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Finding a solution to this equation using Frobenius method |
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| Sep19-11, 04:05 AM | #1 |
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Finding a solution to this equation using Frobenius method
Hi, I have this equation to solve.
y'' + (1/x)y' + [(x^2) + k + (m^2 / x^2)]y = 0 now, I've tried to solve this using frobenius method but cannot formulate a solution. I have that a_(n+4) = [-ka_(n+2) - a_(n)] / [n^2 +/- 2inm] is my recurrence relation, but now I'm stuck and don't know how to proceed, any help greatly appreciated. thanks. |
| Sep19-11, 03:46 PM | #2 |
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Using the standard convention of letting [itex]a_0=1[/itex], I get:
[tex]a_1=0[/tex] [tex]a_2=-\frac{h a_0}{(2+c)(1+c)+1+m^2}[/tex] with the remaining odd a_n=0 and even a_n equal to: [tex]a_n=-\frac{a_{n-2}+a_{n-4}}{(n+c)(n+c+1)+1+m^2}[/tex] with a similar relation for the other root expressed in b_n and so I can write the solution as: [tex]y(x)=K_1 \sum_{n=0}^{\infty} a_n x^{n+mi}+K_2 \sum_{n=0}^{\infty} b_n x^{n-mi},\quad x>0[/tex] and I believe because of the conjugates in the solution, for real initial conditions [itex]y(x_0)=y_0[/itex], and [itex]y'(x_0)=y_1[/itex], the imaginary component is annihilated leaving the desired real solution. If you into it, here's some Mathematica code to check the solution. I believe it's correct but not as close as I would have expected. May be some convergence issues though. Code:
nmax = 75;
x0 = 0.1;
y0 = 0;
y1 = 1.;
h = 5;
m = 4;
mysol = NDSolve[{x^2*Derivative[2][y][x] + x*Derivative[1][y][x] + (x^4 + h*x^2 + m^2)*y[x] == 0, y[x0] == y0,
Derivative[1][y][x0] == y1}, y, {x, x0, 2}];
sol[x_] := Evaluate[y[x] /. mysol];
p1 = Plot[y[x] /. mysol, {x, x0, 2}];
c1 = I*m;
Subscript[a, 0] = 1;
Subscript[a, 1] = 0;
Subscript[a, 2] = ((-h)*Subscript[a, 0])/((2 + c1)*(1 + c1) + 1 + m^2);
Subscript[a, 3] = 0;
Table[Subscript[a, n] = -(Subscript[a, n - 4] + h*Subscript[a, n - 2])/((n + c1)*(n + c1 - 1) + 1 + m^2),
{n, 4, nmax}];
f1[x_] := Sum[Subscript[a, n]*x^(n + c1), {n, 0, nmax}]
f1d[x_] = D[f1[x], x];
c2 = (-I)*m;
Subscript[b, 0] = 1;
Subscript[b, 1] = 0;
Subscript[b, 2] = ((-h)*Subscript[b, 0])/((2 + c2)*(1 + c2) + 1 + m^2);
Subscript[b, 3] = 0;
Table[Subscript[b, n] = -((Subscript[b, n - 4] + h*Subscript[b, n - 2])/((n + c2)*(n + c2 - 1) + 1 + m^2)),
{n, 4, nmax}];
f2[x_] := Sum[Subscript[b, n]*x^(n + c2), {n, 0, nmax}];
f2d[x_] = D[f2[x], x];
myks = First[NSolve[{y0 == k1*f1[x0] + k2*f2[x0], y1 == k1*f1d[x0] + k2*f2d[x0]}, {k1, k2}]];
myy[x_] := k1*f1[x] + k2*f2[x] /. myks;
myd1[x_] = D[myy[x], x];
myd2[x_] = D[myy[x], {x, 2}];
N[x^2*myd2[x] + x*myd1[x] + (x^4 + x^2*h + m^2)*myy[x]] /. x -> 0.334
p2 = Plot[myy[x], {x, x0, 2}, PlotStyle -> Red]
Show[{p1, p2}]
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