Series Solutions to Differential Equations. 

Prof. C. Madigan 

Nova Scotia Agricultural College 

Truro, N.S.    B2N 5E3 

cmadigan@nsac.ca 

Solving linear differential equations with constant coefficients reduces to an algebraic problem.  There is no similar procedure for solving linear differential equations with variable coefficients.  With the exception of special types, such as the Cauchy equations, these will generally require the use of the power series techniques for a solution. 

Initialization. 

 

 

Review of  Series and Power Series. 

 

SERIES 

 

Recall a power series in [powers of]  x - a  is an infinite series of the form 

                    

 

If  a = 0, this is a power series in x 

 

                     

 

 

An important aspect of any series is whether or not  it converges  (IE the infinite sum exists).   (one might say that the only good series is a converging series)  

NOTE:  A converging series can then be approximated by using its nth partial sum. 

Definition 1.  Given an infinite series   Let the sequence of partial sums of the series be .  If this sequence is convergent IE if the series is called convergent and S is called its sum.  If the limit does not exist, the series is called divergent. 

 

Examples.  1.  Geometric series  where a and r are constants.  This series converges to if and diverges for The sum of the first n terms   

                 2.  p series  where p is a constant, converges for ", mathvariant = "normal", fence = "false", separator = "false", stretchy = "false", symmetric = "fal..." align="center" border="0"> and diverges for  

 

Tests for convergence of a series 

 

There are several tests for convergence of a series. 

 

1. Comparison test for series of non-negative terms 

                 Let and suppose converges. Then if   0≤ then converges.  

                 If diverges and if then diverges. 

2. Quotient test for series of non-negative terms. 

               If  and  and if either both converge or both diverge. 

              If A = 0 and if converges then converges 

              If A = ∞ and diverges then diverges 

3.  Integral test.  for series of non-negative terms. 

 

                If f(x) is positive, continuous and monotonic decreasing for x ≥ N and is such that f(n) = , n = N, N+1, N+2, ...then   converges or diverges according as converges or diverges. 

4. Alternating Series test.   

                 

                 An alternating series converges if it satisfies the two conditions 

                                     a) for n ≥ 1 

                             b) or  

 

Definition:  The series is called absolutely convergent if converges.  If converges but diverges, then is called conditionally convergent. 

NOTE: An  absolutely convergent series is convergent. 

 

5. Ratio test.    

               Let   Then the series  

                  a) converges (absolutely) if L < 1 

                  b) diverges if L > 1 

               If L = 1 the test fails. 

 

6. Raabe's test.   

               Let   Then the series  

                  a) converges (absolutely) if L > 1 

                  b) diverges or converges conditionally if L < 1 

               If L = 1 the test fails. 

 

7. Gauss' test. 

            If  where  for all n > N, then the series  

                      a)  converges absolutely if  L > 1 

                      b)  diverges or is conditionally convergent if  L ≤ 1 

 

 

 

POWER SERIES 

 

A power series    converges at the point x = if the infinite series (of real numbers)   converges. 

In general a power series converges absolutely for and diverges for .  R is called the radius of convergence.  

 

Note:  convergence at the endpoints requires separate analysis. 

 

Example 1. Determine the convergence of     

 

              Applying the ratio test we have   =  

 

              Hence  the series converges absolutely for diverges for  

 

              For   IE at we must test separately 

 

                 At our series is the harmonic series known to diverge. 

                 At x = our series is an alternating harmonic series that converges. 

 

The given series converges for   and diverges everywhere else. 

 

Definition   A function is said to be analytic at a point if, in an open interval about this function can be represented as a power series that has a positive radius of convergence. 

 

Taylor Series  If  f is analytic at , then the representation  

 

                           holds in some open interval about and is called the Taylor series for about When , it is also called the Maclaurin series for  

 

 

Power Series Solutions to Linear Differential Equations. 

 

We now consider a method for obtaining a power series solution to a linear differential equation with polynomial coefficients. 

 

Given the differential equation  

 

                             

 

we begin by writing it in the standard form 

 

                            and  

 

Definition 1  A point is called an ordinary point of equation (1) if both p(x) and q(x) are analytic at  If it is not an ordinary point, it is called a singular point of the equation. 

