# Solving heat equation on a circle

Let's consider heat equation $$\frac{\partial u}{\partial t} = \frac{\partial^2u}{\partial x^2}$$ or equivalently $$u_t = u_{xx}.$$ Of course $$t > 0$$ and $$x \in [0, 1]$$.

It's defined on a circle, thus

1. $$u(0, t) = u(1, t)$$
2. $$u_x(0, t) = u_x(1, t)$$

Moreover the initial condition is given by $$f(x) = u(x, 0)$$.

Using the separation of variables I managed to show that $$u(x,t) = X(x)T(t)$$ where $$X(x) = C_1 \sqrt\frac{1}{\lambda} \sin \big(\sqrt \lambda x\big) + C_2 \sqrt \frac{1}{\lambda} \cos \big(\sqrt \lambda x \big)$$ $$T(t) = Ae^{- \lambda t}$$

Using assumptions $$(1)$$ and $$(2)$$ we do get the following system of equations $$\begin{cases} C_2 = C_1 \sin(\sqrt \lambda) + C_2 \cos(\sqrt \lambda)\\C_1 = C_1 \cos(\sqrt \lambda) - C_2 \sin(\sqrt \lambda)\end{cases}$$

How can I solve the system above?

$$C_2(1-\cos u) = C_1\sin u\\ C_1(1-\cos u) =-C_2\sin u$$
so $$\cos u = 1$$ and $$\sin u = 0$$ giving $$\sqrt{\lambda} = 0+2k\pi,\ \ k = 0,1\cdots$$