# Can the PDE $u_{xx} + u_{xy} + u_{yy} = 0$ be separated by variables?

I just started reading Gerald Folland's book Fourier Analysis and Its Applications. I have a question about problem 1 from section 1.3.

The problem is the following.

1. Derive pairs of ordinary differential equations from the following partial differential equations by separation of variables, or show that it is not possible.

$\quad$ (c) $\,$ $u_{xx} + u_{xy} + u_{yy} = 0$

Then if I try to separate variables by letting $u(x, y) = X(x)Y(y)$, I get the following equation

$$X''(x)Y(y) + X'(x)Y'(y) + X(x)Y''(y) = 0$$

For the other three equations corresponding to parts (a), (b) and (d) of the exercise I was able to separate the variables quite easily after doing the above substitution $u(x, y) = X(x)Y(y)$, but for this one I just don't think it's possible.

Question

Can the above equation be separated by variables? And if not, how would one prove that it can't be done?

I haven't been able to find examples where a particular PDE is shown not to be separable by variables.

Thanks for any help.

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Try rotating by 45 degrees? – paul garrett Aug 26 '12 at 19:41
Dear @paul can you please elaborate on that? I don't really have a background in partial differential equations so I can't extract anything from your comment. – Adrián Barquero Aug 26 '12 at 20:00
Ah, sorry, it's not so much about PDEs as about the corresponding quadratic polynomial $x^2+xy+y^2$. Its level curves are ellipses with axes at 45 degrees to the $x$-axis and $y$-axis. (We anticipate that this is relevant by taking Fourier transform, for example.) Thus, use variables $s=x+y$ and $t=x-y$, or similar. – paul garrett Aug 26 '12 at 20:16
@paul I see, thanks a lot for your answer. – Adrián Barquero Aug 26 '12 at 20:23

You've almost answered your question already. If you divide by $XY$ you obtain:

$$\frac{X''}{X}+ \frac{X'Y'}{XY}+ \frac{Y''}{Y}=0$$

Hence,

$$\frac{X''}{X}= \frac{X'Y'}{XY}- \frac{Y''}{Y}$$

Now, if it worked you'd be able to observe that each side of the expression depends only on $x$ (the l.h.s.) and only on $y$ (the r.h.s.). But, that fails here due to the mixed derivative term. Perhaps that is all is expected of you.

Maybe someone has a better answer? The hint about $45^o$ by paul garrett was encouraging you that this PDE is a quadratic form in $\partial_x, \partial_y$ and it can be diagonalized by changing to eigencoordinates. Personally, I would delighted if I had a student employ such a technique.

Note: $Q(x,y) = x^2+xy+y^2$ implies $A = \left[ \begin{array}{cc} 1 & 1/2 \\ 1/2 & 1 \end{array} \right]$ hence $det(A - \lambda I) = (\lambda -1)^2-1/4=0$. We find $\lambda_1 = 1/2$ and $\lambda_2 = 3/2$.This suggests you can choose coordinates $z,w$ such that the PDE takes the form $\frac{1}{2}u_{yy}+\frac{3}{2}u_{zz}=0$.

But, this is not allowed. Otherwise most classification questions in differential equations would be meaningless. Take the question of exactness for first order ODEs. You could always say it is exact by Pfaff's Theorem.

More to the point, rotation by 45 degrees (those are the eigencoordinates) is not a fair play in your game.

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Thanks a lot James. – Adrián Barquero Aug 26 '12 at 20:25
James, why is this not allowed? It takes some smart thinking to get there, but the transformation seems rather linear to me, so in that respect it's very "nice". – akkkk Aug 26 '12 at 20:38
@Auke, it is nice, I totally agree. However, it is not the standard technique of proposing the net-solution is a product of functions of just one-variable. The question is not about how to solve the problem nicely, the problem constrained us to that one technique. – James S. Cook Aug 26 '12 at 22:39
@James S. Cook: -1 for your post. It should be a problem that you answer further about the eigencoordinates. As the separability of a homogeneous linear PDE only depends on the partial derivatives orders combos and their coefficient combos of the terms in the homogeneous linear PDE and should not depends on about the eigencoordinates, you are trying to divert the askers' attentions. – doraemonpaul Aug 26 '12 at 23:05

In fact this question is easy to think that do not rely on books.

In fact the homogeneous linear PDE with two independent variables, the most general separable condition is that if it can rewritable to the form $\sum_{m_1=0}^{n_1}p(x)X^{(m_1)}(x)\sum_{m_2=0}^{n_2}q(y)Y^{(m_2)}(y)=\sum_{m_3=0}^{n_3}r(x)X^{(m_3)}(x)\sum_{m_4=0}^{n_4}s(y)Y^{(m_4)}(y)$ when letting $u(x,y)=X(x)Y(y)$ .

When you try to solve the PDE $u_{xx}+u_{xy}+u_{yy}=0$ by separation of variables, i.e. let $u(x,y) = X(x)Y(y)$ , you will get the key equation $X''(x)Y(y)+X'(x)Y'(y)+X(x)Y''(y)=0$ , and this is obviously fail to rewrite it to the form $\sum_{m_1=0}^{n_1}p(x)X^{(m_1)}(x)\sum_{m_2=0}^{n_2}q(y)Y^{(m_2)}(y)=\sum_{m_3=0}^{n_3}r(x)X^{(m_3)}(x)\sum_{m_4=0}^{n_4}s(y)Y^{(m_4)}(y)$ , so the PDE $u_{xx}+u_{xy}+u_{yy}=0$ should be unseparable.

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Dear doraemon, is it obvious that the equation I obtained cannot be written in that form? It isn't obvious to me at least. – Adrián Barquero Aug 26 '12 at 20:29