ch11: BVPs, Sturm-Liouville, eigenfunction expansions

chapters/ch11_boundary_value_problems.tex

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+% =============================================================================
+% ch11_boundary_value_problems.tex
+% Chapter 11: Boundary Value Problems
+% =============================================================================
+
 \section{Boundary Value Problems}
 \label{ch:boundary_value_problems}
 
 \subsection{BVP vs IVP}
 \label{sec:ch11_bvp_vs_ivp}
 
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+In the preceding chapters we have focused almost exclusively on \textbf{initial value problems (IVPs)}: a differential equation together with conditions specified at a \emph{single} point. For example,
+\[
+  y'' + p(x)y' + q(x)y = f(x),
+  \qquad y(x_0) = y_0,\; y'(x_0) = y_0'.
+\]
+An IVP asks: given the state of a system at one instant, what will it do in the future? This is the natural framework for time-evolution problems.
+
+A \textbf{boundary value problem (BVP)}, by contrast, prescribes conditions at \emph{two different points} of the independent variable:
+\[
+  y'' + p(x)y' + q(x)y = f(x),
+  \qquad y(a) = \alpha,\; y(b) = \beta,
+\]
+where $a \neq b$. Instead of ``initial'' conditions at a starting time, we impose \textbf{boundary conditions} at the endpoints of an interval. The independent variable $x$ typically represents spatial position rather than time.
+
+\paragraph{When BVPs arise.} Boundary value problems appear whenever a physical quantity is constrained at the boundaries of a spatial domain. Classic examples include:
+\begin{itemize}
+  \item The displacement $y(x)$ of a string fixed at both ends: $y(0) = 0$, $y(L) = 0$.
+  \item The temperature $u(x)$ in a rod whose ends are held at prescribed temperatures.
+  \item The electric potential between two conducting plates.
+\end{itemize}
+In all of these, the differential equation describes the internal physics, while the boundary conditions encode the geometry or external constraints.
+
+\paragraph{Key differences from IVPs.} Unlike IVPs, BVPs do \emph{not} always have a solution, and when a solution exists it may not be unique. The three possibilities are:
+\begin{enumerate}
+  \item \textbf{Unique solution}: exactly one function satisfies both the ODE and the boundary conditions.
+  \item \textbf{No solution}: the boundary conditions are incompatible with any solution of the ODE.
+  \item \textbf{Infinitely many solutions}: the boundary conditions are satisfied by a whole family of solutions. This occurs precisely when the associated \textbf{homogeneous BVP} admits nontrivial solutions---a phenomenon we explore in the next subsection.
+\end{enumerate}
+
+\begin{workedexample}
+  Solve the BVP $y'' + y = 0$, $y(0) = 0$, $y(\pi) = 0$.
+
+  \textbf{Solution.} The general solution of the ODE is
+  \[
+    y(x) = c_1 \cos x + c_2 \sin x.
+  \]
+  Apply the first boundary condition:
+  \[
+    y(0) = c_1 \cdot 1 + c_2 \cdot 0 = c_1 = 0,
+  \]
+  so $c_1 = 0$ and $y(x) = c_2 \sin x$.
+
+  Apply the second boundary condition:
+  \[
+    y(\pi) = c_2 \sin \pi = c_2 \cdot 0 = 0.
+  \]
+  This condition is satisfied for \emph{any} $c_2$. Hence we have infinitely many solutions:
+  \[
+    y(x) = c_2 \sin x, \qquad c_2 \in \R.
+  \]
+  The homogeneous BVP admits nontrivial solutions; this is the hallmark of an \textbf{eigenvalue problem}.
+\end{workedexample}
+
+\begin{workedexample}
+  Solve the BVP $y'' + y = 0$, $y(0) = 0$, $y(\pi/2) = 1$.
+
+  \textbf{Solution.} Again the general solution is $y(x) = c_1 \cos x + c_2 \sin x$.
+
+  Apply $y(0) = 0$: we obtain $c_1 = 0$, so $y(x) = c_2 \sin x$.
+
+  Apply $y(\pi/2) = 1$:
+  \[
+    y(\pi/2) = c_2 \sin(\pi/2) = c_2 \cdot 1 = 1 \quad\Longrightarrow\quad c_2 = 1.
+  \]
+  The solution is unique: $y(x) = \sin x$.
+\end{workedexample}
 
 \subsection{Eigenvalue Problems}
 \label{sec:ch11_eigenvalue_problems}
 
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+The phenomenon of infinitely many solutions leads us to one of the most important structures in applied mathematics: the \textbf{eigenvalue problem}.
+
+Consider the second-order equation
+\begin{equation}
+  \label{eq:eigenvalue_problem}
+  y'' + \lambda y = 0, \qquad 0 < x < L,
+\end{equation}
+with Dirichlet boundary conditions
+\begin{equation}
+  \label{eq:dirichlet_bc}
+  y(0) = 0, \qquad y(L) = 0.
+\end{equation}
+Here $\lambda$ is a parameter. For most values of $\lambda$, the only solution is the trivial one $y \equiv 0$. But for certain special values of $\lambda$---called \textbf{eigenvalues}---there exist nontrivial solutions, called \textbf{eigenfunctions}.
+
+\paragraph{Case analysis.} We treat three cases for the sign of $\lambda$.
+
+\medskip
+\noindent\textbf{Case 1: $\lambda < 0$.} Let $\lambda = -\mu^2$ with $\mu > 0$. The general solution is
+\[
+  y(x) = c_1 e^{\mu x} + c_2 e^{-\mu x}
+  = A \cosh(\mu x) + B \sinh(\mu x).
