Ideas on how to solve a PDE with mixed derivatives

Question

I work with PDEs and want to solve a PDE that I come up with by myself. The PDE is given below $$u_{xx}+2u_{xy}+u_{yy}=0, \;\;\;\;u(x,0)=x^2,\;\;\;\;\; u(x,1)=x.$$ In Maple I obtain the solution: $$u(x,y)=F_1(y-x)+F_2(y-x)x$$ and with my conditions, $$u(x,y)=-{y}^{3}+2\,x{y}^{2}+ \left( -{x}^{2}-x+1 \right) y+{x}^{2} $$

Here is my question. Is it possible that I can solve this PDE with separation of variables or maybe method of charateristics or substitution? I can't figure out on how I should set up some equations for my problem. What I think is that I can use seperation of variables, i.e. $u_{xx}=X''$, $u_{xy}=X'Y'$ and $u_{yy}=Y''$. Any hints or suggestion for my problem would be appreciated.

Using the method given in my answer to a similar question results in the substitutions $x=\xi$ and $y=\xi+\eta$ which gives the PDE $u_{\xi \xi}=0$. — projectilemotion, Oct 23 '20 at 16:02
Thank you! I will work it out and read your comment on the link you posted. — NabbKitha, Oct 24 '20 at 10:16

score 3 · Accepted Answer · answered Oct 24 '20 at 15:22

3

So I managed to derive the solution to the PDE, thanks to @projectilemotion. Down below one can see the calculations, correct me if any mistakes!

Transform to canonical form, $a=b=c=1$. Then, $$\frac{dy}{dx}=1$$ Solution yields $y=x+C$ or $C=y-x$. Let $\xi=y-x$ and $\eta=x$ then, $$\begin{align*} &u(x,y)=U(\xi,\eta)\\ &u_{xx}=U_{\xi \xi}-2U_{\xi \eta}+U_{\eta \eta}\\ &u_{yy}=U_{\xi \xi}\\ &u_{xy}=-U_{\xi \xi}+U_{\xi \eta} \end{align*}$$

So the $u_{xx}+2u_{xy}+u_{yy}=0$ can be transformed into $U_{\eta \eta}=0$. Integrating twice, one obtain $U=\eta F_1(\xi)+F_2(\xi)$ and therefore $u(x,y)=xF_1(y-x)+F_2(y-x)$.

Yet, my problem is now to find the solution with $u(x,1)=x$ and $u(1,y)=y$. Any suggestion would be appreciated.

answered Oct 24 '20 at 15:22

NabbKitha

541

2

Nice job! Regarding the particular solution for the conditions given in the bottom of your answer, the change of variable $u=1-x$ and $v=y-1$ and replacing $v$ by $u$ in the second equation leads to the simultaneous equations $$1-u=(1-u)F_1(u)+F_2(u),$$ $$1+u=F_1(u)+F_2(u),$$ for which the solution is $F_1(u)=2$ and $F_2(u)=u-1$. This leads to the solution $u(x,y)=x+y-1$. – projectilemotion Oct 25 '20 at 06:45
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I would like to say thank you for helping me improve my knowledge in PDEs. I understand your method and managed to solve it. I will try to work out some more problems. – NabbKitha Oct 25 '20 at 12:06

Ideas on how to solve a PDE with mixed derivatives

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