How to Find Functions with Given Condition

Question

I am given that $\frac{\partial F}{\partial x}(x) + \frac{\partial F}{\partial y}(-y) = 0$ and that $F(x,-1) = x^2$. I am supposed to find $2$ $F(x,y)$'s that satisfy these conditions.

I have found one of them, which is $F(x,y)=(xy)^2$ but I have no clue how to find the other one. The question also seemed to imply that there are more than two $F(x,y)$'s that are possible, but I am not entirely sure about this. My ultimate goal is to show that any two solutions to this problem differ only when $y \gt 0$ But I am not sure how to go on since I only have one solution at the moment. Can someone help? Thanks

Answer 1

Is your equation in another form of writing $xF_x-yF_y=0$? I was wondering why $F$ in the equation has one and in the boundary condition two arguments.

Then you are correct, the general solution is $F(x,y)=\phi(xy)$ and the initial condition gives $\phi(-x)=x^2\implies F(x,y)=(xy)^2$. [1] This formula is valid on all characteristic curves $xy=c_1$ that contain an initial condition, which means that the part of the hyperbolic curve is to be selected that contains the point $(x_0,-1)$, which can be summarized as $y<0$.

For $y>0$ you can select any function of the for $F(x,y)=\phi_2(xy)$, to make the transition smooth restrict to $F(x,y)=(xy)^2\phi_3(xy)$.

[1] edited from here, see answer of Tsemo Aristide for an alternative parametrisation of the characteristics.

Answer 2

Partial answer.

Let $X$ be a vector field, $dF.X=0$ implies that $F\circ \phi_t$ is constant where $\phi_t$ is the flow of $X$, here $X=(x,-y)$, the flow of $X$ is $(e^tx,e^{-t}y)$, you deduce that $F(e^tx,e^{-t}y)=F(x,y)$, if $y<0$, there exists $t$ such that $e^{t}=-y$, you have $F(-yx,-1)=F(x,y)=(-xy)^2=(xy)^2$.