1.3

Analytic Solution.

The diffusion equation is given by
\begin{equation}
\frac{\partial{u}}{\partial{t}} = \nu \frac{\partial^2{u}}{\partial{x^2}}
\end{equation}
Using separation of variables, let
\begin{equation}
u(x,t) = T(t)X(x)
\end{equation}
then
\begin{equation}
\frac{\partial{(XT)}}{\partial{t}} = \nu\frac{\partial^2{(XT)}}{\partial{x^2}}
\end{equation}
\begin{equation}
\frac{1}{\nu T} \frac{\mathrm{d}T}{\mathrm{d}t} = \frac{1}{X} \frac{\mathrm{d^2}X}{\mathrm{d}x} = -\frac{1}{\lambda^2}
\end{equation}
Separating the two solutions,
\begin{equation}
T(t) = A e^{\frac{-\nu t}{\lambda^2}}
\end{equation}
\begin{equation}
X(x) = B\cos(\frac{x}{\lambda}) + C\sin(\frac{x}{\lambda})
\end{equation}
The general solution is then of the form (after multiplying through by \(A$\))
\begin{equation}
\label{eq:general}
u(x,t) = e^{\frac{-\nu t}{\lambda^2}}\left[D\cos(\frac{x}{\lambda}) + E\sin(\frac{x}{\lambda})\right]
\end{equation}
Next, the boundary conditions are applied to determine the coefficients.
\begin{equation}
u(x=0,t) = e^{\frac{-\nu t}{\lambda^2}}\left[D\right] = 0
\end{equation}
Therefore, $D = 0$.
\begin{equation}
u(x = \pi,t) = e^{\frac{-\nu t}{\lambda^2}}\left[E\sin(\frac{\pi}{\lambda})\right] = 0
\end{equation}
which gives the non-trivial solution \(\lambda = \frac{1}{n}\).
The general solution for a given \(n\) is
\begin{equation}
u_n(x,t) = E_n e^{-\nu n^2t}\sin(nx)
\end{equation}
and for all \(n\)
\begin{equation}
u(x,t) = \sum_{n=1}^\infty E_n e^{-\nu n^2t}\sin(nx)
\end{equation}
Now, the initial condition is applied
\begin{equation}
u(x,t = 0) = \sum_{m=2}^5 \frac{1}{m} \sin(mx) = \sum_{n=1}^\infty E_n \sin(nx)
\end{equation}
giving \(E_n = \frac{1}{m}\) and \(n = m = [2,3,4,5]\). The final form of the solution with all boundary and initial conditions satisfied over the domain \(0 \leq x \leq \pi\) is
\begin{equation}
u(x,t) = \sum_{m=2}^5 \frac{1}{m} e^{-\nu m^2t}\sin(mx)
\end{equation}
The solution at \(t=2\) and \(\nu = 0.05\) is
\begin{equation}
u(x=[0,\pi],t=2)|_{\nu=0.05} = \sum_{m=2}^5 \frac{1}{m} e^{-0.1m^2}\sin(mx)
\end{equation}