4.2. Eigenvalue problem

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4.2. Eigenvalue problems

Problems with explicit solutions

Example 1. (Dirichlet-Dirichlet; from Section 4.1). Consider eigenvalue problem \begin{align} &X'' +\lambda X=0 && 0< x< l,\label{eq-4.2.1}\\[3pt] \label{eq-4.2.2}& X(0)=X(l)=0 \end{align} as eigenvalues and corresponding eigenfunctions \begin{align} & \lambda_n=\bigl(\frac{\pi n}{l}\bigr)^2,&&n=1,2,\ldots \label{eq-4.2.3}\\[3pt] & X_n=\sin \bigl(\frac{\pi n}{l}x\bigr). \label{eq-4.2.4} \end{align}

Visual examples (animation)

Example 2. (Neumann-Neumann). Eigenvalue problem \begin{align} & X'' +\lambda X=0 && 0< x< l,\label{eq-4.2.5}\\[3pt] & X'(0)=X'(l)=0\label{eq-4.2.6} \end{align} has eigenvalues and corresponding eigenfunctions \begin{align} & \lambda_n=\bigl(\frac{\pi n}{l}\bigr)^2,&&n=0,1,2,\ldots \label{eq-4.2.7}\\[3pt] & X_n=\cos \bigl(\frac{\pi n}{l}x\bigr). \label{eq-4.2.8} \end{align}

Visual examples (animation)

Indeed, plugging $X=Ae^{kx}+ Be^{-kx}$, $k=\sqrt{-\lambda}\ne 0$ into (\ref{eq-4.2.6}) $X'(0)=X'(l)=0$ we get \begin{align*} &A \qquad -B\qquad\ = 0,\\ &Ae^{\sqrt{-\lambda}l} - B e^{-\sqrt{-\lambda}l}=0 \end{align*} where we divided the last equation by $\sqrt{-\lambda}$, which leads to the same condition $2\sqrt{-\lambda}l=2\pi in$, $n=1,2,\ldots$ as before but with eigenfunctions (\ref{eq-4.2.8}).

But now, plugging $\lambda=0$ and $X=A+Bx$ we get $B=0$ and $X(x)=1$ is also an eigenfunction and we should add $n=0$.

Example 3. (Dirichlet-Neumann). Consider eigenvalue problem \begin{align} & X'' +\lambda X=0 && 0< x< l,\label{eq-4.2.9}\\[3pt] & X(0)=X'(l)=0\label{eq-4.2.10} \end{align} has eigenvalues and corresponding eigenfunctions \begin{align} & \lambda_n=\bigl(\frac{\pi (2n+1)}{2l}\bigr)^2,&&n=0,1,2,\ldots \label{eq-4.2.11}\\[3pt] & X_n=\sin \bigl(\frac{\pi (2n+1)}{2l}x\bigr).\label{eq-4.2.12} \end{align}

Visual examples (animation)

Indeed, plugging $X=Ae^{kx}+ Be^{-kx}$, $k=\sqrt{-\lambda}\ne 0$ into (\ref{eq-4.2.6}) $X'(0)=X'(l)=0$ we get \begin{align*} &A \qquad +B\qquad\ = 0,\\ &Ae^{\sqrt{-\lambda}l} - B e^{-\sqrt{-\lambda}l}=0 \end{align*} where we divided the last equation by $\sqrt{-\lambda}$, which leads to condition $2\sqrt{-\lambda}l=(2\pi n+1) i$, $n=0,1,2,\ldots$ and with eigenfunctions (\ref{eq-4.2.12}).

Plugging $\lambda=0$ and $X=A+Bx$ we get $B=0$ and $A=0$, so $\lambda=0$ is not an eigenvalue.

The same problem albeit with the ends reversed (i.e. $X'(0)=X(l)=0$) has the same eigenvalues and eigenfunctions $\cos \bigl(\frac{\pi (2n+1)}{2l}x\bigr)$.