 

Definition 2.   A singular point of  (2) is said to be a regular singular point if both and are analytic at  Otherwise is called an irregular singular point. 

 

Solutions about an ordinary point  x =  

 

We assume that a power series solution of the form  exists and our task is to determine the coefficients This task is accomplished by substituting this series into the differential equation, combining the result into a single series by collecting the result in powers of x and then in order for this series to be identically zero, we must have that all of its coefficients must be equal to zero. 

 

We make use of Maples commands , series, series ,coeff, coeff ,convert, convert  collect, collect , etc  to carry out the required work. 

 

Example 1 Solve the equation 2y'' +xy' + y  = 0 about the ordinary point x = 0. 

 

Step 1.  Define the deq 

 

(3.1.1.1)

 

Step 2.  Define the series solution (begin by setting the order to be used for the series) 

 

(3.1.1.2)

 

 

(3.1.1.3)

 

Step 3.  Substitute the series into the deq. 

 

 

(3.1.1.4)

 

Step 4  convert the series into a polynomial and collect in terms of powers of x 

 

 

(3.1.1.5)

 

 

Step 5.  Use the Maple command coeff to select the coefficients of the powers of x.  An example of this command is 

 

 

(3.1.1.6)

 

We then set the coefficients equal to zero and solve for the unknown derivatives i terms of the initial conditions that are required in order to solve a second order DEQ  IE  y(0) = a  and y'(0) = b 

 

 

(3.1.1.7)

 

Step 6.  Finally we substitute these values into our solution and collect the terms in terms of the parameters a and b. 

 

 

(3.1.1.8)

 

We can check our result using Maple's dsolve command 

 

 

(3.1.1.9)

 

 

 

 

Example 2. Solve the differential equation  about the ordinary                                    point x = 0. 

 

Step 1.  Define the deq 

 

 

 

(3.1.2.1)

 

Step 2.  Define the series solution (begin by setting the order to be used for the series) 

 

 

 

(3.1.2.2)

 

Step 3.  Substitute the series into the equation 

 

(3.1.2.3)

 

Step 4  convert the series into a polynomial and collect in terms of powers of x 

 

 

(3.1.2.4)

 

 

Step 5.  Use the Maple command coeff to select the coefficients of the powers of x.  Then set the coefficients equal to zero and solve for the unknown derivatives i terms of the initial conditions that are required in order to solve a second order DEQ  ie  y(0) = a  and y'(0) = b 

 

 

(3.1.2.5)

 

 

Step 6.  Finally we substitute these values into our solution and collect the terms in terms of the parameters a and b. 

 

 

(3.1.2.6)

 

 

 

We can check our result using Maple's dsolve command 

 

 

(3.1.2.7)

 

 

 

 

 

 

NOTE:  One of the difficulties is determining the radius of convergence when we are not able to find a general form for the coefficients and use the ratio test. 

 

EXISTENCE OF ANALYTIC SOLUTIONS. 

 

    There is a theorem that states that if is an ordinary point then equation has two linearly independent analytic solutions of the form    and that the radius of convergence is as least as large as the distance from to the nearest singular point (real or complex-valued) of the deq. 

 

In example 1 above since there are no singular points the radius of convergence is ∞. 

In example 2  there are two singular points  ? i and hence the radius of convergence of our solutions about the point x = 0 is at least 1. 

 

 

Maple also has a power series package PowerSeries  that allows you to work directly with power series. 

 

We will use this package in the following examples. 

 

Example 1.  Solve the IVP   y'' - (x-2)y' + 2y = 0;    y(0) = 1, y'(0) = -1 

 

We will use power series package in  Maple to find the solution 

 First to create the series  solution Ys(x)  =   

   

 

The command tpsform  converts the Powseries created above into a power  series form of the variable stated in the command. 

 

(3.1.3.1)

 

Next we determine the necessary derivatives found in the given deq. 

