Series solution

  • Thread starter Benny
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Can someone please explain some steps of a worked example.

Q. Find the power series solutions about z = 0 of 4zy'' + 2y' + y = 0.

(note: y = y(z))

Writing the equation in standard form:

[tex]
y'' + \frac{1}{{2z}}y' + \frac{1}{{4z}}y = 0
[/tex]

Let [tex]y = z^\sigma \sum\limits_{n = 0}^\infty {a_n z^n }[/tex]

The indicial equation has roots zero and (1/2).

Using the substitution for y above the DE becomes:

[tex]
\sum\limits_{n = 0}^\infty {\left[ {\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right) + \frac{1}{2}\left( {n + \sigma } \right) + \frac{1}{4}z} \right]} a_n z^n = 0
[/tex]...(1)

Demanding that the coefficients of z^n vanish separately in the above equation we obtain the recurrence relation:

[tex]
\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right)a_n + \frac{1}{2}\left( {n + \sigma } \right)a_n + \frac{1}{4}a_{n - 1} = 0
[/tex]

What happened to the factor of z which was multiplied by (1/4) in (1)?

If we choose the larger root, 1/2, then the recurrence relation becomes [tex]a_n = \frac{{ - a_{n - 1} }}{{2n\left( {n + 1} \right)}}[/tex]

Setting a_0 = 1 we find...etc

I am told that the larger of the two roots of the indicial equation always leads to a solution. However, I don't understand why a_0 is set equal to one? Is that an arbitrary selection or is it convenient?

Any help would be good thanks.
 
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Answers and Replies

  • #2
AKG
Science Advisor
Homework Helper
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Benny said:
[tex]\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right)a_n + \frac{1}{2}\left( {n + \sigma } \right)a_n + \frac{1}{4}a_{n - 1} = 0[/tex]

What happened to the factor of z which was multiplied by (1/4) in (1)?
It's there, it's the [itex]\frac{1}{4}a_{n-1}[/itex]. Do you see why?
If we choose the larger root, 1/2, then the recurrence relation becomes

[tex]a_n = \frac{{ - a_{n - 1} }}{{2n\left( {n + 1} \right)}}[/tex]
I've never done problems like this before, but if you are getting this by substituting [itex]\sigma = \frac{1}{2}[/itex] into:

[tex]\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right)a_n + \frac{1}{2}\left( {n + \sigma } \right)a_n + \frac{1}{4}a_{n - 1} = 0[/tex]

then you should get:

[tex]a_n = \frac{{ - a_{n - 1} }}{{2n\left( {\mathbf{2}n + 1} \right)}}[/tex]
Setting a_0 = 1 we find...etc

I am told that the larger of the two roots of the indicial equation always leads to a solution. However, I don't understand why a_0 is set equal to one? Is that an arbitrary selection or is it convenient?
Different choices of a0 will give different solutions, and you should indeed expect a family of solutions, not just one solution. Are there any given boundary conditions that might explain the choice of a0 = 1? If not, then perhaps it is just arbitrary/convenient/conventional, I don't know myself.
 
  • #3
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There aren't any ICs or BCs in the example so a_0 = 1 might have been chosen simply to illustrate the solution method. Thansk for your help.
 

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