%\documentstyle{article} % % USE % TO COMMENT OUT THE FOLLOWING 5 STATEMENTS % AND UNCOMMENT THE FIRST LINE TO MAKE THIS WORK % % \documentclass[epsf]{article} \usepackage[]{times} \usepackage[]{color} \usepackage[]{pstricks,pst-node} \usepackage[]{graphicx} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % MACROS % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \def\eq{\begin{equation}} \def\endeq{\end{equation}} \def\leaderfill{\leaders\hbox to 1em{\hss.\hss}\hfill} \def\reals{{{\rm I}\kern - .15em{\rm R}}} \def\aint#1{{\int \! \! \! \! \! \! -}_{#1}} \def\crossout#1{\setbox\tempbox=\hbox{#1}% \rlap{#1}\raise 2pt\hbox to \wd\tempbox{\hrulefill}} \begin{document} \section{Second Order Boundary Value Problems} \large In this section we consider existence of adjoint operators and Green's functions, homogenization of boundary conditions and solvability of second order boundary value problems. Throughout, we let $u(x)$ be a function in $L^2[a,b]$ and define \eq Lu \equiv a_2(x) u''(x) +a_1(x) u'(x) + a_0(x) u(x) \quad , \quad x \in (a,b) \endeq and $B = [b_{ij}] \in \reals^{2\times 4}$.We use the overbar notation $\bar{u}$ to denote \eq \bar{u} = (u(a),u'(a),u(b),u'(b))^T \in \reals^4 \endeq for any function $u(x)$. Letting $f(x) \in L^2[a,b]$, $\gamma = (\gamma_1,\gamma_2)^T$, a general second order boundary value problem may be posed: \begin{eqnarray} Lu & = & f \label{BVP1} \\ B\bar{u} & = & \gamma \label{BVP2} \end{eqnarray} \subsection{Conversion to Fredholm Integral Equations} Here we demonstrate how to convert \begin{eqnarray} Lu & = & f \label{FRED1} \\ B\bar{u} & = & 0 \label{FRED2} \end{eqnarray} into an equivalent Fredholm integral equation when the boundary conditions are separated. We present two methods. Given \eq Lu \equiv a_2(x) u''(x) +a_1(x) u'(x)+ a_0(x) u(x) \quad , \quad x \in (a,b) \endeq we define \eq L_0u \equiv a_2(x) u''(x) +a_1(x) u'(x) \quad , \quad x \in (a,b) \endeq For the first method, we assume \begin{itemize} \item[(A1)] The boundary conditions are separated with $rank(B)=2$. \item[(A2)] $a_2(x) \ne 0$ for all $x\in[a,b]$. \item[(A3)] The only solution of the homogenous problem \begin{eqnarray} L_0u & = & 0 \label{FRED3} \\ B\bar{u} & = & 0 \label{FRED4} \end{eqnarray} is the trivial solution $u(x) \equiv 0$. \end{itemize} The assumptions (A1)-(A3) assure the existence of a Greens's function $g_0^*(x,t)$ for the adjoint problem \begin{eqnarray} L_0^* g_0^* & = & \delta(x-t) \\ B^*\bar{g}_0^* & = & 0 \end{eqnarray} Conversion of (\ref{FRED1})-(\ref{FRED2}) to a Fredholm integral equation is then accomplished via the calculations \begin{eqnarray} & = & \\ \ & = & + \\ \ & = & + \\ \ & = & + \\ \ & = & u(t) + . \end{eqnarray} This can be written as \eq u(t) + \int_a^b k(x,t) u(x) dx =(I+K)u = F(t) \label{FRED5} \endeq by making the identifications \begin{eqnarray} k(x,t) & = & a_0(x) g_0^*(x,t) \\ F(t) & = & \int_a^b f(x) g_0^*(x,t) dx \end{eqnarray} In short, if assumptions (A1)-(A3) are satisfied, every solution $u(t)$ of (\ref{FRED1})-(\ref{FRED2}) also solves (\ref{FRED5}). To show the converse is true we first note the Green's function $g_0(x,t)$ solving \begin{eqnarray} L_0 g_0 & = & \delta(x-t) \\ B\bar{g}_0 & = & 0 \end{eqnarray} necessarily exists and $g_0^*(x,t)=g(t,x)$. Therefore, (\ref{FRED5}) can