---
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&rsquo;&rsquo; = \lambda X \end{equation}
we have:
\begin{equation} X = \begin{cases} c_1 e^{\sqrt{\lambda}x} + c_2 e^{-\sqrt{\lambda}x}, \lambda &gt; 0 \\ c_1 x + c_2, \lambda =0 \\ c_1 \cos \qty(\sqrt{|\lambda|}x) +c_2 \sin \qty(\sqrt{|\lambda|}x), \lambda &lt; 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&rsquo;&rsquo; = \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&rsquo;t escape).
Solutions For:
\begin{equation} X&rsquo;&rsquo; = \lambda X \end{equation}
in the vanishing Case (\(X&rsquo;(0) = 0 = X&rsquo;(L)\)):
\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.
