial condition as

$$ \begin{cases} x0(s)&=s\ y0(s)&=0\ \hat{u}_0(s)&=s^2 \end{cases} \tag{3} $$

and we obtain the following system:

$$ \begin{cases} \frac{d}{dt}x(t,s) = a(x(t,s), y(t,s))=1\ \frac{d}{dt}y(t,s) = b(x(t,s), y(t,s))=1\ \frac{d}{dt}\hat{u}(t,s) = c0(x(t,s), y(t,s), \hat{u}(t,s) + c1(x(t,s), y(t,s)) =2 \end{cases} \tag{4} $$

using the initial conditions (3).

Therefore the characteristic curves are given by:

$$ \begin{cases} x(t,s) &= t+s\ y(t,s) &= t\ \hat{u}(t,s) &= s^2+2t \end{cases} \tag{5} $$

The parametrized solution surface is then given by the relation

$$u(x(t,s), y(t,s))=\hat{u}(t,s)=s^2+2t \tag{6}$$

Since we are looking for a solution in $(x,y)$ coordinates, we have to find the inverse map $(t,s)\mapsto (x,y)$ to find $u(x,y)$\ In this case it is very easy:

$$ \begin{cases} x(t,s)&=s+t\ y(t,s)&=t \end{cases} \tag{7} $$

so

$$ \begin{cases} y=t\ s=x-t=x-y \end{cases} \tag{8} $$

Hence the solution to the PDE is given by:

$$u(x,y)=s^2+2t=(x-y)^2+2y\tag{9}$$ $$

Saturday, 5 October 2019 11:02 GMT