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Recently, I was got in trouble in solving a fourth order differential equation with Mathematica. And it is written like: $$ -a^2\zeta+\frac{1}{b^2}(\zeta''''+4ia\zeta'''-6a^2\zeta''-4ia^3\zeta'+a^4\zeta)-\frac{c}{d}[-\zeta'-ia\zeta+(1-x)(\zeta''+2ia\zeta'-a^2\zeta)]-\frac{1}{d}(-2f(a)\sqrt{\frac{1-x}{x}}\zeta'+2a^2\sqrt{x(1-x)}\zeta) = a^2x-\frac{1}{b^2}(-4ia^3+a^4x)+\frac{c}{d}[-1-ia x+(1-x)(2ia-a^2 x)] $$ and $$ f(a)=\frac{H{_1^{(2)}}(a/2)}{H{_1^{(2)}}(a/2)+iH{_0^{(2)}}(a/2)} $$ The boundary condition are $$ \zeta(0)=0 $$ $$ \zeta''(0)+2ia\zeta'(0)=-2ia $$ $$ \zeta''(1)+2ia\zeta'(1)-a^2\zeta(1)=-2ia+a^2 $$ $$ \zeta'''(1)+3ia\zeta''(1)-3a^2\zeta'(1)-ia^3\zeta(1)=3a^2+ia^3 $$ where $H{_i^{(2)}}(a/2)$ are Hankel functions of the 2nd kind and $i$th order; $a$, $b$, $c$ and $d$ are known real numbers; $\zeta$ is an unknown complex function of real variable $x$. I really want to figure it out, so I would much appreciate anyone who can do me a favor. Thanks in advance.

Here are my related codes with Dsolve and HankelH2 commands:

a = 8*Pi; b = 5; c = 2.74*10^-3; d = 0.3;  
f = HankelH2[1, a/2]/(HankelH2[1, a/2] + I*HankelH2[0, a/2]);  
eq = -a^2*y + 1/b^2*(y'''' + I*4*a*y''' - 6*a^2*y'' - I*4*a^3*y') - 
    c/d*(-y' - I*a*y + (1 - x)*(y'' + I*2*a*y' - a^2*y)) - 
    1/d*(-2*f*Sqrt[(1 - x)/x]*y' + 2*a^2*Sqrt[(1 - x)*x]*y) == 
    a^2*x - 1/b^2*(-I*4*a^3 + a^4*x) + 
    c/d*(-1 - I*a*x + (1 - x)*(I*2*a - a^2*x));  
DSolve[{eq, y'[0] == 0, y''[0] + I*2*a*y'[0] == -I*2*a, 
    y''[1] + I*2*a*y'[1] - a^2*y[1] == -I*2*a + a^2, 
    y'''[1] + I*3*a*y''[1] - 3*a^2*y'[1] - I*a^3*y[1] == 3*a^2 + I*a^3},
    y[x], x]  

After implementing the codes, I got no answer.

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Welcome to the site! Please edit your question adding the best Mathematica code you think you got at for solving it! –  belisarius Sep 14 '12 at 2:50
    
So, you know about DSolve[] and HankelH2[] already? –  J. M. Sep 14 '12 at 3:08
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When you post a question in this site, be prepared to stay online answering comments and commenting on answers for at least one hour. That way your question will receive more attention and hopefully better answers –  belisarius Sep 14 '12 at 4:06
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To cormullion: I have changed it. Actually, I do believe there is no analytic solution for this equation. So only numerical solution is possible to get, maybe with collocation method. Thank all for your friendly advice and help. –  lesdeux Sep 18 '12 at 3:21
2  
Voting to close as TL since this OP has now given up with the analytical solution and moved on to trying to solve this numerically (see their other question, which could benefit from some attention). –  Oleksandr R. Feb 15 '13 at 12:41
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closed as too localized by Oleksandr R., whuber, m_goldberg, rm -rf Feb 17 '13 at 2:58

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1 Answer

This is not solution but it might suggest that it won't be trivial to find one.

Let us call

eqn = {-a^2 f[x] + 
1/b^2 (f''''[x] + 4 I a f'''[x] + 6  a^2 f''[x] - 4 I a^3 f'[x] + 
   a^4 f[x]) - c/d (-f'[x] - I a f[x] + (1 - x) (f''[x] + 2 I a f'[x] - a^2 f[x])) - 
1/d (-2 F[a] Sqrt[(1 - x)/x] f'[x] + 2 a^2 Sqrt[x (1 - x)] f[x]) == 
 h (a^2 x - 1/b^2 (-4 I a^3 + a^4 x) +  c/d (-1 - I a x + (1 - x) (2 I a - a^2 x))),
f[0] == 0, f''[0] + 2 I a f'[0] == -2  I a,
f''[1] + 2 I a f'[1] - a^2 f[1] == -2 I a + a^2 ,
f'''[1] + 3 I a f''[1] - 3 a^2 f[1] - I a^3 f[1] == 3 a^2 + I a^3}

Let us now simplify the problem: put F to zero make the equation homogeneous and replace variable by numbers

eqn2 = eqn /. 
F[a] -> 0  HankelH2[1, 
    a/2]/(HankelH2[1, a/2] + I HankelH2[0, a/2]) /. 
 h -> 0 /. {a -> 1 , b -> 2, c -> 3, d -> 4}

This does not yield a solution

DSolve[eqn2, f[x], x]

(* returns the same *) while the suggestion of Pragabhava

eqn3 = eqn2 /. f -> Function[x, -x + \[Nu][x]] // Simplify;
DSolve[eqn3, \[Nu][x], x]

does not seem to produce the expected miracle.

So in short, one needs to simplify the problem further and find a solution unless you are happy to find numerical solutions?

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Thanks all the same. –  lesdeux Sep 14 '12 at 8:04
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