Given the biharmonic heat equation on the unit disk (of radius $1$):
$$ u_t = -\nabla^4 u, \qquad t\geq0,$$
together with boundary conditions zero Dirichlet and Neumann boundary conditions at $r=1$. See http://mathworld.wolfram.com/BiharmonicEquation.html
Does anyone know a list of analytical solutions (that are not trivial) to this equation for testing purposes? I am happy for $u(\theta,r,0)$ to be any smooth function.
I suspect that separation of variables will derive solutions, but I wonder if there are already a set of analytical solutions on the internet.
Thank you in advance.
Let $u(\theta, r, t) = T(t) R(r) \Phi(\theta)$ then using this expression of the $\Delta^2$ in polar coodniates and separating the variables gives $$ \frac{T'}{T} + \frac{\frac{1}{r}\left( r\left( \frac{1}{r}\left( rR' \right)' \right)' \right)'}{R} + \frac{2}{r^2} \frac{\Phi''}{\Phi}\frac{R''}{R} + \frac{1}{r^4} \frac{\Phi''''}{\Phi} - \frac{2}{r^3} \frac{\Phi''}{\Phi} \frac{R'}{R} + \frac{4}{r^4} \frac{\Phi''}{\Phi} = 0. $$ Taking $\Phi = \sin m\theta,\cos m\theta$ gives $\frac{\Phi''}{\Phi} = -m^2$, so $$ \frac{T'}{T} + \frac{\frac{1}{r}\left( r\left( \frac{1}{r}\left( rR' \right)' \right)' \right)'}{R} - \frac{2m^2}{r^2}\frac{R''}{R} + \frac{m^4}{r^4} + \frac{2m^2}{r^3} \frac{R'}{R} - \frac{4m^2}{r^4} = 0. $$ Let $$ \lambda^4 = -\frac{T'}{T} = \frac{\frac{1}{r}\left( r\left( \frac{1}{r}\left( rR' \right)' \right)' \right)'}{R} - \frac{2m^2}{r^2}\frac{R''}{R} + \frac{m^4}{r^4} + \frac{2m^2}{r^3} \frac{R'}{R} - \frac{4m^2}{r^4} $$ so we need to solve the following boundary problem $$ \frac{1}{r}\left( r\left( \frac{1}{r}\left( rR' \right)' \right)' \right)' - \frac{2m^2}{r^2}R'' + \frac{2m^2}{r^3} R' + \frac{m^4-4m^2-\lambda^4 r^4}{r^4}R = 0,\\ |R(0)| < \infty,\quad |R'(0)| < \infty,\quad R(1) = 0,\quad R'(1) = 0. $$ Plugging $R(r) = Z(\lambda r), \rho = \lambda r$ gives the same equation, but without $\lambda$ $$ \frac{1}{\rho}\left( \rho\left( \frac{1}{\rho}\left( \rho Z' \right)' \right)' \right)' - \frac{2m^2}{\rho^2}Z'' + \frac{2m^2}{\rho^3} Z' + \frac{m^4-4m^2-\rho^4}{\rho^4}Z = 0,\\ |Z(0)| < \infty,\quad |Z'(0)| < \infty,\quad Z(\lambda) = 0,\quad Z'(\lambda) = 0 $$ Wolfram Mathematica gives $$ Z(\rho) = C_1 I_{-m}(\rho) + C_2 J_{-m}(\rho) + C_3 I_m(\rho) + C_4 J_m(\rho). $$ Assuming that $J_{-m}, I_{-m}$ are the unlimited at $\rho = 0$ solutions to the Bessel equation and to the modified Bessel equation, the only possibility we have is $$ Z(\rho) = C_3 I_m(\rho) + C_4 J_m(\rho) = C \left[\sin \phi\, I_m(\rho) + \cos \phi\, J_m(\rho)\right]. $$ Now we have to find such pair $\phi, \lambda$ so $$ \sin \phi I_m(\lambda) + \cos \phi J_m(\lambda) = 0\\ \sin \phi I_m'(\lambda) + \cos \phi J_m'(\lambda) = 0\\ $$ or, in other words $$ W(\lambda) = \operatorname{det} \begin{pmatrix} I_m(\lambda) & J_m(\lambda)\\ I_m'(\lambda) & J_m'(\lambda) \end{pmatrix} = 0. $$
For example, taking $m = 1$ and solving $W(\lambda) = 0$ numerically gives $$ \lambda_1 \approx 4.6108998790490558273\\ \lambda_2 \approx 7.7992738008112319023\\ \lambda_3 \approx 10.958067191919497803\\ \vdots $$ Let's take $\lambda = \lambda_1 \approx 4.6108998790490558273$. The corresponding solution to the $$ \sin \phi I_m(\lambda) + \cos \phi J_m(\lambda) = 0\\ \sin \phi I_m'(\lambda) + \cos \phi J_m'(\lambda) = 0\\ $$ is $$ \phi = 0.0152149896210699406. $$ Note that $I_m(z)$ tends to infinity very quickly, thus $\phi \ll 1$.
Then the solutions are $$ u(r, \theta, t) = e^{-\lambda^4 t} {\cos m \theta \choose \sin m \theta} \left( \cos \phi J_m(\lambda r) + \sin \phi I_m(\lambda r) \right) $$
The simpler derivation of the $Z(\rho)$ is given by following: let $\Delta^2 u = \lambda^4 u$. Then it is equivalent to the system $$ \Delta v = \lambda^2 u\\ \Delta u = \lambda^2 v. $$ Adding and subtracting gives $$ \Delta (u+v) = \lambda^2 (u + v)\\ \Delta (u-v) = -\lambda^2 (u - v). $$ Thus $u + v$ may be expressed in terms of $(C_1 J_n(\rho) + C_2 Y_n(\rho))e^{in\theta}$ and $u - v$ may be expressed in terms of $(C_3 I_n(\rho) + C_4 K_n(\rho))e^{in\theta}$. Throwing away $K_n$ and $Y_n$ gives the same expression as above.