Example 4. (periodic). Consider eigenvalue problem \begin{align} & X'' +\lambda X=0 && 0< x< l,\label{eq-4.2.13}\\[3pt] & X(0)=X(l), \quad X'(0)=X'(l)\label{eq-4.2.14} \end{align} has eigenvalues and corresponding eigenfunctions \begin{align} & \lambda_0=0,\label{eq-4.2.15}\\[3pt] & X_0=1,\label{eq-4.2.16}\\ & \lambda_{2n-1}=\lambda_{2n}=\bigl(\frac{2\pi n}{l}\bigr)^2,&&n=1,2,\ldots \label{eq-4.2.17}\\[3pt] & X_{2n-1}=\cos \bigl(\frac{2\pi n}{l}x\bigr), && X_{2n}=\sin \bigl(\frac{2\pi n}{l}x\bigr).\label{eq-4.2.18} \end{align}

Visual examples (animation)

Alternatively, as all eigenvalues but $0$ have multiplicity $2$ one can select \begin{align} & \lambda_n=\bigl(\frac{2\pi n}{l}\bigr)^2,&&n=\ldots, -2,-1,0, 1,2,\ldots \label{eq-4.2.19}\\[3pt] & X_{n}=\exp \bigl(\frac{2\pi n}{l}i x\bigr).\label{eq-4.2.20} \end{align}

Indeed, now we get \begin{align*} &A(1-e^{\sqrt{-\lambda} l}) +B(1-e^{-\sqrt{-\lambda} l})= 0,\\ &A(1-e^{\sqrt{-\lambda} l}) - B (1-e^{-\sqrt{-\lambda} l})=0 \end{align*} which means that $e^{\sqrt{-\lambda} l}=1\iff \sqrt{-\lambda}=2\pi ni$ and in this case we get two linearly independent eigenfunctions.

As $\lambda =0$ we get just one $X_0=1$.

Example 5. (quasiperiodic). Consider eigenvalue problem \begin{align} & X'' +\lambda X=0 && 0< x< l,\label{eq-4.2.21}\\[3pt] & X(0)=e^{-ikl}X(l), \quad X'(0)=X'(l)e^{-ikl}X(l)\label{eq-4.2.22} \end{align} with $0< k<\frac{2\pi}{l}$ has eigenvalues and corresponding eigenfunctions \begin{align} & \lambda_{n}=\bigl(\frac{2\pi n}{l}+k \bigr)^2,&& n=0,2,4,\ldots \label{eq-4.2.23}\\[3pt] & X_{n}=\exp \bigl(\bigl[\frac{2\pi n}{l}+k\bigr]i x\bigr), \label{eq-4.2.24}\\[3pt] & \lambda_{n}=\bigl(\frac{2\pi (n+1)}{l}-k \bigr)^2,&& n=1,3,5,\ldots \label{eq-4.2.25}\\[3pt] & X_{n}=\exp \bigl(\bigl[\frac{2\pi (n+1)}{l}-k\bigr]i x\bigr). \label{eq-4.2.26} \end{align} $k$ is called quasimomentum.

1. For $k=0, \frac{2\pi}{l}$ we get periodic solutions, considered in the previous Example 4,
2. for $k=\frac{\pi}{l}$ we get antiperiodic solutions

(Visual examples (animation))

where all eigenvalues have multiplicity $2$ (both $l$-periodic and $l$-antiperiodic functions are $2l$-periodic, and each $2l$-periodic is the sum of $l$-periodic and $l$-antiperiodic functions).
3. For all other $k$ eigenvalues are real and simple but eigenfunctions are not real-valued.

Indeed, we get now \begin{align*} &A(e^{ikl}-e^{\sqrt{-\lambda} l}) +B( -e^{\sqrt{-\lambda} l})= 0,\\ &A(e^{ikl}-e^{\sqrt{-\lambda} l}) - B (e^{ikl}-e^{-\sqrt{-\lambda} l})=0 \end{align*} and we get either $\sqrt{-\lambda} = ik + \pi m i/l$ with $m\in \mathbb{Z}$.

Remark 1. This is the simplest example of problems appearing in the description of free electrons in the crystals; much more complicated and realistic example would be SchrÃ¶dinger equation \begin{equation*} X'' +\bigl(\lambda-V(x)\bigr)X=0 \end{equation*} or its $3D$-analog.

