# Separation of variables in polar form

Will the separation of variable method for solving partial differential equations useful for all equations? or any conditions on PDE? And also please explain about separation constant in polar form equations.

• The main theme of this question is about PDE, why don't this question have pde tag? – doraemonpaul Sep 16 '13 at 8:49

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