+\]
+Apply $y(0) = 0$: $A = 0$, so $y(x) = B \sinh(\mu x)$.
+
+Apply $y(L) = 0$: $B \sinh(\mu L) = 0$. Since $\mu > 0$ and $L > 0$, we have $\sinh(\mu L) > 0$, so $B = 0$. Hence $y \equiv 0$. There are \emph{no} eigenvalues with $\lambda < 0$.
+
+\medskip
+\noindent\textbf{Case 2: $\lambda = 0$.} The equation becomes $y'' = 0$, with general solution
+\[
+  y(x) = c_1 x + c_2.
+\]
+Apply $y(0) = 0$: $c_2 = 0$. Apply $y(L) = 0$: $c_1 L = 0$, so $c_1 = 0$. Again $y \equiv 0$. The value $\lambda = 0$ is \emph{not} an eigenvalue.
+
+\medskip
+\noindent\textbf{Case 3: $\lambda > 0$.} Let $\lambda = \mu^2$ with $\mu > 0$. The general solution is
+\[
+  y(x) = c_1 \cos(\mu x) + c_2 \sin(\mu x).
+\]
+Apply $y(0) = 0$: $c_1 = 0$, so $y(x) = c_2 \sin(\mu x)$.
+
+Apply $y(L) = 0$: $c_2 \sin(\mu L) = 0$. For a nontrivial solution we need $c_2 \neq 0$, which requires
+\[
+  \sin(\mu L) = 0 \quad\Longrightarrow\quad \mu L = n\pi, \qquad n = 1, 2, 3, \dots
+\]
+Thus $\mu_n = \dfrac{n\pi}{L}$ and $\lambda_n = \mu_n^2 = \left(\dfrac{n\pi}{L}\right)^2$.
+
+\begin{keyresult}
+  \textbf{Dirichlet eigenvalue problem.} For
+  \[
+    y'' + \lambda y = 0, \qquad y(0) = 0, \;\; y(L) = 0,
+  \]
+  the eigenvalues and eigenfunctions are
+  \[
+    \lambda_n = \left(\frac{n\pi}{L}\right)^{\!2},
+    \qquad
+    y_n(x) = \sin\!\left(\frac{n\pi x}{L}\right),
+    \qquad n = 1, 2, 3, \dots
+  \]
+  Any positive constant multiple of $y_n(x)$ is also an eigenfunction.
+\end{keyresult}
+
+\paragraph{Neumann boundary conditions.} If instead we impose
+\[
+  y'(0) = 0, \qquad y'(L) = 0,
+\]
+the analysis changes slightly. With $\lambda = \mu^2 > 0$, we have $y(x) = c_1 \cos(\mu x) + c_2 \sin(\mu x)$ and $y'(x) = -c_1 \mu \sin(\mu x) + c_2 \mu \cos(\mu x)$.
+
+Apply $y'(0) = 0$: $c_2 \mu = 0$, so $c_2 = 0$. Then $y(x) = c_1 \cos(\mu x)$ and $y'(x) = -c_1 \mu \sin(\mu x)$.
+
+Apply $y'(L) = 0$: $-c_1 \mu \sin(\mu L) = 0$. For nontrivial solutions ($c_1 \neq 0$), we need $\sin(\mu L) = 0$, so $\mu L = n\pi$ with $n = 0, 1, 2, \dots$.
+
+Note that $n = 0$ gives $\mu = 0$, $\lambda_0 = 0$, and $y_0(x) = \cos(0) = 1$ (a nonzero constant). This means $\lambda_0 = 0$ \emph{is} an eigenvalue for Neumann conditions.
+
+\begin{keyresult}
+  \textbf{Neumann eigenvalue problem.} For
+  \[
+    y'' + \lambda y = 0, \qquad y'(0) = 0, \;\; y'(L) = 0,
+  \]
+  the eigenvalues and eigenfunctions are
+  \[
+    \lambda_n = \left(\frac{n\pi}{L}\right)^{\!2},
+    \qquad
+    y_n(x) = \cos\!\left(\frac{n\pi x}{L}\right),
+    \qquad n = 0, 1, 2, \dots
+  \]
+  Note that $\lambda_0 = 0$ with eigenfunction $y_0(x) = 1$.
+\end{keyresult}
+
+\begin{workedexample}
+  Find all eigenvalues and eigenfunctions of
+  \[
+    y'' + \lambda y = 0, \qquad y(0) = 0, \;\; y(2) = 0.
+  \]
+
+  \textbf{Solution.} Here $L = 2$. From \cref{eq:eigenvalue_problem}, the eigenvalues are
+  \[
+    \lambda_n = \left(\frac{n\pi}{2}\right)^{\!2} = \frac{n^2 \pi^2}{4},
+    \qquad n = 1, 2, 3, \dots
+  \]
+  and the eigenfunctions are
+  \[
+    y_n(x) = \sin\!\left(\frac{n\pi x}{2}\right).
+  \]
+  Listing the first few:
+  \begin{align*}
+    \lambda_1 &= \frac{\pi^2}{4}, & y_1(x) &= \sin\!\left(\frac{\pi x}{2}\right), \\
+    \lambda_2 &= \pi^2, & y_2(x) &= \sin(\pi x), \\
+    \lambda_3 &= \frac{9\pi^2}{4}, & y_3(x) &= \sin\!\left(\frac{3\pi x}{2}\right).
+  \end{align*}
+\end{workedexample}
+
+\begin{workedexample}
+  Find all eigenvalues and eigenfunctions of
+  \[
+    y'' + \lambda y = 0, \qquad y'(0) = 0, \;\; y'(3) = 0.
