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Two coupled Sturm-Liouville Eigenvalue Problems in 2-D

  1. Sep 26, 2010 #1
    Hello everybody,
    I have been trying to solve coupled two eigenvalue (Sturm-Liouville) problems in terms of two (eigen) functions u[x,y] and v[x,y].
    I have been using Mathematica trying to solve the coupled equations analytically in their original form,
    but the Mathematica doesn't seem to give th ecorrect result. Here are the equations in its general form;

    La(u[x,y]) + Lb(v[x,y]) = Es u[x,y]
    Lc(u[x,y]) + Ld(v[x,y]) = Es v[x,y]

    where La,Lb,Lc,Ld are some differential operators which may contain x,y variables, their derivatives or just some constants.
    The actual equations in Mathematica notations are as follows (I move the eigenvalues Es to the left hand side so as to make it
    like a constant to be determined from boundary conditions)


    DSolve[{-I*vf*D[u[x, y], x] + ec/c*vf*H*y*u[x, y] - Es*u[x, y] -
    8*d0/(pf*pf)*D[D[v[x, y], y], y] ==
    0, -8*d0/(pf*pf) D[D[u[x, y], y], y] +
    I*vf*D[v[x, y], x] + (ec/c*vf*H*y - Es)*v[x, y] == 0}, {u[x, y],
    v[x, y]}, {x, y}]

    where variables other than x,y are constants. ('I' is the unit imaginary number).

    The mathematica does give some output but doesn't seem to satisfy the original equations.
    {{u[x, 0] ->
    C[4] + C[5] Sech[x C[1] + C[3]] + C[6] Sech[x C[1] + C[3]]^2 +
    C[10] Tanh[x C[1] + C[3]] +
    C[11] Sech[x C[1] + C[3]] Tanh[x C[1] + C[3]],
    v[x, 0] ->
    C[7] + C[8] Sech[x C[1] + C[3]] + C[9] Sech[x C[1] + C[3]]^2 +
    C[12] Tanh[x C[1] + C[3]] +
    C[13] Sech[x C[1] + C[3]] Tanh[x C[1] + C[3]]},

    {u[x, y] ->
    C[4] + C[5] Sech[x C[1] + y C[2] + C[3]] +
    C[6] Sech[x C[1] + y C[2] + C[3]]^2 +
    C[10] Tanh[x C[1] + y C[2] + C[3]] +
    C[11] Sech[x C[1] + y C[2] + C[3]] Tanh[x C[1] + y C[2] + C[3]],
    v[x, y] ->
    C[7] + C[8] Sech[x C[1] + y C[2] + C[3]] +
    C[9] Sech[x C[1] + y C[2] + C[3]]^2 +
    C[12] Tanh[x C[1] + y C[2] + C[3]] +
    C[13] Sech[x C[1] + y C[2] + C[3]] Tanh[x C[1] + y C[2] + C[3]]}}

    Not sure what the u[x,0] and v[x,0] mean but I know they are probably some sort of boundary conditions that are assumed/suggested by the Mathematica for the problem.
    Well unfortunately I have not had the boundary conditions yet, I just hope to find some general solution with some constants/coefficients that can be determined from
    the boundary condition to be fixed or 'chosen' later. By naive counting, may be wrong though, the equations are 2nd order in y and 1st order in x and there are two functions to solve,
    we need 6 boundary conditions, but Mathematica solution requires 13 constants here, a mismatch.

    I try to another approach;

    I rewrite the two equations into one PDE for u[x,y] and obtain the following 4th order in y and 2nd order in x DE, in Mathematica notations


    DSolve[vf*vf*D[D[u[x, y], x], x] + (el*vf*H*y/c - Es)^2*u[x, y] -
    64.0*d0*d0/pf^4*D[D[D[D[u[x, y], y], y], y], y] == 0,
    u[x, y], {x, y}]


    I assume (guess) solution of the form u[x,y] = Exp[i*k*x] * P(y) Exp(-(y-y0)^2/(2 s*s))

    where P(y) is some polynomial in y (not yet known, to be determined)
    The reason is, the equation is 2nd order in x (a typical Schrodinger equation for free particle) and 4th order in y with dependence on (y-y0)^2.
    Substituting this guess solution, I obtain ODE for P(y)

    DSolve[(D[D[D[D[P[y],y],y],y],y]-4.0/s^4*(y-y0)*D[D[D[P[y],y],y],y]-6.0*(1/s^2-1/s^4*(y-y0)^2)*D[D[P[y],y],y]+(12.0/s^4*(y-y0)-4.0/s^6*(y-y0)^3)*D[P[y],y]+(3.0/s^4-6.0/s^6*(y-y0)^2+(y-y0)^4/s^8)*P[y])*64.0*d0*d0/pf^4+(k^2*vf^2-(ec*vf*H/c)^2*(y-y0)^2)*P[y]
    0,P[y],y]

    Again I try using Mathematica to solve this but it gives up, it doesn't give answer, just returning the original equation instead.

    So, I still couldn't solve this problem after several days of effort. Anyone very expert in differential equations who can see what the solution should be, perhaps involving some special or oher non-trivial functions?
    Please suggest solution.

    Thanks in advance,

    Urkel
     
  2. jcsd
  3. Sep 26, 2010 #2
    Looks too messy to me. Also, don't use capital letters in Mathematica as the first letter for a symbol name since it may conflict with built-in commands which are all capitalized. This is what I'd do: strip away all the constants or set to one or whatever, then trying to solve that. This is what a simpler version looks like to me:

    [tex]\frac{\partial u}{\partial x}-\frac{\partial^2 v}{\partial y^2}+(y-1)u=0[/tex]

    [tex]\frac{\partial v}{\partial x}+\frac{\partial^2 u}{\partial y^2}+yv=0[/tex]

    Suppose that's it. Then I'd try and solve that. I tried DSolve and it can't do it. I'm pessimistic it would be able to do so for a more complicated version but it's possible. I'm not familiar with any technique to solve it by hand but perhaps others here know or the literature would be helpful.

    If not, I'd try NDSolve for a numerical solution and just for now, use very simple boundary and initial conditions.

    If numerical is acceptable, then I'd get a simple solution "working", meaning I have high confidence it's right, then I would begin adding the original formulation step-wise, get that working, then add more, and hopefully build my way to the problem I want.
     
    Last edited: Sep 26, 2010
  4. Sep 27, 2010 #3
    I have many different versions of ODE/PDE that describe this same problem.

    A simple version but which is not yet solved is this 4th order nonlinear ODE

    a*f(y) + b*(y-y0)^2*f(y) + c*d^4 f(y)/ dy^4 = 0

    where a, b, c and y0 are all constants.

    I hope this is nicer to look at than the lengthy problem description above.

    I find Mathematica couldn't give analytical solution to this though.
    I also Fourier transform this equation and get

    p*F(ky) + q * d^2 F(ky)/dky^2 + r * d F(ky)/dky + s ky^4 F(ky) = 0

    where p, q, r and s are constants. Unfortunately, it's still not analytically solvable by Mathematica. I don't have Maple or Matlab or other software. I hope someone who does have them can solve this problem. In this case, I indeed expect analytical rather than numerical solution.
     
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