%% 6. Eigenvalues of linear BVPs %% % % \setcounter{page}{60} % %% % The first BVP of the last chapter, equation (5.1), can be written % $$y'' + b\kern .3pt y = 0, \quad x \in [\kern .3pt 0,60\kern.3pt], ~ y(0) = 1, ~y(60)=0 % \eqno (6.1)$$ % with $b=1$, and the solution had amplitude % about $3.3$. Look what happens % if $b$ is reduced by just $1\%$, to $0.99$. ODEformats, b = 0.99; L = chebop(0,60); L.op = @(x,y) diff(y,2) + b*y; L.lbc = 1; L.rbc = 0; y = L\0; plot(y,CO,bvp) title(['Fig.~6.1.~~Solution of (6.1) for $b=0.99$, ' ... 'near an eigenvalue'],FS,11) %% % \vskip 1.02em %% % % \noindent % The amplitude has increased to $111.3\kern .5pt!$\ \ With % $b = 0.9897$, it is about $16606$, and % with $b = (19\pi/60)^2\approx 0.98970199,$ % it reaches infinity'': the BVP has no solution. % %% % What is happening is that the solution depends % on a division by a number that is close to zero. % In this chapter we will explore the mathematics of situations % like this. For a simple problem like (6.1), the algebra is elementary, % but the implications, including the phenomenon % of resonance for time-dependent partial differential equations % (see Chapter 22), are far-reaching. %% % Let's begin with the elementary algebra. From equation % (4.5), we know that % the general solution of (6.1) is % $$y(x) = A \sin(kx) + B\cos(kx)$$ % with $k = \sqrt b$. % The boundary conditions in (6.1) correspond to the equations % $$B = 1, \quad A\sin(60\sqrt b\kern 1pt) % + B\cos(60\sqrt b\kern 1pt) = 0.$$ % This can be regarded as a $2\times 2$ matrix problem for % the unknowns $A$ and $B$, an especially easy one % since the first equation is trivial. The solution is % $$B = 1, \quad A = - \cot(60\sqrt b\kern 1pt) ,$$ % where $\cot x = \cos x / \sin x$ as always. % Here is the value of $A$ for $b=0.99$. b = 0.99; A = -cot(60*sqrt(b)) %% % % \noindent % And here it is for $b=0.9897$. % b = 0.9897; A = -cot(60*sqrt(b)) %% % % \noindent % With $b = (19\pi/60)^2$, we get $60\sqrt b = 19\pi$, % and the cotangent is infinite. Likewise it would be % infinite with $b = (\kern 1pt j\pi/60)^2$ for any positive integer $j$. % %% % % We are seeing here the phenomenon of eigenvalues of a % differential operator with boundary conditions. % To frame the matter more generally, % let $L$ be the linear operator $L: y\mapsto y''$ on % $[\kern .3pt 0,60\kern .3pt]$ and let us consider the problem % $$Ly = y'' = \lambda\kern .3pt y + f, \quad x \in [\kern .3pt 0,60\kern.3pt], % ~ y(0) = \alpha , ~y(60)=\beta , \eqno (6.2)$$ % where $\lambda$ is a given number and $f$ is a function of $x$. % It is clear that regardless of the choices of $f$, $\alpha$, % and $\beta$, there can be no unique solution of (6.2) % if there exists a nonzero function $v(t)$ that satisfies % the homogeneous equation % $$Lv = v''= \lambda v , \quad v(0) = v(60) = 0. \eqno (6.3)$$ % The reason is that any multiple of $v$ could be added to $y$, % and the result would still satisfy (6.2). % Such a function is called an {\bf eigenfunction} % of $L$, and $\lambda$ is the corresponding % {\bf eigenvalue.