+  \]
+
+  \textbf{Solution.} Here $L = 3$ with Neumann boundary conditions. The eigenvalues are
+  \[
+    \lambda_n = \left(\frac{n\pi}{3}\right)^{\!2} = \frac{n^2 \pi^2}{9},
+    \qquad n = 0, 1, 2, \dots
+  \]
+  and the eigenfunctions are
+  \[
+    y_n(x) = \cos\!\left(\frac{n\pi x}{3}\right).
+  \]
+  The first few are:
+  \begin{align*}
+    \lambda_0 &= 0, & y_0(x) &= 1, \\
+    \lambda_1 &= \frac{\pi^2}{9}, & y_1(x) &= \cos\!\left(\frac{\pi x}{3}\right), \\
+    \lambda_2 &= \frac{4\pi^2}{9}, & y_2(x) &= \cos\!\left(\frac{2\pi x}{3}\right).
+  \end{align*}
+  Note that $y_0(x) = 1$ corresponds to the constant equilibrium state.
+\end{workedexample}
 
 \subsection{Sturm--Liouville Form}
 \label{sec:ch11_sturm_liouville}
 
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+The eigenvalue problems we just studied are special cases of a broad and powerful framework: the \textbf{Sturm--Liouville (SL) problem}. This theory unifies eigenvalue problems, orthogonality, and series expansions.
+
+\paragraph{General SL form.} A second-order linear ODE is in \textbf{Sturm--Liouville form} when it can be written as
+\begin{equation}
+  \label{eq:sl_form}
+  \bigl(p(x)\,y'\bigr)' + \bigl(q(x) + \lambda\,r(x)\bigr)y = 0,
+  \qquad a \leq x \leq b,
+\end{equation}
+where $p(x)$, $q(x)$, and $r(x)$ are given coefficient functions and $\lambda$ is the eigenvalue parameter. The function $r(x)$ is called the \textbf{weight function}.
+
+\paragraph{Converting to SL form.} Any equation of the form
+\begin{equation}
+  \label{eq:general_second_order}
+  P(x)\,y'' + Q(x)\,y' + R(x)\,y + \lambda\,S(x)\,y = 0
+\end{equation}
+can be converted to SL form by dividing by $P(x)$ and multiplying by an \textbf{integrating factor}. First divide by $P(x)$:
+\[
+  y'' + \frac{Q(x)}{P(x)}\,y' + \frac{R(x)}{P(x)}\,y + \lambda\,\frac{S(x)}{P(x)}\,y = 0.
+\]
+Multiply by the integrating factor
+\[
+  \mu(x) = \exp\!\left(\int \frac{Q(x)}{P(x)}\,\diff x\right).
+\]
+Then the first two terms combine as a derivative:
+\[
+  \mu(x)\,y'' + \mu(x)\,\frac{Q(x)}{P(x)}\,y' = \bigl(\mu(x)\,y'\bigr)',
+\]
+which is exactly the $(py')'$ structure of the SL form with $p(x) = \mu(x)$, $q(x) = \mu(x)\,\dfrac{R(x)}{P(x)}$, and $r(x) = \mu(x)\,\dfrac{S(x)}{P(x)}$.
+
+\paragraph{Regular SL problem.} We restrict attention to the \textbf{regular} case:
+
+\begin{definition}[Regular Sturm--Liouville Problem]
+  \label{def:regular_sl}
+  A Sturm--Liouville problem is \textbf{regular} on $[a,b]$ if:
+  \begin{enumerate}
+    \item $p(x)$, $p'(x)$, $q(x)$, and $r(x)$ are continuous on $[a,b]$.
+    \item $p(x) > 0$ and $r(x) > 0$ on $[a,b]$.
+    \item The boundary conditions are of the form
+      \[
+        \alpha_1 y(a) + \alpha_2 y'(a) = 0,
+        \qquad
+        \beta_1 y(b) + \beta_2 y'(b) = 0,
+      \]
+      where $\alpha_1^2 + \alpha_2^2 > 0$ and $\beta_1^2 + \beta_2^2 > 0$.
+  \end{enumerate}
+\end{definition}
+
+\begin{theorem}[SL Existence Theorem]
+  \label{thm:sl_existence}
+  A regular Sturm--Liouville problem has:
+  \begin{enumerate}
+    \item An infinite sequence of real eigenvalues
+      \[
+        \lambda_1 < \lambda_2 < \lambda_3 < \cdots \to \infty.
+      \]
+    \item Each eigenvalue $\lambda_n$ has exactly one corresponding eigenfunction $\phi_n(x)$ (up to a constant multiple).
+    \item The $n$-th eigenfunction $\phi_n(x)$ has exactly $n-1$ zeros in the open interval $(a,b)$.
+  \end{enumerate}
+\end{theorem}
+
+This theorem guarantees that eigenvalue problems arising from well-behaved physical systems always have a rich structure of eigenvalues and eigenfunctions.
+
+\begin{workedexample}
+  Convert the equation $x\,y'' + 2y' + \lambda\,x\,y = 0$ to Sturm--Liouville form on $[0,1]$, and identify $p(x)$, $q(x)$, and $r(x)$.
+
+  \textbf{Solution.} Here $P(x) = x$, $Q(x) = 2$, $R(x) = 0$, and $S(x) = x$.
+
+  Divide by $P(x) = x$:
+  \[
+    y'' + \frac{2}{x}\,y' + \lambda\,y = 0.
+  \]
+
+  The integrating factor is
+  \[
+    \mu(x) = \exp\!\left(\int \frac{2}{x}\,\diff x\right)
+            = \exp(2\ln x) = x^2.
