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The differential equation $y''+xy=0$ is given.

Find the solution of the differential equation, using the power series method.

That's what I have tried:

We are looking for a solution of the form $y(x)= \sum_{n=0}^{\infty} a_n x^n$ with radius of convergence of the power series $R>0$.

Then:

$$y'(x)= \sum_{n=1}^{\infty} n a_n x^{n-1}= \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n$$

$$y''(x)= \sum_{n=1}^{\infty} (n+1) n a_{n+1} x^{n-1}= \sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n$$

Thus:

$$\sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n+ x \sum_{n=0}^{\infty} a_n x^n=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+ \sum_{n=0}^{\infty} a_n x^{n+1}=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+ \sum_{n=1}^{\infty} a_{n-1} x^n=0 \\ \Rightarrow 2a_2+\sum_{n=1}^{\infty} \left[ (n+2) (n+1) a_{n+2}+ a_{n-1}\right] x^n=0$$

So it has to hold:

$$a_2=0 \\ (n+2) (n+1) a_{n+2}+a_{n-1}=0, \forall n=1,2,3, \dots$$

For $n=1$: $3 \cdot 2 \cdot a_3+ a_0=0 \Rightarrow a_3=-\frac{a_0}{6}$

For $n=2$: $4 \cdot 3 \cdot a_4+a_1=0 \Rightarrow a_4=-\frac{a_1}{12}$

For $n=3$: $5 \cdot 4 \cdot a_5+a_2=0 \Rightarrow a_5=0$

For $n=4$: $6 \cdot 5 \cdot a_6+a_3=0 \Rightarrow 30 a_6-\frac{a_0}{6}=0 \Rightarrow a_6=\frac{a_0}{6 \cdot 30}=\frac{a_0}{180}$

For $n=5$: $7 \cdot 6 \cdot a_7+ a_4=0 \Rightarrow 7 \cdot 6 \cdot a_7-\frac{a_1}{12}=0 \Rightarrow a_7=\frac{a_1}{12 \cdot 42}$

Is it right so far? If so, how could we find a general formula for the coefficients $a_n$?

EDIT: Will it be as follows:

$$a_{3k+2}=0$$

$$a_{3k}=(-1)^k \frac{a_0}{(3k)!} \prod_{i=0}^{k-1} (3i+1)$$

$$a_{3k+1}=(-1)^k \frac{a_1}{(3k+1)!} \prod_{i=0}^{k-1} (3i+2)$$

If so, then do we have to write seperately the formula for the coefficients of $x^0, x^1$, because otherwhise the sum would be from $0$ to $-1$ ?

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  • $\begingroup$ The coefficients aren't too nice. Check out the Wolfram Alpha result here and go to the documentation on the Airy functions here. It'll help you figure out what the general form of the coefficients is. Particularly, see equations $(9)$ and $(10)$ on the Wolfram MathWorld page. $\endgroup$ – Cameron Williams Mar 19 '15 at 21:09
  • $\begingroup$ You can find them through recursion by rewriting it as $$(n+2)(n+1)a_{n+3} + a_{n} = 0 $$ $$ a_{n+3} = -\frac{a_{n}}{(n+2)(n+1)}$$ Since $a_2 = 0$, you have $a_{2+3k} = 0$ for all integer $k$ $\endgroup$ – Dylan Mar 19 '15 at 21:20
  • $\begingroup$ @Dylan Shouldn't it be $(n+3)(n+2) a_{n+3}+a_n=0$ ? $\endgroup$ – evinda Mar 19 '15 at 22:43
  • $\begingroup$ @evinda You're right, I screwed up. But you get my point, right? $\endgroup$ – Dylan Mar 20 '15 at 2:09
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    $\begingroup$ What are the initial conditions for $y(0)$ and $y'(0)$ ? $\endgroup$ – Han de Bruijn Jun 4 '15 at 9:27
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Follow-up from my comment, every $a_{n+3}$ can be expressed in terms of $a_n$. $$a_{n+3} = -\frac{a_{n}}{(n+3)(n+2)}$$