} Another way to say it is that an eigenvalue % of a linear operator $L$ is a number $\lambda$ such that % the operator $L-\lambda I$, acting on functions % satisfying the homogeneous boundary conditions, % has a nontrivial nullspace. (Here $I$ denotes the % identity operator, mapping a function $y$ to itself.) % An eigenfunction $v$ of $L$ % is a nonzero function in this nullspace. % Note that any nonzero multiple of an eigenfunction % is also an eigenfunction for the same $\lambda$. % %% % Let's do an experiment to scan systematically for eigenvalues of the % operator $L$ of (6.2). Since $\lambda$ is the % negative of $b$ in (6.1), the eigenvalues will be negative numbers, and % the smallest will be quite close to zero. Here we arbitrarily pick % boundary conditions $y(0)=e$ and $y(60)=\pi$ % and the forcing function $f(x) = \exp(-(x-10)^2)$, and we % solve the ODE successively with $\lambda = -.0300, -.02998, % -.02996,\dots, 0$. For each value of $\lambda$ we plot a % dot representing $\max_{x\in [\kern .3pt 0,60\kern.3pt]} |y(x)|$. f = chebfun('exp(-(x-10)^2)',[0 60]); L.lbc = exp(1); L.rbc = pi; for lambda = -0.03:.0002:0 L.op = @(x,y) diff(y,2) - lambda*y; y = L\f; plot(lambda,norm(y,inf),'.k',MS,7), hold on, drawnow end axis([-.03 0 0 400]), hold off, xlabel('$\lambda$',FS,10,IN,LT) title(['Fig.~6.2.~~$\max_x |y(x)|$ for solutions to (6.2)' ... ' with various values of $\lambda$'],FS,11) %% % \vskip 1.02em %% % % \noindent % The evidence is clear: there are three eigenvalues of $L$ % in this range, lying near $-0.003$, $-0.01$, and $-0.025$. % Here are plots of the solutions $y(x)$ for these values of $\lambda$. % f = chebfun('exp(-(x-10)^2)',[0 60]); llam = [-.003 -.01 -.025]; for j = 1:3 lambda = llam(j); L.op = @(x,y) diff(y,2) - lambda*y; subplot(1,3,j), plot(L\f,CO,bvp), axis([0 60 -115 115]) text(30,95,['$\lambda = ' num2str(lambda) '$'],FS,9,HA,CT) end subplot(1,3,2) title('Fig.~6.3.~~Three large solutions of (6.2)',FS,11) %% % \vskip 1.02em %% % % \noindent % The amplitudes of all three functions are on the order of $100$, and we could % have made them larger by estimating the eigenvalues more % accurately. Just as the three values of $\lambda$ approximate the % first three eigenvalues of $L$ (i.e., the three % smallest eigenvalues in absolute % value), the three functions approximate multiples of the corresponding % eigenfunctions. Note that these curves have 1, 2, and 3 % humps, respectively, separated by 0, 1, and 2 interior zeros, called % {\bf nodes}. The % property that the $j$th eigenfunction of $L$ has exactly $j-1$ nodes % is typical of eigenfunctions for a wide range of differential % operators.\footnote{In particular this holds for the % self-adjoint equation $(\kern .6pt p(x)y')' + q(x) y = \lambda\kern .3pt y$ % on an interval $[a,b]$ with homogeneous Dirichlet, Neumann, % or Robin boundary conditions, where $p'$ and $q$ are continuous % real functions and $p(x) > 0$. Such problems are the % subject of Sturm--Liouville theory.} % %% % Eigenvalues and eigenfunctions are much too important % for us to have to find them by % scanning values of $\lambda$''! % Chebfun computes these quantities directly % with the |eigs| command. We can find the first five eigenvalues of $L$ % like this. L.op = @(x,y) diff(y,2); sort(eigs(L,5),'descend')' %% % % \noindent % To get eigenfunctions as well as eigenvalues we can type % [V,D] = eigs(L,5); llam = diag(D); [llam,ii] = sort(llam,'descend'); V = V(:,ii); %% % % \noindent % A plot of the first three eigenfunctions shows convincing approximations % of multiples of the curves above.\footnote{As mentioned % earlier, eigenfunctions are only defined % up to multiplicative constants. Those returned by Chebfun are % scaled to have square-integral equal to 1, a % normalization that determines their % shape apart from an arbitrary choice of sign.} % \label{chap5foot} % for j = 1:3 subplot(1,3,j), plot(V(:,j),CO,bvp), axis([0 60 -.3 .3]) text(30,.24,['$\lambda = ' num2str(llam(j)) '$'],HA,CT,FS,9) end subplot(1,3,2) title('Fig.~6.4.