+  \]
+
+  Multiply the equation by $x^2$:
+  \[
+    x^2\,y'' + 2x\,y' + \lambda\,x^2\,y = 0.
+  \]
+  The first two terms combine as
+  \[
+    x^2 y'' + 2x y' = (x^2 y')'.
+  \]
+  Hence the SL form is
+  \[
+    (x^2 y')' + \lambda\,x^2\,y = 0.
+  \]
+  Reading off the coefficients:
+  \[
+    p(x) = x^2, \qquad q(x) = 0, \qquad r(x) = x^2.
+  \]
+  On $(0,1]$, we have $p(x) = x^2 > 0$ and $r(x) = x^2 > 0$. Note that $p(0) = 0$, so strictly speaking this is a \emph{singular} SL problem (not regular) at $x = 0$, since the regularity condition requires $p > 0$ on the \emph{closed} interval. Singular SL problems require additional care at the singular endpoint.
+\end{workedexample}
 
 \subsection{Orthogonality Theorem}
 \label{sec:ch11_orthogonality_theorem}
 
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+The most powerful consequence of the Sturm--Liouville framework is the \textbf{orthogonality} of eigenfunctions belonging to distinct eigenvalues.
+
+\begin{theorem}[SL Orthogonality Theorem]
+  \label{thm:sl_orthogonality}
+  Let $\phi_n(x)$ and $\phi_m(x)$ be eigenfunctions corresponding to distinct eigenvalues $\lambda_n \neq \lambda_m$ of a regular Sturm--Liouville problem on $[a,b]$. Then $\phi_n$ and $\phi_m$ are \textbf{orthogonal with respect to the weight function} $r(x)$:
+  \[
+    \int_a^b \phi_n(x)\,\phi_m(x)\,r(x)\,\diff x = 0, \qquad n \neq m.
+  \]
+\end{theorem}
+
+\begin{proof}
+  The eigenfunctions satisfy the SL equations
+  \[
+    \begin{cases}
+      \bigl(p\,\phi_n'\bigr)' + \bigl(q + \lambda_n\,r\bigr)\phi_n = 0, \\[4pt]
+      \bigl(p\,\phi_m'\bigr)' + \bigl(q + \lambda_m\,r\bigr)\phi_m = 0.
+    \end{cases}
+  \]
+  Multiply the first equation by $\phi_m(x)$ and the second by $\phi_n(x)$:
+  \[
+    \begin{cases}
+      \phi_m\,(p\,\phi_n')' + \phi_m\,q\,\phi_n + \lambda_n\,\phi_m\,r\,\phi_n = 0, \\[4pt]
+      \phi_n\,(p\,\phi_m')' + \phi_n\,q\,\phi_m + \lambda_m\,\phi_n\,r\,\phi_m = 0.
+    \end{cases}
+  \]
+  Subtract the second from the first:
+  \[
+    \phi_m\,(p\,\phi_n')' - \phi_n\,(p\,\phi_m')'
+    + (\lambda_n - \lambda_m)\,\phi_n\,\phi_m\,r = 0.
+  \]
+  The terms involving $q(x)$ cancel. Rearrange:
+  \[
+    (\lambda_n - \lambda_m)\,\phi_n(x)\,\phi_m(x)\,r(x)
+    = \phi_n\,(p\,\phi_m')' - \phi_m\,(p\,\phi_n')'.
+  \]
+  The right-hand side can be written as a total derivative:
+  \[
+    \phi_n\,(p\,\phi_m')' - \phi_m\,(p\,\phi_n')'
+    = \frac{\diff}{\diff x}\Bigl[p\bigl(\phi_n\,\phi_m' - \phi_m\,\phi_n'\bigr)\Bigr].
+  \]
+  To verify this, differentiate the expression inside the brackets:
+  \[
+    \frac{\diff}{\diff x}\Bigl[p(\phi_n\phi_m' - \phi_m\phi_n')\Bigr]
+    = p'(\phi_n\phi_m' - \phi_m\phi_n') + p(\phi_n'\phi_m' + \phi_n\phi_m'' - \phi_m'\phi_n' - \phi_m\phi_n'')
+    = \phi_n(p\phi_m')' - \phi_m(p\phi_n')',
+  \]
+  as claimed. Now integrate both sides from $a$ to $b$:
+  \[
+    (\lambda_n - \lambda_m)\int_a^b \phi_n(x)\,\phi_m(x)\,r(x)\,\diff x
+    = \Bigl[p(x)\bigl(\phi_n(x)\,\phi_m'(x) - \phi_m(x)\,\phi_n'(x)\bigr)\Bigr]_{x=a}^{x=b}.
+  \]
+  The right-hand side consists of \textbf{boundary terms}. For a regular SL problem, the boundary conditions are of the form
+  \[
+    \alpha_1 y(a) + \alpha_2 y'(a) = 0, \qquad \beta_1 y(b) + \beta_2 y'(b) = 0.
+  \]
+  Both $\phi_n$ and $\phi_m$ satisfy these boundary conditions. It is a standard verification that under any such separated boundary conditions, the boundary expression
+  \[
+    p\bigl(\phi_n\phi_m' - \phi_m\phi_n'\bigr)
+  \]
+  vanishes at both $x = a$ and $x = b$. (For example, if Dirichlet conditions $y(a) = 0$ apply, then $\phi_n(a) = \phi_m(a) = 0$ makes the expression zero immediately. For mixed conditions, a short algebraic argument shows cancellation.)
