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I am trying to solve the following differential equation: $$ (1-x^2)^2\frac{d^4f}{dx^4} + a f(x) = 0,$$ where $a \neq 0$ and $|x| \leq 1$.

When I run DSolve, Mathematica gives me this monster of a solution: $$ \begin{align} f(x) &= c_1 \,_2F_1\left(-\frac{\alpha+1}{4},\frac{\alpha-1}{4},\frac{1}{2};x^2\right)\,_2F_1\left(-\frac{\beta+1}{4},-\frac{\beta-1}{4},\frac{1}{2};x^2\right)\\ &\quad + x \left[c_2 \, _2F_1\left(-\frac{\alpha+1}{4},\frac{\alpha-1}{4},\frac{1}{2};x^2\right)\,_2F_1\left(-\frac{\beta-1}{4},\frac{\beta+1}{4},\frac{3}{2};x^2\right) \right.\\&\quad\left. + c_3 \,_2F_1\left(-\frac{\alpha-1}{4},\frac{\alpha+1}{4},\frac{3}{2};x^2\right)\,_2F_1\left(-\frac{\beta+1}{4},-\frac{\beta-1}{4},\frac{1}{2};x^2\right) \right] \\ &\quad + c_4 x^2 \,_2F_1\left(-\frac{\alpha-1}{4},\frac{\alpha+1}{4},\frac{3}{2};x^2\right) \,_2F_1\left(-\frac{\beta-1}{4},\frac{\beta+1}{4},\frac{3}{2};x^2\right), \end{align} $$ where $$\alpha = \sqrt{1-2\sqrt{a}}, \qquad \beta = \sqrt{1+2\sqrt{a}}.$$ However, plugging this output back into the differential equation does not yield zero, even with FullSimplify. This is true even for just the $c_1$ term. Doing a series expansion around $x=0$ shows that I either must set all the $c_i$ coefficients to zero or set $a=0$, in which case the solution becomes a cubic polynomial as expected. I can also plot the differential equation evaluated on the solution given above for particular nonzero values of $a$, and it appears to be a smooth non-zero function.

What's going on here? For reference, the exact line of code I'm running is

DSolve[(1 - x^2)^2 * D[f[x],{x,4}] + a*f[x] == 0, f[x], x]

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    $\begingroup$ Welcome to Mathematica.SE! I suggest that: 1) You take the introductory Tour now! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! 3) As you receive help, try to give it too, by answering questions in your area of expertise. $\endgroup$ Commented Feb 20 at 19:01
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    $\begingroup$ Are there any boundary conditions which should be considered? $\endgroup$ Commented Feb 21 at 8:36
  • $\begingroup$ No. This is one small part of a much larger problem I'm solving, so I want the general solution. $\endgroup$ Commented Feb 21 at 19:22
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    $\begingroup$ It is possible that Mathematica has given you a solution that is valid somewhere in the complex plane, but not in the region of interest to you. Since your equation is linear, you could solve it for specific starting values (e.g. `f[0]==1&&f'[0]==0...') and sum the results. When I try this I get a message "For some branches of the general solution, the given boundary conditions lead to an empty solution". $\endgroup$ Commented Feb 23 at 8:59

1 Answer 1

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I could not find the trick to do it. May be someone will.

So translated Maple's solution to Mathematica code, and verified in Mathematica the solution is correct. Here is the solution

enter image description here

Code

ode=(1-x^2)^2*D[f[x],{x,4}]+a*f[x]==0 sol = f -> Function[{x}, C[1]*(x - 1)*(x + 1)* LegendreP[Sqrt[5 - 4*Sqrt[-a + 1]]/2 - 1/2, 2, x] + C[2]*(x - 1)*(x + 1)* LegendreP[Sqrt[5 + 4*Sqrt[-a + 1]]/2 - 1/2, 2, x] + C[3]*(x - 1)*(x + 1)* LegendreQ[Sqrt[5 - 4*Sqrt[-a + 1]]/2 - 1/2, 2, x] + C[4]*(x - 1)*(x + 1)* LegendreQ[Sqrt[5 + 4*Sqrt[-a + 1]]/2 - 1/2, 2, x]] ode /. sol // FullSimplify (* True after 1-2 minutes *)
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  • $\begingroup$ Thanks! I'll mark this as the answer unless someone else can explain how to get around this bug in Mathematica. $\endgroup$ Commented Feb 22 at 3:05
  • $\begingroup$ Given the linearity of the ODE, the Loewy decomposition could be used for reducing it to general Legendre differential equations. I have tried some experiments, so far with only modest indications that it might work. It seems that the Loewy decomposition is not yet implemented in either WL or the Wolfram Function Repository. $\endgroup$ Commented Mar 12 at 21:45

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