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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.

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  • $\begingroup$ The main theme of this question is about PDE, why don't this question have pde tag? $\endgroup$ – doraemonpaul Sep 16 '13 at 8:49
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At the lowbrow level, & as far as I can see, separation of variables has two major limitations: first, the domain must be rectangular; & second, the PDE cannot contain mixed derivatives. Thus, for example, to apply the technique to Laplace's eq. (no mixed derivatives) on a disk (which is not a rectangle), we first pass to polar coordinates. This introduces no mixed derivatives (check the formula for the Laplacian in polar coordinates) and turns the disk onto a rectangle.

As for your question re: polar coordinates: (a) what exactly you need is beyond me & (b) why not consult the tons of web resources (e.g., OCW) on PDEs + polar coordinates?

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

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