tg

tg
tgt

Wednesday, 27 March 2013

Series Solutions: Airy's Equation


The general form of a homogeneous second order linear differential equation looks as follows:

y''+p(ty'+q(ty=0.



The series solutions method is used primarily, when the coefficients p(t) or q(t) are non-constant.

One of the easiest examples of such a case is Airy's Equation

y''-t y=0,


which is used in physics to model the defraction of light.
We want to find power series solutions for this second-order linear differential equation.

The generic form of a power series is

\begin{displaymath}y(t)=\sum_{n=0}^\infty a_n t^n.\end{displaymath}


We have to determine the right choice for the coefficients (an).
As in other techniques for solving differential equations, once we have a "guess" for the solutions, we plug it into the differential equation. Recall that

\begin{displaymath}y''(t)=\sum_{n=2}^\infty n(n-1) a_n t^{n-2}.\end{displaymath}



Plugging this information into the differential equation we obtain:

\begin{displaymath}\sum_{n=2}^\infty n(n-1) a_n t^{n-2}-t\sum_{n=0}^\infty a_n t^n=0, \end{displaymath}


or equivalently

\begin{displaymath}\sum_{n=2}^\infty n(n-1) a_n t^{n-2}-\sum_{n=0}^\infty a_n t^{n+1}=0. \end{displaymath}



Our next goal is to simplify this expression such that (basically) only one summation sign "$\sum$" remains. The obstacle we encounter is that the powers of both sums are different, tn-2 for the first sum and tn+1 for the second sum. We make them the same by shifting the index of the first sum up by 2 units and the index of the second sum down by one unit to obtain



\begin{displaymath}\sum_{n=0}^\infty (n+2)(n+1) a_{n+2} t^{n}-\sum_{n=1}^\infty a_{n-1} t^{n}=0.\end{displaymath}



Now we run into the next problem: the second sum starts at n=1, while the first sum has one more term and starts at n=0. We split off the 0th term of the first sum:

\begin{displaymath}\sum_{n=0}^\infty (n+2)(n+1) a_{n+2} t^{n}=2\cdot 1\cdot a_2+
\sum_{n=1}^\infty (n+2)(n+1) a_{n+2} t^{n}.\end{displaymath}



Now we can combine the two sums as follows:

\begin{displaymath}2 a_2 +\sum_{n=1}^\infty \left( \phantom{\int}(n+2)(n+1) a_{n+2} t^{n}- a_{n-1} t^n\phantom{\int}\right)=0,\end{displaymath}


and factor out tn:

\begin{displaymath}2 a_2+\sum_{n=1}^\infty \left( \phantom{\int}(n+2)(n+1) a_{n+2} - a_{n-1} \phantom{\int}\right)t^n=0.\end{displaymath}




The power series on the left is identically equal to zero, consequently all of its coefficients are equal to 0:

\begin{displaymath}\left\{\begin{array}{cl}2a_2&=0\\ (n+2)(n+1) a_{n+2} - a_{n-1} &=0 \mbox{ for all } n=1,2,3,\ldots\end{array}\right.\end{displaymath}



We can slightly rewrite as

\begin{displaymath}\left\{\begin{array}{ll}a_2&=0\\ a_{n+2}& =\displaystyle{\fra...
...{(n+1)(n+2)}} \mbox{ for all } n=1,2,3,\ldots\end{array}\right.\end{displaymath}


These equations are known as the "recurrence relations" of the differential equations. The recurrence relations permit us to compute all coefficients in terms of a0 and a1.
We already know from the 0th recurrence relation that a2=0. Let's compute a3 by reading off the recurrence relation for n=1:

\begin{displaymath}a_3=\frac{a_0}{2\cdot 3}.\end{displaymath}


Let us continue:

\begin{eqnarray*}a_4&=&\frac{a_1}{3\cdot 4}\\
a_5&=&\frac{a_2}{4\cdot 5}=0\\
a...
...&\frac{a_6}{8\cdot 9}=\frac{a_0}{(2\cdot 3)(5\cdot 6)(8\cdot 9)}
\end{eqnarray*}



The hardest part, as usual, is to recognize the patterns evolving; in this case we have to consider three cases:

1. All the terms $a_2,a_5,a_8,\ldots$ are equal to zero. We can write this in compact form as

\begin{displaymath}a_{3k+2}=0\mbox{ for all }k=0,1,2,3,\ldots\end{displaymath}



2. All the terms $a_3,a_6,a_9,\ldots$ are multiples of a0. We can be more precise:

\begin{displaymath}a_{3k}=\frac{1}{(2\cdot 3)(5\cdot 6)\cdots((3k-1)\cdot (3k))}\cdot a_0
\mbox{ for all }k=1,2,3,\ldots\end{displaymath}


(Plug in k=1,2,3,4 to check that this works!)
3. All the terms $a_4,a_7,a_{10},\ldots$ are multiples of a1. We can be more precise:

\begin{displaymath}a_{3k+1}=\frac{1}{(3\cdot 4)(6\cdot 7)\cdots((3k)\cdot (3k+1))}\cdot a_1
\mbox{ for all }k=1,2,3,\ldots\end{displaymath}


(Plug in k=1,2,3,4 to check that this works!)
Thus the general form of the solutions to Airy's Equation is given by

\begin{eqnarray*}y(t)&=&a_0\left(1+\sum_{k=1}^\infty \frac{t^{3k}}{(2\cdot 3)(5\...
...{t^{3k+1}}{(3\cdot 4)(6\cdot 7)\cdots((3k)\cdot (3k+1))}\right).
\end{eqnarray*}



Note that, as always, y(0)=a0 and y'(0)=a1. Thus it is trivial to determine a0 and a1 when you want to solve an initial value problem.

