# The Biharmonic Eigenvalue Problem on a Rectangle with Dirichlet Boundary Conditions

I am interested in solving the following biharmonic eigenvalue problem.

$$\begin{array}{cccc} & \Delta ^2 \Psi (x,y) = \lambda \Psi (x,y), & - a \le x \le a & - b \le y \le b \\ & x = \phantom{-}a & \Psi = 0 & \dfrac{\partial \Psi }{\partial x} = 0 \\ & x = - a & \Psi = 0 & \dfrac{\partial \Psi }{\partial x} = 0 \\ & y = \phantom{-}b & \Psi = 0 & \dfrac{\partial \Psi }{ \partial y} = 0 \\ & y = - b & \Psi = 0 & \dfrac{\partial \Psi }{ \partial y} = 0 \end{array}$$

where

$$\Delta ^2 \Psi = \frac{\partial ^4 \Psi }{\partial x^4} + 2 \frac{\partial^4 \Psi }{\partial x^2 \partial y^2} + \frac{\partial ^4 \Psi }{\partial y^4}$$

$$\Psi \in {{\bf{C}}^{\infty}}\left( {[ - a,a] \times [ - b,b]} \right)$$

To describe the problem in words, we are looking for the eigenfunctions of the biharmonic operator over a rectangular domain where all its derivatives are continuous. The boundary conditions are of Dirichlet type, i.e., the function and it's normal derivative are prescribed over the boundary of the rectangular domain.

Facts and Motivations

1) This problem occurs in many physical areas. One of the most famous ones is the vibration of a rectangular isotropic elastic clamp plate.

2) It is believed between the engineers that the problem doesn't have a closed form solution. It may be asked that even the problem has a solution or not. Numerical evidence shows that such a solution may exists. However, I am looking for some strong theoretical basis to prove the existence of the solution so I planned to ask this question in a society of mathematicians.

Questions

1) Is there any non-zero solution for this problem? In other words, I am asking an existence or non-existence theorem for this problem.

This question is completely answered by TKS. According to TKS, it is an old result firstly proved by K. Friedrichs. Maybe the reason that many people are unaware of this is that the paper by K. Friedrichs is written in German entitled as

Die Randwert- und Eigenwertprobleme aus der Theorie der elastischen Platten. (Anwendung der direkten Methoden der Variationsrechnung)

The translation in English is

The boundary value and eigenvalue problems in the theory of elastic plates. (Application of direct methods of variational calculus)

Another short answer to this question is given by Jean Duchon on Math Over Flow.

2) Assuming the existence, how can one compute these eigenvalues and eigenfunctions? Is there a closed form solution for this purpose?

This question remained unanswered!

• Even if it is not a homework question, please include what you have tried in your question. Thank you. Sep 23, 2015 at 18:08
• @HoseinRahnama I'm think this question is quite interesting, and since you can't afford a bounty, I added one for you. (This means that your question will show up for 7 days in the featured tab, and that the person with the best answer will get a +50 rep bonus) Sep 26, 2015 at 6:49
• I will award half the bounty to his answer and then start a new one. Oct 3, 2015 at 10:48
• @H.R. There appears to be a problem with letting the system half-awarding the bounty, so I decided to give the whole bounty to Michael's answer. I will now start a new bounty. Oct 4, 2015 at 7:42
• For $\lambda=\mu^{2}\ne 0$, your problem is equivalent to finding solutions of $$\nabla^{2}\Psi_{+}=\mu \Psi_{+} \\ \nabla^{2}\Psi_{-}=-\mu \Psi_{-}$$ that have matching function values and normal derivative data on the boundary of the your rectangular region. The difference $\Psi=\Psi_{+}-\Psi_{-}$ is a solution of $(\nabla^{2})^{2}\Psi=\mu^{2}\Psi$ with the required conditions on the boundary. Conversely, you can show that a solution of your equation gives such solutions $\Psi_{\pm}$. Trying to match function values & normal derivatives seems tough, but necessary. Oct 9, 2015 at 8:46

You will find what you are looking for in Chapter 3.1 of Gazzola, F., Grunau, H.-Ch., Sweers, G.: Polyharmonic Boundary Value Problems. Lecture Notes 1991. Springer, Berlin (2010).

