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Consider

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

We have the initial curve

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

We parametrize the initial condition as

$$
\begin{cases}
x_0(s)&=s\\
y_0(s)&=0\\
\hat{u}_0(s)&=s^2
\end{cases} \tag{3}
$$

and we obtain the following system:

$$
\begin{cases}
\frac{d}{dt}x(t,s) = a(x(t,s), y(t,s))=1\\
\frac{d}{dt}y(t,s) = b(x(t,s), y(t,s))=1\\
\frac{d}{dt}\hat{u}(t,s) = c_0(x(t,s), y(t,s), \hat{u}(t,s) + c_1(x(t,s), y(t,s)) =2
\end{cases} \tag{4}
$$

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:

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

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

Hence the solution to the PDE is given by:

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

Sat, 05 Oct 2019 11:02 GMT