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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.

Dr. belisarius
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dearN
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  • According to the documentation of `DSolve`, the function can "solve many linear equations up to second order with nonconstant coefficients". So my guess is that `DSolve` fails because the biharmonic equation is a fourth order PDE. – Heike Oct 18 '11 at 16:15
  • @Heike Looks to be the case. How am I to solve this equation in mathematica? I have solved fourth order non-linear pdes before, but with NDSolve... – dearN Oct 18 '11 at 16:23
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    See http://mathworld.wolfram.com/BiharmonicEquation.html – Dr. belisarius Oct 18 '11 at 16:48
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    Shouldn't your second term be `D[f[x, y], {x, 2}, {y, 2}]` or `D[D[f[x, y], {x, 2}], {y, 2}]`? – Meng Lu Oct 18 '11 at 16:59
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    See also http://www.ekayasolutions.com/UCDMath/PapkovitchEigenFunctions.nb – Dr. belisarius Oct 18 '11 at 16:59
  • @belisarius Yes, I did check that previously. Although they don't really attempt at solving it through DSolve. – dearN Oct 18 '11 at 18:04
  • @MengLu Hmmm... well what do you know? It didn't make much of a difference, mathematica still just spits my equation back at me... – dearN Oct 18 '11 at 18:04
  • I had a meeting with a math faculty and I realized that the biharmonic equation (cartesian) doesn't probably have an analytical solution. There is some substitution and separation of variables involved which MATHEMATICA doesn't quite figure out on its own. – dearN Oct 19 '11 at 17:37

2 Answers2

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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]]
Meng Lu
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  • This helps a lot! Thanks! :) however, I'd love to be able to do it in cartesian coordinates. – dearN Oct 19 '11 at 17:39
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DSolve[D[f[x, y], {x, 4}] + 2 D[f[x, y], {x, 2}, {y, 2}] + D[f[x, y], {y, 4}] == 0, f, {x, y}]

This ought to be the actual syntax

fuller
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