In particular

\begin{displaymath}y_1(t)= 1+\sum_{k=1}^\infty \frac{t^{3k}}{(2\cdot 3)(5\cdot 6)\cdots((3k-1)\cdot (3k))}\end{displaymath}


and

\begin{displaymath}y_2(t)=t+\sum_{k=1}^\infty\frac{t^{3k+1}}{(3\cdot 4)(6\cdot 7)\cdots((3k)\cdot (3k+1))}\end{displaymath}


form a fundamental system of solutions for Airy's Differential Equation.
Below you see a picture of these two solutions. Note that for negative t, the solutions behave somewhat like the oscillating solutions of y''+y=0, while for positive t, they behave somewhat like the exponential solutions of the differential equation y''-y=0.



Reduction of Order Technique


This technique is very important since it helps one to find a second solution independent from a known one. , in order to find the general solution to y'' + p(x)y' + q(x)y = 0, we need only to find one (non-zero) solution, tex2html_wrap_inline143 .
Let tex2html_wrap_inline143 be a non-zero solution of
displaymath147
Then, a second solution tex2html_wrap_inline149 independent of tex2html_wrap_inline143 can be found as
displaymath135
Easy calculations give
displaymath136,
where C is an arbitrary non-zero constant. Since we are looking for a second solution one may take C=1, to get
displaymath157
Remember that this formula saves time. But, if you forget it you will have to plug tex2html_wrap_inline159 into the equation to determine v(x) which may lead to mistakes !
The general solution is then given by
displaymath137

Example: Find the general solution to the Legendre equation
displaymath163,
using the fact that tex2html_wrap_inline165 is a solution.
Solution: It is easy to check that indeed tex2html_wrap_inline165 is a solution. First, we need to rewrite the equation in the explicit form
displaymath169
We may try to find a second solution tex2html_wrap_inline171 by plugging it into the equation. We leave it to the reader to do that! Instead let us use the formula
displaymath173
Techniques of integration (of rational functions) give
displaymath175,
which gives
displaymath177
The general solution is then given by
displaymath179

Series Solutions: Taking Derivatives and Index Shifting


Throughout these pages I will assume that you are familiar with power series and the concept of the radius of convergence of a power series.

Given a power series

\begin{displaymath}y(t)=\sum_{n=0}^\infty a_n t^n=a_0+a_1 t+a_2 t^2 +a_3 t^3+\ldots,\end{displaymath}


we can find its derivative by differentiating term by term:

\begin{displaymath}y'(t)=\sum_{n=1}^\infty n a_n t^{n-1}=a_1+2 a_2 t + 3 a_3 t^2+\ldots\end{displaymath}


Here we used that the derivative of the term an tn equals an n tn-1. Note that the start of the summation changed from n=0 to n=1, since the constant term a0 has 0 as its derivative.
The second derivative is computed similarly:

\begin{displaymath}y''(t)=\sum_{n=2}^\infty n(n-1) a_n t^{n-2}=2 a_2 +6 a_3 t+\ldots.\end{displaymath}



Taking the derivative of a power series does not change its radius of convergence, so $y(t), y'(t),y''(t),\dots$ will all have the same radius of convergence.



The rest of this section is devoted to "index shifting". Consider the example

\begin{displaymath}\int_2^5 (x+1)^5\,dx.\end{displaymath}


Using a simple substitution u=x+1, we can rewrite this integral as

\begin{displaymath}\int_3^6 u^5 \,du,\end{displaymath}


or changing the dummy variable u back to x, we get:

\begin{displaymath}\int_2^5 (x+1)^5\,dx=\int_3^6 x^5\, dx.\end{displaymath}


The expression (x+1) is "shifted down" by one unit to x, while the limits of integration are "shifted up" by one unit from 2 to 3, and 5 to 6.
Summation is just a special case of integration, so an analogous "index shifting" will work:

\begin{displaymath}\sum_2^5 (n+1)^5=\sum_3^6 n^5.\end{displaymath}


You should convince yourself that both of these expressions are indeed the same, by writing out explicitly the four terms of each of the two formulas!
Let's try this for our derivative formulas:

\begin{eqnarray*}y'(t)&=&\sum_{n=1}^\infty n a_n t^{n-1}\\
&=&\sum_{n=0}^\infty...
..._{n+1} t^{((n+1)-1)}\\
&=&\sum_{n=0}^\infty (n+1) a_{n+1} t^{n}
\end{eqnarray*}


We shifted each occurrence of n in the expression up by one unit, while the limits of summation were shifted down by one unit, from 1 to 0, and from $\infty$ to $\infty-1=\infty$.
You should once again convince yourself that the first and the last formula are indeed the same, by writing out explicitly the first few terms of each of the two formulas!

As a last example, let's shift the formula for the second derivative by 2 units:

\begin{eqnarray*}y''(t)&=&\sum_{n=2}^\infty n(n-1) a_n t^{n-2}\\
&=&\sum_{n=0}^...
...2} t^{((n+2)-2)}\\
&=&\sum_{n=0}^\infty (n+2)(n+1)a_{n+2} t^{n}
\end{eqnarray*}