Frobenius Method - Roots differ by integer

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The discussion focuses on the Frobenius method for solving differential equations, specifically regarding the linear independence of solutions y1 and y2 when the roots σ1 and σ2 of the indicial equation differ by an integer. The original poster seeks clarification on why the condition of non-integer difference implies linear independence. A participant explains that linear dependence occurs when a linear combination of y1 and y2 equals a constant, which can only happen if σ1 and σ2 differ by an integer, allowing terms to cancel. Thus, for y1 and y2 to be linearly independent, the difference σ1 - σ2 must be a fraction. This highlights the critical relationship between the roots of the indicial equation and the nature of the solutions.
asras
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I'm reading up on some methods to solve differential equations. My textbook states the following:

"y_{1} and y_{2} are linearly independent ... since \sigma_{1}-\sigma_2 is not an integer."

Where y_{1} and y_{2} are the standard Frobenius series and \sigma_1 and \sigma_2 are the roots of the indicial equation.

I'm having trouble seeing how the above follows and would appreciate some input. I'm using "Essential Mathematical Methods for the Physical Sciences" and the quote is (albeit slightly paraphrased) from page 282, for reference.

Incidentally this is my first post. Looking forward to participating in this forum.
 
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y1 and y2 are linearly dependent if ay1 + by2 = c
a,b,c= const.
let, y1=\sumnxn+σ1
y1=\sumnxn+σ1
then ay1\sumnxn+σ1 + by1\sumnxn+σ1=c

Since right hand side is const. , all the coefficients of the powers of x are zero. This is possible if one term arising in the first summation cancels the other. This is possible only when σ1 and σ2 differ by integer. then n can assume different values and cancel the coefficients.
So, for independentness, σ1-σ2=fraction
 
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