Solve differential equation using "Power Series"

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July 23rd 2010, 11:13 AM
Ackbeet

No, you didn't deal with the coefficients right. The sum sign is for addition. The same distributive rules apply to that kind of addition as to any other kind of addition. I would actually move the $1+x^{2}$ in the $y''$ term into the sum, as well as the $x$ into the $y'$ term thus:

$\displaystyle{\sum_{k=0}^{\infty}(k+2)(k+1)c_{k+2} (x^{k}-x^{k+2})-6\,\sum_{k=0}^{\infty}(k+1)c_{k+1}x^{k+1}-4\,\sum_{k=0}^{\infty}c_{k}x^{k}=0}.$

Now, I think you can add everything together. How does that look?
July 23rd 2010, 11:21 AM
EmilyL

ah, right, makes sense. So:

$\displaystyle{\sum_{k=0}^{\infty}(k+2)(k+1)c_{k+2} (x^{k}-x^{k+2})-(6k+6)c_{k+1}x^{k+1}-4c_{k}x^{k}=0}.$

Unfortunately I have to leave for work now, but I'll be back online in 6-7 hours. Hopefully you'll still be around to help me through this. Thanks so much for all the help so far!
July 23rd 2010, 11:23 AM
Ackbeet

I understand what you mean, but you need to be more careful with parentheses. You really ought to have:

$\displaystyle{\sum_{k=0}^{\infty}\Big[(k+2)(k+1)c_{k+2}(x^{k}-x^{k+2})-(6k+6)c_{k+1}x^{k+1}-4c_{k}x^{k}\Big]=0}.$

I'll wait for your next step.
July 23rd 2010, 07:05 PM
EmilyL

$\displaystyle{\Big[(k+2)(k+1)c_{k+2}(x^{k}-x^{k+2})-(6k+6)c_{k+1}x^{k+1}-4c_{k}x^{k}\Big]=0, k=0,1,2,3...}}.$

I'm not sure this is right, but I'm trying to follow what the example in the book did. I took out the sum symbol, and then wrote k = 0, 1, 2...

From here, according to the example, I should be isolating $\displaystyle C_{k+2}$ on one side. I can do this on paper, but not in LaTeX
July 24th 2010, 03:15 AM
Ackbeet

I was just doing a bit of research, and I thought for a minute there that we were going to have to re-do this whole thing using the method of Frobenius. However, I needn't have worried. Your original DE is very much like the Legendre DE, which does not require the method of Frobenius.

We skipped a step. Before shifting indices, we need to break off terms from the sums so that the powers of x all match up. If we start from post # 8's equation:

$\displaystyle{\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2}-<br /> \sum_{n=2}^{\infty}n(n-1)c_{n}x^{n}<br /> -6\,\sum_{n=1}^{\infty}nc_{n}x^{n}-4\,\sum_{n=0}^{\infty}c_{n}x^{n}=0},$

we break off the terms we need so that all the powers of x are the same. The way you do this is to ensure that the first term of each sum always has the same power of x. In order to do that, by inspection, I can see that the smallest power of x common to all the sums is $x^{2}$ . So, for any sum beginning with a power of x lower than that, I split off terms until I can start the sum at $x^{2}.$ From the first sum, I split off two terms. From the second sum, none. From the third sum, one term, and from the fourth sum, two terms. I get the following:

$\displaystyle{(2c_{2}x^{0}+6c_{3}x^{1}-6c_{1}x^{1}-4c_{0}x^{0}-4c_{1}x^{1})}$
$\displaystyle{+\sum_{n=4}^{\infty}n(n-1)c_{n}x^{n-2}-<br /> \sum_{n=2}^{\infty}n(n-1)c_{n}x^{n}<br /> -6\,\sum_{n=2}^{\infty}nc_{n}x^{n}-4\,\sum_{n=2}^{\infty}c_{n}x^{n}=0}.$

Now, by looking at this expression, and especially the sums, I can tell that the only series we need to shift is the first one: $k=n-2$ . For all the other sums, just substitute $k=n.$ I'll also gather like terms in the broken-off terms. We get:

$\displaystyle{(2c_{2}-4c_{0})+(6c_{3}-6c_{1}-4c_{1})x^{1}}$
$\displaystyle{+\sum_{k=2}^{\infty}(k+2)(k+1)c_{k+2 }x^{k}-<br /> \sum_{k=2}^{\infty}k(k-1)c_{k}x^{k}<br /> -6\,\sum_{k=2}^{\infty}kc_{k}x^{k}-4\,\sum_{k=2}^{\infty}c_{k}x^{k}=0}.$

Now we can add all the series together. You really have to have the indices starting at the same place, and the powers of x all matching up in order to combine the series together. Doesn't this make more sense now? It certainly is making more sense to me! We get the following:

$\displaystyle{(2c_{2}-4c_{0})+(6c_{3}-10c_{1})x}$
$\displaystyle{+\sum_{k=2}^{\infty}\Big[(k+2)(k+1)c_{k+2}-<br /> k(k-1)c_{k}<br /> -6kc_{k}-4c_{k}\Big]x^{k}=0}.$

Now, the big question: what happens next?
July 24th 2010, 09:19 PM
bondesan

I will not add more ideas about the mechanics required to solve this question because Ackbeet already did it, instead, I should mention that makes more sense to work for the recurrence relation if you think that you're in a vectorial space (actually, a space of functions) with polinomials as the basis and you want to find the coefficients that satisfies the linear combination.