+
+  Therefore the right-hand side is zero. Since $\lambda_n \neq \lambda_m$, we divide by $(\lambda_n - \lambda_m)$ to obtain
+  \[
+    \int_a^b \phi_n(x)\,\phi_m(x)\,r(x)\,\diff x = 0,
+  \]
+  completing the proof.
+\end{proof}
+
+\paragraph{Orthogonality in action.} The orthogonality of $\{\sin(n\pi x/L)\}$ on $[0,L]$ is a direct consequence of \cref{thm:sl_orthogonality}. For the Dirichlet problem $y'' + \lambda y = 0$ on $[0,L]$, we have $p = 1$, $q = 0$, $r = 1$, and $\phi_n(x) = \sin(n\pi x/L)$. The theorem guarantees
+\[
+  \int_0^L \sin\!\left(\frac{n\pi x}{L}\right)\sin\!\left(\frac{m\pi x}{L}\right)\diff x = 0, \qquad n \neq m,
+\]
+which is precisely the sine orthogonality relation used throughout Fourier analysis (\cref{ch:fourier_series}).
+
+\begin{workedexample}
+  Verify the orthogonality of $\sin(\pi x)$ and $\sin(2\pi x)$ on $[0,1]$ by direct computation.
+
+  \textbf{Solution.} We compute
+  \[
+    I = \int_0^1 \sin(\pi x)\,\sin(2\pi x)\,\diff x.
+  \]
+  Use the product-to-sum identity $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$:
+  \[
+    \sin(\pi x)\sin(2\pi x) = \frac{1}{2}\bigl[\cos(-\pi x) - \cos(3\pi x)\bigr]
+    = \frac{1}{2}\bigl[\cos(\pi x) - \cos(3\pi x)\bigr].
+  \]
+  Integrate:
+  \[
+    I = \frac{1}{2}\left[\frac{\sin(\pi x)}{\pi} - \frac{\sin(3\pi x)}{3\pi}\right]_0^1
+      = \frac{1}{2}\left[\frac{0}{\pi} - \frac{0}{3\pi} - 0\right] = 0.
+  \]
+  The integral vanishes, confirming orthogonality.
+\end{workedexample}
+
+\begin{workedexample}
+  Show that the eigenfunctions of $y'' + \lambda y = 0$ with Neumann conditions $y'(0) = 0$, $y'(1) = 0$ on $[0,1]$ are orthogonal with respect to the weight $r(x) = 1$.
+
+  \textbf{Solution.} The eigenfunctions are $\phi_n(x) = \cos(n\pi x)$ for $n = 0, 1, 2, \dots$. We need to verify
+  \[
+    \int_0^1 \cos(n\pi x)\,\cos(m\pi x)\,\diff x = 0, \qquad n \neq m.
+  \]
+  Use $\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$:
+  \[
+    \cos(n\pi x)\cos(m\pi x) = \frac{1}{2}\bigl[\cos\bigl((n-m)\pi x\bigr) + \cos\bigl((n+m)\pi x\bigr)\bigr].
+  \]
+  Integrate:
+  \[
+    \int_0^1 \cos(n\pi x)\cos(m\pi x)\,\diff x
+    = \frac{1}{2}\left[\frac{\sin((n-m)\pi x)}{(n-m)\pi} + \frac{\sin((n+m)\pi x)}{(n+m)\pi}\right]_0^1.
+  \]
+  Since $n \neq m$ are integers, both $(n-m)\pi$ and $(n+m)\pi$ are nonzero multiples of $\pi$, so $\sin(k\pi) = 0$ for any integer $k$. The integral evaluates to $0$, confirming orthogonality.
+\end{workedexample}
 
 \subsection{Eigenfunction Expansions}
 \label{sec:ch11_eigenfunction_expansions}
 
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+The orthogonality of eigenfunctions allows us to expand an arbitrary function as a series, much like a Taylor series or a Fourier series.
+
+\begin{keyresult}
+  \textbf{Eigenfunction expansion.} Let $\{\phi_n(x)\}_{n=1}^\infty$ be the eigenfunctions of a regular SL problem on $[a,b]$ with weight $r(x)$. Any piecewise smooth function $f(x)$ on $[a,b]$ can be expanded as
+  \[
+    f(x) = \sum_{n=1}^\infty c_n\,\phi_n(x),
+  \]
+  where the \textbf{expansion coefficients} are
+  \begin{equation}
+    \label{eq:expansion_coefficient}
+    c_n = \frac{\displaystyle\int_a^b f(x)\,\phi_n(x)\,r(x)\,\diff x}
+               {\displaystyle\int_a^b \phi_n(x)^2\,r(x)\,\diff x}.
+  \end{equation}
+  The series converges to $f(x)$ at points of continuity, and to the average $\frac{1}{2}[f(x^+) + f(x^-)]$ at points of discontinuity.
+\end{keyresult}
+
+\paragraph{Derivation of the coefficient formula.} Multiply the expansion $f(x) = \sum_{k=1}^\infty c_k \phi_k(x)$ by $\phi_n(x)\,r(x)$ and integrate over $[a,b]$:
+\[
+  \int_a^b f(x)\,\phi_n(x)\,r(x)\,\diff x
+  = \sum_{k=1}^\infty c_k \int_a^b \phi_k(x)\,\phi_n(x)\,r(x)\,\diff x.
+\]
+By orthogonality, all terms in the sum vanish except $k = n$:
+\[
+  \int_a^b f(x)\,\phi_n(x)\,r(x)\,\diff x
+  = c_n \int_a^b \phi_n(x)^2\,r(x)\,\diff x.