Since $a_2 = 0$, all coefficients with $n = 3k + 2$ will be $0$ The rest will have to based on the initial conditions of $a_0 = y(0)$ and $a_1 = y'(0)$

For $n = 3k$ $$a_3 = -\frac{a_0}{3\cdot2}$$ $$a_6 = -\frac{a_3}{6\cdot5} = \frac{a_0}{6\cdot5\cdot3\cdot2} $$

This can be extrapolated to $$ a_{n} = \pm\frac{a_0}{n(n-1)(n-3)(n-4)\cdots\cdot6\cdot5\cdot3\cdot2} $$ Basically, the denominator is a product of all integers from $1$ to $n$, skipping every 3rd number and alternating signs. An alternate notation is $$a_{3k} = (-1)^k\frac{a_0}{(3k)!}\prod_{i=0}^{k-1} (3i+1) $$

The same can be done for $n = 3k+1$ $$a_{3k+1} = (-1)^k\frac{a_1}{(3k+1)!}\prod_{i=0}^{k-1} (3i+2)$$

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  • $\begingroup$ Isn't it as follows, Dylan? $$$$ $$a_{n+3}=-\frac{a_n}{(n+3)(n+1)}$$ $$a_3=-\frac{a_0}{3 \cdot 1}$$ $$a_6=\frac{a_0}{3 \cdot 4 \cdot 6}$$ $$a_9=\frac{-a_0}{3 \cdot 4 \cdot 6 \cdot 7 \cdot 9}$$ So: $$a_{3k}=(-1)^k \frac{a_0}{n(n-2)(n-3) \cdots 3}$$ $$a_4=-\frac{a_1}{2 \cdot 4}$$ $$a_7=\frac{a_1}{2 \cdot 4 \cdot 5 \cdot 7}$$ $$a_{3k+1}=(-1)^k \frac{a_1}{n(n-2)(n-3) \cdots 4 \cdot 2}$$ Don't we also have to check the case $n=3k+2$? $$$$ Also what can we deduce for the radius of convergence? $\endgroup$ – evinda Mar 22 '15 at 23:06
  • $\begingroup$ I fixed the calculation. Also, as I said above, $a_{3k+2} = 0$ since $a_2 = 0$ $\endgroup$ – Dylan Mar 23 '15 at 23:49
  • $\begingroup$ I think that it should be as follows: $$$$ For $n=3k$: $$a_{n}=(-1)^k \frac{a_0}{(3k)!} \prod_{i=0}^{k-1} (3i+1)$$ and for $n=3k+1$: $$a_n=(-1)^k \frac{a_1}{(3k+1)!} \prod_{i=0}^{k-1} (3i+2)$$ Or am I wrong? $\endgroup$ – evinda Mar 24 '15 at 22:39
  • $\begingroup$ Also taking the limit of the ratios $\frac{a_{3(k+1)}}{a_{3k}}$ and $\frac{a_{3(k+1)+1}}{a_{3k+1}}$, we will get $0$ and so the radii of convergence of $\sum_k a_{3k}x^{3k}$ and $\sum_k a_{3k+1} x^{3k+1}$ will be $+\infty$, so the series converge . $$$$ So the solution of the differential equation will be: $$\sum_k a_{3k} x^{3k}+ \sum_k a_{3k+1} x^{3k+1}$$ where $a_{3k}, a_{3k+1}$ have the above formulas and the radius of convergence of the solution will be $+\infty$. Right? $\endgroup$ – evinda Mar 24 '15 at 22:56
  • $\begingroup$ Looks right to me :) $\endgroup$ – Dylan Mar 25 '15 at 23:18
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$$y''+xy=0$$ $$y=\sum_{n=0}^{\infty} c_n x^n, y'=\sum_{n=1}^{\infty} nc_n x^{n-1}, y''=\sum_{n=2}^{\infty} n(n-1)c_n x^{n-2}$$ $$\therefore y''+xy=0=\underbrace{\sum_{n=2}^{\infty} n(n-1)c_n x^{n-2}}_{k=n-2\Rightarrow n=k+2}+\underbrace{\sum_{n=0}^{\infty} c_n x^{n+1}}_{k=n+1\Rightarrow n=k-1}=\sum_{k=0}^{\infty} (k+2)(k+1)c_{k+2} x^{k}+\sum_{k=1}^{\infty} c_{k-1} x^k$$ $$=2c_2+\sum_{k=1}^{\infty} (k+2)(k+1)c_{k+2} x^{k}+\sum_{k=1}^{\infty} c_{k-1} x^k=2c_2+\sum_{k=1}^{\infty} [(k+2)(k+1)c_{k+2}+ c_{k-1}] x^k=0$$

