Using the Frobenius method on a 2D Laplace

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The discussion focuses on the application of the Frobenius method to solve a 2D Laplace equation. A key issue identified is the dual use of the symbol ##\ell##, which complicates the analysis. Participants suggest using a different summation variable, such as ##j##, in the Frobenius ansatz to avoid confusion. The correct approach involves plugging the ansatz into the ordinary differential equation (ODE) and comparing coefficients to determine ##\lambda## and the coefficients ##a_j##. Ultimately, clarity in notation leads to successful problem resolution.
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.
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Frobenius method of solving ode
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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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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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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?
 
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
 
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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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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