Pointers and help - solving an equation

**liquidFuzz** · December 20th 2010, 11:42 AM

I want to solve a differential equation, the exercise being, find two different independent solutions. (I don't know the proper English terms for this.)

$z^2 w'' - (z^2 + z)w' + (z+1)w = 0$

I start out with $\displaystyle w = \sum_0^\infty a_n z^{n+m}$ . Differentiate to match each term in the equation and shove it in.

$\displaystyle \sum_0^\infty (n+m)(n+m-1)a_n z^{n+m} - \sum_0^\infty (n+m)a_n z^{n+m+1}-\sum_0^\infty (n+m)a_n z^{n+m} + \sum_0^\infty a_n z^{n+m+1} + \sum_0^\infty a_n z^{n+m} = 0$

$\displaystyle \sum_0^\infty [(n+m)(n+m-1) -(n+m) - 1]a_n z^{n+m} = \sum_0^\infty [(n+m) -1]a_n z^{n+m+1}$

Recursive series.

$\displaystyle (m+n-1)^2 a_n = (n+m -2)a_{n-1}$

Now I'm starting to get wobbly legs. I set $n = 0, a_0 = 1$ to see if I can pinpoint $m$ . $(m-1)^2 = 0 \implies m = \pm 1$

$m = -1 \implies a_n = \frac{(n-3)}{(n-2)^2} a_{n-1}$ results in a series $a_n = \infty, n \geq 2$ . Can I render a solution from that even if it diverges?

$\displaystyle m = 1 \implies a_n = \frac{(n-1)}{n^2} a_{n-1}$ results in a series $a_n = 0, n \geq 1$ that would imply that one solution is w = z. Or..?

Now I've sort of played all my cards here. What if I state a solution w = zu and solve the equation..? Or is there a simpler way to render an other solution here?

**chisigma** · December 20th 2010, 01:28 PM

Given the linear second order 'incomplete' DE...

$\displaystyle z^{2}\ w^{''} -(z + z^{2})\ w^{'} + (1+z)\ w=0$ (1)

... we are able to solve it from the knowledge of two independent solutions u and v. Because u and v are solutions of (1) You can write...

$\displaystyle z^{2}\ u^{''} -(z + z^{2})\ u^{'} + (1+z)\ u=0$

$\displaystyle z^{2}\ v^{''} -(z + z^{2})\ v^{'} + (1+z)\ v=0$ (2)

... so that subtracting with some easy step You obtain...

$\displaystyle (v\ u^{''} - u\ v^{''}) - \frac{1+z}{z}\ (v\ u^{'} - u\ v^{'}) =0$ (3)

Now setting $\varphi = v\ u^{'} - u\ v^{'}$ the (3) becomes...

$\displaystyle \varphi^{'} = \frac{1+z}{z}\ \varphi$ (4)

... the solution of which is very easy...

$\varphi= c_{1}\ z\ e^{z}$ (5)

Deviding both terms of (3) by $v^{2}$ and taking into account (5) You obtain...

$\displaystyle \frac{v\ u^{''} - u\ v^{''}}{v^{2}} = \frac{d}{dz} \frac{u}{v} = c_{1}\ \frac{z\ e^{z}}{v^{2}} \implies \frac{u}{v}= c_{2} + c_{1} \int \frac{z\ e^{z}}{v^{2}}\ dz \implies$

$\displaystyle \implies u= c_{2}\ v + c_{1}\ v\ \int \frac{z\ e^{z}}{v^{2}}\ dz$ (6)

Now it is easy to verify that $v=z$ is solution of (1) so that the general solution of (1) is...

$\displaystyle w(z)= c_{2}\ z + c_{1}\ z\ \int \frac{e^{z}}{z}\ dz$ (7)

The integral is (7) is a non elementary function known as 'exponential integral function'...

Merry Christmas from Italy

$\chi$ $\sigma$

**liquidFuzz** · December 20th 2010, 10:31 PM

Ok, the part where you solve the de i did slightly different.

With my approach, $w = zu$ I get $u''+ (\frac{1}{z} + z) u' = 0$ . Multiply with $e^{ln z - z} = z e ^{-z}$ to get $u' = C z^{-1} e ^z$ .

$C = 1$ should work as I need one solution, I already have $w = z$ .

$\displaystyle u = \int \frac{e^z}{z} dz = \int \sum_{n=0}^\infty \frac{z^{n-1}}{n!} dz = \log z + \sum_{n=1}^\infty \frac{z^{n}}{n*n!} dz$

$w_1 = z$

$\displaystyle w_2 = zu = z \log z + \sum_{n=1}^\infty \frac{z^{n+1}}{n*n!} dz$

Edit: Back to my questions, is there a simpler way to get a solution, using the recursive formula? Can I render a solution if it diverges?

**chisigma** · December 21st 2010, 08:30 AM

Originally Posted by liquidFuzz

...

$\displaystyle u = \int \frac{e^z}{z} dz = \int \sum_{n=0}^\infty \frac{z^{n-1}}{n!} dz = \log z + \sum_{n=1}^\infty \frac{z^{n}}{n*n!} dz$

The definition of 'exponential integral function' is a little different as You can see in...

Exponential Integral -- from Wolfram MathWorld

Pratically is...

$\displaystyle Ei(z)= \ln z + \gamma + \sum_{n=1}^{\infty} \frac{z^{n}}{n\ n!}$ (1)

... where the constant $\gamma$ is the Euler's constant...

$\displaystyle \gamma= \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k} - \ln n$ (2)

That is not a marginal detail because the 'particular solution' is given by the product...

$\displaystyle z\ Ei(z)= z\ \ln z + \gamma\ z + \sum_{n=2}^{\infty} \frac{z^{n}}{(n-1)\ (n-1)!}$ (3)

Merry Christmas from Italy

$\chi$ $\sigma$

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