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Find the general solution of the PDE $z_{xx}+z_{xy}+z_y=z$.

By separation of variable method, $z=XY$ gives, $X''-\lambda X'+(\lambda-1)X=0$ and $Y'+\lambda Y=0$.

Case 1. If $\lambda>1$, $X=c_1e^{(\lambda-1)x}+c_2e^x$, $Y=c_3e^{-\lambda y}$ and hence $z$.

Case 2. If $\lambda=1$, $z(x,y)=(A+Be^x)e^{-y}$.

Case 3. If If $\lambda<1$, $X=c_1e^{(\lambda-1)x}+c_2e^x$, $Y=c_3e^{-\lambda y}$ i.e. $z(x,y)=\left(Ae^{(\lambda -1)x}+Be^x\right)e^{-\lambda y}$.

My question:

  • If this answer correct, should I have to write all three cases? If not suggest.

  • If the question is solved by canonical form, what would be the answer? (Please solve). Is there any possibility that answers are same for both method?

  • Is there other suitable method to solve this problem?

Thank you.

math131
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    A problem involving a partial differential equation typically has a domain and boundary conditions associated with it. Those tend to clarify, in an approach like yours, the appropriate simple solutions. – A rural reader Feb 20 '23 at 16:13
  • I think equation for $X$ should be $X''-\lambda X' - (\lambda+1)X=0$. – Abolfazl Chaman motlagh Feb 20 '23 at 20:22
  • @AbolfazlChamanmotlagh Thanks. would you suggest better solution – math131 Feb 21 '23 at 06:10
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    I don't know why you divide the problem into cases for different $\lambda$. in all of them $(A e^{(\lambda-1)x}+Be^x)e^{-\lambda y}$ works! also considering the typo i mention in above comment, it should be $(A e^{(\lambda+1)x}+Be^{-x})e^{-\lambda y}$. – Abolfazl Chaman motlagh Feb 21 '23 at 07:32

1 Answers1

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Separation of variables is not suitable if there are no boundary conditions. What works is a change of variables: let $a=x-y$ and $b=y$, then $$z_x=z_a,z_y=-z_a+z_b,z_{xx}=z_{aa}$$ $$z_{xy}=z_{aa}a_y+z_{ab}b_y=-z_{aa}+z_{ab}$$ $$z_{xx}+z_{xy}+z_y=z_{ab}-z_a+z_b=z$$ Hence the corresponding linear operator factors, leaving two ODEs: $$(z_a+z)_b=z_a+z\implies z_a+z=F_1(a)e^b$$ $$e^a(z_a+z)=(ze^a)_a=F_1(a)e^ae^b\implies ze^a=F_2(a)e^b+G_1(b)$$ $$z=e^bF_3(a)+e^{-a}G_1(b)=e^yF_3(x-y)+e^{y-x}G_1(y)$$ $$=e^xF(x-y)+e^{-x}G(y)$$ where the $F$'s and $G$'s are arbitrary functions.

Parcly Taxel
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