# Find solutions of a PDE (by substitution?)

Find the solutions of the PDE $$2v_{\xi\eta}+v_{\eta}=0.~~~~~(+)$$

Hello, with this question I refer to Transforming hyperbolic PDE into normal form. I sum it up what happened there:

The original PDE was $$x^2u_{xx}-y^2u_{yy}=0~~~~\text{in}~~~~\Omega:=\left\{(x,y)\in\mathbb{R}^2: x>0,y>0\right\}~~~~~(++).$$ Using the transformation $$\xi:=\ln(y/x),~~~~~\eta:=-\ln(xy),$$ the PDE (++) was transformed into the PDE (+).

Now to find the solutions of (++), it is easier to find the solutions of (+). And that's what I want to do now:

I use $$v_{\xi\eta}=v_{\eta\xi}$$ to write $$2v_{\xi\eta}+v_{\xi}=2\cdot (v_{\xi})_{\eta}+v_{\xi}=0$$ and then I substitute $z:=v_{\xi}$, getting the ODE $$z_{\eta}=-\frac{1}{2}z.$$ Using "separation of the variables", I get $$z=\exp\left(-\frac{1}{2}\eta+C_1(\xi)\right).$$ Re-substituting, that means $$v_{\xi}=\exp\left(-\frac{1}{2}\eta+C_1(\xi)\right)$$ and integrating to $\xi$ gives me $$v(\xi,\eta)=\int_{\xi_0}^{\xi}\exp\left(-\frac{1}{2}\eta+C_1(\tau)\right)\, d\tau+C_2(\eta),$$ which means for the original PDE (++) in my opinion, that the solutions are given by $$u(x,y)=\int_{z_0}^{\ln(y/x)}\exp\left(\frac{1}{2}\ln(xy)+C_1(\tau)\right)\, d\tau+C_2(-\ln(xy)).$$

Can anybody tell me if this is right, i.e. if this are the right solutions?

With kind regards and thank for your help!

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It is looking correct. I did not check with pen and paper. – Dutta Oct 16 '13 at 12:32
Unfortunately there is a difference between looking correct and being correct, hehe. – math12 Oct 16 '13 at 13:05
I would like to verify that this are indeed solutions. But I do not know how to do so. – math12 Oct 16 '13 at 18:15
Try substituting. By hand it appears to be pain enough, so I suggest you get $u(x,y)$ and $(++)$ into Maple and check. The PDETools is fantastic. If it is not correct you can always recheck each step on Maple. – D G Jan 7 '14 at 1:24