SU-MATH53 FEB232024

Boundary Value Problem A BVP for an ODE is defined at two different points x_0 and x_1 at two different values of l, whereby we are given:

\begin{equation} X_0 = a, X(L) = b \end{equation}

which we use to further specify a PDE. BVPs can either have no or lots of solutions.
To aid in the discovery of solutions, for:

\begin{equation} X’’ = \lambda X \end{equation}

we have:

\begin{equation} X = \begin{cases} c_1 e^{\sqrt{\lambda}x} + c_2 e^{-\sqrt{\lambda}x}, \lambda > 0 \\ c_1 x + c_2, \lambda =0 \\ c_1 \cos \left(\sqrt{|\lambda|}x\right) +c_2 \sin \left(\sqrt{|\lambda|}x\right), \lambda < 0 \end{cases} \end{equation}

Which specific solution arises out of which initial condition you use.
Dirichlet Conditions Initial conditions:

\begin{equation} \begin{cases} u(t,0) = 0 \\ u(t, l) = 0 \end{cases} \end{equation}

This tells us that we are holding the ends of the rod at a constant temperature.
Solutions For:

\begin{equation} X’’ = \lambda X \end{equation}

in the vanishing Case (X(0) = 0 = X(L)):

\begin{equation} X = c \sin \left( \frac{k \pi x}{L}\right) \end{equation}

where c \neq 0, and the solutions quantized k = 1, 2, 3, \ldots.
which gives rise to:

\begin{equation} \lambda = \frac{-n^{2}\pi^{2}}{L^{2}} \end{equation}

Neumann Conditions \begin{equation} \begin{cases} \pdv{u}{x}(t,0) = 0 \ \pdv{u}{x}(t, l) = 0 \end{cases} \end{equation}
this tells us there is no heat flux across the boundary (i.e. heat doesn’t escape).
Solutions For:

\begin{equation} X’’ = \lambda X \end{equation}

in the vanishing Case (X’(0) = 0 = X’(L)):

\begin{equation} X = c \cos \left( \frac{k \pi x}{L}\right) \end{equation}

where c \neq 0, and the solutions quantized k = 1, 2, 3, \ldots.
which gives rise to:

\begin{equation} \lambda = \frac{-n^{2}\pi^{2}}{L^{2}} \end{equation}

Examples See Heat Equation, and its worked solution.