Helmholtz and Biharmonic equation examples with exact solution I'm looking for examples of Helmholtz and Biharmonic equations in Cartesian co-ordinates with exact solutions, in order to compare results of my numerical solutions with it.
I was able to find quite a few examples on the internet, where the problem with boundary conditions were precisely defined. Those were, unfortunately only illustrative examples and exact solutions were not shown.
Any particular example, or a useful link to a web-page or a paper is appreciated.
 A: For $u_{xx}+u_{yy}=m^2u$ ,
Let $p=mx$ ,
Then $m^2u_{pp}+u_{yy}=m^2u$
Let $q=my$ ,
Then $u_{qq}=u-u_{pp}$
Similar to Deducing the existence of a PDE by constructing it inductively via its Taylor series expansion,
Consider $u(p,a)=f(p)$ and $u_q(p,a)=g(p)$ ,
Let $u(p,q)=\sum\limits_{n=0}^\infty\dfrac{(q-a)^n}{n!}\dfrac{\partial^nu(p,a)}{\partial q^n}$ ,
Then $u(p,q)=\sum\limits_{n=0}^\infty\dfrac{(q-a)^{2n}}{(2n)!}\dfrac{\partial^{2n}u(p,a)}{\partial q^{2n}}+\sum\limits_{n=0}^\infty\dfrac{(q-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(p,a)}{\partial q^{2n+1}}$
$u_{qqqq}=u_{qq}-u_{ppqq}=u-u_{pp}-u_{pp}+u_{pppp}=u-2u_{pp}+u_{pppp}$
$u_{qqqqqq}=u_{qq}-2u_{ppqq}+u_{ppppqq}=u-u_{pp}-2u_{pp}+2u_{pppp}+u_{pppp}-u_{pppppp}=u-3u_{pp}+3u_{pppp}-u_{pppppp}$
Similarly, $\dfrac{\partial^{2n}u}{\partial q^{2n}}=\sum\limits_{k=0}^n(-1)^kC_k^n\dfrac{\partial^{2k}u}{\partial p^{2k}}$
$u_{qqq}=u_q-u_{ppq}$
$u_{qqqqq}=u_{qqq}-u_{ppqqq}=u_q-u_{ppq}-u_{ppq}+u_{ppppq}=u_q-2u_{ppq}+u_{ppppq}$
$u_{qqqqqqq}=u_{qqq}-2u_{ppqqq}+u_{ppppqqq}=u_q-u_{ppq}-2u_{ppq}+2u_{ppppq}+u_{ppppq}-u_{ppppppq}=u_q-3u_{ppq}+3u_{ppppq}-u_{ppppppq}$
Similarly, $\dfrac{\partial^{2n+1}u}{\partial q^{2n+1}}=\sum\limits_{k=0}^n(-1)^kC_k^n\dfrac{\partial^{2k+1}u}{\partial p^{2k}\partial q}$
$\therefore u(p,q)=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kC_k^nf^{(2k)}(p)(q-a)^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kC_k^ng^{(2k)}(p)(q-a)^{2n+1}}{(2n+1)!}$
For $u_{xx}+u_{yy}=-m^2u$ ,
Let $p=mx$ ,
Then $m^2u_{pp}+u_{yy}=-m^2u$
Let $q=my$ ,
Then $u_{qq}=-u-u_{pp}$
Similar to Deducing the existence of a PDE by constructing it inductively via its Taylor series expansion,
Consider $u(p,a)=f(p)$ and $u_q(p,a)=g(p)$ ,
Let $u(p,q)=\sum\limits_{n=0}^\infty\dfrac{(q-a)^n}{n!}\dfrac{\partial^nu(p,a)}{\partial q^n}$ ,
Then $u(p,q)=\sum\limits_{n=0}^\infty\dfrac{(q-a)^{2n}}{(2n)!}\dfrac{\partial^{2n}u(p,a)}{\partial q^{2n}}+\sum\limits_{n=0}^\infty\dfrac{(q-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(p,a)}{\partial q^{2n+1}}$
$u_{qqqq}=-u_{qq}-u_{ppqq}=u+u_{pp}+u_{pp}+u_{pppp}=u+2u_{pp}+u_{pppp}$
$u_{qqqqqq}=u_{qq}+2u_{ppqq}+u_{ppppqq}=-u-u_{pp}-2u_{pp}-2u_{pppp}-u_{pppp}-u_{pppppp}=-u-3u_{pp}-3u_{pppp}-u_{pppppp}$
Similarly, $\dfrac{\partial^{2n}u}{\partial q^{2n}}=\sum\limits_{k=0}^n(-1)^nC_k^n\dfrac{\partial^{2k}u}{\partial p^{2k}}$
$u_{qqq}=-u_q-u_{ppq}$
$u_{qqqqq}=-u_{qqq}-u_{ppqqq}=u_q+u_{ppq}+u_{ppq}+u_{ppppq}=u_q+2u_{ppq}+u_{ppppq}$
$u_{qqqqqqq}=u_{qqq}+2u_{ppqqq}+u_{ppppqqq}=-u_q-u_{ppq}-2u_{ppq}-2u_{ppppq}-u_{ppppq}-u_{ppppppq}=-u_q-3u_{ppq}-3u_{ppppq}-u_{ppppppq}$
Similarly, $\dfrac{\partial^{2n+1}u}{\partial q^{2n+1}}=\sum\limits_{k=0}^n(-1)^nC_k^n\dfrac{\partial^{2k+1}u}{\partial p^{2k}\partial q}$
$\therefore u(p,q)=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^nf^{(2k)}(p)(q-a)^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^ng^{(2k)}(p)(q-a)^{2n+1}}{(2n+1)!}$
