Solution for the One-dimensional Diffusion Equation
To satisfy the boundary conditions we require
B=0 and
\lambda=n\pi/L . One solution is therefore found to be
\begin{equation*}
u(x,t)=A_n\sin(n\pi x/L)e^{-n^2\pi^2 t/L^2}.
\end{equation*}
But there are infinitely many possible
n values (infinite number of solutions). Moreover,
the diffusion equation is linear and because of this we know that a superposition of solutions
will also be a solution of the equation. We may therefore write
\begin{equation*}
u(x,t)=\sum_{n=1}^{\infty} A_n \sin(n\pi x/L) e^{-n^2\pi^2 t/L^2}.
\end{equation*}