power series solutions at ordinary points

7/29/2019 Power Series Solutions at Ordinary Points

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Copyright Ren Barrientos Page 1

MAP2302 Lecture 10

2011-3

Series Solutions at ordinary Points

The idea behind series methods is this: well-behaved functions are analytic in some subset of

. It stands to

reason that if the equation 1has well-behaved coefficients and forcing function, then its solutions should be also well-behaved analyticfunctions. Hence, we should expect two linearly independent solutions which have power series representationsin some common subsetIof the reals. Hence, we assume a solution of the form

This solution involves an infinite number of arbitrary constants (namely, the coefficients), but we know that asecond order equation can only have two. Therefore, the quantities must be related andthe goal of the methodis to find that relationship. Before we continue we need to state a few fundamental facts and definitions.

Clearly if 0 then is a singular point. We will not treat singular points in these notes, but you shouldbe able to distinguish between the following two types of singularities:

Regular and Irregular Singular Points

A point

is a regular singularpoint if it is a singular point but both

and

are analytic at

.

Otherwise we say it is a irregular singular point. Methods to solve equations at regular singular points are treated in theappendix.

Existence of Power Series Solutions

Example 1The equation

1

0has

1as a singularity since it is clear that

and are not analytic at 1. On the other hand, both 1 1 and 1 1 are analytic at 1. Therefore, 1 is a regular singular point.Observe that on the other hand, 0 is an ordinary point. Therefore, we are guaranteed to have twolinearly independent solutions of the form

Suppose is an ordinary point of equation (1). Then there exist two linearly independent soluti ons of theform

which converge on the interval | | where is at least as large as the distance from to thenearest singularity of the equation.

and Ordinary Points of an Equation

A point is anordinary pointof equation (1) if both

areanalyticat. On the contrary, If is not an ordinary point, we call it asingular point or a singularity.


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which converge on the interval | 0| where is at least 1, the distance from the ordinary point 0 to the closest singularity 1. Notice that is not the only ordinary point of this equation; anyother point other than

1would have done.

Example 2 Determine the singular points of the equation 9 2 4 0. What interval ofconvergence is expect for the series solution about 1?Solution 9 so 3 are the a singular points of the equation. 2/ 3 and 4/ 3 are both analytic at 1. Therefore, 1 is an ordinary point and we are

guaranteed to have two linearly independent solutions of the form

1 Which converges on | 1| , where R is at least 2, the distance from 1 to the nearestsingularity

3. Thus

2.

The Method

Suppose is an ordinary point of the differential equation (1). ThenStep 1 Let Step 2 Substituteyin the differential equation and adjust the terms so that all sums can be combined into

a single sum.

Step 3 Obtain a recursive formulaand generate as many terms as needed in order to determine theirgeneral form, or if not possible enough terms to obtain the pattern of the two linearly independentsolutions.

Step 4 If possible, write the solution in summation form and apply the initial conditions if they are given.

As a quick review, term-by-term differentiation can be done within a series interval of convergence and the

result is as follows: If Then 2 3 and 2 3 2 4 3 2 1 and so on.

However, notice that nothing is altered if we start the last two sums at 0. This will be an option that may beexercised and both methods will be illustrated in the examples below.

Example 3 Find the series solution of the equation 0.SolutionWe already well know that this equation has as solution wherek is a constant. However, letus see what the series method gives us. Observe that 0 is an ordinary point.Step 1 let . Then


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Step 2 Substituting the sums in the equation 0: 0

In order to combine these, we must re-index the sums. Let

1 in the first sum. Then

1

and substitution gives us

1 0Since the indeces are just dummy variables, we can express both sums in terms of or; let us use,which is a totally arbitrary choice:

1 0These may be combined into a single sum:

1 0Step 3 This equation is true for all

if and only if

1 0 ; 0 ,1 ,2 ,or 1 1 ; 0 ,1 ,2 ,

This relation, called therecursive relation or formula, may be used to generate the terms of the series.Since we have a first order equation, there can only be one arbitrary constant. Therefore, thecoefficients must reduce to a single one. Let us generate these terms in order to establish thisconnection. 0 :

1 :

2

2 : 3 3 : 4 4 : 5 ...

It is clear that

![recall that ! 1 2 3 2 1]Look at what the series solution has become: Substituting the coefficients,

!


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Factoring: !

Thus, we reach the pleasant but anticipated conclusion that .Let us take this opportunity to show another technique for obtaining the recursive relation which, as you haveseen, is at the heart of this method. Observe that

because the term corresponding to 0 in the second sum is 0 anyway. So nowStep 2 above looks like this:

0In order to write the left-hand side as a single sum, let in the first sum. Then and 0 1

. Thus,

0

becomes

1 0Renaming to the index in the second sum:

1

0Now notice that when 1, the first term of the first sum is 0. Therefore, we may start that sumwith 0 as well: 1 0

Thus, we can express this as a single sum:

1 0Hence, 1 0

or

, , , ,

This is exactly the same recursive formula we obtained earlier.