~~First three eigenfunctions of (6.3)',FS,11) %% % \vskip 1.02em %% % % \noindent % From these images it appears that the $j$th % eigenfunction consists of the sine function scaled so % that exactly $j$ lobes, i.e.\ $j/2$ periods, fit in the % interval $[\kern .3pt 0,60\kern .3pt]$, % $$v_j^{}(t) = \sin(\kern 1pt j\pi t/60) , \quad j = 1,2,\dots . % \eqno (6.4)$$ % Substituting this trial solution in (6.3) confirms that it is % indeed an eigenfunction. The associated eigenvalue comes out as % $$\lambda_j^{} = -(\kern 1pt j\pi/60)^2, % \quad j = 1,2,\dots, \eqno (6.5)$$ % and this formula enables us to confirm the five numbers computed above, % -((1:5)*pi/60).^2 %% % % The eigenfunctions of $L$ are {\bf orthogonal}, which % means that the number % $$a_{ij}^{} = \int_0^{60} v_i^{} (x) v_j^{}(x) dx \eqno (6.6)$$ % is zero whenever $i\ne j$.\footnote{Eigenvectors of real symmetric matrices % are also orthogonal when they correspond to distinct % eigenvalues. The proofs are closely related.} % To see why, consider two arbitrary functions % $v_i^{}$ and $v_j^{}$ satisfying the homogeneous boundary conditions. % Using integration by parts, we compute % $$\int_0^{60}\kern -2pt (L v_i^{}(x)) v_j^{}(x) dx % = \int_0^{60}\kern -2pt v_i'' (x) v_j^{}(x) dx % = -\kern -2pt \int_0^{60}\kern -2pt v_i' (x) v_j'(x) dx,$$ % where we have discarded boundary terms that are % zero because of the boundary conditions on $v_j^{}$. % A second integration by parts exploiting the zero boundary % conditions on $v_i^{}$ shows that this is equal to % $$\int_0^{60}\kern -2pt v_i^{} (x) v_j''(x) dx % = \int_0^{60}\kern -2pt v_i^{}(x) (L v_j^{}(x))dx.$$ % Thus we have % $$\int_0^{60} (Lv_i^{})v_j^{}dx = \int_0^{60} % \kern -2pt v_i^{} (L v_j^{}) dx. \eqno (6.7)$$ % An operator satisfying this property for arbitrary functions % $v_i^{}$ and $v_j^{}$ is said to be {\bf self-adjoint}. Now suppose % $v_i^{}$ and $v_j^{}$ are eigenfunctions of a self-adjoint % operator $L$ corresponding to % eigenvalues $\lambda_i^{}$ and $\lambda_j^{}$ with % $\lambda_i^{}\ne \lambda_j^{}$. Then (6.7) becomes % $$\lambda_i^{} \kern -2pt \int_0^{60} \kern -2pt v_i^{} v_j^{} dx = % \lambda_j^{}\kern -2pt \int_0^{60} \kern -2pt v_i^{} v_j^{} dx,$$ % and if $\lambda_i^{}\ne \lambda_j^{}$, this implies that % $v_i^{}$ and $v_j^{}$ must be orthogonal. % We shall repeat this calculation in greater generality % for Theorem~7.1 in the next chapter. % %% % % Let us verify orthogonality numerically for the operator $L$ % of (6.2). The object $\bf V$ computed above by % {\tt eigs} is a {\em quasimatrix\/} of dimensions $\infty\times 5$, % whose five columns'' are the functions $v_1^{}, \dots,v_5^{}$. % The object ${\bf V}^T {\bf V}$ is the product of the $5\times \infty$ transpose % of ${\bf V}$ with ${\bf V}$ itself, that is, the $5\times 5$ matrix of entries % $a_{ij}^{}$ defined % by (6.6), $1\le i,j \le 5$, known as a {\em Gram matrix}. % Orthogonality of the % eigenfunctions is confirmed by the fact % that this matrix has zero entries off the diagonal. % format short, A = V'*V %% % % \noindent % The fact that the entries on the diagonal are equal % to $1$, making $\bf A$ the identity matrix, % indicates that the eigenfunctions are not just orthogonal but % {\bf orthonormal}, a consequence of the % normalization mentioned in the footnote on p.