Problems with "almost" explicit solutions

Example 6. (Robin boundary conditions). Consider eigenevalue problem \begin{align} & X'' +\lambda X=0 && 0< x< l,\label{eq-4.2.27}\\[3pt] & X'(0)=\alpha X(0), \quad X'(l)=-\beta X(l)\label{eq-4.2.28} \end{align} with $\alpha\ge 0$, $\beta\ge 0$ ($\alpha+\beta>0$). Then \begin{align} \lambda \int_0^l X^2\,dx=&-\int_0^l X''X\,dx\notag \\ =&\int_0^l X'^2\,dx-X'(l)X(l) +X'(0)X(0)\notag\\ =&\int_0^l X'^2\,dx+\beta X(l)^2 +\alpha X(0)^2\qquad \label{eq-4.2.29} \end{align} and $\lambda_n=\omega_n^2$ where $\omega_n>0$ are roots of \begin{align} & \tan (\omega l)= \frac{(\alpha+\beta)\omega}{\omega^2-\alpha\beta}; \label{eq-4.2.30}\\ & X_n= \omega_n \cos (\omega_n x) +\alpha \sin (\omega_n x);\label{eq-4.2.31} \end{align} ($n=1,2,\ldots$).

Indeed, looking for $X=A\cos(\omega x) + B\sin(\omega x)$ with $\omega=\sqrt{\lambda }$ (we are smarter now!) we find from the first equation of (\ref{eq-4.2.28}) that $\omega B= \alpha A$ and we can take $A=\omega$, $B= \alpha$, $X= \omega \cos(\omega x) -\alpha\sin( \omega x)$.

Then plugging to the second equation we get \begin{gather*} -\omega ^2\sin(\omega l)+\alpha \omega \cos(\omega l) =-\beta \omega \cos(\omega l) - \alpha \beta \sin(\omega l) \end{gather*} which leads us to (\ref{eq-4.2.30})--(\ref{eq-4.2.31}).

Observe that

1. $\alpha,\beta\to 0^+\implies \omega_n \to \frac{\pi (n-1)}{l}$.
2. $\alpha,\beta\to +\infty \implies \omega_n \to \frac{\pi n}{l}$.
3. $\alpha\to 0^+,\beta\to +\infty\implies \omega_n \to \frac{\pi (n-\frac{1}{2})}{l}$.

Example 7. (Robin boundary conditions (negative eigenvalues)). However if $\alpha$ and/or $\beta$ are negative, one or two negative eigenvalues $\lambda_n=-\gamma_n^2$ can also appear where \begin{align} & \tanh (\gamma_n l )= {-\frac{(\alpha + \beta)\gamma_n }{\gamma ^2 + \alpha\beta}}, \label{eq-4.2.32}\\ & X_n(x) = \gamma \cosh (\gamma_n x) + \alpha \sinh (\gamma_n x). \label{eq-4.2.33} \end{align}

Indeed, looking for $X=A\cosh(\gamma x) + B\sinh(\gamma x)$ with $\gamma=\sqrt{-\lambda }$ we find from the first equation of (\ref{eq-4.2.28}) that $\gamma B= \alpha A$ and we can take $A=\gamma$, $B= \alpha$, $X= \gamma \cosh(\gamma x) -\alpha \sinh( \gamma x)$.

Then plugging to the second equation we get \begin{gather*} \gamma ^2\sinh(\gamma l)+\alpha \gamma \cosh(\gamma l) =-\beta \gamma \cosh(\gamma l) - \alpha \beta \sinh(\gamma l) \end{gather*} which leads us to (\ref{eq-4.2.32})--(\ref{eq-4.2.33}).

To investigate when it happens, consider the threshold case of eigenvalue $\lambda=0$: then $X=cx+d$ and plugging into boundary conditions we have $c=\alpha d$ and $c=-\beta (d+lc)$; this system has non-trivial solution $(c,d)\ne 0$ iff $\alpha+\beta+\alpha\beta l =0$. This hyperbola divides $(\alpha,\beta)$-plane into three zones:

To calculate the number of negative eigenvalues one can either apply the general variational principle or analyze the case of $\alpha=\beta$; for both approaches see Appendix 4.A.