EDIT:

• To be more specific, you are looking for Theorem 3.8 at page 69. The original proof goes back to Friedrichs, K. Die Randwert- und Eigenwertprobleme aus der Theorie der elastischen Platten. (Anwendung der direkten Methoden der Variationsrechnung). Math. Ann. 98, 205–247 (1927) (p. 233-240). Theorem 3.8 states precisely: Let $\Omega=(0,1)\times (0,1)$ denote the unit square. Then there exist $\lambda_1>0,u_1\not\equiv 0$ that solve \begin{aligned} {}&\Delta^2 u =\lambda\,u {}& \text{ in }\Omega\\ {}&u=|\nabla u|=0 {}& \text{ on } \partial \Omega. \end{aligned} (the problem has an eigenfunction). Moreover, for $\lambda<\lambda_1$ the above problem has no non-zero solution $u$ (it is a first eigenfunction). Finally, $u_1(x,y)$ changes sign as $(x,y)$ varies in $\Omega$.

• As for computing the solution, I would not be very optimistic: there is no known Green's function for the clamped plate, except in the case of a ball or a half-space (see Chapter 2.6, p. 47-49, same book as above). Computing an explicit solution is far stronger than finding a Green's function, that is, a solution in closed form, that is, you would achieve a better result than what people have been trying for more than 100 years.

• Using an abstract result (Theorem 7.22 in Folland's Introduction to partial differential equations). The abstract theorem states: Let $\Omega\subset \mathbb R^2$ be an open bounded and connected domain. Let $X$ be a closed subspace of $H^m(\Omega)$ containing $H^m_0(\Omega)$ and $D$ be a self-adjoint and coercive (there exist $C>0$ and $\mu\geq0$ such that $\mathrm{Re}\,D(u,u)\geq C\|u\|_{H^m}^2-\mu\|u\|_{L^2}^2$) Dirichlet form defined in $X$. Then there exists an orthonormal basis $u_j$ of $L^2(\Omega)$ consisting of eigenfunctions of $D$ on $X$, that is, for each $j$ we have $u_j\in X$ and there is a real constant $\lambda_j$ such that $D(u_j,v)=\lambda_j(u_j,v)_{L^2}$ for all $v\in X$. Moreover $\lambda_j>\mu$ for all $j$, $\lim_{j\rightarrow\infty}\lambda_j=+\infty$. In this case define the bilinear form $D:H_0^2(\Omega)\times H_0^2(\Omega)\rightarrow \mathbb R$, by $D(u,v)=\int_\Omega \Delta u\,\Delta v\;dxdy$. First note the the $L^2$ norm of the Laplacian is an equivalent norm in $H_0^2$, which makes $D$ continuous: $|D(u,v)|\leq \|\Delta u\|_{L^2} \|\Delta v\|_{L^2}$ (use Cauchy-Schwartz inequality) and coercive: $|D(u,u)|= \|\Delta u\|_{L^2}^2$. Apply the referenced theorem to obtain a sequence of eigenpairs $\lambda_j,u_j$ satisfying $$\int_\Omega \Delta u_j\,\Delta v\;dxdy=\lambda_j\int_\Omega u_j\,v\;dxdy,$$ for all $v\in H_0^2(\Omega)$. Regularity theory then gives you $u_j\in H^4(\Omega')$ for all $\Omega'\subset\subset \Omega$ (away from the boundary) so, taking $v\in C_0^\infty(\Omega')$ and integrating by parts (use the so-called second Green's identity) you get $\Delta ^2u_j=\lambda_j u_j$ in $\Omega'$.

• You can get more regularity by applying a trick for the square: reflecting at all sides (like unfolding a folded piece of paper) and using regularity results for the interior of the unfolded domain you get that $u_j\in H^4(square)$. To get $C^4$ smoothness you then need to bootstrap: $u_j\in H^4$ implies $\Delta^2u_j\in H^4$ and doing the whole thing again you get $u_j\in H^8$. Iterating the process you get $u_j\in C^\infty$. But this holds only for the square or for smooth domains!! (at least with $C^4$ boundary).

Short answers: no, there isn't a separable solution; yes, there are eigenfunctions of the biharmonic.

Unfortunately, the biharmonic isn't separable like the Laplacian. Off the top of my head, I don't know of a nice closed-form solution for the eigenfunctions of the biharmonic. The image below is an approximation of the eigenfunction for the smallest eigenvalue. I generated it using P3 Hermite finite elements (ignoring the extra data from the gradient). You can see the data here.

Yes, of course the biharmonic has eigenfunctions. A paper by Pereira and Pereira shows results for general domains in $R^n$ for $n\geq 2$. It's positive definite. The only reference I found in Evans' PDE book was in an exercise; dun't know what else to suggest.

