Question

Solve the Partial differential equation

x² d²z/dx² - y² d²z/dy² + x dz/dx - y dz/dy = ln(x)

Answer 1

Let $x=e^t$, then

$ln(x)=t$

Using chain rule:

$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{dx}{dt}=\frac{\partial z}{\partial x}e^t$

$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial t}e^{-t}$

$\frac{\partial ^2z}{\partial t^2}=(\frac{\partial z}{\partial x}e^t)'_t=\frac{\partial ^2z}{\partial x^2}\frac{dx}{dt}e^t+\frac{\partial z}{\partial x}e^t=\\ \frac{\partial ^2z}{\partial x^2}e^{2t}+\frac{\partial z}{\partial t}$

$\frac{\partial ^2z}{d\partial ^2}=(\frac{\partial ^2z}{\partial t^2}-\frac{\partial z}{\partial t})e^{-2t}$

$x^2\frac{\partial ^2z}{\partial x^2}+x\frac{\partial z}{\partial x}=e^{2t}(\frac{\partial ^2z}{\partial t^2}-\frac{\partial z}{\partial t})e^{-2t}+e^t\frac{\partial z}{\partial t}e^{-t}=\\ \frac{\partial ^2z}{\partial t^2}-\frac{\partial z}{\partial t}+\frac{\partial z}{\partial t}=\frac{\partial ^2z}{\partial t^2}$

Similarly after substitution $y=e^s$ we have

$y^2\frac{\partial ^2z}{d\partial ^2}+y\frac{\partial z}{\partial y}=\frac{\partial ^2z}{\partial s^2}$

And equation becomes:

$\frac{\partial ^2z}{\partial t^2}-\frac{\partial ^2z}{\partial s^2}=t$

Now introduce new variables $\xi=t+s,\ \eta=t-s$

$z=z(\xi,\eta)$

Using chain rule again:

$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial \xi}\frac{\partial \xi}{\partial t}+\frac{\partial z}{\partial \eta}\frac{\partial \eta}{\partial t}= \frac{\partial z}{\partial \xi}+\frac{\partial z}{\partial \eta}$

$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial \xi}\frac{\partial \xi}{\partial s}+\frac{\partial z}{\partial \eta}\frac{\partial \eta}{\partial s}= \frac{\partial z}{\partial \xi}-\frac{\partial z}{\partial \eta}$

$\frac{\partial ^2z}{\partial t^2}=\frac{\partial ^2z}{\partial \xi^2}\frac{\partial \xi}{\partial t}+\frac{\partial ^2z}{\partial \xi \partial \eta}\frac{\partial \eta}{\partial t}+\frac{\partial ^2z}{\partial \xi \partial \eta}\frac{\partial \xi}{\partial t}+\frac{\partial ^2z}{\partial \eta^2}\frac{\partial \eta}{\partial t}=$

$\frac{\partial ^2z}{\partial \xi^2}+2\frac{\partial ^2z}{\partial \xi \partial \eta}+\frac{\partial ^2z}{\partial \eta^2}\frac{\partial \eta}{\partial t}$

$\frac{\partial ^2z}{\partial s^2}=\frac{\partial ^2z}{\partial \xi^2}\frac{\partial \xi}{\partial s}+\frac{\partial ^2z}{\partial \xi \partial \eta}\frac{\partial \eta}{\partial s}-(\frac{\partial ^2z}{\partial \xi \partial \eta}\frac{\partial \xi}{\partial s}+\frac{\partial ^2z}{\partial \eta^2}\frac{\partial \eta}{\partial s})=$

$\frac{\partial ^2z}{\partial \xi^2}-\frac{\partial ^2z}{\partial \xi \partial \eta}-(\frac{\partial ^2z}{\partial \xi \partial \eta}-\frac{\partial ^2z}{\partial \eta^2})= \frac{\partial ^2z}{\partial \xi^2}-2\frac{\partial ^2z}{\partial \xi \partial \eta}+\frac{\partial ^2z}{\partial \eta^2})$

$\frac{\partial ^2z}{\partial t^2}-\frac{\partial ^2z}{\partial s^2}=4\frac{\partial ^2z}{\partial \xi \partial \eta}$

$t=(t+s)/2+(t-s)/2=\frac{\xi+\eta}{2}$

$4\frac{\partial ^2z}{\partial \xi \partial \eta}=\frac{\xi+\eta}{2}$

$\frac{\partial ^2z}{\partial \xi \partial \eta}=\frac{\xi+\eta}{8}$

Integrating:

$\frac {\partial z}{\partial \eta}=\int \frac{\partial ^2z}{\partial \xi \partial \eta}\partial \xi=\int \frac{\xi+\eta}{8}d\xi=\xi^2/16+\eta\xi/8+g(\eta)$

$z=\int (\xi^2/16+\eta\xi/8+g(\eta))d\eta=\\ \xi^2\eta/16+\eta^2\xi/16+G(\eta)+F(\xi)$,

where F and G are arbitrary twice differentiable functions,

and G is antiderivative of g

Now use reverse substitutions $t=ln(x),\ s=ln(y)$

we have:

$\xi=t+s=ln(x)+ln(y)=ln(xy),\\ \eta=t-s=ln(x)-ln(y)=ln(x/y)$

$z=\frac{ln(xy)ln(x/y)}{16}(ln(xy)+ln(x/y))+G(ln(x/y)+F(ln(xy))=\\ \frac{ln(xy)ln(x/y)}{8}(ln(x))+G(ln(x/y)+F(ln(xy))$

Answer: $z=\frac{ln(xy)ln(x/y)}{8}(ln(x))+G(ln(x/y)+F(ln(xy))$, where F and G are arbitrary twice differentiable functions.