Series Solutions of Differential Equations
5.1 Review of Power Series
Not all differential equations have solutions that can be expressed in terms of elementary functions such as polynomials, exponentials, trigonometric functions, etc. Even when they do, finding these solutions explicitly can be complex or impossible. Series solutions offer a way to represent the solution as an infinite sum of terms. They can provide insights into the behavior of solutions, such as their convergence, oscillation, or growth properties when an explicit solution is unknown. In practical applications, an exact solution may not be necessary, and a finite series (a truncation of the infinite series) can serve as an approximate solution. This method is especially useful in computational methods and simulations.
Before delving into power series solutions of differential equations, let’s review the concept of a power series and its relevant properties.
A. Power Series
A power series is an infinite series of the form
[asciimath]sum_(n=0)^oo\ a_n(x-x_0)^n[/asciimath]
where [asciimath]n[/asciimath] is the index of summation, [asciimath]a_n[/asciimath] represents the coefficient of the nth term, [asciimath]x_0[/asciimath] is the center of the series, and [asciimath]x[/asciimath] is the variable. The series can be expressed as
[asciimath]a_0+a_1(x-x_0)+[/asciimath] [asciimath]a_2(x-x_0)^2+ a_3(x-x_0)^3+ ...[/asciimath]
This allows us to approximate functions in regions where the series converges, which is essential for understanding and solving differential equations. We may sometimes be interested in the pattern or form of initial terms in the series or manipulating terms such as re-indexing or combining terms. Thus, we can ‘strip out’ these terms from the general series notation.
[asciimath]sum_(n=0)^oo\ a_n(x-x_0)^n[/asciimath] [asciimath]=a_0+a_1(x-x_0) +sum_(n=2)^oo\ a_n(x-x_0)^n[/asciimath]
Here, the first two terms are stripped out of the general series notation, and the summation index now starts from [asciimath]n=2[/asciimath].
B. Shifting the Index of a Power Series
Shifting the index of a power series changes the starting point of the summation and reindexes the terms of the series. This is particularly useful for aligning terms for addition or subtraction of series. Consider a power series
[asciimath]sum_(n=n_0)^oo\ a_nx^n[/asciimath]
Shifting Right (Increasing index)
To shift the series right by [asciimath]k[/asciimath] unit, replace [asciimath]n[/asciimath] with [asciimath]n-k[/asciimath] in the general term and add [asciimath]k[/asciimath] to the original lower limit of the summation.
[asciimath]sum_(n=n_0+k)^oo\ a_(n-k)x^(n-k)[/asciimath]
Shifting Left (Decreasing index)
To shift the series left by [asciimath]k[/asciimath] unit, replace [asciimath]n[/asciimath] with [asciimath]n+k[/asciimath] in the general term and subtract [asciimath]k[/asciimath] from the original lower limit of the summation.
[asciimath]sum_(n=n_0-k)^oo\ a_(n+k)x^(n+k)[/asciimath]
C. Linear Combination of Power Series
When solving differential equations using series often we need to add or subtract series. When adding or subtracting series, we ensure the terms being added or subtracted correspond to the same power of the variable. This means ensuring both series have the same power of [asciimath]x-x_0[/asciimath] and their summation indices are aligned properly to start from the same lower limit. Consider two power series
[asciimath]f(x)=sum_(n=0)^oo\ a_n(x-x_0)^n[/asciimath] and [asciimath]g(x)=sum_(n=0)^oo\ b_n(x-x_0)^n[/asciimath]
Since the power of [asciimath]x-x_0[/asciimath] term is the same in both series and the index in both start from the same value, they can be linearly combined as
[asciimath]c_1f(x)+c_2g(x)=sum_(n=0)^oo\ (c_1a_n+c_2b_n)(x-x_0)^n[/asciimath]
where [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath] are constants.
If there is an [asciimath]x-x_0[/asciimath] term in front of the summation in a series, we move it inside the summation and combine it with [asciimath](x-x_0)^n[/asciimath] term there. For example,
[asciimath](x-x_0)^c sum_(n=0)^oo\ a_n(x-x_0)^n[/asciimath]
[asciimath]=sum_(n=0)^oo\ a_n(x-x_0)^(n+c)[/asciimath]
This simplifies the handling and manipulation of series in differential equation solutions.
Write the following as a single series in terms of [asciimath](x+2)^n[/asciimath].
