Separation of variables for partial differential equations

Question

What class of Partial Differential Equations can be solved using the method of separation of variables?

Answer 1

There is an extremely beautiful Lie-theoretic approach to separation of variables, e.g. see Willard Miller's book [1] (freely downloadable). I quote from his introduction:

This book is concerned with the relationship between symmetries of a linear second-order partial differential equation of mathematical physics, the coordinate systems in which the equation admits solutions via separation of variables, and the properties of the special functions that arise in this manner. It is an introduction intended for anyone with experience in partial differential equations, special functions, or Lie group theory, such as group theorists, applied mathematicians, theoretical physicists and chemists, and electrical engineers. We will exhibit some modem group-theoretic twists in the ancient method of separation of variables that can be used to provide a foundation for much of special function theory. In particular, we will show explicitly that all special functions that arise via separation of variables in the equations of mathematical physics can be studied using group theory. These include the functions of Lam6, Ince, Mathieu, and others, as well as those of hypergeometric type.

This is a very critical time in the history of group-theoretic methods in special function theory. The basic relations between Lie groups, special functions, and the method of separation of variables have recently been clarified. One can now construct a group-theoretic machine that, when applied to a given differential equation of mathematical physics, describes in a rational manner the possible coordinate systems in which the equation admits solutions via separation of variables and the various expansion theorems relating the separable (special function) solutions in distinct coordinate systems. Indeed for the most important linear equations, the separated solutions are characterized as common eigenfunctions of sets of second-order commuting elements in the universal enveloping algebra of the Lie symmetry algebra corresponding to the equation. The problem of expanding one set of separable solutions in terms of another reduces to a problem in the representation theory of the Lie symmetry algebra.

[1] Willard Miller. Symmetry and Separation of Variables.
Addison-Wesley, Reading, Massachusetts, 1977 (out of print)

Answer 2

For example the linear homogeneous PDEs with dependent variable $u$ and independent variables $x$ and $y$ , the separable condition is that the PDEs can rewrite to the form $\dfrac{\sum\limits_{a_1=0}^{b_1}M_{a_1}(x)X^{[a_1]}(x)}{\sum\limits_{a_2=0}^{b_2}N_{a_2}(x)X^{[a_2]}(x)}=\dfrac{\sum\limits_{a_3=0}^{b_3}P_{a_3}(y)Y^{[a_3]}(y)}{\sum\limits_{a_4=0}^{b_4}Q_{a_4}(y)Y^{[a_4]}(y)}$ when letting $u(x,y)=X(x)Y(y)$ .

For example, the PDE $x^2u_{xy}-yu_{yy}+u_x-4u=0$ mentioned in The canonical form of a nonlinear second order PDE is an unseparable example while the PDE $u_{xy}-yu_{yy}+u_x-4u=0$ is a separable example.

Start from the PDEs with three independent variables, the separable conditions are more difficult to described, since for example the linear homogeneous PDEs with dependent variable $u$ and independent variables $x$ , $y$ and $z$ , the PDEs are separable when the PDEs not only can rewrite to the form $\dfrac{\sum\limits_{a_1=0}^{b_1}M_{1,a_1}(x)X^{[a_1]}(x)}{\sum\limits_{a_2=0}^{b_2}M_{2,a_2}(x)X^{[a_2]}(x)}+\dfrac{\sum\limits_{a_3=0}^{b_3}M_{3,a_3}(y)Y^{[a_3]}(y)}{\sum\limits_{a_4=0}^{b_4}M_{4,a_4}(y)Y^{[a_4]}(y)}+\dfrac{\sum\limits_{a_5=0}^{b_5}M_{5,a_5}(z)Z^{[a_5]}(z)}{\sum\limits_{a_6=0}^{b_6}M_{6,a_6}(z)Z^{[a_6]}(z)}=0$ when letting $u(x,y,z)=X(x)Y(y)Z(z)$ , but also when the PDEs can rewrite to the form $\dfrac{\sum\limits_{a_1=0}^{b_1}M_{1,a_1}(x)X^{[a_1]}(x)}{\sum\limits_{a_2=0}^{b_2}M_{2,a_2}(x)X^{[a_2]}(x)}+\dfrac{\sum\limits_{a_3=0}^{b_3}M_{3,a_3}(y)Y^{[a_3]}(y)}{\sum\limits_{a_4=0}^{b_4}M_{4,a_4}(y)Y^{[a_4]}(y)}+\dfrac{\sum\limits_{a_3=0}^{b_3}N_{3,a_3}(y)Y^{[a_3]}(y)\sum\limits_{a_5=0}^{b_5}M_{5,a_5}(z)Z^{[a_5]}(z)}{\sum\limits_{a_4=0}^{b_4}N_{4,a_4}(y)Y^{[a_4]}(y)\sum\limits_{a_6=0}^{b_6}M_{6,a_6}(z)Z^{[a_6]}(z)}=0$ when letting $u(x,y,z)=X(x)Y(y)Z(z)$ .