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