Example 4 (Another familiar equation)Find the series solution of the equation 0.Solution

Once again 0 is an ordinary point. Therefore, we can take as our tentative solution.What we expect is to obtain the two already known solutions s i n and c o s . Let us see whatsolutions the method of series produces.

1 1 0 1


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Computing the derivatives and using the idea outlined in the discussion following example 3,

1

1

Substituting in the equation, 1 0We need to do some adjustments:

1 0

With this re-indexing we have:

2 1 0Observe that the first two terms of the first series corresponding to 2 and 1 are 0.

Therefore, we may start it at 0 as well. 2 1 0

Thus, 2 1 0The recursive formula is:

2 1 0 ; 0 ,1 ,2 ,Solving for : ; , , , Generating some of the coefficients will give us an idea of what they look like. Remember, in thisproblem there can only be two arbitrary constants since we have a second order equation.

0: 2 1 1: 3 2

2:

4 3

4 3 2 1 3: 5 4 5 4 3 2 4: 6 5 6 5 4 3 2 1

5: 7 6 7 6 5 4 3 2and so on.

Thus, the solution becomes


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!

!

Grouping the common constants,

! ! ! ! We could not be happier. The terms in parentheses correspond to the cosine and the sine series,respectively. Thus, as expected.

Example 5 (non-constant coefficients) find the series solution of the equation 0. Express the answer insummation form.

Solution

This equation can be solved by separation of variables and you should do so in order to compare theanswers. 0 is an ordinary point of this equation. In fact, since the leading function 1 for allx, everyreal number is an ordinary point. This is a first order equation so we expect only one arbitrary constant.

Let be the solution. Substituting in the equation,

0Absorbingx in the second sum,

0We need to fix the powers in both sums:

0

With these substitutions, the equation becomes

1

0

Once again, the term corresponding to 1 in the first sum is0. Therefore, we may write1

0In order to align the sums, we expand one term of the first sum (this procedure of expanding terms of someof the sums will be necessary from time to time).

1 0 1 1 0 1


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1 1

0

We now can combine the sums:

1 1 0This equality holds for allx if and only the coefficients of all the powers ofx on one side of the equationare equal to those on the other: Therefore, 0 and 1 0 ; 1 ,2 ,3 ,

The recursive relation is,

We can now generate some of the constants and hopefully discern a pattern.

1 : 2 : 3 :

4 : 5 : 6 :

The pattern indicates that 0 and for 1 , 2 , 3 , A few moreterms will verify this.

Thus, the odd powers of are absent from the sum because their coefficients are0: Substituting the coefficients,

Factoring,

This is another of the few instances when we can write the solution in summation (closed) form:

First termof the seriescorresponding to

0

; , , , ,


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More often than not one will not be able to express the solution in closed form. In those cases, it will benecessary to generate enough terms of the series so that at least the first three terms of each of the two solutionsare apparent.

Example 6 Find the series solution of the equation 2 0.Solution

Since 0 is an ordinary point of this equation we seek solutions at 0 of the form Then,

1

Substituting:

1

0Absorb all coefficients into the sums: 1 2 0

Now re-adjust the indices so that all powers of are the same. This can be accomplished by letting 2 2 in the first sumand renaming the indices in the other two: 2 1 2 0

The first two terms of the first series corresponding to 2 and 1 are0 so we can start it at 0 as well: 2 1

2

0Combining these in a single sum:

2 1 2 0Thus, 2 1 2 0is the recursive relation that generates the terms of the solution.

Solving for: 2 2 1

0,1,2,or

, , , Remember: we have a second order equation; hence only two arbitrary constants can survive.

0 : 22 1


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1 : 3 2 2 : 0

3 :

5 4

5 4 3 2 4 : 26 5 0 5 : 37 6 37 6 5 4 3 2 6 : 48 7 0...

.

Let us see what the solution

becomes. Substituting the coefficients,

3 2 0 5 4 3 2 0 If we group together terms of like coefficients we obtain:

1 13 2 15 4 3 2 37 6 5 4 3 2 Interestingly, one of the sums is finite (rather it is infinite but the even powers ofx beyond the secondpower have0 coefficients) and, as expected, only two arbitrary constants remain.

Exercise: verify that 1 is a solution of the equation.Example 7 Find the first five non-zero terms of the series solution of the differential equation.

