# Solving biharmonic equation with Mathematica

I would like to solve a biharmonic equation in polar coordinates of the form:

$\Delta \Delta \Phi[r,\theta] = r^i cos(j \, \theta) \quad i,j \, \epsilon \, \mathbb{N}_0$

I know that a solution for the homogeneous problem

$\Delta \Delta \Phi[r,\theta] = 0$

was found quite a while ago (100 years+), see here. I am wondering why Mathematica is not able to solve even the simplified case via

$DSolve[ \Delta[ \Delta \phi [r, \theta], \{r, \theta\}, "Polar"], \{r, \theta\}, "Polar"] == 0, \phi, \{r, \theta\}]$

Code to copy:

  DSolve[Simplify[
Laplacian[
Laplacian[ϕ[r, θ], {r, θ}, "Polar"], {r, θ}, "Polar"]] == 0, ϕ, {r, θ}]


because the problem is quite common in elasticity.

I know a similar question has been asked before, but the recommended solution is just valid for rotational symmetric case for which the PDE simplifies to an ODE.

I would be very thankful for any suggestion.

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• You can format inline code and code blocks by selecting the code and clicking the {} button above the edit window. The edit window help button ? is also useful for learning how to format your questions and answers. You may also find this meta Q&A helpful – Michael E2 Jun 22 '16 at 17:50

If you accept that the general solution can be constructed as an (infinite) Fourier series,

$$\sum_{m=-\infty}^{\infty}\phi_m(r) \exp(i m \theta),$$

then you can obtain the expression for $\phi_m(r)$ as follows:

eqn =
0 == Simplify@
Laplacian[ Laplacian[ ϕ[r] Exp[I m θ], {r, θ}, "Polar"], {r, θ}, "Polar"];

DSolve[eqn, ϕ, r]

(*
==> {{ϕ ->
Function[{r},
r^-m C + r^(2 - m) C + r^(2 + m) C + r^m C]}}
*)


The integration constants in this result can then be chosen differently for each $m$. This solves the homogeneous problem (if the constants are chosen to satisfy the given boundary conditions).

You can do the same for the inhomogeneous problem with specific $i, j$, e.g., $i=2$, $j=3$:

eqn1 =
r^2 Exp[I 3 θ] ==
Simplify@
Laplacian[Laplacian[ ϕ[r] Exp[I 3 θ], {r, θ}, "Polar"], {r, θ}, "Polar"];

DSolve[eqn1, ϕ, r]

(*
==> {{ϕ ->
Function[{r}, r^6/189 + C/r^3 + C/r + r^3 C + r^5 C]}}
*)

eqn2 =
r^2 Exp[-I 3 θ] ==
Simplify@
Laplacian[ Laplacian[ ϕ[r] Exp[-I 3 θ], {r, θ}, "Polar"], {r, θ}, "Polar"];

DSolve[eqn2, ϕ, r]

(*
==> {{ϕ ->
Function[{r}, r^6/189 + C/r^3 + C/r + r^3 C + r^5 C]}}
*)


Here, I broke the $\cos(3\theta)$ up into two parts containing $\exp(\pm 3 i\theta)$ and solved the two separately. Then you have to add both solutions and divide by two to get the solution to with $r^2 \cos(3 \theta)$ as the inhomogeneous term.

Finally, using the same superposition principle as above for some special choices of $i,j$, I tried the general solution for arbitrary $i,j$ in the following form:

eqn3 =
r^i Cos[j θ] ==
Simplify@
Laplacian[ Laplacian[ ϕ[r] Cos[j θ], {r, θ}, "Polar"], {r, θ}, "Polar"];

DSolve[eqn3, ϕ, r]

(*
==> {{ϕ ->
Function[{r},
r^(4 + i)/((-4 - i + j) (-2 - i + j) (2 + i + j) (4 + i + j)) +
r^-j C + r^(2 - j) C + r^(2 + j) C + r^j C]}}
*)


So for the inhomogeneous equation, all we had to do is choose the same $\theta$ dependence for the unknown solution as appears in the given right-hand side.

• Thank you so much, Jens! That's exactly what I was looking for and it works perfectly fine. – Paul Saturday Jun 24 '16 at 6:42