# How to Find Functions with Given Condition

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

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.

• 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 $x^2sin(-x)$ – dmsj djsl Oct 24 '18 at 14:22
• 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 $C^2$. – LutzL Oct 24 '18 at 14:40

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 $$F(-yx,-1)=F(x,y)=(-xy)^2=(xy)^2$$.