Using the Frobenius method on a 2D Laplace

In summary, the problem is you're using the symbol ##\ell## in two different ways: (1) as part of the constant that appears on the righthand side of the differential equation and (2) as the index of the summation. This can make the equation simpler, but you have to be careful because you can get different results depending on which notation you use.
  • #1
jkthejetplane
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Homework Statement
I have some supplemental hmwk that I need to use a method I am not familiar with and I am not sure I have applied it correctly bc if i solve for the missing variable i will not achieve the right answer.
Relevant Equations
Frobenius method of solving ode
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  • #2
The problem is you're using the symbol ##\ell## in two different ways: (1) as part of the constant that appears on the righthand side of the differential equation and (2) as the index of the summation.
 
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  • #3
Hint:
$$\frac{\mathrm{d}}{\mathrm{d} r} \left (r^2 \frac{\mathrm{d} R}{\mathrm{d} r} \right)=r \frac{\mathrm{d}^2}{\mathrm{d} r^2} (r R)$$
helps to make the equation simpler.

Then you have to choose another summation variable in your Frobenius ansatz than ##l##, e.g.,
$$R(r)=\sum_{j=0}^{\infty} a_j r^{j+\lambda},$$
and you can assume ##a_0 \neq 0## for uniqueness. Then plug this ansatz into your ODE and determine first ##\lambda## and then the ##a_j## by comparing the coefficients in the generalized power series on both sides of your equation.
 
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  • #4
vanhees71 said:
Hint:
$$\frac{\mathrm{d}}{\mathrm{d} r} \left (r^2 \frac{\mathrm{d} R}{\mathrm{d} r} \right)=r \frac{\mathrm{d}^2}{\mathrm{d} r^2} (r R)$$
helps to make the equation simpler.

Then you have to choose another summation variable in your Frobenius ansatz than ##l##, e.g.,
$$R(r)=\sum_{j=0}^{\infty} a_j r^{j+\lambda},$$
and you can assume ##a_0 \neq 0## for uniqueness. Then plug this ansatz into your ODE and determine first ##\lambda## and then the ##a_j## by comparing the coefficients in the generalized power series on both sides of your equation.
Ok so as far as your hint goes, I should be using that whole expression by getting it equal to zero so I have a y" term or that I should only use the right side equal to 0?
 
  • #5
vela said:
The problem is you're using the symbol ##\ell## in two different ways: (1) as part of the constant that appears on the righthand side of the differential equation and (2) as the index of the summation.
Oh yeah i see that now. I got caught up in our notation and other notation when looking up the method. I'll try again using the standard n
 
  • #6
jkthejetplane said:
Ok so as far as your hint goes, I should be using that whole expression by getting it equal to zero so I have a y" term or that I should only use the right side equal to 0?
Just plug in the ansatz into the equation and compare the coefficients of the series. You'll get an equation for ##\lambda## and then a recursion relation for the ##a_j##.
 
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  • #7
Update: I got it all correct after just trying again without the confusion of l terms. Did it a couple ways to make sure. Thank you
 
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1. What is the Frobenius method and how is it used in solving the 2D Laplace equation?

The Frobenius method is a technique used to solve linear differential equations with variable coefficients. It involves assuming a solution in the form of a power series and then solving for the coefficients by substituting the series into the equation. In the case of the 2D Laplace equation, the Frobenius method is used to find a series solution for the equation in two variables, typically x and y.

2. What are the advantages of using the Frobenius method over other methods for solving the 2D Laplace equation?

The Frobenius method allows for the solution of equations with variable coefficients, making it a more versatile method compared to other techniques such as separation of variables. Additionally, it can be used to find series solutions which may be more accurate than other methods when dealing with complex boundary conditions or irregular geometries.

3. Are there any limitations to using the Frobenius method on the 2D Laplace equation?

One limitation of the Frobenius method is that it can only be applied to linear differential equations. Additionally, it may not always yield a convergent solution, especially when dealing with singular points or complex boundary conditions. In these cases, other methods may need to be used.

4. How do I know if the series solution obtained using the Frobenius method is valid?

The series solution obtained using the Frobenius method will be valid if it satisfies the boundary conditions of the 2D Laplace equation. This can be checked by substituting the series solution into the equation and verifying that it satisfies the given boundary conditions.

5. Can the Frobenius method be used to solve the 2D Laplace equation in any domain?

Yes, the Frobenius method can be used to solve the 2D Laplace equation in any domain as long as the boundary conditions are well-defined and the equation is linear. However, for more complex geometries, it may be more efficient to use other numerical methods such as finite difference or finite element methods.

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