be written \eq u(x) = \int_a^b \left( f(t) - a_0(t) u(t)\right) g_0(x,t) dt \label{FRED6} \endeq It is clear from this expression since $g_0(x,t)$ satisfies the homogenous boundary conditions (in $x$) so does $u(x)$. Moreover, the calculations \begin{eqnarray} L_0u(x) & = & \int_a^b \left( f(t) - a_0(t) u(t)\right) L_0 g_0(x,t) dt \\ \ & = & \int_a^b \left( f(t) - a_0(t) u(t)\right) \delta(x-t) dt \\ \ & = & f(x) - a_0(x) u(x) \end{eqnarray} demonstrate \footnote{ Note how we have used the fact that $\delta(x)$ is an even ``function'' so that $\delta(x-t)=\delta(t-x)$. To see this, note that for any $\phi \in C_0^{\infty}(\reals)$ \[ \int_{-\infty}^{\infty} \delta(-x) \phi(x) dx = \int_{-\infty}^{\infty} \delta(y) \phi(-y) dy = \phi (0) \] } $Lu=f$. We remark that a certain degree of smoothness in the solution $u$ and the functions $a_0, a_1, a_2$ and $f$ has been assumed. Next we determine conditions for which assumption (A3) is satisfied. We do this by explicitly solving \eq a_2(x) u''(x) + a_1(x) u'(x) = 0 \endeq which is first order in $u'(x)$. The general solution of $L_0u=0$ is \begin{eqnarray} u(x) & = & c_1 u_1(x) + c_2 u_2(x) \\ \ & = & c_1 + c_2 \int_a^x exp\left(-\int_a^s \frac{a_1(t)}{a_2(t)}dt \right) ds \end{eqnarray} Defining \eq U= \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 1 & u_2(b) \\ 0 & u_2'(b) \end{array} \right] \endeq the boundary conditions are equivalent to the system \eq BU c = 0 \quad , \quad c=(c_1,c_2)^T \endeq Trivial solutions exists {\sl iff} $c=0$. Thus, assumption (A3) is satisfied if and only if \eq det(BU) \ne 0 \quad . \endeq Clearly, this condition will not always be satisfied for all $a_1,a_2$ and separated boundary conditions. One merely has to consider Neumann boundary conditions for which it is easily verified that $det(BU)=0$ regardless of what $a_1(x)$ or $a_2(x)$ equal. A second method for conversion to a Fredholm integral equation depends on {\em apriori} knowledge of the eigenvalues of $L$. We assume (A1)-(A2) and replace assumption (A3) by \begin{itemize} \item[(A3')] Assume there is a $\lambda \in \reals$ such that the only solution of the homogenous problem \begin{eqnarray} L_{\lambda}u & \equiv & Lu - \lambda u = 0 \label{FRED3X} \\ B\bar{u} & = & 0 \label{FRED4X} \end{eqnarray} is the trivial solution $u(x) \equiv 0$. \end{itemize} In other words, suppose $\lambda$ is not an eigenvalue of $L$. For the same reasons described previously, there is a Greens function $g_{\lambda}(x,t)$ (and corresponding adjoint Greens function $g_{\lambda}^*(x,t)$) solving \eq L_{\lambda}g_{\lambda} = \delta (x-t) \quad , \quad B\bar{g}_{\lambda} = 0 \endeq Conversion to an integral equation is then accomplished by the calculations \begin{eqnarray} F(t) = & = & \\ \ & = & + \lambda \\ \ & = & + \lambda \\ \ & = & u(t) + \lambda \int_a^b g_{\lambda}(t,x) u(x) dx \\ \ & = & (I + \lambda K) u \end{eqnarray} Again, in a similar fashion, it can be verified this integral equation is equivalent to the boundary value problem from which it arose. From a theoretical point of view the second approach has many advantages. For instance, if $L$ is self adjoint then it is clearly evident that $K$ is. Furthermore, if the boundary conditions are Neumann, assumption (A3) cannot be satisfied. From a practical