Thus, $$2c_2=0\Rightarrow c_2=0$$ $$(k+2)(k+1)c_{k+2}+ c_{k-1}=0\Rightarrow c_{k+2}=-\frac{c_{k-1}}{(k+2)(k+1)}\forall k=1,2,3,\ldots$$

Choosing $c_0=1$ and $c_1=0$, we find $$c_2=0,c_3=-1/6,c_4=0, c_5=0, c_6=1/180\ldots$$ and so on.

Choosing $c_0=0$ and $c_1=1$, we find $$c_2=0,c_3=0,c_4=-1/12, c_5=0, c_6=0,c_7=1/504\ldots$$ and so on.

Thus, the two solutions are $$y_1=1-\frac{1}{6}x^3+\frac{1}{180}x^6+\ldots$$ $$y_2=x-\frac{1}{12}x^4+\frac{1}{504}x^7+\dots$$

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  • $\begingroup$ @evinda: Does this solve your problem? $\endgroup$ – Yagna Patel May 31 '15 at 16:07
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Disclaimer. This is not an answer, but rather a too long comment with some graphics in it.
The equation looks like the common ODE for a harmonic oscillator: $y'' + \omega^2 y = 0$ with the square of the frequency varying proportional to "time" $x$ . Numerical simulation with the initial conditions $y(0)=0$ and $y'(0)=1$ reveals that the solution indeed looks like that (I hate solutions without a picture, you know):
enter image description here
The viewport is $\,0 < x < 40\,$ and $\,-1.5 < y < +1.5\,$.
Program (Delphi Pascal) snippet for doing the calculations and the drawing (not optimized at all):

{
The equations of motion are solved numerically as follows.
Start with: y(0) = 0 ; x = 0 ;
(y(dx) - y(0))/dx = v ==> y(dx) = y(0) + v.dx
y(x + dx) - 2.y(x) + y(x - dx) ------------------------------ + x.y(x) = 0 ==> dx^2
y(x + dx) = 2.y(x) - y(x - dx) - dx^2.x.y(x) ==>
y(0.dx) = 0 y(1.dx) = y(0.dx) + v.dx y(2.dx) = 2.y(1.dx) - y(0.dx) - dx^2.x.y(1.dx) y(3.dx) = 2.y(2.dx) - y(1.dx) - dx^2.x.y(2.dx) .............................................. y(k+1).dx) = 2.y(k.dx) - y((k-1).dx) - dx^2.x.y(k.dx) } procedure bereken; const N : integer = 40000; v : double = 1; var x,dx,y0,y1,y2 : double; k : integer; begin x := 0; dx := 0.001; y0 := 0; y1 := y0+v*dx; Form1.Image1.Canvas.MoveTo(x2i(0),y2j(0)); for k := 0 to N-1 do begin x := x + dx; y2 := 2*y1 - y0 - dx*dx*x*y1; Form1.Image1.Canvas.LineTo(x2i(x),y2j(y2)); y0 := y1; y1 := y2; end; end;

Note that the solution becomes very oscillatory (i.e. singular) for $x\to\infty$ . However, the frequency only varies with the square root of the distance: $\omega=\sqrt{x}$ , therefore it doesn't happen immediately.

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