[asciimath](x+2)^2 sum_(n=3)^oo na_n (x+2)^(n-4)-sum_(n=1)^oo na_n (x+2)^(n+1)[/asciimath]
Show/Hide Solution
1. First, we multiply [asciimath](x+2)^2[/asciimath] term into the first summation.
[asciimath]=sum_(n=3)^oo na_n (x+2)^(n-4+2)-sum_(n=1)^oo na_n (x+2)^(n+1)[/asciimath]
[asciimath]=sum_(n=3)^oo na_n (x+2)^(n-2)-sum_(n=1)^oo na_n (x+2)^(n+1)[/asciimath]
2. We shift the indexes in both series to make the exponent of [asciimath](x+2)[/asciimath] be [asciimath]n[/asciimath]. Thus we need to shift the first series two units to left and the second series one unit to right.
[asciimath]=sum_(n=3-2)^oo (n+2)a_(n+2) (x+2)^(n+2-2)-sum_(n=1+1)^oo (n-1)a_(n-1) (x+2)^(n-1+1)[/asciimath]
[asciimath]=sum_(n=1)^oo (n+2)a_(n+2) (x+2)^(n)-sum_(n=2)^oo (n-1)a_(n-1) (x+2)^(n)[/asciimath]
3. Finally, we ensure both series start from the same lower limit. Depending on the series, we can sometimes strip out terms or adjust the index if the preceding terms are already zero. Notice that if the second series starts from [asciimath]n=1[/asciimath], the initial term will be zero because of the factor [asciimath](n-1)[/asciimath]. Therefore, initiating the index from [asciimath]n=1[/asciimath] does not alter its overall value.
[asciimath]=sum_(n=1)^oo (n+2)a_(n+2) (x+2)^(n)-sum_(n=1)^oo (n-1)a_(n-1) (x+2)^(n)[/asciimath]
Now we can combine the series to obtain the final answer.
[asciimath]=sum_(n=1)^oo [(n+2)a_(n+2)-(n-1)a_(n-1)] (x+2)^(n)[/asciimath]
Note: Generally, whenever a series contains a factor of [asciimath](n-a)[/asciimath], the term at [asciimath]n=a[/asciimath] (where a is the starting index) will be zero.
Try an Example
D. Convergence of Power Series
The convergence of power series is essential for ensuring that the series represents the function accurately over some interval. A series converges at a particular point if the sum approaches a finite limit as [asciimath]n[/asciimath] approaches infinity. In other words, a power series converges for a given [asciimath]x[/asciimath] if the following limit exists.
[asciimath]lim_(N->oo)sum_(n=0)^Na_n (x-x_0 )^n[/asciimath]
For any power series, any of the three cases can be true:
- Converges only at [asciimath]x=x_0[/asciimath]: Here, the sum of the series equals to [asciimath]a_0[/asciimath].
- Converges for all values of [asciimath]x[/asciimath].
- Converges within a radius of convergence [asciimath]R[/asciimath] : The series converges if [asciimath]|x-x_0 |ltR[/asciimath] and diverges if [asciimath]|x-x_0 |gtR[/asciimath] . [asciimath]R[/asciimath] is called the radius of convergence, and the interval [asciimath](-R+x_0,R+x_0)[/asciimath] is the interval of convergence.
To determine the radius and interval of convergence for a given power series, the Ratio Test is often used. The Ratio Test involves taking the limit
[asciimath]lim_(n->oo)|(b_(n+1))/(b_n)|=L[/asciimath]
where [asciimath]b_n[/asciimath] denotes the nth term of the series. If [asciimath]Llt1[/asciimath], the series converges, and the radius of convergence is [asciimath]R=1//L[/asciimath].
E. Differentiation of Power Series
Differentiation and integration of power series within their interval of convergence can be performed term-by-term. For a given power series centered at [asciimath]x_0[/asciimath]
[asciimath]f(x)=sum_(n=0)^oo\ a_n(x-x_0)^n[/asciimath]
The first derivative of [asciimath]f(x)[/asciimath] is
[asciimath]f'(x)=sum_(n=1)^oo\ a_nd/(dx) (x-x_0)^n[/asciimath]
[asciimath]=sum_(n=1)^oo\ na_n(x-x_0)^(n-1)[/asciimath]
Note that the index of the first derivative starts at [asciimath]n=1[/asciimath] because the first term in the original series is constant ([asciimath]a_0[/asciimath]) and disappears upon differentiation. The interval of convergence for the derivative series is at least as large as that of the original series, but careful attention should be paid to the endpoints.
