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As the title says, the task is

Find the common solution of the PDE $$ u_{xy}+2u_x+u_y+2u=0.~~~~(*) $$

What I have to mention here is that this is the second part of the task which I asked here: Which PDE does v fullfill?. Maybe one does need that here, therefore I mention it.

My idea is the following:

Concerning to the link, $u$ is a solution of (*) exactly then, when $v(x,y):=u(x,y)\exp(x+2y)$ is a solution of the PDE $$ v_{xy}=0. $$

Now the question is which possibilities there are for $v$, in order to get $v_{xy}=0$.

$v_{xy}=0$ exactly then, when $u_y$ is constant or only depends on $y$ or is a combination of that, i.e. $v=const+g(y)$ for any $g\in C^2(\mathbb{R})$?

Then to my thoughts, $$ u(x,y)=v(x,y)\exp(-bx-ay), v(x,y):=\mbox{const}+g(y), g\in C^2(\mathbb{R}) $$

is the common solution of (*).

What do you think about that?

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1 Answers1

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Integrating $v_{xy} = 0$ with respect to $x$ should give you $v_y = a(y)$ for an arbitrary function $a$. Now integrate with respect to $y$ and you should get $v = A(y) + B(x)$ where $A$ and $B$ are arbitrary functions.

Robert Israel
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  • Thanks! What is about adding a constant $c$, i.e. $v=A(y)+B(x)+c, c\in\mathbb{R}$ - or is this to be thought of contained already in the functions $A$ and $B$? --- So the common solution is $u(x,y)=(A(y)+B(x))\exp(-x-2y)$ with arbirtrary functions $A,B\colon\mathbb{R}\to\mathbb{R}$? Or is it necessary/ better to suppose $A,B$ being in $C^2(\mathbb{R})$? –  Oct 14 '13 at 17:48
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    Yes, a constant can be considered as a function of $x$. Differentiability is a bit tricky: if you consider $v_{xy}$ as meaning $\dfrac{\partial}{\partial y} \dfrac{\partial}{\partial x} v$, then for a solution of $v_{xy} = 0$, $A(y)$ could be literally anything but $B(x)$ must be differentiable. However, for your $u$ equation to be defined you do need $A(y)$ and $B(x)$ to be differentiable. – Robert Israel Oct 14 '13 at 18:40