+\]
+Solving for $c_n$ gives \cref{eq:expansion_coefficient}. The denominator
+\[
+  \|\phi_n\|^2 = \int_a^b \phi_n(x)^2\,r(x)\,\diff x
+\]
+is the squared \textbf{norm} of the eigenfunction with respect to the weight $r(x)$.
+
+\paragraph{Connection to Fourier series.} The Fourier sine series is a special case of eigenfunction expansion. Consider the SL problem with $p = 1$, $q = 0$, $r = 1$ on $[0,L]$ and Dirichlet boundary conditions. The eigenfunctions are $\phi_n(x) = \sin(n\pi x/L)$, and
+\[
+  \|\phi_n\|^2 = \int_0^L \sin^2\!\left(\frac{n\pi x}{L}\right)\diff x = \frac{L}{2}.
+\]
+The expansion formula then gives the familiar Fourier sine coefficients:
+\[
+  c_n = \frac{2}{L}\int_0^L f(x)\,\sin\!\left(\frac{n\pi x}{L}\right)\diff x.
+\]
+Similarly, the Fourier cosine series arises from the Neumann SL problem on $[0,L]$.
+
+\begin{workedexample}
+  Expand $f(x) = x$ on $[0,1]$ as a series of eigenfunctions of $y'' + \lambda y = 0$ with Dirichlet boundary conditions $y(0) = 0$, $y(1) = 0$.
+
+  \textbf{Solution.} The eigenfunctions are $\phi_n(x) = \sin(n\pi x)$ with weight $r(x) = 1$. We need
+  \[
+    c_n = \frac{\displaystyle\int_0^1 x\,\sin(n\pi x)\,\diff x}
+               {\displaystyle\int_0^1 \sin^2(n\pi x)\,\diff x}.
+  \]
+  The denominator is
+  \[
+    \int_0^1 \sin^2(n\pi x)\,\diff x = \frac{1}{2}.
+  \]
+
+  For the numerator, use integration by parts with $u = x$ and $\diff v = \sin(n\pi x)\,\diff x$:
+  \[
+    \int_0^1 x\,\sin(n\pi x)\,\diff x
+    = \left[-\frac{x}{n\pi}\cos(n\pi x)\right]_0^1 + \frac{1}{n\pi}\int_0^1 \cos(n\pi x)\,\diff x.
+  \]
+  The boundary term gives
+  \[
+    -\frac{1}{n\pi}\cos(n\pi) + 0 = -\frac{(-1)^n}{n\pi} = \frac{(-1)^{n+1}}{n\pi}.
+  \]
+  The remaining integral is
+  \[
+    \frac{1}{n\pi}\left[\frac{\sin(n\pi x)}{n\pi}\right]_0^1 = 0.
+  \]
+  So the numerator is $\dfrac{(-1)^{n+1}}{n\pi}$.
+
+  Therefore
+  \[
+    c_n = \frac{(-1)^{n+1}/(n\pi)}{1/2} = \frac{2(-1)^{n+1}}{n\pi}.
+  \]
+  The eigenfunction expansion is
+  \[
+    x = \sum_{n=1}^\infty \frac{2(-1)^{n+1}}{n\pi}\,\sin(n\pi x),
+    \qquad 0 < x < 1.
+  \]
+  Writing out the first few terms:
+  \[
+    x \approx \frac{2}{\pi}\sin(\pi x) - \frac{1}{\pi}\sin(2\pi x) + \frac{2}{3\pi}\sin(3\pi x) - \cdots
+  \]
+\end{workedexample}
+
+\begin{workedexample}
+  Expand $f(x) = 1$ on $[0,1]$ as a series of eigenfunctions of $y'' + \lambda y = 0$ with Neumann boundary conditions $y'(0) = 0$, $y'(1) = 0$.
+
+  \textbf{Solution.} The eigenfunctions are $\phi_n(x) = \cos(n\pi x)$ for $n = 0, 1, 2, \dots$, with weight $r(x) = 1$.
+
+  For $n = 0$: $\phi_0(x) = 1$. The norm is
+  \[
+    \|\phi_0\|^2 = \int_0^1 1^2\,\diff x = 1.
+  \]
+  The coefficient is
+  \[
+    c_0 = \frac{\int_0^1 1 \cdot 1\,\diff x}{1} = 1.
+  \]
+
+  For $n \geq 1$:
+  \[
+    c_n = \frac{\displaystyle\int_0^1 1 \cdot \cos(n\pi x)\,\diff x}
+               {\displaystyle\int_0^1 \cos^2(n\pi x)\,\diff x}
+        = \frac{\left[\frac{\sin(n\pi x)}{n\pi}\right]_0^1}{1/2}
+        = \frac{0}{1/2} = 0.
+  \]
+
+  The expansion is simply
+  \[
+    1 = 1,
+  \]
+  i.e., only the $n = 0$ term survives. This makes sense: the constant function is itself the $n = 0$ eigenfunction.
+
+  As a more instructive exercise, expand $f(x) = x$ with Neumann conditions. Then for $n = 0$:
+  \[
+    c_0 = \frac{\int_0^1 x \cdot 1\,\diff x}{1} = \frac{1}{2}.
+  \]
+  For $n \geq 1$:
+  \[
+    c_n = \frac{\int_0^1 x\,\cos(n\pi x)\,\diff x}{1/2}.
+  \]
+  Integrate by parts with $u = x$, $\diff v = \cos(n\pi x)\,\diff x$:
+  \[
+    \int_0^1 x\cos(n\pi x)\,\diff x
+    = \left[\frac{x}{n\pi}\sin(n\pi x)\right]_0^1 - \frac{1}{n\pi}\int_0^1 \sin(n\pi x)\,\diff x.
