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Separation of variables is explained in my notes to solve certain types of partial differential equations. This method requires the assumption that the solution, say $u(x,t)$, is in the form

$$u(x,t)=X(x)T(t).$$

Then you solve some ordinary differential equations and consider the sum:

$$\sum b_k X_k(x) T_k(t)$$

So provided the series converges and with the conditions given for the pde you can obtain the coefficients $b_{k}$ and this is the/a solution of the PDE.

This is pretty clear for me. What I want to know now is how can you ensure $u(x,t)=X(x)T(t)$, or more precisely, How can you assure you found every solution for the PDE? It is unclear to me why sometimes the method is used and it is not.

These are some examples from my notes:

The heat equation: (Can be solved using separation of variables) $$ \begin{cases} u_{t}-u_{xx}=0 & 0<x<\pi, t>0\\ u(x,0)=f(x) & 0<x<\pi & \\ u(0,t)=u(\pi,t)=0 & t>0 \end{cases} $$

Also,

The wave equation: (Can be solved using separation of variables) $$ \begin{cases} u_{tt}-c^2u_{xx}=0 & 0<x<\pi, t>0\\ u(0,t)=u(\pi,t)=0 & 0<x<\pi \\ u(x,0)=f(x) & t>0 \\ u_{t}(x,0)=g(x) & 0<x<\pi \end{cases} $$

Also, the Laplace equation: $$ \begin{cases} \Delta u=0 & \text{in }B \\ u=f & \text{in } \partial(B) \end{cases} $$

for the circle $B$, can be solved by separation of variables if you consider the change of variables $u(r,\theta)=(r\cos\theta,r\sin\theta)$, which reduces the ecuation to:

$$ \begin{cases} u_{rr}+ \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta \theta}=0 & \text{in } [- \pi, \pi] \times [0,1) \\ u=f & \text{in } [- \pi, \pi] \end{cases} $$

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    $\begingroup$ This question is related, along with those under "Linked" in the sidebar on that page. $\endgroup$ – Antonio Vargas Jun 11 '15 at 21:50
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For linear PDE, the PDE can be solved by separation of variables if the homogeneous part of the PDE can rewrite to the form $\dfrac{\sum\limits_{a_1=0}^{b_1}M_{a_1}(x)X^{[a_1]}(x)}{\sum\limits_{a_2=0}^{b_2}N_{a_2}(x)X^{[a_2]}(x)}=\dfrac{\sum\limits_{a_3=0}^{b_3}P_{a_3}(t)T^{[a_3]}(t)}{\sum\limits_{a_4=0}^{b_4}Q_{a_4}(t)T^{[a_4]}(t)}$ when letting $u(x,t)=X(x)T(t)$ .

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  • $\begingroup$ Is that a necessary condition for the method? $\endgroup$ – Raúl Astete Dec 18 '19 at 3:28

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