# Solving an eigenvalue problem with a 4th-order linear differential operator

Function $$\phi_n$$ satisfies the differential equation of this form

$$\mathcal{L}_z\phi_n=-\nu \frac{\mathrm{d}^4 \phi_n}{\mathrm{d} z^4}-\frac{\mathrm{d}^2 \phi_n}{\mathrm{d} z^2}=\mu_n \phi_n, \quad n=1,2,...,\infty \quad (1)$$

where the parameter $$\nu>0$$ and $$z\in[-\pi,\pi]$$, $$\phi_n$$ denotes an eigenfunction and $$\mu_n$$ denotes an eigenvalue, subject to $$4$$ periodic boundary conditions

$$\frac{\mathrm{d}^j \phi_n}{\mathrm{d} z^j}(-\pi)=\frac{\mathrm{d}^j \phi_n}{\mathrm{d} z^j}(\pi), \quad j=0,1,2,3 \quad (2)$$

I can solve a simpler version of the eigenvalue problem, that is,

$$\frac{\mathrm{d}^2 \phi_n}{\mathrm{d} z^2}=\mu_n \phi_n, \quad n=1,2,...,\infty \quad (3)$$

subject to 2 boundary conditions

$$\frac{\mathrm{d}^j \phi_n}{\mathrm{d} z^j}(-\pi)=\frac{\mathrm{d}^j \phi_n}{\mathrm{d} z^j}(\pi), \quad j=0,1 \quad (4)$$

where the eigenvalue $$\mu_n$$ is a real constant.

To solve the eigenvalue problem (3) and (4), first, letting $$\mu_n=-\lambda_n^2$$, the characteristic equation of (3) is $$r^2+\lambda_n^2=0$$, whose roots are $$r=\pm \mathrm{i}\lambda_n$$. Then (3) has the general solution $$\phi_n(z)=a \sin\lambda_n z+b\cos\lambda_n z$$ with constant $$a$$ and $$b$$. By applying the boundary conditions (4), the pair of algebraic equations for $$a$$ and $$b$$ can be written as

$$\left[ \begin{array}{cc} \sin\lambda_n\pi&0\\ 0&\lambda_n\sin\lambda_n\pi \end{array} \right]\left[ \begin{array}{c} a\\ b \end{array} \right]=\left[ \begin{array}{c} 0\\ 0 \end{array} \right]$$

For there to be nontrivial solutions to this set of equations for $$a$$ and $$b$$, the determinant of the coefficient matrix must be zero, that is, $$\lambda_n\sin^2\lambda_n\pi=0$$, which determines the values of $$\lambda_n$$. It is $$\lambda_n=0$$ or $$\sin\lambda_n\pi=0$$. The roots of the latter one is $$\lambda_n=n$$, where $$n=0, \pm1, \pm2, ...$$. The resulting eigenfunctions are $$\phi_0=b$$ corresponding to the eigenvalue $$\mu_0=0$$ and $$\phi_n(z)=a \sin n z+b\cos n z$$ corresponding to the eigenvalue $$\mu_n=-n^2=-1,-4,-9,...$$, in which $$\mu_0=0$$ has multiplicity two.

Question: I have difficulty in solving the problem with a higher-order differential operator, say, 4th-order in my problem of Eqs.(1) and (2), in which the eigenvalue $$\mu_n$$ should be a complex number.

Can anybody give me some suggestions for solving the eigenvalue problem of Eqs.(1) and (2)? Thank you in advance!

The eigenvalues: $$\mu_0=0$$ and $$\mu_n=-\nu n^4+n^2$$ ($$\mu_n$$ has multiplicity two)

The eigenfunctions $$\psi_0=\frac{1}{\sqrt{2\pi}}$$, $$\phi_n(z)=\frac{1}{\sqrt{\pi}}\sin{(nz)}$$ and $$\psi_n(z)=\frac{1}{\sqrt{\pi}}\cos{(nz)}$$ with $$n=1,2,...,\infty$$.

The equation can be written as $$(\nu D^4+D^2+\mu)\phi=0$$ Its characteristic roots are $$\pm\alpha$$, $$\pm\beta$$, which can be written in terms of $$\nu$$ and $$\mu$$ (and can be complex). $$\alpha^2,\beta^2=\frac{-1\pm\sqrt{1-4\nu\mu}}{2\nu}$$ In order to satisfy the periodic boundary conditions, it is necessary that these roots are purely imaginary, $$\alpha=it$$, $$\beta=is$$. (The cases $$\alpha^2>0$$, $$\beta^2\le0$$ or $$\alpha^2,\beta^2>0$$ need to be eliminated by showing that they cannot satisfy the boundary conditions; this part is omitted here.) The solutions are therefore of the form $$\phi(z)=a\cos(t z)+b\sin(t z)+c\cos(s z)+d\sin(s z)$$ where $$a,b,c,d$$ are real constants. Substituting the four boundary conditions gives four equations in $$a,b,c,d$$. $$\begin{pmatrix}0&2\sin(t\pi)&0&2\sin(s\pi)\\ -2t\sin(t\pi)&0&-2s\sin(s\pi)&0\\ 0&-2t^2\sin(t\pi)&0&-2s^2\sin(s\pi)\\ 2t^3\sin(t\pi)&0&2s^3\sin(s\pi)&0 \end{pmatrix}\begin{pmatrix}a\\b\\c\\d\end{pmatrix}=0$$ For non-trivial solutions (eigenvector) the determinant must be zero, and this gives the condition for the eigenvalues, namely (after Gaussian reduction etc.) $$st\sin^2(s\pi)\sin^2(t\pi)(s^2-t^2)^2=0$$ Thus either $$t=0$$, $$s=0$$, $$s=t$$, $$s=-t$$, or $$t=n$$, or $$s=n$$,.
The first two imply $$\mu=0$$ with eigenfunction $$\phi=constant$$; the next non-integer cases reduce to these as well.
The non-trivial cases are for $$t=n$$ or $$s=n$$. Then $$\frac{1-\sqrt{1-4\mu\nu}}{2\nu}=-n^2\implies \mu=-\nu n^4+n^2$$
Edit: I just realised that the working can be simplified by noticing that the characteristic equation is equivalent to $$(D^2+\tfrac{1}{2\nu})^2=\tfrac{1-4\mu\nu}{4\nu^2}$$ so it is enough to solve the eigenvalue problem $$(D^2+\tfrac{1}{2\nu})\phi=\lambda\phi$$.
• Thank you. I need some time to investigate and then accept. The original DE should be equivalent to $(D^2+\frac{1}{2\nu})^2\phi=\frac{(1-4\mu\nu)\phi}{4\nu^2}$. But why it is enough to solve the eigproblem $(D^2+\frac{1}{2\nu})\phi=\lambda \phi$? Sep 5, 2020 at 7:13
• @user55777 The same eigenfunctions apply, giving $\lambda^2$. Sep 5, 2020 at 9:31
• Thanks! I understand $t=0$ gives $\alpha=0$ so $-1+\sqrt{1-4\mu\nu}=0$, which gives $\mu=0$, but not understand why $s=0$ gives $\mu=0$? Test: if $s=0$ then $\beta=0$ so $-1-\sqrt{1-4\mu\nu}=0$ which cannot give $\mu=0$. Sep 5, 2020 at 11:01
• @user55777 What I meant was that $s=0$ implies, from the matrix, that $a=0=b$ and so $\phi=$constant, thus $\mu=0$. But I should word it better; there are no solutions with $s=0$. Sep 5, 2020 at 13:15