+  \]
+  The boundary term vanishes ($\sin(n\pi) = 0$). The remaining integral is
+  \[
+    -\frac{1}{n\pi}\left[-\frac{\cos(n\pi x)}{n\pi}\right]_0^1
+    = \frac{1}{(n\pi)^2}\bigl(\cos(n\pi) - \cos 0\bigr)
+    = \frac{(-1)^n - 1}{(n\pi)^2}.
+  \]
+  This is zero for even $n$ and $-\dfrac{2}{(n\pi)^2}$ for odd $n$. Therefore
+  \[
+    c_n = \begin{cases}
+      0, & n \text{ even}, \\[6pt]
+      -\dfrac{4}{(n\pi)^2}, & n \text{ odd}.
+    \end{cases}
+  \]
+  The expansion is
+  \[
+    x = \frac{1}{2} - \frac{4}{\pi^2}\cos(\pi x) - \frac{4}{9\pi^2}\cos(3\pi x) - \frac{4}{25\pi^2}\cos(5\pi x) - \cdots
+  \]
+  This converges to $x$ on $[0,1]$.
+\end{workedexample}
+
+\subsection{Applications}
+\label{sec:ch11_applications}
+
+Eigenfunction expansions provide a powerful method for solving \textbf{nonhomogeneous} boundary value problems.
+
+\paragraph{The nonhomogeneous SL problem.} Consider
+\begin{equation}
+  \label{eq:nonhomogeneous_sl}
+  \bigl(p(x)\,y'\bigr)' + q(x)\,y + \lambda\,r(x)\,y = f(x),
+  \qquad a < x < b,
+\end{equation}
+subject to homogeneous boundary conditions at $x = a$ and $x = b$. Suppose we already know the eigenfunctions $\{\phi_n(x)\}$ and eigenvalues $\{\lambda_n\}$ of the associated homogeneous problem
+\[
+  \bigl(p(x)\,\phi_n'\bigr)' + q(x)\,\phi_n + \lambda_n\,r(x)\,\phi_n = 0.
+\]
+
+\paragraph{Eigenfunction expansion method.} Expand both the solution $y(x)$ and the source term $f(x)$ in the eigenfunction basis:
+\[
+  y(x) = \sum_{n=1}^\infty c_n\,\phi_n(x), \qquad
+  f(x) = \sum_{n=1}^\infty f_n\,\phi_n(x).
+\]
+Substitute the series for $y(x)$ into \cref{eq:nonhomogeneous_sl}:
+\[
+  \sum_{n=1}^\infty c_n\,\Bigl[\bigl(p\,\phi_n'\bigr)' + q\,\phi_n + \lambda\,r\,\phi_n\Bigr] = f(x).
+\]
+Using the homogeneous eigenvalue equation $\bigl(p\,\phi_n'\bigr)' + q\,\phi_n = -\lambda_n\,r\,\phi_n$:
+\[
+  \sum_{n=1}^\infty c_n\,\bigl(-\lambda_n\,r\,\phi_n + \lambda\,r\,\phi_n\bigr) = f(x).
+\]
+\[
+  \sum_{n=1}^\infty c_n\,(\lambda - \lambda_n)\,\phi_n(x)\,r(x) = f(x).
+\]
+Multiply by $\phi_m(x)$ and integrate, using orthogonality:
+\[
+  c_m\,(\lambda - \lambda_m)\,\|\phi_m\|^2 = f_m.
+\]
+Therefore, provided $\lambda \neq \lambda_n$ for any $n$:
+\begin{equation}
+  \label{eq:nonhomogeneous_coefficient}
+  c_n = \frac{f_n}{(\lambda - \lambda_n)\,\|\phi_n\|^2}
+       = \frac{\displaystyle\int_a^b f(x)\,\phi_n(x)\,r(x)\,\diff x}
+              {(\lambda - \lambda_n)\displaystyle\int_a^b \phi_n(x)^2\,r(x)\,\diff x}.
+\end{equation}
+
+\paragraph{Resonance.} If $\lambda = \lambda_n$ for some $n$ and the corresponding $f_n \neq 0$, there is no solution---this is the phenomenon of \textbf{resonance}. If $f_n = 0$ for that particular $n$, the coefficient $c_n$ is undetermined (infinite solutions), as discussed in the context of homogeneous BVPs.
+
+\begin{workedexample}
+  Solve the nonhomogeneous BVP
+  \[
+    y'' + \pi^2\,y = x, \qquad 0 < x < 1,
+    \qquad y(0) = 0, \;\; y(1) = 0.
+  \]
+
+  \textbf{Solution.} The associated homogeneous problem $y'' + \lambda y = 0$ with Dirichlet BCs has eigenfunctions $\phi_n(x) = \sin(n\pi x)$ and eigenvalues $\lambda_n = n^2\pi^2$.
+
+  Here $\lambda = \pi^2 = \lambda_1$---we are at the first eigenvalue. This raises the question of whether a solution exists. We must check the solvability condition.
+
+  Expand the source term $f(x) = x$ in the eigenfunction basis. From the previous worked example:
+  \[
+    x = \sum_{n=1}^\infty \frac{2(-1)^{n+1}}{n\pi}\,\sin(n\pi x).
+  \]
+  So $f_n = \dfrac{2(-1)^{n+1}}{n\pi}$.
+
+  For $n = 1$: $\lambda = \lambda_1 = \pi^2$, so the denominator $(\lambda - \lambda_1) = 0$. The coefficient $f_1 = \dfrac{2(-1)^2}{\pi} = \dfrac{2}{\pi} \neq 0$.
