Airy's equation bounces back

[logistic_guy]

Find a series solution of Airy's equation $\displaystyle y'' - xy = 0$ in the form $\displaystyle \sum_{n=0}^{\infty}a_n(x - 1)^n$.

 

[khansaheb]

Find a series solution of Airy's equation $\displaystyle y'' - xy = 0$ in the form $\displaystyle \sum_{n=0}^{\infty}a_n(x - 1)^n$.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[logistic_guy]

A long time ago I solved the Airy's equation from scratch by using the form $\displaystyle \sum_{n=0}^{\infty}a_nx^n$. I was the only one on the earth to solve it like that, and I will always be the only one. Why? Because it is very difficult for others to understand the art of differential equations!

Now I am challenging myself to solve it again but with the new form $\displaystyle \sum_{n=0}^{\infty}a_n(x - 1)^n$. There's a baker who works in the bakery where I always go to buy bread. He is a nice guy and is very interested in my posts in this forum. He follows my solutions in a daily basis. He was once a professor of mathematics, but got bored of teaching and decided to work in the bakery. He advised me to change the form the Airy's equation to $\displaystyle y'' - (x - 1)y - y = 0$ before I solve it. He did not tell the reason and I did not ask, but I am guessing that the calculations will be easier.

So let us listen to the baker and beat this Airy again.

We have the differential equation:
$\displaystyle y'' - (x - 1)y - y = 0$

And
I will let $\displaystyle y = \sum_{n=0}^{\infty}a_n(x - 1)^n$

In the next post, we will take the first and the second derivatives of this function $\displaystyle y$. Be in touch with this thread if you wanna enjoy the show!

 

[logistic_guy]

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

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

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

 

[khansaheb]

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

 

[logistic_guy]

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

You're stealing my brand

$\displaystyle y'' - (x - 1)y - y = 0$

Now we substitute what we got in the differential equation above.

$\displaystyle \sum_{n=2}^{\infty}a_n n(n-1)(x - 1)^{n-2} - (x - 1)\sum_{n=0}^{\infty}a_n(x - 1)^n - \sum_{n=0}^{\infty}a_n(x - 1)^n = 0$

 

[logistic_guy]

Now we will do trick $\displaystyle 47$. What's that? It's the hitman style $\displaystyle \rightarrow$ we change any index $\displaystyle n$ that is not starting at $\displaystyle 0$ to $\displaystyle 0$.

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

The fun has just started!

 

[logistic_guy]

We have solved the index problem, but a new problem popped up. The first and third terms contain $\displaystyle (x - 1)^n$ while the second term contains $\displaystyle (x - 1)^{n+1}$. They don't match.

We can do this for now:

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

 

[logistic_guy]

$\displaystyle \sum_{n=0}^{\infty}a_{n+2} (n + 2)(n + 1)(x - 1)^{n} = 2a_2 + \sum_{n=1}^{\infty}a_{n+2} (n + 2)(n + 1)(x - 1)^{n}$

And

$\displaystyle \sum_{n=0}^{\infty}a_n(x - 1)^n = a_0 + \sum_{n=1}^{\infty}a_n(x - 1)^n$

Then, we have:

$\displaystyle 2a_2 + \sum_{n=1}^{\infty}a_{n+2} (n + 2)(n + 1)(x - 1)^{n} - \sum_{n=1}^{\infty}a_{n-1}(x - 1)^{n} - a_0 - \sum_{n=1}^{\infty}a_n(x - 1)^n = 0$

 

[logistic_guy]

Let us simplify it further.

$\displaystyle 2a_2 - a_0 + \sum_{n=1}^{\infty}\left[a_{n+2} (n + 2)(n + 1) - a_{n-1} - a_n\right](x - 1)^n = 0$

 

[fresh_42]

Why do you think $y'$ is incorrect?

 

[khansaheb]

Why do you think $y'$ is incorrect?

Differentiate term by term and you will see.

 

[fresh_42]

Differentiate term by term and you will see.

I don't see any mistakes in either derivation. The factor $n=0$ eliminates the negative power.

 

[logistic_guy]

Let us simplify it further.

$\displaystyle 2a_2 - a_0 + \sum_{n=1}^{\infty}\left[a_{n+2} (n + 2)(n + 1) - a_{n-1} - a_n\right](x - 1)^n = 0$

This gives us:

$\displaystyle 2a_2 - a_0 = 0$

And

$\displaystyle a_{n+2} (n + 2)(n + 1) - a_{n-1} - a_n = 0$

Or

$\displaystyle a_2 = \frac{a_0}{2}$

And

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

It seems that we have a recursive formula in here

 

[logistic_guy]

Let us use the recursion formula and calculate some arbitrary constants.

