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--- title: "SU-MATH53 FEB232024" source: https://www.jemoka.com/posts/kbhsu_math53_feb232024/ date: 2024-02-23 --- 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 \qty(\sqrt{|\lambda|}x) +c_2 \sin \qty(\sqrt{|\lambda|}x), \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 \qty( \frac{k \pi x}{L}) \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 \qty( \frac{k \pi x}{L}) \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.