Chapter 2 Eigenvalues and eigenvectors/eigenfunctions

2.1 Revision of MA20220

Consider the initial value problem for the ordinary differential equation

\[ \dot {u}(t)=Au(t),\qquad u(0)=u^0, \]

where \(A\in \mC ^{n\times n}\) and \(u^0\in \mC ^n\) are given and \(u:[0,\infty )\to \mC ^n\) is to be found. Here \(\dot {u}\) is notation for the derivative of \(u\) with respect to \(t\) (which we interpret as time).

We first consider only the differential equation and ignore the initial condition. We look for a solution \(u\) of the following form

\[ u(t)=\e ^{\lambda t}v, \]

where \(v\in \mC ^n\). Substituting this Ansatz into the differential equation gives

\[ \lambda \e ^{\lambda t}v=A\e ^{\lambda t}v. \]

Cancelling (the always positive) \(\e ^{\lambda t}\) on both sides gives

\[ \lambda v=Av. \]

We therefore need that \(v\) is an eigenvector of \(A\) with eigenvalue \(\lambda \). The above implications reverse, so this is not only necessary but also sufficient. We therefore obtain the solutions

\[ u_k(t)=C_k\e ^{\lambda _k t}v_k, \]

for arbitrary constants \(C_k\) where \((v_k)\) are linearly independent eigenvectors (which we assume we have \(n\) of).

To satisfy the initial condition we look at a sum of these solutions:

\[ u(t)=\sum _{k=1}^n C_k\e ^{\lambda _k t}v_k. \]

The initial condition \(u(0)=u^0\) then becomes

\begin{equation} \label {eq:v} u^0=\sum _{k=1}^n C_kv_k, \end{equation}

which means that we should choose the \(C_k\) as the coefficients of \(u^0\) in the basis \((v_k)\). If we assume that the \((v_k)\) are orthonormal (i.e. \(\ipd {v_k}{v_m}\) equals 1 if \(k=m\) and 0 if \(k\neq m\)), then we can obtain a more explicit formula for the \(C_k\). In (2.1) we take the inner-product with \(v_m\) and use orthonormality to conclude that

\[ \ipd {u^0}{v_m}=C_m. \]

Note that above we assumed that \(A\) has an orthonormal basis of eigenvectors. This is not always true, but it is true if \(A=A^*\) (here \(A^*\) denotes the complex conjugate transpose of \(A\)) as follows from the spectral theorem from Algebra 1B (or Algebra 2A, depending on which year you started your studies...).

2.2 The connection

We see that the solution method for the heat equation on the interval and the ODE \(\dot {u}=Au\) is very similar (especially when \(A=A^*\)). We have the following correspondences (here \(\delta _{km}\) is the Kronecker delta which equals 1 if \(k=m\) and \(0\) if \(k\neq m\))

From the correspondence table we see that the inner-product relevant for the heat equation on an interval is

\[ \ipd {g}{h}:=\int _0^L g(x)h(x)\,dx. \]

This inner-product makes sense for functions for which the corresponding norm is finite, i.e.

\[ \|f\|^2:=\int _0^L |f(x)|^2\,dx<\infty , \]

i.e. functions which are square-integrable. This gives the natural generalization of the Euclidean space \(\mC ^n\) to functions defined on the spatial domain \((0,L)\). We will denote this space of square-integrable functions by \(L^2(0,L)\) (the \(L\) in \(L^2\) stands for Lebesgue).

From the correpondence table we see that the generalization of the matrix \(A\) is the second derivative operator \(X\mapsto X''\). However, we also have to deal with the boundary conditions. The way to deal with these is to include them in the domain of the operator, i.e. to define

\begin{multline} \label {eq:OpDirichlet} AX=X'',\quad A:\dom (A)\subset L^2(0,L)\to L^2(0,L), \\ \dom (A)=\{X\in L^2(0,L): X',X''\in L^2(0,L),~X(0)=0,~X(L)=0\}, \end{multline} (here we impose the conditions \(X',X''\in L^2(0,L)\) to ensure that \(A\) indeed maps \(\dom (A)\) into \(L^2(0,L)\)). The eigenvalue equation \(AX=\lambda X\) implicitly contains the condition that \(X\in \dom (A)\) and therefore in particular that an eigenfunction \(X\) satisfies the boundary conditions \(X(0)=X(L)=0\). Therefore the ODE boundary value problem (1.2) is simply the eigenvalue problem for the operator \(A\) given by (2.2).

As in the matrix case, the crucial thing in the operator case is that the eigenvectors of \(A\) form an orthonormal basis of our inner-product space (\(\mC ^n\) in the matrix case, \(L^2(0,L)\) in the operator case).

2.3 Background: the Spectral Theorem

The following theorem which will be proven in some form in MA40256 Analysis in Hilbert Spaces (and can be considered the main result in that module).

Differential operators are never compact, but often they have compact resolvent. Applying Theorem 2.5 to the resolvent, we obtain the following (using that \(\lambda \) is an eigenvalue of \(T\) if and only if \(\mu =\frac {1}{z-\lambda }\) is an eigenvalue of \((zI-T)^{-1}\) and that the eigenvectors are the same).