When $\displaystyle n = 1$

$\displaystyle a_3 = \frac{a_0 - a_1}{6}$


When $\displaystyle n = 2$

$\displaystyle a_4 = \frac{a_1 - a_2}{12} = \frac{a_1 - \frac{a_0}{2}}{12} = \frac{2a_1 - a_0}{24}$


When $\displaystyle n = 3$

$\displaystyle a_5 = \frac{a_2 - a_3}{20} = \frac{\frac{a_0}{2} - \frac{a_0 - a_1}{6}}{20} = \frac{2a_0 + a_1}{120}$


When $\displaystyle n = 4$

$\displaystyle a_6 = \frac{a_3 - a_4}{30} = \frac{\frac{a_0 - a_1}{6} - \frac{2a_1 - a_0}{24}}{30} = \frac{5a_0 - 6a_1}{720}$

 

[logistic_guy]

Let us calculate the first $\displaystyle 7$ terms.

$\displaystyle y(x) = \sum_{n=0}^{\infty}a_n(x - 1)^n = a_0 + a_1(x - 1) + a_2(x - 1)^2 + a_3(x - 1)^3 + a_4(x - 1)^4 + a_5(x - 1)^5 + a_6(x - 1)^6 + \cdots$

 

[logistic_guy]

Let us now replace some of the constants with what we got from the recursion formula.

$\displaystyle y(x) = a_0 + a_1(x - 1) + \frac{a_0}{2}(x - 1)^2 + \left(\frac{a_0 - a_1}{6}\right)(x - 1)^3 + \left(\frac{2a_1 - a_0}{24}\right)(x - 1)^4 + \left(\frac{2a_0 + a_1}{120}\right)(x - 1)^5 + \left(\frac{5a_0 - 6a_1}{720}\right)(x - 1)^6 + \cdots$

 

[fresh_42]

I have found a recursion $b_{n+2}=b_n+nb_{n-1}$ where $b_n=n!a_n$ and $b_0=y(1)\, , \,b_1=y'(1)\, , \,b_2=y(1).$ I haven't found a closed formula for it since the lack of a $b_{n+1}$ term together with the asymmetric weight make the values jump a bit.

Here is my calculation:
$$y''=xy\, , \,y=\sum_{n=0}^\infty a_n(x-1)^n\, , \,a_0=y(1)\, , \,a_1=y'(1)$$$$\begin{array}{lll} y'' &=\displaystyle{ \sum_{n=0}^{\infty}a_n n(n-1)(x - 1)^{n-2}=\sum_{k=2}^\infty a_k k(k-1)(x - 1)^{k-2}} \\[18pt] &=\displaystyle{\sum_{n=0}^\infty a_{n+2}(n+2)(n+1)(x-1)^n=x\sum_{n=0}^{\infty}a_n(x-1)^n}\\[18pt] &=\displaystyle{(x-1+1)\sum_{n=0}^{\infty}a_n(x-1)^n=\sum_{n=0}^{\infty}a_n(x-1)^n+\sum_{n=0}^\infty a_n(x-1)^{n+1}}\\[18pt] &=\displaystyle{\sum_{n=0}^{\infty}a_n(x-1)^n+\sum_{k=1}^\infty a_{k-1}(x-1)^{k}} \end{array}$$
$$\begin{array}{lll} 2a_2=a_0=y(1)\, , \,a_1=y'(1)\\[12pt] a_{n+2}=\dfrac{a_{n-1}+a_n}{(n+1)(n+2)}=\dfrac{n!}{(n+2)!}(a_{n-1}+a_n)\, , \,n\geq 1\\[18pt] (n+2)!a_{n+2}=n!a_n+n\cdot (n-1)!a_{n-1}\, , \,n\geq 1\\[12pt] b_n:=n!a_n\, , \,b_0=a_0\, , \,b_1=a_1\, , \,b_2=2a_2=a_0\, , \,n\geq 0\\[12pt] b_{n+2}=b_n+nb_{n-1}\, , \,n\geq 1 \end{array}$$
However, $b_{n-1}=\dfrac{b_{n+2}-b_n}{n}$ goes like $f(x)=f'(x)$ and is thus exponential.

 

[fresh_42]

I got

$$\begin{array}{lll} \begin{bmatrix} b_{2n-1}\\b_{2n} \\b_{2n+1} \\ \end{bmatrix}& =\underbrace{\left(\prod_{k=1}^{n-1}\begin{bmatrix}0&0&1\\2k&1&0\\0&2k+1&1\end{bmatrix}\right)}_{k=n-1\text{ most left}} \begin{bmatrix} b_{1}\\b_0 \\b_0+b_1 \\ \end{bmatrix} \end{array}$$

 

[fresh_42]

The matrix formula allows you to calculate all the $a_n$ but you should use a computer program. Already small values become inconvenient:
$$\begin{pmatrix}b_9\\b_{10}\\b_{11}\end{pmatrix}=\begin{pmatrix}9!\,a_9\\10!\,a_{10}\\11!\,a_{11}\end{pmatrix}= \begin{pmatrix}24&15&29\\82&83&18\\42&186&119\end{pmatrix} \begin{pmatrix}a_1\\a_0\\a_0+a_1\end{pmatrix}$$