Characteristics of semilinear PDE $xu_x+yu_y+xyu_z=0$

Question

Remark: Initially, I have incorrectly copied over the problem. The edit below shows the correct one and how I checked that the function really does satisfy the "new" pde.

I have a pde of the following form

$$xu_{x}+yu_{y}+xyu_{z}=0$$

with the boundary condition $$u(x,y,0)=x^2+y^2.$$

I solved the pde with the method of characteristic and got the solution $$u(x,y,z)=x^2+y^2-2(\frac{x}{y}+\frac{y}{x})z.$$

However, the other part of this problem asks me to explain the following.

We know that $u$ is constant along characteristics and we know that $u(x,y,0)\geq0$ then why is $u(1,1,1)=-2$.

What I was thinking is that even though we know that $u$ is going to be constant on each characteristic the fact that $u(1,1,1)=-2$ means that no characteristic curve begging on $(x,y,0)$ will pass through the point $(1,1,1)$. Am I correct in thinking this is the answer or is there something else that I forgot about.

EDIT: Just to check that the solution really does satisfy the pde:

$$x(2x-2\frac{z}{y}+2\frac{yz}{x^2})+y(2y-2\frac{z}{x}+2\frac{xz}{y^2})-2xy(\frac{x}{y}+\frac{y}{x})=$$ $$=x^2-\frac{xz}{y}+\frac{yz}{x}+y^2-\frac{yz}{x}+\frac{xz}{y}-x^2-y^2=0$$ Which really does satisfy the equation.

Answer 1

$$xu_x+yuy+xyu_z=0$$ Charpit-Lagrange system of characteristic ODEs : $$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{xy}=\frac{du}{0}$$ A first characteristic equation comes from $\frac{du}{0}\:$finite $\implies\:du=0$. $$u=c_1$$ A second characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{y}$ : $$\frac{y}{x}=c_2$$ A third characteristic equation comes from $\frac{dx}{x}=\frac{dy}{y}=\frac{ydx+xdy}{yx+xy}=\frac{dz}{xy}\quad\implies\quad \frac12 d(xy)=dz$ $$xy-2z=c_3$$ The general solution of the PDE expressed on the form $c_1=F(c_2,c_3)$ is : $$\boxed{u(x,y,z)=F\left(\frac{y}{x}\:,\:xy-2z\right)}$$ $F(X,Y)$ is an arbitrary function of two variables.

CONDITION : $u(x,y,0)=x^2+y^2$ $$x^2+y^2=F\left(\frac{y}{x}\: , \: xy\right)$$ With $X=\frac{y}{x}$ and $Y=xy$ $\quad\implies\quad$ $x^2=\frac{Y}{X}$ and $y^2=XY$. $$F(X,Y)=x^2+y^2=\frac{Y}{X}+XY$$ So the function $F(X,Y)$ is known now. We put it into the above geheral solution where $X=\frac{y}{x}$ and $Y=xy-2z$. $$u=\frac{xy-2z}{\frac{y}{x}}+\frac{y}{x}(xy-2z)$$ $$u=x^2+y^2-2z\left(\frac{x}{y}+\frac{y}{x} \right)$$ This is exactly the result that you obtained.

Answer 2

One way to approach the second half of the problem is to repeat the method of characteristics, but restricting to the characteristic which starts at $(1,1,1)$. Then the characteristic ODEs

$$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{xy}=\frac{du}{0}=ds$$

yield the solution $$x(s) =y(s)= e^s,\quad z(s)=\frac12 (e^{2s}+1)$$ with $u$ constant along the characteristic. Note that $z(s)\geq 1/2$ for all $s$, so this characteristic can never reach the $z=0$ plane on which the initial condition is prescribed. So requiring that $u$ is positive on the $z=0$ plane does not mean that $u$ must be positive everywhere.

This does raise the question of what points $(x,y,z)$ are connected to the $z=0$ plane by characteristics. If we generalize to arbitrary initial $(x_0,y_0,z_0)$ then the solution for $z(s)$ becomes $$z(s) =\frac12 x_0 y_0(e^{2s}-1)+z_0$$ which is bounded below by $z(-\infty)=z_0-x_0y_0/2$. Hence for $z_0> x_0 y_0/2$ the characteristic does not reach the $z=0$ plane. For comparison, the solution obtained by the OP can be expressed as

$$u(x,y,z) = (x/y+y/x)(xy-2z)$$ which is indeed negative only if $z >xy/2$. (In the present case, $z>xy/2$ is a sufficient condition for $u<0$ rather than just a necessary one. This appears specific to the initial condition $u(x,y,0)=x^2+y^2$. For instance, if $u(x,y,0)=x^2+y^2+a$ for some $a>0$, I expect that there would exist some nonnegative values of $u(x,y,z)$ in the region $z>xy/2$.)