- #1

Urkel

- 15

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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