Consider $$(\Delta^2-\mu^4)u=(\Delta-\mu^2)(\Delta+\mu^2)u=0$$ coupled with $u=0$ and $\frac{\partial u}{\partial {n}}=0$ at the rectangular boundary $\Gamma$. $\frac{\partial }{\partial {n}}$ denotes the normal derivatives. Assume also $\mu\ne 0$.

Inspired by the comment of TrialAndError, one can look instead for $$\begin{cases} (\Delta+μ^2)v = 0 & v = 0 \text{ on } \Gamma \\ (\Delta-\mu^2)u = v & \frac{\partial u}{\partial {n}}=0 \text{ on } \Gamma \end{cases}$$ Note that the boundary conditions here aren't an unique choice, however the following discussion won't change.

Note that $$(\Delta+μ^2)v = 0$$ is Helmholtz equation, which is in some domains may have resonance, i.e. non unique solution. This hints that the original problem may have non unique solution, which would mean that there is a non zero solution to the original problem. But again, this is depends on the domain.

For example in one dimensional case, $u''''=\lambda u$ in $[0,1]$ coupled with $u=0$, $u'=0$ at the boundary has a solution $$u = (\cosh(k)− \cos(k))(\sin(k x)−\sinh(k x))−(\cosh(kx)− \cos(kx))(\sin(k)−\sinh(k))$$ where $k^4=\lambda$

• It's really a nice Idea. Voted up. Just as a correction you can write your system in the most general form $\left\{ \matrix{ \left( {\Delta + {\mu ^2}} \right)v = 0\,\,\,\,\,\,\,\,\,\,\,\,v = \Delta u\,\,\,on\,\,\Gamma \hfill \cr \left( {\Delta - {\mu ^2}} \right)u = v\,\,\,\,\,\,\,\,\,\,\,{{\partial u} \over {\partial n}} = 0\,\,\,on\,\,\,\Gamma \hfill \cr} \right.$. But it seems we are again at nowhere since we don't know what is $\Delta u\,$ on $\Gamma$. So we can use a trial and error scheme to go through. Oct 14, 2015 at 9:47
• sorry, but $v\ne\Delta u$ unless $\mu=0$ Oct 14, 2015 at 9:51
• you forgot to use the condition that $u = 0\,\,on\,\,\Gamma$! :) Oct 14, 2015 at 9:52
• you cannot require $u=0$ and $\frac{\partial u}{\partial n} =0$ together on all $\Gamma$. not in this system Oct 14, 2015 at 9:56
• Sorry, I didn't get you. It's the boundary conditions of our biharmonic eigenvalue problem! but you just tried to decouple the problem into two Laplace and Poisson BVPs with proper boundary conditions and the one I wrote is the most general form for decoupling. :) Oct 14, 2015 at 9:59

The work done earlier as below is similar to what you are attempting (except may be the coupling of different orders). Features are non-orthogonal eigen-functions $$F_k(y)$$satisfying boundary conditions of Biharmonic Equation, supplying characteristic equation yielding eigen values $$\rho_k$$ developed in $$k$$ Fadle- Popkovich expansion procedure using auxiliary clamped beam functions $$Y_m$$ to successfully orthogonalize and evaluate all coefficients including $$Y_m$$ numerically.

$$\sum A_k a_{km} = C_m ;\qquad \sum A_k a_{km} \rho_k = D_m ;$$

The satisfaction of zero deflection at corners is found accurate to the order $$10^{-6}$$. Coefficient $$a_{km}$$ evaluated is useful for your work too hopefully. Full details are in the following paper Elsevier published original research.

Orthotropic plate Eigen function Expansions

• If I succeed sure shall let you know.. Oct 22, 2015 at 19:19
• At present I have something on hand. I could however attend to some portions . Please send me a copy for a re-look.The availability of evaluated $a_{km}$ between eigenfunction(k) and clamped beam orthogonalization (m) yielded like $E-10$ numerical accuracy. Oct 23, 2015 at 13:45
• I have no strategy for closed form. The above work is fiercely numerical. However finding $a_{km}$ tanamounted to $almost$ closed form. Attempts to integrate / couple. the two is ambitious it appears. Later I shall attempt to send a reference in Timoshenko plates and shells text book where Von Karman formulated stess function and plate deformation together, when deriving compatibility of deformations and stress resultants. The second source I can may be suggest is the classical Elasticity book by AEH Love. Oct 24, 2015 at 14:37
• OK, I shall try. The other author was my guide, not around now otherwise I would provide the contact Oct 24, 2015 at 14:44
• Shall chat tomorrow. Oct 25, 2015 at 20:21