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Solution of the PDE?

bkarpuz

New member
Jan 27, 2012
11
Dear MHB members,

I have the following equation
$xy(z_{xx}-z_{yy})+(x^{2}-y^{2})z_{xy}=yz_{x}-xz_{y}-2(x^{2}-y^{2})$.
When I transform this into the canonical form via $\xi=2xy$ and $\eta=x^{2}-y^{2}$, I obtain
$z_{\xi\eta}+\frac{\eta}{\xi^{2}+\eta^{2}}z_{\xi}=-\frac{\eta}{2(\xi^{2}+\eta^{2})}$.
This is the point I stuck at.
It follows from here that
$z_{\xi}=\frac{f_{1}(\xi)}{\sqrt{\xi^{2}+\eta^{2}}}-\frac{1}{2}$,
which yields
$z=\int^{\xi}\frac{f_{1}(u)}{\sqrt{u^{2}+\eta^{2}}}{\rm d}u+f_{2}(\eta)-\frac{\xi}{2}$.
Therefore, the given equation has the general solution
$z=\int^{2xy}\frac{f_{1}(u)}{\sqrt{u^{2}+(x^{2}-y^{2})^{2}}}{\rm d}u+f_{2}(x^{2}-y^{2})-xy$.

My question is how to modify the part
$\int^{\xi}\frac{f_{1}(u)}{\sqrt{u^{2}+\eta^{2}}}{\rm d}u$ in a better way
and get rid of the integral if possible.

Thank you very much.
bkarpuz
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Dear MHB members,

I have the following equation
$xy(z_{xx}-z_{yy})+(x^{2}-y^{2})z_{xy}=yz_{x}-xz_{y}-2(x^{2}-y^{2})$.
When I transform this into the canonical form via $\xi=2xy$ and $\eta=x^{2}-y^{2}$, I obtain
$z_{\xi\eta}+\frac{\eta}{\xi^{2}+\eta^{2}}z_{\xi}=-\frac{\eta}{2(\xi^{2}+\eta^{2})}$.
This is the point I stuck at.
It follows from here that
$z_{\xi}=\frac{f_{1}(\xi)}{\sqrt{\xi^{2}+\eta^{2}}}-\frac{1}{2}$,
which yields
$z=\int^{\xi}\frac{f_{1}(u)}{\sqrt{u^{2}+\eta^{2}}}{\rm d}u+f_{2}(\eta)-\frac{\xi}{2}$.
Therefore, the given equation has the general solution
$z=\int^{2xy}\frac{f_{1}(u)}{\sqrt{u^{2}+(x^{2}-y^{2})^{2}}}{\rm d}u+f_{2}(x^{2}-y^{2})-xy$.

My question is how to modify the part
$\int^{\xi}\frac{f_{1}(u)}{\sqrt{u^{2}+\eta^{2}}}{\rm d}u$ in a better way
and get rid of the integral if possible.

Thank you very much.
bkarpuz
Hi bkarpuz, :)

I don't have any idea about how you obtained \(z_{\xi}=\frac{f_{1}(\xi)}{\sqrt{\xi^{2}+\eta^{2}}}-\frac{1}{2}\). However if you integrate, \(z_{\xi\eta}+\frac{\eta}{\xi^{2}+\eta^{2}}z_{\xi}=-\frac{\eta}{2(\xi^{2}+\eta^{2})}\) with respect to \(\eta\) we can obtain,

\[z_{\xi}+\int\frac{\eta}{\xi^{2}+\eta^{2}}z_{\xi}\,d\eta=-\frac{1}{4}\ln(\xi^{2}+\eta^{2})+g(\xi)\]

Kind Regards,
Sudharaka.