If the basis are $\{1=x^0,x^1,x^2,\ldots,x^n\}$ and the coefficients $a_0, a_1,\ldots, a_n$ , then you can write the linear combination as $\sum_{n=0}^{\infty} a_n x^n$ .

So if you have $\sum_{n=0}^{\infty} a_n x^n = 0$ , it follows that $a_0,a_1,\ldots,a_n$ should be all zero given that the basis's vectors are linearly independent.

You can also write like

$\displaystyle{\sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + \sum_{n=2}^{\infty} a_n x^n = 0 \Longleftrightarrow a_0 = a_1 = \ldots = a_n = 0$

See that you can break the sum in as many as terms you wish, and you will often have to do this, because you must match both the index and the power of every sum. So suppose that you took off the terms corresponding to n=0 and n=1 of two distintict sums. Then you should write

$\displaystyle{a_0 x^0 + a_1 x^1 + b_0 x^0 + b_1 x^1 + \sum_{n=2} a_n x^n + \sum_{n=2} b_n x^n =}$

$= \displaystyle{(a_0 + b_0)x^0 + (a_1 + b_1)x^1 + \sum_{n=2} (a_n + b_n)x^n}$ .

So, when you have to strip terms in more than one sum, pay attention to the power of x so you can simplify by gathering the coefficients with same power of x, like Ackbeet did.

I hope it helps.
July 25th 2010, 05:20 AM
EmilyL

Thank you both very much. I was able to contact a student who took this class semester, and he said to do it a completely different way. I'm not sure how it relates to the way we were originally going. Would I end up getting the same answer either way? This is the way he showed me to do it:

Put it in the form $\displaystyle{(1-x^{2})\,\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2}-6\,x\,\sum_{n=1}^{\infty}nc_{n}x^{n-1}-4\,\sum_{n=0}^{\infty}c_{n}x^{n}=0}.$

Then substitute different values for n (0, 1, 2,...) into each part and write them down. The order I'm writing it below is: what n equals, first term, second term, third term.. each seperated by a "||" (i apologize for the confusing notation, its easier on paper where you can space out the columns and rows)

n=0 || 0 || 0 || -4c(0)
n=1 || 0 || -6c(1) || -4c(1)x
n=2 || 2(1-x^2)c(2) || -12x^2c(2) || -4c(2)x^2
n=3 || 6c(3)x - 6c(3)x^3) || -18x^3c(3) || -4c(3)x^3
n=4 || 12c(4)x^2 - 12c(4)x^4 || -24x^4c(4) || -4c(4)x^(4)

then you look at the powers of x in each of those, and set everything with equal powers of x to 0. For example, -6xc(1), -4xc(1) and 6xc(3) are all the terms that involve x to the first power, so you add those terms and set it equal to zero. by doing this for each value of x, you get a bunch of equations, from which you can get all c values in terms of c(1) and c(2).

The main equation is apparently:

$\displaystyle{\sum_{n=0}^{\infty}c_0 + xc_0 + x^2c_2 + x^3c_3....\,}.$

since we can get c(2), c(3), etc in terms of c(0) and c(1), we substituted those values into the above equation for a final answer of:

$\displaystyle{\{y}={c_0}(1 + 2x^2 + 3x^4....) + {c_1}(x +( 5x^3)/3 + (7x^5)/3....)}.$
July 25th 2010, 07:47 AM
bondesan

What your friend taught you was, mutatis mutandi, what we did, but in a less formal way. It works too, but I find more secure doing how Ackbeet did.

If you look closely, first we assumed that the power series $\sum_{n=0}^{\infty} a_n x^n$ would converge to a function y(x), and this y(x) would satisfy the diff. equation. Second, we plugged our guess (or ansatz) in the equation and arranged the terms that was outside the sums into them. By doing this, we sealed the relation of the coefficients.

Right now, you can either plug values of n and write them in a sheet like you did or you can change the indexes and the powers of the sums to match them. By doing this you'll have to strip some terms out (that will have the same power of x), but this is essentially what your friend did.

Now, we are in a situation that an entire expression is equal to zero, but you know that some terms of this expression are not. More than that, you know that this is a linear combination of some functions that are not linearly dependent each other, so it implies that our coefficients should all be equal to zero. Since you do not know the coefficients, but you can find a relation between each other (the recurrence relation), it follows that the series will be dependet of two of them, and this is natural - when you solve a second order diff.

So, the coefficients with even index will necessarily be written in terms of $a_0$ and the odd index as a function of $a_1$ , for example. This will generate two series: one with even powers of x whose coefficients deppends on $a_0$ and one with odd powers with coefficients dependent on $a_1$ . You should realize that this is the same when we solve second order diff. equations without power series - we always end with two functions and two coefficients wich we can determine according to the intial conditions. By the way, you should try to solve with power series some well known second order diff. equations and see that it gives you the same answer (but in taylor series).

Pay attention to your notation too, you wrote $\sum_{n=0}^{\infty} c_0 + xc_0 + x^2 c_2 + ...$ and this is wrong. Either you use the sigma or you write term by term. All right?

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