 

 

(3.1.3.2)

 

 

(3.1.3.3)

 

 

Note the answer is in the form of a procedure. You can see the power series by using the tpsform command.We also need to make the coeff of our eqn into a power series in order to use the powseries commands. 

 

 

 

You can see the power series by using the tpsform command.  For example   

 

 

(3.1.3.4)

 

Next we combine these series to form the lhs of the equation given using the commands for multiplying and adding power series. 

 

 

(3.1.3.5)

 

 

(3.1.3.6)

 

Using the op command to identify the coefficients , setting them equal to zero, and solving for them in terms of a[0] and a[1] 

 

(3.1.3.7)

 

Substituting the coeff into Y(x) 

 

 

(3.1.3.8)

 

 

Converting our solution into a polynomial and collecting in terms  of   

 

 

(3.1.3.9)

 

 

(3.1.3.10)

 

Finally, substituting the initial conditions and solving for a[0] and a[1] 

 

 

(3.1.3.11)

 

 

(3.1.3.12)

 

 

 

 

Using Maples dsolve command to solve the given deq. 

 

 

 

 

(3.1.3.13)

 

NOTE:   Since this deq has NO singular points the radius of convergence for this solution is ∞.  ie it is valid for all values of x. 

 

 

Example 2. Solve the DEQ  y'' + xy' +(2x-3)y = 0  near x = -1. 

              

Note  x = -1 is an ordinary point for this deq.  

 

We first make the substitution  t = x - (-1) = x + 1. The resulting deq can then be solved near  t = 0 using the above techniques. 

ie  Since  y(x) = y(t-1)  and   and  the eqn becomes  

 

 

First create the power series solution 

 

To calculate the derivative of Y(x) 

 

(3.1.4.1)

 

 

(3.1.4.2)

 

converting the coefficients into power series 

 

 

 

Combining these series as defined by the equation 

 

 

(3.1.4.3)

 

Using the op command to identify the coefficients , setting them equal to zero, and solving for them in terms of a[0] and a[1] 

 

(3.1.4.4)

 

Substituting the coeff into Y(x) and converting the solution into a polynomial  

 

(3.1.4.5)

 

Substituting that t = x+1 we obtain the solution to the original problem 

 

 

(3.1.4.6)

 

Again since this eqn has NO singular points this solution is valid for all values of x. 

Given initial conditions we could then proceed to determine the values for and   

Suppose that y(-1) = 2 and y'(-1) = -2   

 

 

(3.1.4.7)

 

 

(3.1.4.8)

 

 

 

Using Maple's dsolve command. 

 

 

 

(3.1.4.9)

 

 

 

 

 

 

Solutions about a regular singular point  --- Method of Frobenius. 

 

Definition 3.  If is a regular singular point of y'' + p(x) y' +q(x) y = 0, then the indicial equation for this point is  

 

                                                 

 

                   where  and      

 

                   The roots of the indicial equation are called the exponents(indices) of the singularity  

 

 

Frobenius method  of solving ordinary differential equations near a regular singular point, , by positing a solution of the form  

 

 

 the values of r and the coefficients are then found by iteration by substituting the potential solution into the equation. 

 

NOTE  The first step in this method is to find the roots and (Re ) of the indicial equation.  Then utilizing the larger root Frobenius's theorem assures us that our deq has a series solution of the form above and that this series converges for all x such that  where R is the distance from to the nearest other singular point (real or complex).  

           

          If   is not an integer, then there exist two linearly independent solutions of the form  

                                         

                                         

         If  then there exist two linearly independent solutions of the form 

                                         

                                          

 

        If   is a positive integer, then there exist two linearly independent solutions  of the form 

                                           

                                          where C is a constant that could be zero. 

 

Example 1.   Find a series solution for the differential equation   

 

Now x = 0 and x = -2 are both singular points for this deq. 

Also x = 0 is a regular singular point since and are analytic at  

 

We are looking for a solution of the form  

Step 1.  Define our deq 

 

 

(3.2.1.1)

 

 

 

Step 2.  Define our series solution 

 

(3.2.1.2)

 

 

(3.2.1.3)

 

 

Step 3.  Substitute our series into the deq. 