point of view, however, the first method may be superior. We illustrate these issues by way of two examples. \vspace{0.1in} \noindent {\sc Example 1}: Let $Lu = u''+\sin(x) u$ and suppose the boundary conditions are $u(0)=u(1)=0$. Because explicit solutions of the homogenous problem \eq L_{\lambda}u = u''+\sin(x) u - \lambda u = 0 \quad, u(0)=u(1)=0 \endeq cannot be found for any $\lambda$, $g_{\lambda}(x,t)$ cannot be found. However, the Greens function $g_0(x,t)$ solving \eq L_0 g_0 = \delta (x-t) \quad , u(0)=u(1)=0 \endeq is easily found as \eq g_0(x,t) = \left\{ \begin{array}{ll} t(x-1) & 1 \ge x > t \ge 0\\ x(t-1) & 1 \ge t > x \ge 0 \end{array} \right. \endeq so that an equivalent integral equation can explicity be found. \vspace{0.1in} \noindent {\sc Example 2}: Let $Lu = u''$ and $u'(0)=u'(1)=0$. No Greens function for $L_0=L$ exists. However, so long as $\lambda \ne (n \pi)^2, n=0,1,\ldots$ the Greens function $g_{\lambda}$ satisfying the Neumann boundary conditions and \eq L_{\lambda} g_{\lambda} = g_{\lambda}'' + \lambda g_{\lambda} = \delta (x-t) , \endeq is easily found as ($\lambda > 0$) \eq g_{\lambda}(x,t) = \left\{ \begin{array}{ll} \frac{\sin \sqrt{\lambda} (x-1) \sin \sqrt{\lambda} t} {\sqrt{\lambda} \sin \sqrt{\lambda}} & 1 \ge x > t \ge 0\\ \frac{\sin \sqrt{\lambda} (t-1) \sin \sqrt{\lambda} x} {\sqrt{\lambda} \sin \sqrt{\lambda}} & 1 \ge t > x \ge 0 \end{array} \right. \endeq \subsection{Fredholm Alternative} Suppose (\ref{FRED1})-(\ref{FRED2}) can be converted into an equivalent integral equation \eq (I+ \lambda K)u(x) = F(x) \label{F3} \endeq where \eq (Ku)(x) = \int_a^b k(x,t) u(x) dx \endeq Continuity (on $[a,b]^2$) of the Greens function used to make the conversion is sufficient to guarantee that $K$ is compact. Therefore, the Fredholm alternative for compact operators applies and (\ref{F3}) has a solution if and only if $F$ is orthogonal to $N(I+K^*)$ where the adjoint \eq (K^*v)(x) = \int_a^b k(t,x) v(t) dt \quad . \endeq Thus, to prove the following theorem: \vspace{0.1in} \noindent {\em Theorem: The problem defined in (\ref{FRED1})-(\ref{FRED2}) has a solution if and only if} \[ = 0 \quad \forall v \in N(L^*) \] \vspace{0.2in} \noindent it suffices to show that $v(x)$ solves \eq (I+\lambda K^*) v(x) = 0 \label{F4} \endeq if and only if $L^*v=0$ and $B^*\bar{v}=0$. Since the same assumptions (A1)-(A3') imply the existence of the adjoint Green's function $g_{\lambda}^*(x,t)$ and $g_{\lambda}^*(t,x)=g_{\lambda}(x,t)$, this result follows trivially by writing down the equivalent integral equation for $L^*v=0$ and comparing it to(\ref{F4}). The same proof holds for the first method involving $g_0$. Still, a natural question to ask is when is assumption (A3') satisfied? In Coddington and Levinson (Theory of Ordinary Differential Equations), it is shown that for any self adjoint $n$-th order boundary value problem with general (nonseparated) boundary conditions having $a_n(t)\ne 0$ on $[a,b]$, the associated operator $L$ has at most an enurable number of eigenvalues. Therefore, for these problems (A3') is satisfied and the Fredholm alternative holds. The necessity of the alternative theorem is true for a much broader class of problems. Defining some operator $L:D(L)\subset H \rightarrow H$ for some Hilbert space $H$ if we know the adjoint $L^*$ exists on some domain $D(L^*)\subset H$ then necessity follows from \eq = = = 0 \quad \forall v \in N(L^*) \endeq Alternative theorems are often stated in two parts. One part involves the existence of solutions (we have already discussed this). The second part addresses uniqueness. For completeness we state this second part here. \vspace{0.1in} \noindent {\em Theorem: If $L:D(L)\subset H \rightarrow H$ for some Hilbert space $H$ then} \[ \exists ! \ u \in D(L) \ such \ that \ Lu=f \quad \Leftrightarrow \quad N(L) = \{ 0 \} \] \vspace{0.1in} \noindent This result follows from theory on linear transformations. Namely, a linear transformation $T:X \rightarrow Y$ on two linear spaces is one-to-one if and only if $N(T)=0$. Last, we demonstrate the application of these theorems through several examples: \vspace{0.1in} \noindent {\sc Example 1}: The problem \eq u'' = f(x) \quad , \quad u'(0)=u'(1) = 0 \label{ex1} \endeq has a solution if and only if $f$ is orthononal to all solutions $v$ of \eq v'' = 0 \quad , \quad v'(0)=v'(1) = 0 \endeq Since $N(L^*) = span \{ 1 \}$ we have \eq \int_0^1 f(x) dx = 0 \endeq as both a necessary and sufficient condition for the existence of solutions $u(x)$ of (\ref{ex1}). \vspace{0.1in} \noindent {\sc Example 2}: Consider the nonlinear boundary value problem \eq Lu \equiv u''+\pi^2 u = \varepsilon f(u,u',\mu_1) \quad , \quad u(0)=u(1) = 0 \endeq where $0 \le \varepsilon \ll 1$ and $\mu_1\in \reals$. For $\varepsilon = 0$ the solution of this problem is not unique. However, for $\varepsilon >0$ solutions may not even exist. If we seek solutions $u(x,\varepsilon)$ which vary smoothly in $\varepsilon$ the Fredholm alternative can be used to deduce necessary conditions for the existence of solutions which perturb from $u(x,0)$. In a linear analysis of this problem one assumes the expansion \eq u(x,\varepsilon) = u_0(x) + \varepsilon u_1(x) + O(\varepsilon^2) \endeq where where $u_0(x) = \mu_0 \sin(\pi x)$ solves the $\varepsilon = 0$ problem and the $O(\varepsilon)$ problem is \eq Lu_1 = f(u_0,u_0',\mu_1) \quad , \quad u_1(0)=u_1(1)=0 \label{order1} \endeq Since $\mu_0 \sin(\pi x)$ spans the nullspace of the adjoint problem (here self-adjoint), solutions of (\ref{order1}) exist if and only if \eq F(\mu) = F(\mu_0,\mu_1) \equiv \int_0^1 \sin(\pi x) f(\mu_0 \sin(\pi x), \mu_0 \pi \cos(\pi x) ) dx = 0 \endeq Notice that $F=0$ is (in general) some curve in the $(\mu_0,\mu_1)$-plane. Only for these values can a solution $u$ smoothly perturb from a solution of the $\varepsilon = 0$ problem. Notice that, for instance, if $\mu_1$ is fixed there may be only one amplitude $\mu_0$ of $u_0(x)$ for which solutions may perturb. This type of argument is often used to determine leading behavior of solutions. \vspace{0.1in} \noindent {\sc Example 3}: Let $H=L^2[1,2]$ and define \begin{eqnarray} Lu & \equiv & u''(x)+\frac{1}{x} u'(x)+\lambda^2 u(x) \\ D(L) & = & \{u\in H : Lu \in H, u(1)=0, u(2)=0 \} \end{eqnarray} The adjoint for this problem is given by \begin{eqnarray} L^*v & \equiv & v- \frac{1}{x} v'(x) +(\frac{1}{x^2}+\lambda^2) v(x) \\ D(L^*) & = & \{v\in H : L^*v \in H, v(1)=0, v(2)=0 \} \end{eqnarray} Note that even the boundary conditions and the adjoint boundary conditions are the same, $L$ is not