+
+  Since $f_1 \neq 0$ and $\lambda = \lambda_1$, the BVP has \textbf{no solution}. The source term $f(x) = x$ has a component along the first eigenfunction, and at resonance the system cannot respond.
+
+  To see this directly: suppose a solution $y(x)$ exists. Multiply the ODE by $\phi_1(x) = \sin(\pi x)$ and integrate:
+  \[
+    \int_0^1 (y'' + \pi^2 y)\sin(\pi x)\,\diff x = \int_0^1 x\sin(\pi x)\,\diff x.
+  \]
+  The right-hand side is $\dfrac{2}{\pi} \neq 0$. But integrating the left side by parts twice and using the boundary conditions yields zero (this is the Fredholm alternative). Contradiction: no solution exists.
+\end{workedexample}
+
+\begin{workedexample}
+  Solve the nonhomogeneous BVP
+  \[
+    y'' + 2\pi^2\,y = \sin(\pi x), \qquad 0 < x < 1,
+    \qquad y(0) = 0, \;\; y(1) = 0.
+  \]
+
+  \textbf{Solution.} Eigenvalues: $\lambda_n = n^2\pi^2$. Here $\lambda = 2\pi^2$, which is \emph{not} an eigenvalue ($\sqrt{2}\pi$ is not a multiple of $\pi$).
+
+  Expand $f(x) = \sin(\pi x)$ in eigenfunctions. Since $\sin(\pi x) = \phi_1(x)$, the expansion is simply
+  \[
+    f(x) = \sin(\pi x) = 1 \cdot \phi_1(x).
+  \]
+  So $f_1 = \|\phi_1\|^2 = \dfrac{1}{2}$ and $f_n = 0$ for $n \neq 1$.
+
+  The coefficients are
+  \[
+    c_n = \frac{f_n}{(\lambda - \lambda_n)\,\|\phi_n\|^2}.
+  \]
+  For $n = 1$:
+  \[
+    c_1 = \frac{1/2}{(2\pi^2 - \pi^2)\cdot(1/2)} = \frac{1}{\pi^2}.
+  \]
+  For $n \neq 1$, $f_n = 0$ so $c_n = 0$.
+
+  The solution is
+  \[
+    y(x) = \frac{1}{\pi^2}\,\sin(\pi x).
+  \]
+  \textbf{Verification.} Compute $y'' = -\sin(\pi x)$. Then
+  \[
+    y'' + 2\pi^2 y = -\sin(\pi x) + 2\pi^2 \cdot \frac{1}{\pi^2}\sin(\pi x) = -\sin(\pi x) + 2\sin(\pi x) = \sin(\pi x).
+  \]
+  $\checkmark$ The solution satisfies the ODE and the boundary conditions.
+\end{workedexample}
 
 \subsection{Summary}
 \label{sec:ch11_summary}
 
-% Summary table at end of chapter
 \begin{table}[htbp]
 \centering
-\caption{Chapter Summary}
+\caption{Eigenvalue problems and Sturm--Liouville theory}
 \label{tab:ch11_summary}
-\begin{tabular}{l l}
+\begin{tabular}{l l p{5.5cm}}
 \toprule
-\textbf{Concept} & \textbf{Key Formula/Method} \\
+\textbf{Concept} & \textbf{Key formula/result} \\
 \midrule
-TBD & TBD \\
+BVP vs IVP & BVP: conditions at $x=a$ and $x=b$; IVP: conditions at one point \\
+Dirichlet eigenvalues & $\lambda_n = (n\pi/L)^2$, $y_n(x) = \sin(n\pi x/L)$, $n = 1, 2, \dots$ \\
+Neumann eigenvalues & $\lambda_n = (n\pi/L)^2$, $y_n(x) = \cos(n\pi x/L)$, $n = 0, 1, 2, \dots$ \\
+SL form & $(py')' + (q + \lambda r)y = 0$ \\
+Integrating factor & $\mu(x) = \exp\!\left(\int \frac{Q(x)}{P(x)}\,\diff x\right)$ \\
+SL existence & Infinite real eigenvalues $\lambda_1 < \lambda_2 < \cdots \to \infty$ \\
+Orthogonality & $\displaystyle\int_a^b \phi_n \phi_m\, r\,\diff x = 0$ for $n \neq m$ \\
+Expansion coefficients & $\displaystyle c_n = \frac{\int_a^b f\phi_n\,r\,\diff x}{\int_a^b \phi_n^2\,r\,\diff x}$ \\
+Nonhomogeneous solution & $\displaystyle c_n = \frac{f_n}{(\lambda - \lambda_n)\,\|\phi_n\|^2}$ (if $\lambda \neq \lambda_n$) \\
+Resonance & If $\lambda = \lambda_n$ and $f_n \neq 0$, no solution exists \\
 \bottomrule
 \end{tabular}
 \end{table}
+
+\begin{hintbox}
+  \textbf{Problem-solving checklist for BVPs.}
+  \begin{enumerate}
+    \item Identify the boundary conditions and classify (Dirichlet, Neumann, mixed).
+    \item Determine the eigenvalues and eigenfunctions of the associated homogeneous problem.
+    \item Verify orthogonality of eigenfunctions with respect to the weight $r(x)$.
+    \item For nonhomogeneous problems, expand $f(x)$ in the eigenfunction basis.
+    \item Check whether $\lambda$ coincides with any eigenvalue (resonance).
+    \item If $\lambda \neq \lambda_n$ for all $n$, compute coefficients using \cref{eq:nonhomogeneous_coefficient}.
+    \item If $\lambda = \lambda_n$, check the solvability condition ($f_n = 0$).
+  \end{enumerate}
+\end{hintbox}