Finding a solution to this equation using Frobenius method

In summary, the equation given is a second-order differential equation with variable coefficients. The speaker has attempted to solve it using the Frobenius method and has derived a recurrence relation for the coefficients. They have also written a solution using the initial conditions and provided some code for checking the solution.
  • #1
climbon
18
0
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.
 
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  • #2
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}]
 
Last edited:

FAQ: Finding a solution to this equation using Frobenius method

1. What is the Frobenius method?

The Frobenius method is a technique used to find solutions to linear differential equations with variable coefficients. It involves assuming a solution in the form of a power series and using algebraic manipulations to determine the coefficients of the series.

2. When is the Frobenius method used?

The Frobenius method is typically used when the coefficients of a differential equation cannot be expressed as a finite series, making it difficult to solve using other methods such as separation of variables or substitution.

3. How do you apply the Frobenius method to an equation?

To apply the Frobenius method, you first assume a solution in the form of a power series, substitute it into the equation, and then solve for the coefficients of the series. This often involves solving a recurrence relation or using the ratio test to determine the convergence of the series.

4. What are the limitations of the Frobenius method?

The Frobenius method can only be used to find solutions for certain types of differential equations, such as those with regular singular points. It also may not always result in a solution that converges for all values of the independent variable.

5. Are there any alternative methods to the Frobenius method?

Yes, there are other methods for solving differential equations with variable coefficients, such as the Laplace transform method and the method of undetermined coefficients. However, the Frobenius method is often preferred for its simplicity and applicability to a wide range of equations.

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