self adjoint since $L \ne L^*$. Now consider the solvability of the problem \eq Lu = f \quad , \quad u \in D(L) \label{ex3-1} \endeq It is possible to explicitly verify that this problem can be converted into an equivalent integral equation. We omit this step and simply note that the alternative theorem holds. Therefore, (\ref{ex3-1}) has a solution if and only if $f \in N(L^*)^{\perp}$. That is $=0$ for every $v \in D(L^*)$ solving $L^*v=0$. The general solution of the homogenous adjoint equation is \eq v(x) = c_1 x J_0(\lambda x) + c_2 x Y_0(\lambda x) \endeq where $c_1,c_2$ are constants and $J_0, Y_0$ are the zeroth order Bessel functions of first and second kind, respectively. $N(L^*)$ will be trivial unless $\lambda$ solves the eigenvalue equation \eq \Lambda(\lambda) \equiv J_0(\lambda) Y_0(2 \lambda) - J_0(2\lambda) Y_0(\lambda) = 0 \endeq A plot of the function $\Lambda$ reveals several roots $\lambda_n, n=1,2,\ldots$. If $\lambda=\lambda_n$ then \[ N(L^*) = span \{ v_n(x) \} \] where \[ v_n(x) = x (J_0(\lambda_n x) - J_0(\lambda_n) Y_0(\lambda_n x)) \] If $\lambda \ne \lambda_n$, $N(L^*)=\{0\}$. Therefore, (\ref{ex3-1}) has a solution for all $\lambda \ne \lambda_n$. Furthermore, if $\lambda = \lambda_n$ a solution exists if and only if \[ < f, v_n > = 0 \] It can be shown that $N(L)$ is nontrivial for the same $\lambda_n$ so that if $\lambda=\lambda_n$, should a solution of (\ref{ex3-1}) exist, it is not unique. However, it should be noted that the $u_n(x) \in D(L)$ solving \eq u_n''(x)+\frac{1}{x} u_n'(x)+\lambda_n^2 u_n(x) = 0 \endeq are not equal to $v_n$. In particular, $=0$ is not the correct solvability condition for this problem. \vspace{0.2in} \subsection{Some comments on singular boundary value problems} Much of the theory in the preceding sections depended heavily on the requirement that $a_2(x) \ne 0$ on $[a,b]$. If $a_2$ vanishes on $[a,b]$ the problem is said to be singular. In much literature, if $a_2$ vanishes at some $x_0$ in the interior $(a,b)$, the problem is referred to as a ``turning point problem''. For singular problems, many of the conclusions that are true for regular problems are no longer true. We present a few examples to illustrate these points. \vspace{0.1in} {\sc Example 1}: Let $H=L^2[0,1]$ and define \begin{eqnarray} Lu & \equiv & x u''(x) \\ D(L) & = & \{ u \in H: Lu \in H, u(0)=u'(0)=0 \} \end{eqnarray} Define \begin{eqnarray} L^*v & \equiv & x v''(x)+ 2 v'(x) \\ D(L^*) & = & \{ v \in H: Lv \in H, v(1)=v'(1)=0 \} \end{eqnarray} The operator $L$ is singular since $a_2(x)=x$ vanishes at $x=0$. Also, there are certainly a great many functions for which $=$. The general solution of $L^*v=0$ is \[ v(x) = \frac{c_1}{x}+c_2 \] From this we see $N(L^*)=\{ 0\}$. If the Fredholm alternative did hold, the conclusion would be that there is always a $u \in D(L)$ solving $Lu=f$. To see this, note the general solution of $Lu=1$ is \eq u(x) = x(ln(x) -1) + c_1 + c_2 x \endeq Clearly the boundary conditions cannot be satisfied since \[ u'(x) = ln(x)+c_2 \] This example is one illustration of how the Fredholm alternative can fail in singular problems. \vspace{0.1in} \noindent {\sc Example 2}: Consider the singular boundary value problem \[ Lu+\lambda u = x^2 u''(x)+x u(x)+\lambda u(x) = 0 \] where \[ u(1)=0, u,u' \ bounded \ at \ x=0 \] For all $\lambda > 0$ this problem has nontrivial solutions: \[ u(x) = \sin ( \sqrt{\lambda} ln(x)) \] Thus, eigenvalues of singular problems may not be countable. \vspace{0.2in} \subsection{Eigenvalues for Invertible Self-Adjoint Differential Operators} If a differential operator $L$ defined on $D(L)\subset L^2[a,b]$ is self-adjoint and $Lu=f$ is equivalent to a Fredholm integral equation \[ (I-\lambda K) u = F \] with $K=K^*$ and $K$ compact, then the existence of eigenvalues and complete sets of eigenfunctions can be deduced from spectral theory for compact self-adjoint Fredholm operators. Due to the result in Coddington and Levinson, there is always a $\lambda \in \reals$ such that for all self-adjoint $n$-th order differential operators defined with separated boundary conditions, \[ Lu = \lambda u \quad , \quad u\in D(L) \Leftrightarrow u(x) \equiv 0 \] For the theory we have developed for 2nd order $L$ this implies the existence of a symmetric Green's function $g=g(x,t)$ solving \[ Lg = \delta (x-t) \quad , \quad g \in D(L) \] such that \[ (Ku)(x) = \int_a^b g(x,t) u(t) dt \] is self-adjoint since $g(x,t)=g(t,x)$ and compact because $g(x,t)$ is continuous making $K$ Hilbert-Schmidt. For the following Theorem, one also needs to know that $N(L) \ne {0}$ or equivalently that $L^{-1}$ exists on the range $R(L)$ of $L$. \vspace{0.1in} \noindent {\em Theorem: Let $L$ be the second order differential operator defined on $H=L^2[a,b]$ as follows: \begin{eqnarray} Lu & = & a_2(x) u''(x) + a_1(x) u'(x) +a_0(x) \quad , x\in(a,b) \\ a_0(x) & \ne & 0 \quad \forall x \in [a,b] \\ B_1u & = & b_{11}u(a) + b_{12} u'(a) \\ B_2u & = & b_{23}u(b) + b_{24} u'(b) \\ D(L) & = & \{ u \in H: Lu \in H, B_iu=0, i=1,2 \} \end{eqnarray} so that $L:D(L)\rightarrow H$ and \[ R(L) = \{ f \in H: \exists u \in D(L) \ni Lu=f\} \] Furthermore, assume $N(L)=\{0\}$. Then \begin{itemize} \item[1)] If $\lambda$ is an eigenvalue of $L$ then $dim E_{\lambda}(L) \le 2$. \item[2)] All the eigenvalues of $L$ are real. \item[3)] Eigenfunctions corresponding to different eigenfunctions are orthogonal with respect to the $L^2[a,b]$ inner product. \item[4)] The eigenfunctions of $L$ form a complete set on $L^2[a,b]$ \item[5)] The eigenvalues $\lambda_n$ of $L$ have $|\lambda_n| \rightarrow \infty$ as $n\rightarrow \infty$. \end{itemize} } \vspace{0.1in} {\em Proof:} 1) follows since $L$ is second order. 2) follows from \[ = <\phi,L\phi> = \lambda \parallel \phi \parallel^2 = \bar{\lambda} \parallel \phi \parallel^2 \] If $\lambda_1 \ne \lambda_2$ are eigenvalues of $L$ with respective eigenfunctions $\phi_1$, $\phi_2$ then \[ = <\phi_1,L\phi_2> = \lambda_1 <\phi_1,\phi_2> = \lambda_2 <\phi_1,\phi_2> \] implies $<\phi_1,\phi_2> \ne 0$ demonstrating 3). To show 4) let $g(x,t)$ be the symmetric Green's function for $L$. Then, $L\phi = \lambda \phi$ is equivalent to the integral equation \eq \phi=\lambda K\phi \equiv \int_a^b g(x,t) \phi (t) dt \endeq Since $N(L)=\{0\}$, $\lambda$=0 is not an eigenvalue of $L$ so that every eigenvalue $\mu$ of $K$ satisfying \[ K\phi = \mu \phi \] corresponds to an eigenvalue $\lambda = \mu^{-1}$ of $L$ and vice-versa. Now note the eigenfunctions of $K$ are complete over the range of $K$. But, $R(K)=D(L)$ since $K=L^{-1}$ (existence