When is separation of variables an acceptable assumption to solve a PDE?

Question

We know that one of the classical methods for solving some PDEs is the method of separation of variables. It works for known types of PDEs and many examples of physical phenomena are successfully represented in PDE systems where an assumption that the functions are separable in variables seems to work just fine, and we get correct solutions.

My question is how do we arrive at the conclusion that the assumption that our function is 'separable in variables' is valid/correct for a particular equation in the first place? Also, I suspect that it will rely on the given scenario of boundary conditions too (not only the type of equation), such as how symmetric the boundaries are to the coordinates, whether they are curved or separable in variables themselves, etc.

Here is an example of what I mean: suppose I have the Helmholtz equation, say $\nabla^{2}f(x,y)+k^{2}f(x,y)=0$ and the boundary values are defined for some lines, like $f=0$ for $x=0$ and $x=x_{0}$ and for $y=0$ and $y=y_{0}$. Suppose that separation of variables worked here and we got the solution. Can we solve the same equation using the same method (i.e. separation of variables) for a new domain of boundaries, say that $f=0$ over the line segments $|y|=ax$ for $x\in[0,x_{0}]$ with $a$ being a scalar, or over a curved boundary like $y=ax^{2}$?

The point is, if the boundary conditions are not seperable in the given coordinate variables, wouldn't the method of separation of varaibles fail? This begs the question whether its utility is also subject to the nature of BVs, and not only to the form of the PDEs.

Thanks for any help.

Answer 1

In fact the separable conditions (conditions than can apply the separation of variables method) of the PDEs are very depending on about the partial derivatives combinations and the coefficients of the terms in the PDE, symmetries and types of the bounded domain are basically independent.

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 $u(x,y,z)=X(x)Y(y)Z(z)$ .