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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

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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.

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  • $\begingroup$ I guess my real problem is trying to find those $\phi_{2}, \phi_{3}$. I know that any expression with $xy$ satisfies the pde looking thingy but I am not sure how to find those $\phi$. For example, I could do $(xy)^2sin(xy)$ but this would give $x^2sin(-x)$ $\endgroup$ – dmsj djsl Oct 24 '18 at 14:22
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    $\begingroup$ No to the last, this new function is only for $y\ge0$. The curves through $y=-1$ do not reach positive $y$, there is an almost complete disconnect. So $F=(xy)^2$ for $y<0$ and $F=(xy)^2\cos(xy)$ for $y\ge 0$ is completely correct and $C^1$ and may be even $C^2$. $\endgroup$ – LutzL Oct 24 '18 at 14:40
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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 $F(-yx,-1)=F(x,y)=(-xy)^2=(xy)^2$.

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