~\pageref{chap5foot}. % %% % % Eigenvalue problems are associated with % homogeneous boundary conditions, because they are all about % identifying nontrivial solutions of the homogeneous problem. % It does not make sense, for example, to specify $v(0) = 1$ % for an eigenvalue problem. A variation that does make % sense, however, is to change a homogeneous boundary % condition from Dirichlet to Neumann form. For example, here is % what happens if we change the right-hand boundary % condition of (6.3) from $v(60)=0$ to $v'(60) = 0$. % L.rbc = 'neumann'; sort(eigs(L,5),'descend')' %% % % \noindent % The eigenfunctions have zero slope at the right boundary. % [V,D] = eigs(L,5); llam = diag(D); [llam,ii] = sort(llam,'descend'); V = V(:,ii); for j = 1:3 subplot(1,3,j), plot(V(:,j),CO,bvp), axis([0 60 -.3 .3]) text(30,.24,['$\lambda = ' num2str(llam(j)) '$'],HA,CT,FS,9) end subplot(1,3,2) title('Fig.~6.5.~~Eigenfunctions with a Neumann BC',FS,11) %% % \vskip 1.02em %% % % \noindent % The images reveal $1/2$, $3/2$, $5/2,\dots$ % lobes of the sine function scaled to $[\kern .3pt 0,60\kern .3pt]$, suggesting % the general formula % $$v_j^{}(x) = \sin((\kern 1pt j-\textstyle{1\over 2})\pi x/60) , % \quad j = 1,2,3,\dots . \eqno (6.8)$$ % As before, substitution confirms that these are eigenfunctions, with % eigenvalues % $$\lambda_j^{} = -((\kern 1pt j-\textstyle{1\over 2})\pi/60)^2, % \quad j = 1,2,\dots. \eqno (6.9)$$ % The numbers match properly, % -((0.5:4.5)*pi/60).^2 %% % % \noindent % Again the eigenfunctions are orthonormal, % V = chebfun(V); A = V'*V %% % Eigenfunctions of differential operators are not always orthogonal. % For example, here is an advection-diffusion problem adapted % from equation (4.9): % $$Ly = y'' +0.1 y' , \quad x\in [\kern .3pt 0,60\kern .3pt], ~y(0)=1, ~ y'(0) = 0. % \eqno (6.10)$$ % Unlike the other operators of this chapter, this one is not self-adjoint. % Here are the first four eigenfunctions. % Note that the nonzero boundary value specified in (6.10) % is ignored by Chebfun in solving the eigenvalue problem, which % is defined in terms of homogeneous boundary conditions as always. L = chebop(0,60); L.op = @(x,y) diff(y,2) + 0.1*diff(y); L.lbc = 1; L.rbc = 0; [V,D] = eigs(L,5); llam = diag(D); [llam,ii] = sort(llam,'descend'); V = V(:,ii); clf for j = 1:4 subplot(2,2,j), plot(V(:,j),CO,bvp), axis([0 60 -.5 .5]) text(57,.30,['$\lambda = ' num2str(llam(j)) '$'],FS,9,HA,RT) end subplot(2,2,1) title('\kern 2.5in Fig.~6.6.~~Eigenfunctions of (6.10)',FS,11) %% % \vskip 1.02em %% % % \noindent % The matrix of inner products confirms that % the eigenfunctions are nonorthogonal. % V = chebfun(V); A = V'*V %% % % \noindent % Here are some higher eigenfunctions, corresponding to % eigenvalues $\lambda_8$, $\lambda_{16}$, $\lambda_{24}$, % and $\lambda_{32}$. The higher the % eigenvalue, the more oscillations. % [V,D] = eigs(L,32); llam = diag(D); [llam,ii] = sort(llam,'descend'); V = V(:,ii); for j = 1:4 subplot(2,2,j), plot(V(:,8*j),CO,bvp), axis([0 60 -.5 .5]) text(57,.30,['$\lambda = ' num2str(llam(8*j)) '$'],FS,9,HA,RT) end subplot(2,2,1) title('\kern 2.45in Fig.~6.7.~~Some higher eigenfunctions',FS,11) %% % \vskip 1.02em %% % % \begin{center} % \hrulefill\\[1pt] % {\sc Application: eigenstates of the Schr\"odinger equation}\\[-3pt] % \hrulefill % \end{center} % %% % % One of the biggest scientific discoveries of all time was % the Schr\"odinger equation of quantum mechanics, % in 1926, which in principle % reduces much of the subject of chemistry to self-adjoint eigenvalue calculations. % It is from this date that eigenvalues and eigenfunctions became % a universally known tool in the physical sciences. % %% % % The steady-state (time-reduced) 1D Schr\"odinger % equation on a finite or infinite % interval $[-d,d\kern .5pt]$ takes the form % $$-h^2 y'' + V(x) y = \lambda\kern .3pt y, \quad x\in [-d,d\kern .5pt ], ~ % y(-d\kern .5pt ) = y(d\kern .5pt ) = 0, \eqno (6.11)$$ % where $V(x)$ is a fixed potential function and % $h$ is Planck's constant. % Equation (6.11) describes what states a particle may occupy, % in the quantum theory, if it is confined to the given interval % and subject to an energy potential $V(x)$ there. % (More precisely, $h^2$ in (6.11) should be % $h^2/8\pi^2 \mu$, where $h$ is the true % Planck constant and $\mu$ is the mass of the particle.) % The eigenvalue $\lambda$ corresponds to the energy of a particle % in that state, and its probability density of being % at each point $x\in [-d,d\kern .5pt]$ is equal to % $|y(x)|^2$, where $y$ is the corresponding normalized % eigenfunction.\footnote{This is an ODE problem % because there is just one particle in one dimension. % For $n$ particles in three dimensions, % it would be a PDE in $3n$ independent variables, far too difficult % in most cases to be solved numerically. That is why we say % that Schr\"odinger's equation reduces chemistry to mathematical % calculations in principle.'' % In practice, chemists have developed powerful methods for % approximating the equations for multiple-particle systems, % including those that led to the Nobel Prize for Kohn and % Pople in 1998.} % %% % % The simplest choice of $V(x)$ would be a constant; the % corresponding eigenfunctions are sines and cosines. Next % simplest is a parabola such as $V(x) = x^2$, corresponding % to a linear restoring force as in a Hookean spring; this is % the {\em quantum harmonic oscillator.} % Here we use Chebfun to compute the first ten eigenstates % of this problem with $h=0.1$, taking $d=3$ as a good % approximation to $d=\infty$. Following a format standard in physics, % we show the potential function as a heavy black curve and % plot each eigenfunction raised up by an amount corresponding % to its eigenvalue. Higher curves correspond to higher % energies. % clf, x = chebfun('x',[-3 3]); V = x^2; h = 0.1; L = chebop(-3,3); L.bc = 0; L.op = @(x,y) -h^2*diff(y,2) + V*y; [W,D] = eigs(L,10); diag(D)' for k = 1:10, plot(D(k,k)+0.06*W{k},LW,1), hold on, end xlabel('$x$',FS,10,IN,LT), ylabel('eigenfunctions',FS,9,IN,LT) title(['Fig.~6.8.~~First ten eigenstates of ' ... '(6.11) for harmonic oscillator'],FS,11) plot(V,'k',LW,1), axis([-1.5 1.5 0 2.3]), hold off %% % \vskip 1.02em %% % % \noindent % This problem can % be solved exactly with Hermite polynomials (for $d=\infty$); % the eigenvalues are $1,3,5,\dots$ times % $h$, hence $0.1, 0.3, 0.5,\dots$ for our choice $h=0.1$. % %% % The energy levels of systems like this % determine the states of atomic and molecular systems and % of the photons they emit or absorb. A photon emitted by % a radiating system will have a wavelength corresponding to % the difference between two energies, and much of what we % know about objects outside our solar system, including % the expanding universe and the Big Bang, comes from % analysing such wavelengths. % Closer to home, the glowing color of a neon'' fabric comes % from the phenomenon of fluorescence, in which the fabric % absorbs light at ultraviolet wavelengths and radiates it away % at a wavelength corresponding to the difference between % two quantum energy eigenstates. %% % % Chebfun has a command {\tt quantumstates} for % automating such calculations (which sets $h=0.1$ by default). % We illustrate this here with a modified function~$V$ % by adding a small peak in the middle, a % {\em potential barrier,} % making $V$ what is known as a {\em double-well potential.