Example 8. (Oscillations of the rod (beam)) Small oscillations of the rod are described by

$$u_{tt} + Ku_{xxxx}=0 \label{eq-4.2.34}$$ with the boundary conditions \begin{align} &u|_{x=0}=u_x|_{x=0}=0\label{eq-4.2.35},\\ &u|_{x=l}=u_x|_{x=l}=0\label{eq-4.2.36} \end{align} (both ends are clamped) or \begin{align} &u|_{x=0}=u_{x}|_{x=0}=0\label{eq-4.2.37},\\ &u_{xx}|_{x=l}=u_{xxx}|_{x=l}=0\label{eq-4.2.38} \end{align} (left end is clamped and the right one is free); one can consider different and more general boundary conditions.

Separating variables we get \begin{align} &X^{IV}-\lambda X=0, \label{eq-4.2.39}\\ &T'' +\omega^2 T=0, && \omega=\sqrt[4]{K\lambda} \label{eq-4.2.40} \end{align} with the boundary conditions \begin{align} &X(0)=X'(0)=0\label{eq-4.2.41},\\ &X(l)\,=X'(l)\,=0\label{eq-4.2.42} \end{align} or \begin{align} &X(0)\, =X'(0)\,=0\tag{41},\\ &X''(l)=X'''(l)=0.\label{eq-4.2.43} \end{align} Eigenvalues of the both problems are positive (we skip the proof). Then \begin{gather} X(x)= A\cosh(kx) + B\sinh (kx) +C\cos(kx) +D\sin(kx). \label{eq-4.2.44} \end{gather} and (\ref{eq-4.2.41}) implies that $C=-A$, $D=-B$ and $$X(x)= A\bigl(\cosh(kx) -\cos(k x)\bigr)+ B\bigl(\sinh (kx) -\sin(kx)\bigr). \label{eq-4.2.45}$$ Plugging into (\ref{eq-4.2.43}) we get \begin{align*} &A(\cosh(kl) -\cos(kl))+ B(\sinh (kl) -\sin(kl))=0,\\ &A(\sinh(kl)\ +\sin (kl))+ B(\cosh (kl) -\cos(kl))=0. \end{align*} Then determinant must be $0$: \begin{gather*} (\cosh(kl) -\cos(kl))^2 -(\sinh^2 (kl) -\sin^2(kl))=0 \end{gather*} which is equivalent to \begin{gather} \cosh(kl)\cdot\cos(kl)=1. \label{eq-4.2.46} \end{gather}

On the other hand, plugging \begin{equation*} X(x)= A\bigl(\cosh(kx) -\cos(k x)\bigr)+ B\bigl(\sinh (kx) -\sin(kx)\bigr). \tag{45} \end{equation*} into (\ref{eq-4.2.43}) $X''(l)=X'''(l)=0$ leads us to \begin{align*} &A(\cosh(kl) +\cos(kl))+ B(\sinh (kl) +\sin(kl))=0,\\ &A(\sinh(kl)\ -\sin (kl))+ B(\cosh (kl) + \cos(kl))=0. \end{align*} Then determinant must be $0$: \begin{gather*} (\cosh(kl) +\cos(kl))^2 -(\sinh^2 (kl) -\sin^2(kl))=0 \end{gather*} which is equivalent to \begin{gather} \cosh(kl)\cdot\cos(kl)=-1. \label{eq-4.2.47} \end{gather}

We solve (\ref{eq-4.2.46}) and (\ref{eq-4.2.47}) graphically:

Case of both ends free, results in the same eigenvalues $\lambda_n =k_n^4$ as when both ends are clumped, but with eigenfunctions \begin{equation*} X(x)= A\bigl(\cosh(kx) +\cos(k x)\bigr)+ B\bigl(\sinh (kx) +\sin(kx)\bigr). \end{equation*} and also in double eigenvalue $\lambda=0$ and eigenfunctions $1$ and $X$.
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