Transform the partial differential equation $(y-z)\frac{∂z}{∂x}+(y+z)\frac{dz}{dy}=0$

Question

Transform the P.D.E.

$(y-z)\frac{\partial z}{\partial x}+(y+z)\frac{\partial z}{\partial y}=0$

so that the new equation contains $x$ as a new function, and $u=y-z, v=y+z$ are new independent variables.

Answer 1

$$u=y-z\quad\implies\quad du=dy-dz$$ $$v=y+z\quad\implies\quad dv=dy+dz$$ $$dx=\frac{\partial x(u,v)}{\partial u}du+\frac{\partial x(u,v)}{\partial v}dv$$ $$dx=\frac{\partial x(u,v)}{\partial u}(dy-dz)+\frac{\partial x(u,v)}{\partial v}(dy+dz)$$ $$dx=\left(\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}\right)dy+\left(-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}\right)dz$$

$$dz=\frac{1}{-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}dx-\frac{\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}{-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}dy\quad\implies$$

$$\implies\quad\begin{cases} \frac{\partial z(x,y)}{\partial x}=\frac{1}{-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}\\ \frac{\partial z(x,y)}{\partial y}=-\frac{\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}{-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}} \end{cases}$$

We put them into the original PDE where $y-z=u$ and $y+z=v$ :

$$(y-z)\frac{\partial z}{\partial x}+(y+z)\frac{\partial z}{\partial y}=0$$

$$u\frac{1}{-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}-v\frac{\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}{-\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}}=0$$

$$u-v\left(\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}\right)=0$$ $$\boxed{\frac{\partial x(u,v)}{\partial u}+\frac{\partial x(u,v)}{\partial v}=\frac{u}{v}}$$

This is the transformed PDE we were looking for.

This PDE can be solved thanks to the method of characteristics. The result is :

$$\boxed{x(u,v)=v+(u-v)\ln|v|+F(u-v)}$$ $F$ is an arbitrary function until some boundary condition be specified.

Comming back to the original variables : $$x=y+z(x,y)-2z(x,y)\ln|y+z(x,y)|+F\big(-2z(x,y)\big)$$ or equivalenttly $$\boxed{z\ln|y+z|=\frac{y-x}{2}+G(z)}$$ $G$ is an arbitrary function until some boundary condition be specified.

The last equation cannot be solve for $z$ in terms of standard functions. Thus one have to be satisfied for the solution expressed on the form of implicit equation.

NOTE : The original PDE is even easier to solve without the change of variables. The direct solving leads to the solution consistent with the above solution. This is a manner to check that the above result is correct.