of $L^{-1}$ on $R(L)$ is guaranteed by $N(L)=0$). Since $D(L)$ is dense in $L^2[a,b]$ the eigenfunctions of $K$ (equivalently the eigenfunctions of $L$) form a complete set over $L^2[a,b]$. Lastly, 5) follows from the fact that self adjoint compact $K$ have $\mu_n \rightarrow 0$ as $n\rightarrow \infty$ for its eigenvalues $\mu_n$. Q.E.D. \vspace{0.2in} Clearly, 2)-3) are true even if $N(L) \ne 0$ and $L$ is an operator on Hilbert space $H$ for which an adjoint exists. Furthermore, if $L$ is some arbitrary $n$th order differential operator for which a symmetric Green's function exists then the results in 4)-5) hold if $N(L) = 0$. Even if $N(L)=0$, the eigenfunctions of $L$ may form a complete set. For example, the eigenfunctions of the self adjoint problem \[ Lu = u''(x) \] \[ u(0)=u(2\pi) \quad , \quad u'(0)=u'(2\pi) \] are $\phi_n(x)=\cos(nx),\sin(nx)$, $n=0,1,\ldots$. These are complete over $L^2[0,2\pi]$ even though $N(L)=span\{1\}$. Suppose now that the assumptions of the theorem hold for some $L$ and we wish to solve the problem \eq Lu=f \quad , \quad u \in D(L) \label{eqa} \endeq The solution necessarily exists and is unique since $N(L)=N(L^*)=0$. Let $\{\phi_n(x)\}_{n=1}^{\infty}$ be the complete set of mutually orthonormal eigenfunctions of $L$. Then \[ f(x) = \sum_{n=1}^{\infty} \phi_n(x) \] where equality is in the sense of the $L^2[a,b]$ norm \footnote{ For example, suppose $D(L)$ is defined using Dirichlet boundary conditions $u(0)=u(1)=0$, then $\phi_n(0)=0$ for all $n$. Pointwise evaluation of the series suggests $f(0)=0$. Clearly not all $f \in L^2[0,1]$ have $f(0)=0$. However, it can be shown that \[ \lim_{x\rightarrow 0^+} \sum_{n=1}^{\infty} \phi_n(x) = \tilde{f}(0) \] for all $f$ equivalent to a continuous function $\tilde{f}$ on $[a,b]$.}. Assuming a solution $u$ of the form \[ u(x) = \sum_{n=1}^{\infty} u_n \phi_n(x) \] from \[ Lu = \sum_{n=1}^{\infty} u_n \lambda_n \phi_n(x) = \sum_{n=1}^{\infty} \phi_n(x) \] one deduces \[ u(x) = \sum_{n=1}^{\infty} \frac{}{\lambda_n} \phi_n(x) \] where we note $\lambda_n \ne 0$ for an invertible $L$. Alternately we could write this as \[ u(x) = \int_0^1 \tilde{g}(x,t) f(t) dt \] where \[ \tilde{g}(x,t) = \sum_{n=1}^{\infty} \frac{\phi_n(x) \phi_n(t)}{\lambda_n} \] This is different expression for the Green's function $g(x,t)$ of $L$. Again, $g$ and $\tilde{g}$ may differ pointwise but they are equal in the distributional sense. A separate way of seeing this connection is through the following informal calculations. Suppose you wish to solve \[ Lg = \delta (x-t) \quad , \quad g \in D(L) \] using the eigenfunction expansion \[ g(x,t) = \sum_{n=1}^{\infty} g_n(t) \phi_n(x) \] Next expand $\delta(x-t)$ as follows \[ \begin{array}{lcl} \delta (x-t) & = & \sum_{n=1}^{\infty} \delta_n(t) \phi_n(x) \\ \ & = & \sum_{n=1}^{\infty} <\delta_n(t), \phi_n(x)> \phi_n(x) \\ \ & = & \sum_{n=1}^{\infty} \phi_n(t) \phi_n(x) \end{array} \] Then \[ Lg = \sum_{n=1}^{\infty} \lambda_n g_n(t) \phi_n(x) = \sum_{n=1}^{\infty} \phi_n(t) \phi_n(x) \] from which we deduce the same series as in $\tilde{g}(x,t)$ for $g$. Note that formally, the series for the expansion of $\delta (x-t)$ does not converge. Authors refer to such nonconvergent series as a ``representation'' of the $\delta$ function. \end{document}