} % V = x^2 + 1.5*exp(-(x/.25)^4); [U,D] = quantumstates(V); axis([-1.5 1.5 0 2.3]) eigenvalues = diag(D)' title('Fig.~6.9.~~Eigenstates of a double well potential',FS,11) xlabel('$x$',FS,10,IN,LT), ylabel('eigenfunctions',FS,9,IN,LT) %% % \vskip 1.02em %% % % \noindent % The effect on the lower eigenvalues is dramatic if you % look at the numbers printed above: the lower ones now % fall in {\em nearly degenerate pairs.} This is because the potential % barrier nearly decouples % the left and right halves, so that it is approximately as if % we had two identical, uncoupled copies of a single-well % eigenvalue problem. % The lowest eigenvalue % $0.4436$ corresponds to an even eigenfunction, whereas the next % lowest, only slightly higher at $0.4459$, corresponds to an % odd eigenfunction that is almost equal to the same function multiplied % by $\hbox{sign\kern .5pt}(x)$. The third and fourth % eigenvalues and eigenfunctions have similar behavior, but % not as pronounced; at this energy level the two % wells are less perfectly decoupled. % This kind of close but inexact agreement of eigenvalues % as a result of approximate % but imperfect symmetry, known as {\em line splitting,} % is of tremendous importance in physics and led to Nobel Prizes for % the Zeeman effect in 1902 (line splitting in a % magnetic field) and the Stark effect in 1919 (line splitting in % an electric field). % %% % % Further remarkable things happen if we change the % potential of Figure~6.9 slightly, moving % the barrier to the right a distance $0.01$. % The eigenvalues, still in nearly degenerate pairs, do % not change very much. The lower eigenfunctions, however, % change completely. Now they are {\em localized,} % with the lowest eigenfunction being concentrated % in the bigger well, the % next-lowest in the narrower well, and so on. % In Figure 6.10 we plot just these two lowest % eigenfunctions in the two cases of the symmetric-well % potential and the slightly asymmetric one. % V2 = x^2 + 1.5*exp(-((x-.01)/.25)^4); [U2,D] = quantumstates(V2,'noplot'); eigenvalues = diag(D)' subplot(1,2,1), plot(U(1:2)), axis([-1.4 1.4 -1.9 1.9]) text(.2,-.7,'symmetric:',FS,9) text(.2,-1.05,'mode 1 even',FS,9) text(.2,-1.4,'mode 2 odd',FS,9) subplot(1,2,2), plot(U2(1:2)), axis([-1.4 1.4 -1.9 1.9]) text(.2,-.7,'asymmetric:',FS,9) text(.2,-1.05,'mode 1 left',FS,9) text(.2,-1.4,'mode 2 right',FS,9) subplot(1,2,1) title('\kern 2.4in Fig.~6.10.~~Eigenfunction localization',FS,11) %% % \vskip 1.02em %% % % More advanced phenomena of localization are related % to transparency of glass, insulating properties of % ceramics, and further Nobel Prizes. In fact, it can % be argued that eigenvalue problems % are related to the Physics Nobel Prizes of % 1902 (Zeeman), % 1903 (Becquerel, Curie \& Curie), % 1907 (Michelson), % 1911 (Wien), % 1917 (Barkla), % 1918 (Planck), % 1919 (Stark), % 1921 (Einstein), % 1922 (Bohr), % 1923 (Millikan), % 1924 (Siegbahn), % 1927 (Compton), % 1929 (de Broglie), % 1930 (Raman), % 1932 (Heisenberg), % 1933 (Schr\"odinger \& Dirac), % 1943 (Stern), % 1944 (Rabi), % 1945 (Pauli), % 1952 (Bloch \& Purcell), % 1954 (Born), % 1955 (Lamb), % 1956 (Schockley, Bardeen \& Brattain), % 1961 (M\"ossbauer), % 1962 (Landau), % 1963 (Wigner, Mayer \& Jensen), % 1964 (Townes, Basov \& Prokhorov), % 1965 (Tomonago, Schwinger \& Feynman), % 1966 (Kastler)$,\dots.$ % %% % % \smallskip % {\sc History.