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\begin{document}


\begin{center}
Inhomogeneous data
\end{center}

\begin{align*}
(\ast )\qquad u_{t}& =ku_{xx}\qquad 0<x<l,\quad t>0 \\
u(x,0)& =0\quad \qquad \text{initial condition} \\
u(0,t)& =h(t)\qquad \text{boundary condition} \\
u(l,t)& =j(t)\qquad \text{boundary condition}
\end{align*}

\bigskip

\noindent Assume:%
\begin{align*}
(\ast \ast )\qquad u(x,t) &= \tsum\limits_{n=1}^{\infty }u_{n}(t)\sin (\frac{%
n\pi x}{l}) \\
u_{n} &= \frac{2}{l}\tint\limits_{0}^{l}u(x,t)\sin (\frac{n\pi x}{l})dx
\end{align*}

\bigskip

\noindent \textbf{Note: \ }Since $u(x,t)$ is nonzero at the boundaries the
convergence \textbf{cannot} be uniform but either pointwise or else least
square !!

\noindent Substitute (**) into (*) giving%
\begin{equation*}
0=u_{t}-ku_{xx}=\tsum\limits_{n=1}^{\infty }\left[ \frac{du_{n}}{dt}%
+ku_{n}(t)\left( \frac{n\pi x}{l}\right) ^{2}\right] \sin (\frac{n\pi x}{l})
\end{equation*}

\noindent This \textbf{can't} match the boundary conditions:

\bigskip

\noindent \textbf{Lesson}: Differentiating (**) is illegal !!

\bigskip

\noindent Method \#1

\noindent Instead we avoid differentiation os a nonuniform converging
series. Assume

\bigskip

\begin{align*}
u(x,t) &= \tsum\limits_{n=1}^{\infty }u_{n}(t)\sin (\frac{n\pi x}{l}) \\
\frac{\partial u}{\partial t} &= \tsum\limits_{n=1}^{\infty }v_{n}(t)\sin (%
\frac{n\pi x}{l}) \\
v_{n}(t) &= \frac{2}{l}\tint\limits_{0}^{l}\frac{\partial u}{\partial t}\sin
(\frac{n\pi x}{l})dx=\frac{du_{n}}{dt}\quad \text{since new integrand is
continuous} \\
\frac{\partial ^{2}u}{\partial x^{2}} &= \tsum\limits_{n=1}^{\infty
}w_{n}(t)\sin (\frac{n\pi x}{l}) \\
w_{n}(t) &= \frac{2}{l}\tint\limits_{0}^{l}\frac{\partial ^{2}u}{\partial
x^{2}}\sin (\frac{n\pi x}{l})dx \\
&= -\frac{2}{l}\tint\limits_{0}^{l}\left( \frac{n\pi }{l}\right)
^{2}u(x,t)\sin (\frac{n\pi x}{l})dx+\frac{2}{l}\left[ \frac{\partial u}{%
\partial x}\sin (\frac{n\pi x}{l})-\frac{n\pi }{l}\cos (\frac{n\pi x}{l})%
\right] _{0}^{l} \\
&= -\left( \frac{n\pi }{l}\right) ^{2}u_{n}(t)-\frac{2n\pi }{l^{2}}\left[
(-1)^{n}j(t)-h(t)\right] \qquad \text{let }\lambda _{n}= \left( \frac{n\pi }{%
l}\right) ^{2}
\end{align*}

\noindent So%
\begin{align*}
v_{n}(t)-kw_{n}(t) &= 0 \\
\frac{du_{n}}{dt} &= kw_{n}(t)
\end{align*}

\noindent Then%
\begin{align*}
\frac{du_{n}}{dt} &= -k\lambda _{n}u_{n}(t)+\frac{2n\pi }{l^{2}}\left[
(-1)^{n}j(t)-h(t)\right] \\
u_{n}(0) &= 0
\end{align*}

\noindent Solving the ODE we get%
\begin{equation*}
u_{n}(t)=Ce^{-\lambda _{0}kt}-\frac{2n\pi }{l^{2}}k\tint\limits_{0}^{t}e^{-%
\lambda _{n}k\left( t-s\right) }\left[ (-1)^{n}j(s)-h(s)\right] ds
\end{equation*}

\bigskip

\noindent Consider the PDE

\begin{align*}
(\ast )\qquad u_{tt}& =c^{2}u_{xx}+f(x,t)\qquad 0<x<l,\quad t>0 \\
u(x,0)& =\varphi (x)\quad \qquad \text{initial condition} \\
\frac{\partial u}{\partial t}(x,0)& =\psi (x)\quad \qquad \text{initial
condition} \\
u(0,t)& =h(t)\qquad \text{boundary condition} \\
u(l,t)& =k(t)\qquad \text{boundary condition}
\end{align*}

\noindent Method \#2: Define a new variable $v$ using $g(x,t)$that satisfies
the homogenous boundary conditions

\begin{equation*}
v(x,t)=u(x,t)-\frac{\left( l-x\right) h(t)+xk(t)}{l}
\end{equation*}

\noindent Now%
\begin{align*}
v_{tt}& =c^{2}v_{xx}+f(x,t)-\frac{\left( l-x\right) h^{\prime \prime
}(t)+xk^{\prime \prime }(t)}{l}=c^{2}v_{xx}+F(x,t) \\
v(x,0)& =\varphi (x)-\frac{\left( l-x\right) h(0)+xk(0)}{l} \\
\frac{dv}{dt}(x,0)& =\psi (x)-\frac{\left( l-x\right) h^{\prime
}(0)+xk^{\prime }(0)}{l} \\
v(0,t)& =v(l,t)=0
\end{align*}

