wolfram-mathematicapde

Solving the biharmonic equation in mathematica


I am attempting to solve the linear biharmonic equation in mathematica using DSolve. I think this issue is not just limited to the biharmonic equation but MATHEMATICA just spits out the equation when I attempt to solve it.

I've tried solving other partial differential equations and there was no trouble.

The biharmonic equation is just:

Laplacian^2[f]=0

Here is my equation:

DSolve[
 D[f[x, y], {x, 4}] + 2 D[D[f[x, y], {x, 2}, {y, 2}]] + 
   D[f[x, y], {y, 4}] == 0,
 f,
 {x, y}]

The solution is spit out as

DSolve[(f^(0,4))[x,y]+2 (f^(2,2))[x,y]+(f^(4,0))[x,y]==0,f,{x,y}]

That is obviously not the solution. What gives? What am I missing? I've solved other PDEs without boundary conditions.


Solution

  • How about try it in polar coordinates? If f(r, \[Theta]) is symmetric with respect to azimuth \[Theta], the biharmonic equation reduces to something Mathematca can solve symbolically (c.f. http://mathworld.wolfram.com/BiharmonicEquation.html):

    In[22]:= eq = D[r D[D[r D[f[r],r],r]/r,r],r]/r;
    eq//FullSimplify//TraditionalForm
    
    Out[23]//TraditionalForm= f^(4)(r) + (2 r^2 f^(3)(r) - r f''(r)
                               + f'(r))/r^3
    
    In[24]:= DSolve[eq==0,f,r]
    Out[24]= {{f -> Function[{r}, 
                     1/2 r^2 C[2] - 1/4 r^2 C[3] + C[4] + C[1] Log[r] 
                       + 1/2 r^2 C[3] Log[r]
                    ]}}
    
    In[25]:= ReplaceAll[
        1/2 r^2 C[2]-1/4 r^2 C[3]+C[4]+C[1] Log[r]+1/2 r^2 C[3] Log[r],
        r->Sqrt[x^2+y^2]
    ]
    Out[25]= 1/2 (x^2+y^2) C[2]-1/4 (x^2+y^2) C[3]+C[4]+C[1] Log[Sqrt[x^2+y^2]]+ 
    1/2 (x^2+y^2) C[3] Log[Sqrt[x^2+y^2]]