Solving a DE: Finding Power Series Solutions about z = 0

[Benny]

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:

$$y'' + \frac{1}{{2z}}y' + \frac{1}{{4z}}y = 0$$

Let $$y = z^\sigma \sum\limits_{n = 0}^\infty {a_n z^n }$$

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

Using the substitution for y above the DE becomes:

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

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

$$\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right)a_n + \frac{1}{2}\left( {n + \sigma } \right)a_n + \frac{1}{4}a_{n - 1} = 0$$

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 $$a_n = \frac{{ - a_{n - 1} }}{{2n\left( {n + 1} \right)}}$$

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.

 

[AKG]

$$\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right)a_n + \frac{1}{2}\left( {n + \sigma } \right)a_n + \frac{1}{4}a_{n - 1} = 0$$

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

It's there, it's the $\frac{1}{4}a_{n-1}$. Do you see why?

If we choose the larger root, 1/2, then the recurrence relation becomes

$$a_n = \frac{{ - a_{n - 1} }}{{2n\left( {n + 1} \right)}}$$

I've never done problems like this before, but if you are getting this by substituting $\sigma = \frac{1}{2}$ into:

$$\left( {n + \sigma } \right)\left( {n + \sigma - 1} \right)a_n + \frac{1}{2}\left( {n + \sigma } \right)a_n + \frac{1}{4}a_{n - 1} = 0$$

then you should get:

$$a_n = \frac{{ - a_{n - 1} }}{{2n\left( {\mathbf{2}n + 1} \right)}}$$

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.

 

[Benny]

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.