 

 

 

(3.2.1.4)

 

 

 

(3.2.1.5)

 

 

Step 4. Determine the coefficient of the  term  (  it in effect is the indicial eqn needed to determine the values of r in Frobenius' method ) 

 

 

(3.2.1.6)

 

From the above we see that r = 1  or  r  =  1/2 

Step 5.  We  use the larger value of r = 1 first.  We now substitute  that  y1(0) = a ,  D(y1)(0) = b  and r =1  into the coefficients of  the lhs our eqn1  

 

 

(3.2.1.7)

 

 

We now want to solve the above coefficients = 0  for the values of the derivatives in each of themTo do this we must drop the first and 2nd items.  NOTE :  the 2nd  item   a + 3b =0  says that  b = -1/3 a 

 

(3.2.1.8)

 

 

(3.2.1.9)

 

 

 

 

Step 6. Substitute these values into slna minus the last three terms since we have not calculated their correct coefficients 

 

 

(3.2.1.10)

 

 

It may be possible to obtain the second independent series solution by repeating the above using the other value of r  ( This will work in this example since the two r values  do not differ by an integer). 

 

 

(3.2.1.11)

 

We now want to solve the above coefficients = 0  for the values of the derivatives in each of themTo do this we must drop the first and 2nd items.  NOTE :  2nd item  says    b=-3/4a 

 

(3.2.1.12)

 

 

(3.2.1.13)

 

 

Finally we substitute these values into slna minus the last three terms since we have not calculated their correct coefficients 

 

 

(3.2.1.14)

 

 

Hence the general solution to the given differential equation will be   

 

 

(3.2.1.15)

 

 

MAPLE dsolve solution  

 

 

(3.2.1.16)

 

 

(3.2.1.17)

 

 

 

Example 2. Find a series solution for the differential equation about the point x = 0 

 

 

x = 0 is a regular singular point for this deq. 

 

 

(3.2.2.1)

 

 

(3.2.2.2)

 

 

(3.2.2.3)

 

 

 

(3.2.2.4)

 

 

(3.2.2.5)

 

We see that r = 1 is a double root  and proceed to find the first solution using the  procedures outlined above. 

 

 

(3.2.2.6)

 

 

 

(3.2.2.7)

 

 

(3.2.2.8)

 

 

(3.2.2.9)

 

 

We now proceed to find the second solution using the fact that it will be of the form   

                            

 

(3.2.2.10)

 

 

 

(3.2.2.11)

 

 

(3.2.2.12)

 

Since y1(x) is a solution of the deq the factor on ln(x) is zero  (We also convert our series into a finite sum in order to deal with the summations...) 

 

 

(3.2.2.13)

 

 

(3.2.2.14)

 

 

(3.2.2.15)

 

 

(3.2.2.16)

 

 

(3.2.2.17)

 

 

(3.2.2.18)

 

 

(3.2.2.19)

 

MAPLE dsolve solution..... 

 

(3.2.2.20)

 

 

(3.2.2.21)

 

 

Example 3. Find a series solution for the differential equation                     

 

Note x = 0 is a regular singular point for this deq. 

 

 

(3.2.3.1)

 

 

 

(3.2.3.2)

 

 

(3.2.3.3)

 

 

(3.2.3.4)

 

roots are 0 and -3     In this case they differ by a positive integer   

 

 

roots are  r = 0 and -3. Using the larger root  r = 0 

 

 

(3.2.3.5)

 

 

(3.2.3.6)

 

 

(3.2.3.7)

 

 

(3.2.3.8)

 

A second solution of the form   and where C is a constant that may be zero. 

 

 

(3.2.3.9)

 

 

(3.2.3.10)

 

Again since y1(x)  is a solution the factor on ln(x) is 0. 

 

 

 

(3.2.3.11)

 

 

 

 

(3.2.3.12)

 

 

(3.2.3.13)

 

 

(3.2.3.14)

 

NOTE   The factor on for another independent solution we must choose to be nonzero. 

             In effect sln2aff is a general solution to the given deq. 

 

 

 

 

 

(3.2.3.15)

 

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