Solve the PDE

$$y\frac{\partial u^2}{\partial x^2} + (y-x)\frac{\partial u^2}{\partial x \partial y} -x \frac{\partial u^2}{\partial y^2} = \frac{y-x}{y+x}\bigg(\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}\bigg) \qquad \text{for } x+y \neq 0$$

with initial data

$$u(x,1)=x \qquad \text{and} \qquad \frac{\partial u}{\partial y}(x,1) = -1 \qquad \qquad \text{for } x\geq 0$$

State where your solution is uniquely determined by this initial data.


So the two sets of characteristics are

$$x^2+y^2 = \text{constant} \qquad \text{and} \qquad x-y = \text{constant}$$

Using the change of variables $\xi = x^2+y^2$ and $\eta = x-y$, the equation reduces to

$$\frac{\partial^2 u}{\partial \xi \partial \eta} = 0$$

Integrating, changing back to $x,y$ variables, and using the initial data, we find the solution

$$u(x,y) = x-y+1$$

However, where is this solution uniquely determined by the initial data? To determine this, do you simply plot the set of characteristics that touch the data curve then take the region crossed by both sets of characteristics?

I think there was also something about the Jacobian having to be non-zero?

So in this case, the data curve is $y=1$ for $x \geq 0$. The characteristics that touch this line are given by

\begin{align} x^2+y^2 = c \qquad & c \geq 1 \\ x-y = d \qquad & d \geq -1 \end{align}

Does this mean that the solution is uniquely determined in the region

$$\{ (x,y) \in \Bbb R^2: x^2+y^2 \geq 1 \text{ and } x-y \geq -1 \}$$

shown in yellow below:

enter image description here

And also, what happens on the line $x+y=0$?


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