1

0

Solution 0 is an ordinary point [verify]. Thus, let . Substituting, 1 1

0

Distributing 1 and absorbing all coefficients, 1 1

0

Let

in the first sum,

in the second sum, and rename

to

in the last sum. Then,

1 2 1 0The first term of the first sum is 0 so we may start it at 0. The first two terms of the second sum arealso 0 so we may also star it at 0. Thus,

1 2 1

0or


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1 2 1 0Thus, ; , , The recursive relation is ; , , Generating terms:

0 : 1 : 1 2 3 2 2 : 2 3 4 3 2 3

3 2 24 3

3 : 3 4 5 4 3 4 12 4 3 3 25 4 and so on. Thus,

2 3 2 12 4 3

13 56 4 3 Grouping like coefficients:

1 12 16 124 16 112 572 Finding a closed form for the last series is not easy. However, as was stated earlier, it is often not necessary touse all the terms of a series since just a few terms may be enough to approximate the solution in the interval ofconvergence. For example, the figure below shows the graph of the function as well as the graphs ofits Taylor Polynomials of first, third, and fifth degree.

Notice that provides us with a very good approximation for values ofx in the interval 1,3. If moreprecision is required we may expand more terms or expand the function at a different point.

- -1 1 2

10

20

30

40

50

1

1 12! 13! 1 12! 13! 14! 15!


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Example 8 (transformations) Find the series solution of the equation 6 1 0 4 3 6 0; 3 2 , 3 0.Solution

The coefficient 6 1 0 has no real zeros. Therefore, every real number is an ordinary point sinceboth

4 3 6 1 0 and 6 6 1 0are analytic throughout the real line. Thus, we may seek series solutions at 0. However, thecomplexity of the leading coefficient suggests that we try to simplify the equation and this can beaccomplished via a substitution.

Since 6 1 0 3 1let . Then

and

Also, 3 0. Therefore, the new initial conditions are0 2, 0 0 and we havetransformedthe original initial value problems into a new one in the variable:

; , Let . Substituting,

1 1 4 6 0Distribute the coefficients:

1 1 4 6 0As before, without loss of generality we may start the three sums at 0:

1 1 4 6 0In order to align the powers, let in the second sum and rename the index in the others:

1 2 1 4 6 0Observe that the terms corresponding to 2 and 1 in the second sum are 0. Therefore, wecan also start that sum at

0:

1 2 1 4 6 0Thus, 1 2 1 4 6 0This equation is true for allx if, only if 1 2 1 4 6 0; 0


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Solving for the leading term,

4 1 6 2 1 ; 0 , 1 , 2 , Cleaning up:

Generating terms: 0 : 62 3 1 : 26 13 2 : 0 6 0 3 : 0

6 0

4 : 2 30 0 5 : 6 35 0...So the solution is:

3 13 0 Notice that all coefficients beyond the fourth term are

0, therefore:

3 13 0

Applying the initial conditions:0 2 0 0 The particular solution is 21 3 Finally, since :

Exercise: In the previous example, verify that 1 3 and are linearly independent solutions ofthe differential equation and the particular solution is in fact a solution of the initial value problem.

The Method o f Series and Non-homogeneous Equations

Example 9 Find a series solution of the equation 1.

23 2 1 ; 0


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SolutionThis is a non-homogeneous equation so care must be exercised when identifying the series coefficients.Once again 0 is an ordinary point and we may let

be the tentative solution. Differentiating and substituting in the equation,

1 1Absorbing 1 1Re-indexing the first sum with 2 and relabeling the index on the second:

2 1 1or

2 1 1Thus, 2 1 1Observe that the term corresponding to 0 is the constant term of the infinite series. Therefore, itmust be equal to the constant term on the right-hand side, namely1. Equating coefficients:

2 0 1 All the other terms correspond to nontrivial powers of. Therefore their coefficients must be0:

2 1 0or , , , Generating some terms: 1 : 3 2

2 : 24 3 14 3 3 : 35 4 35 4 3 2

4 : 46 5 46 5 4 3 5 : 57 6 5 3 7 6 5 4 3 2...Here are the first few terms of the series:


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12 3 2 14 3 35 4 3 2 46 5 4 3 Collecting terms according to like coefficients,

1

3 2

3

5 4 3 2

12

1

4 3

4

6 5 4 3

Observe the magic: 1) we have exactly two arbitrary constants, as we should, and 2) we have aparticular solution given by which is itself an infinite series1.

Exercise: Investigate this particular solution. Does it look like it will satisfy the non-homogeneous equation?You might want to expend a few more terms.

The Spring-mass System Revisited

Sprigs wear out with the passage of time. Hence, the spring constant does not really remain constant. We mayassume thatkis a decreasing function of time, but that it decreases very slowly:

where is a very small positive number. Thus, in the short run, the spring constant is the constant but as timegoes by its value decreases because of metal fatigue; the spring gets weaker. The equation of motion of thespring-mass system without damping is Exercise: Solve the equation 0 ; 0 , 0 0 where is assumed to be asmall positive real number.

1To what function could this series correspond? Investigate.

power series solutions at ordinary points

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