Do I get ALL solutions for the heat equation by using the separation of variables?

Question

Let's assume the heat equation as: $$\frac{\partial \omega_1}{\partial t}=\nu(\frac{\partial^2 \omega_1}{\partial x_1^2}+\frac{\partial^2 \omega_1}{\partial x_2^2})$$ Using the separation of variables as $\omega_1=X_1(x_1)X_2(x_2)T(t)$ one can find a solution as: $$\omega_1=(C_1 \cos(\sqrt{a}x_1)+C_2 \sin(\sqrt{a}x_1))(C_3 \cos(\sqrt{b}x_2)+C_4 \sin(\sqrt{b}x_2))e^{-\nu(a+b)t}$$ Let's now assume a second function $\omega_2$ which also needs to solve the heat equation, thus can be written as: $$\omega_2 = (K_1 \cos(\sqrt{a}x_1)+ K_2 \sin(\sqrt{a}x_1))(K_3 \cos(\sqrt{b}x_2)+ K_4 \sin(\sqrt{b}x_2))e^{-\nu(a+b)t}$$ Both functions need to fulfill the boundary conditions $\omega_i(x_2=0)=0$ and $\omega_i(x_2=L)=0$ and the continuity equation $\frac{\partial \omega_1}{\partial x_1}+\frac{\partial \omega_2}{\partial x_2}=0$. How can I assure that by using the separation of variables I get ALL the solutions needed to solve this set of conditions?

Answer 1

There are some uniqueness theorems that tell you that certain initial-boundary value problems for the heat equation cannot have more than one solution. The goal of separation of variables is only to find a solution to the problem. Once you have found a solution, then you invoke a uniqueness theorem to show that it is the only solution.