Convection-diffusion equation is:
$\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} = 0.01\frac{\partial^2 u}{\partial x^2}$
Inital conditon is:
$u(x,0) = sin(x)$ over the domain 0 to $2\pi$ with periodic boundary conditon that is $u(0,t) =u(2\pi,t)$.
Second boundary condition is clamped at $x = 0 $ that is $\frac{du}{dx} = 0$
I would like to know
1) Whether this problem has an exact solution? if so please prove the solution.
2) Can any symbolic computing software like Maple, Mathematica, Matlab can solve this problem analytically?
3) Please provide some good tutorial (external links) for finding the analytical solution of the advection-diffusion equation.
Please don't provide a numerical solution because this problem is a toy problem in numerical methods.
Best Answer
Let's rewrite
$$ 2ku_t = u_{xx} - 2ku_x $$
where $k=50$. Separation of variables $u(x,t) = X(x)T(t)$ gives
$$ 2k\frac{T'}{T} = \frac{X''-2kX}{X} = -\lambda $$
The $X$ part has the characteristic polynomial
$$ r^2 - 2kr + \lambda = 0 $$
or $r = k \pm \sqrt{k^2-\lambda} $
Due to the periodicity, we need the roots to have an imaginary part, therefore $k^2-\lambda = -\mu^2$, or $\lambda = \mu^2+k^2$. This makes
$$ X(x) = e^{kx}\big[A\sin(\mu x) + B\cos(\mu x) \big] $$
The periodicity condition $X(x) = X(x+2\pi)$ requires $\mu = n$ where $n=1,2,3$.
The other boundary condition is $X'(0)=0$. Solving this gives $\mu A + kB = 0$, we can write
$$ X_n(x) = e^{kx}\big[k\sin(nx) - n\cos(nx)\big] $$
up to a multiplicative constant.
Solving for $T(t)$ and using superposition, we have the general solution
$$ u(x,t) = \sum_{n=1}^\infty c_n e^{-\frac{n^2+k^2}{2k}t}e^{kx}\big[k\sin(nx) - n\cos(nx)\big] $$
The initial condition is equivalent to solving for $c_n$ such that
$$ \sum_{n=1}^\infty c_n\big[k\sin(nx) - n\cos(nx)\big] = e^{-kx}\sin x $$
Using orthogonality, we have
$$ c_n = \frac{\int_0^{2\pi} e^{-kx}\sin x \big[k\sin(nx) - n\cos(nx)\big] dx}{\int_0^{2\pi} \big[k\sin(nx) - n\cos(nx)\big]^2 dx} $$