Why is the Separation of Variables method valid?

1. Mar 20, 2013

Zag

Why is the "Separation of Variables" method valid?

Hey guys,

Lately I have been focusing on some question that have annoyed me for some time. One of these questions is: Why is the method of separation of variables valid when solving some PDE?

Usually smmetry arguments are presented, and in some physical systems you can guess that it will be the case so that the separation of variables will work. However, is there a rigorous treatment of this method? Could anyone suggest me a reading on this topic or share what kind of analysis is necessary to treat this question?

Thank you very much! ;)

2. Mar 20, 2013

phyzguy

It doesn't work most of the time. But if it finds a solution, then it finds a solution. Any method that finds a solution is 'valid' in that sense.

3. Mar 20, 2013

Ben Niehoff

Take a simple example, the PDE

$$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$$
Then just postulate $f(x,y) = X(x) Y(y)$. Not ALL solutions have this form (in fact, most do not), but we're only interested if some solutions have this form. Plugging this in, you get

$$Y(y) \frac{d^2 X}{dx^2} + X(x) \frac{d^2 Y}{dy^2} = 0$$
which can be rearranged to give

$$\frac{1}{X(x)} \frac{d^2 X}{dx^2} = - \frac{1}{Y(y)} \frac{d^2 Y}{dy^2}$$
but now, everything on the left-hand side is a function of $x$ only, and everything on the right-hand side is a function of $y$ only. I.e., we have

$$g(x) = h(y)$$
But the catch is that the equation must hold for ALL $x$ and $y$. So, in fact, the functions $g(x)$ and $h(y)$ must be constants! So we now we can break this up into two equations,

$$\frac{1}{X(x)} \frac{d^2 X}{dx^2} = m^2, \qquad \frac{1}{Y(y)} \frac{d^2 Y}{dy^2} = - m^2$$
for some constant $m^2$, which we call the "separation constant".

The key point here is that if we guess a solution of the form $f(x,y) = X(x) Y(y)$, we are able to rearrange the equation so that all functions of $x$ are on one side, and all functions of $y$ are on the other side. If it is possible to do this, the PDE is called "separable".

Not all PDEs are separable! But if a PDE is separable, then separation of variables will produce a solution. In fact, it will produce a whole family of solutions, labelled by the separation constants.

Once you have a family of solutions of the form $f_m(x,y)=X_m(x) Y_m(y)$ with separation constant m, you can then use linearity to consider series solutions of the form

$$f(x,y) = \sum_m a_m X_m(x) Y_m(y)$$
for some constants $a_m$. The next question you should be concerned about is whether you can get ALL solutions this way. This depends on whether the family $f_m(x,y)=X_m(x) Y_m(y)$ of separable solutions is "complete" or not. It requires a bit more work to show whether this is true, but there are theorems that tell you when it is true for certain kinds of equations.

4. Mar 21, 2013

NegativeDept

There are rigorous group-theoretic treatments, but I'm afraid I will confuse everyone (and/or myself) if I attempt to explain them in detail. Roughly speaking: look for coordinate transformations which don't change the form of the equation. Those transformations are symmetries of the PDE, and the set of all of them forms a group. Solutions should be invariant with respect to that group.

In practice, I think most of us use something more like Ben Niehoff's pragmatic explanation or the intuitive symmetry argument you mentioned.

5. Mar 28, 2013

alissca123

6. Mar 29, 2013

Zag

Thank you very much, guys! All the replies helped me a lot, and I think now I understand the context of separation of variables much better.

Thank you also for the references and links, I am definitely going to read them! ;)

7. Mar 29, 2013

Coelum

Zag,
it's very simple: if the differential operator associated to the PDE is the sum of operators that act on different set of variables, then the method works. Ben's example is one of those cases. In some cases, you may find a coordinate transformation that "separates" the operator into parts that operates on different variables. Then, you apply the method to the PDE expressed in the new coordinates.