\noindent More generally 
\begin{equation*}
v(x,t)=u(x,t)-U(x,t)
\end{equation*}

\noindent then

\begin{align*}
v_{tt} &= c^{2}v_{xx}+f(x,t)-\left( U_{tt}-c^{2}U_{xx}\right) \\
v(x,0) &= \varphi (x)-U(x,0) \\
\frac{dv}{dt}(x,0) &= \psi (x)-U_{t}(x,0) \\
v(0,t) &= h(t)-U(0,t) \\
v(l,t) &= k(t)-U(l,t)
\end{align*}

\noindent So we choose $U$ so that%
\begin{align*}
U_{tt} &= c^{2}U_{xx}+f(x,t) \\
U(0,t) &= h(t) \\
U(l,t) &= k(t)
\end{align*}

\noindent e.g. if $\ f$,h,k are independent of $t$ then choose $U$ also
independent of $t$ and%
\begin{align*}
U_{xx} &= -\frac{1}{c^{2}}f(x) \\
U(0) &= h \\
U(l) &= k
\end{align*}

\newpage

\begin{center}
Inhomogeneous right hand side
\end{center}

\bigskip \noindent Consider

\begin{equation*}
(\ast )\qquad -\nabla \cdot (p(\mathit{x})\nabla u)+q(\mathit{x})u(\mathit{x}%
)=a\,m(\mathit{x})u(\mathit{x})+f(\mathit{x})\qquad \mathit{x}\in D
\end{equation*}

\noindent with homogeneous Dirichlet or Neumann (or more generally
symmetric) boundary conditions.

\begin{theorem}
(a) If \textbf{a} is not an eigenvalue (f=0) then there exists a unique
solution for all $f$ with $||f||^{2}<\infty $

(b) if \textbf{a} is an eigenvalue then either there is no solution or else
an infinite number of solutions (Fredholm alternative) depending on $f(x)$
\end{theorem}

\begin{proof}
(a) Denote the eigenvalues by $\lambda _{n}$and the eigenfunctions by $v_{n}(%
\mathit{x}),n=1,2,3...$

Let $\delta $ be the distance between \textbf{a} and the nearest eigenvalue.
By completeness we have%
\begin{equation*}
(\ast \ast )\qquad u(\mathit{x})=\tsum\limits_{n=1}^{\infty }\frac{(u,v_{n})%
}{(v_{n},v_{n})}v_{n}(\mathit{x})\quad (u,v)\equiv \tiiint\limits_{D}muvdV
\end{equation*}

and the sum converges in $L^{2}$.

Multiply (*) by $v_{n}$ and integrate to get%
\begin{equation*}
-\int \nabla \cdot (p\nabla u)v_{n}dV+\int quv_{n}dV=a\int muv_{n}dV+\int
fv_{n}dV
\end{equation*}

By Green's second identity we replace the first integral by (using the
symmetric homogenous boundary conditions on $u$ and $v$)%
\begin{equation*}
-\int \nabla \cdot (p\nabla v_{n})udV+\int quv_{n}dV= a\int muv_{n}dV+\int
fv_{n}dV
\end{equation*}

But $v_{n}$ satisfies the homogenous version of (*)

and so%
\begin{equation*}
-\nabla \cdot (p(\mathit{x})\nabla v_{n})+q(\mathit{x})v_{n}(\mathit{x}%
)=\lambda _{n}m(\mathit{x})v_{n}(\mathit{x})
\end{equation*}

Hence, subtracting we get%
\begin{equation*}
\left( \lambda _{n}-a\right) \int muv_{n}dV=\int fv_{n}
\end{equation*}

or%
\begin{equation*}
(u,v_{n})=\frac{\int fv_{n}}{\left( \lambda _{n}-a\right) }
\end{equation*}

\noindent Putting into (**) we finally arrive at%
\begin{equation*}
u(\mathit{x})=\tsum\limits_{n=1}^{\infty }\frac{\int fv_{n}}{\left( \lambda
_{n}-a\right) (v_{n},v_{n})}v_{n}(\mathit{x})
\end{equation*}

\noindent To show convergence we have%
\begin{equation*}
|u(\mathit{x})|\leq \tsum\limits_{n=1}^{\infty }\frac{|\int fv_{n}|}{\delta
(v_{n},v_{n})}v_{n}(\mathit{x})
\end{equation*}

\noindent For simplicity assume the eigenfunctions are orthonormal so $%
(v_{n},v_{n})=1$. Then%
\begin{equation*}
||u||^{2}\overset{Schwarz}{\leq }\frac{1}{\delta ^{2}}|\tsum\limits_{n=1}^{%
\infty }|\int fv_{n}(\mathit{x})|^{2}\overset{Bessel}{\leq }\frac{1}{\delta
^{2}}\int f^{2}\frac{1}{m}dV<\infty
\end{equation*}

\noindent So the series converges

\bigskip

(\noindent b) Suppose $a=\lambda _{n}$ . Then

\begin{equation*}
\left( \lambda _{n}-a\right) \int muv_{n}dV=\int fv_{n}
\end{equation*}

\noindent implies that%
\begin{equation*}
\int fv_{n}=0
\end{equation*}

\noindent If this is not true then there are no solutions. If it is true
then we can add $C v_n$ to any solution to get an infinite set of solutions.
\end{proof}

\end{document}
