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!

Here is the answer for your reference:

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 $(D^2+\tfrac{1}{2\nu})\phi=\lambda\phi$.

  • $\begingroup$ 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 $(D^2+\frac{1}{2\nu})\phi=\lambda \phi$? $\endgroup$ – user55777 Sep 5 '20 at 7:13
  • $\begingroup$ @user55777 The same eigenfunctions apply, giving $\lambda^2$. $\endgroup$ – Chrystomath Sep 5 '20 at 9:31
  • $\begingroup$ 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 $\mu=0$. $\endgroup$ – user55777 Sep 5 '20 at 11:01
  • $\begingroup$ @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 $s=0$. $\endgroup$ – Chrystomath Sep 5 '20 at 13:15

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