Separation of variables and heat

Question

Consider the equation:$$u_t-ku_{xx}-\alpha u=0, x>0, x<2\pi , t>0$$ $$u(0,t)=u(2\pi,t), t\geq 0,$$ $$u_x(0,t)=u_x(2\pi,t), t\geq 0,$$ $$u(x,0)=f(x)$$ $f$ is a smooth periodic function and $k>0$ and \alph is a real number. I want to solve this equation using separations. I did this: Let $u(x,t)=X(x)T(t)$ them we have:$$u_t=XT^\prime$$ $$u_{xx}= X^{\prime\prime}T$$. Then plugging these into the original equation we conclude: $$XT^\prime-kX^{\prime\prime}T-\alpha XT=0$$ Hence: $$\frac{T^\prime}{T} = \frac{kX^{\prime\prime}+\alpha X}{X}= - \lambda$$ Therefore: $$T^\prime + \lambda T =0$$ $$and$$ $$X^{\prime\prime} + (\frac{\alpha + \lambda}{k}) X =0$$ Can anyone please help me how can I proceed from here? Thanks.

Answer 1

You asked the same question before. Consider editing it instead of asking a new one.

Starting from here

$$ XT' -kX''T - \alpha XT = 0 $$

Divide through by $XT$ to get

$$ \frac{T'}{T} - k\frac{X''}{X} = \alpha $$

Let $\frac{X''}{X} = -\lambda$ then

\begin{cases} X'' + \lambda X = 0 \\ T' + (k\lambda-\alpha)T = 0 \end{cases}

We want the solution to be $2\pi$-periodic in $X$, so $\lambda = n^2$ and the general solution is

$$ u(x,t) = A_0e^{\alpha t} + \sum_{n=1}^\infty [A_n\cos(nx) + B_n\sin(nx)]e^{-(kn^2-\alpha)t} $$