} The starting ideas of eigenvalue analysis % go back about 200 years to Fourier, Poisson, Sturm, and Liouville. % Sylvester and Cayley were diagonalizing % matrices in the 1850s, and by the 1880s the analysis of % eigenfunctions for vibrating membranes % and the theory of sound was well established. The terms % eigenvalue'' and spectral theory,'' however, seem to have come % later, coined by Hilbert around the turn of the 20th century. % %% % % \smallskip % {\sc Our favorite reference.} Although eigenvalues are everywhere % in the mathematical sciences, it is hard to find % discussions of {\em why} they matter so much. What is it about % the seemingly arbitrary algebraic property $L y = \lambda\kern .3pt y$ that % demands our attention? Such a discussion can be found in Chapter 1 of % Trefethen and Embree, % {\em Spectra and Pseudospectra: The Behavior of Nonnormal Matrices % and Operators,} Princeton U. Press, 2005, where four answers to % the question are proposed. % \smallskip % %% % % \begin{displaymath} % \framebox[4.7in][c]{\parbox{4.5in}{\vspace{2pt}\sl % {\sc Summary of Chapter 6.} % A second-order scalar linear ODE\/ $Ly = y'' + a(x) y' + b(x) y = f(x)$ % has a two-dimensional affine space % of solutions, assuming $a$, $b$, and $f$ are continuous or % piecewise continuous. For an IVP\/ $Ly=f$ % with two initial conditions, it follows that % there exists a unique solution. For a BVP with two boundary % conditions, however, there exists a unique solution if and only if % there is no nonzero solution of the homogeneous problem $Lv = 0$ with % homogeneous boundary conditions. Such a solution is % called an eigenfunction of $L$ with eigenvalue $0$. % More generally, if\/ $\lambda$ is a number and $v$ is a nonzero % solution of\/ $Lv=\lambda v$ with homogeneous boundary % conditions, then $v$ is an eigenfunction of\/ $L$ with % eigenvalue~$\lambda$. \vspace{2pt}}} % \end{displaymath} % %% % % \small\parindent=0pt\parskip=1pt % {\em \underline{Exercise $6.1$}. Generalized eigenfunctions and an approximate % delta function.} Consider the generalized eigenvalue problem % $y''(x) + \lambda F_\varepsilon^{}(x) y(x) = 0$ with $y(-1) = y(1) = 0$, % where $F_\varepsilon^{}(x)$ is $(2\kern .3pt\varepsilon)^{-1}$ for % $|x|<\varepsilon$ and $0$ otherwise. The first $k$ eigenvalues % $D(1,1),\dots,D(k,k)$ and % eigenfunctions $V\{1\},\dots, V\{k\}$ can be obtained by executing % {\tt [V,D] = eigs(L,M,k)} with % {\tt x = chebfun('x'),} % \verb|F = (abs(x)0$. (This is actually % a generalized eigenvalue problem, as in Exercise 6.1, % so you will need to define % two chebops and call e.g.\ \verb|eigs(L,M,8)|.) % Which is the first eigenfunction % of this problem that is not positive on$(0,5\kern .3pt]\kern .5pt?$% \par % {\em \underline{Exercise$6.5$}. Davies's complex harmonic oscillator.} % {\em (a)} Use Chebfun to compute the first four % eigenvalues of the problem %$-y''+i\kern.3pt x^2y = \lambda\kern .3pt y$on$(-\infty,\infty)$with %$u(\pm \infty) = 0$. For the computation it will suffice to replace %$\infty$by$8$. % {\em (\kern .7pt b)} The eigenvalues of the usual harmonic oscillator, % without the factor$i$, are % well known to be$1,3,5,7,\dots .$Use this fact and a suitable change % of variables to explain the result of part {\em (a)} analytically. % \par % {\em \underline{Exercise$6.6$}. Robin boundary condition.} % {\em (a)} Use Chebfun to compute the first six eigenvalues of % the problem$y'' = - \lambda\kern .3pt y$on$[\kern .3pt 0,\pi]$with %$y(0) = 0$and$y(\pi) = y'(\pi)\$. Plot the eigenfunctions. % {\em (\kern .7pt b)} One of the eigenfunctions is quite different from the % others. Explain this by working out the form of the eigenfunctions % analytically. %