Similarly, the second derivative of [asciimath]f(x)[/asciimath] is
[asciimath]f''(x)=sum_(n=1)^oo\ na_nd/dx(x-x_0)^(n-1)[/asciimath]
[asciimath]=sum_(n=2)^oo\ n(n-1)a_n(x-x_0)^(n-2)[/asciimath]
Note that the index of the second derivative starts at [asciimath]n=2[/asciimath] as the first term of the first derivative is constant ([asciimath]a_1[/asciimath]) and disappears upon differentiation.
Suppose [asciimath]y[/asciimath] can be expressed as a power series [asciimath]y=sum_(n=0)^oo\ a_nx^n[/asciimath] . Write the following as a single series in terms of [asciimath]x^n[/asciimath].
[asciimath](1+x^2)y''+2xy'-2y[/asciimath]
Show/Hide Solution
1. First, we find [asciimath]y'[/asciimath] and [asciimath]y''[/asciimath] :
[asciimath]y'=d/dxsum_(n=0)^oo\ a_nx^n[/asciimath] [asciimath]=sum_(n=1)^oo\ n a_n x^(n-1)[/asciimath]
[asciimath]y''=d/dxsum_(n=1)^oo\ n a_nx^(n-1)[/asciimath] [asciimath]=sum_(n=2)^oo\ n(n-1) a_n x^(n-2)[/asciimath]
2. Next, we substitute [asciimath]y,\ y',[/asciimath] and [asciimath]y''[/asciimath] into the expression:
[asciimath](1+x^2)sum_(n=2)^oo\ n(n-1) a_n x^(n-2)[/asciimath] [asciimath]+2xsum_(n=1)^oo\ n a_n x^(n-1)[/asciimath] [asciimath]-2sum_(n=0)^oo\ a_nx^n[/asciimath]
3. Multiplying the [asciimath]x[/asciimath] term in front of the summation by the [asciimath]x[/asciimath] term in the general term of each series, we obtain
[asciimath]sum_(n=2)^oo\ n(n-1) a_n x^(n-2)[/asciimath] [asciimath]+sum_(n=2)^oo\ n(n-1) a_n x^n[/asciimath] [asciimath]+2sum_(n=1)^oo\ n a_n x^n[/asciimath] [asciimath]-2sum_(n=0)^oo\ a_nx^n[/asciimath]
4. Note that the exponent of [asciimath]x[/asciimath] is the same in all but the first series. Therefore, we only need to shift the index of the first summations by 2 to the left:
[asciimath]sum_(n=0)^oo\ (n+2)(n+1) a_(n+2) x^n[/asciimath] [asciimath]+sum_(n=2)^oo\ n(n-1) a_n x^n[/asciimath] [asciimath]+2sum_(n=1)^oo\ n a_n x^n[/asciimath] [asciimath]-2sum_(n=0)^oo\ a_nx^n[/asciimath]
5. Finally, we ensure all series start from the same lower limit. Notice that the second series is zero at [asciimath]n=0,1[/asciimath] because there are factors [asciimath]n[/asciimath] and [asciimath]n-1[/asciimath] in the general term of the series. Thus, its index can start at [asciimath]n=0[/asciimath] without changing its value. Likewise, the third series is zero at [asciimath]n=0[/asciimath] so it can too start at [asciimath]n=0[/asciimath] . Then we rewrite the indices and we have
[asciimath]sum_(n=0)^oo\ (n+2)(n+1) a_(n+2) x^n[/asciimath] [asciimath]+sum_(n=0)^oo\ n(n-1) a_n x^n[/asciimath] [asciimath]+2sum_(n=0)^oo\ n a_n x^n[/asciimath] [asciimath]-2sum_(n=0)^oo\ a_nx^n[/asciimath]
Combining the series yields
[asciimath]sum_(n=0)^oo\ [(n+2)(n+1) a_(n+2)+(n(n-1)+2n-2)a_n] x^n[/asciimath]
Try an Example
F. Properties of Power series
Equality of Series
If two power series are equal for all [asciimath]x[/asciimath] in an open interval that contains [asciimath]x_0[/asciimath], then their coefficients must be equal. That is
[asciimath]sum_(n=0)^oo\ a_n(x-x_0)^n=sum_(n=0)^oo\ b_n(x-x_0)^n[/asciimath]
Implies [asciimath]a_n=b_n[/asciimath] for all [asciimath]n[/asciimath].
Power Series Vanishing on an Interval
If a power series equals zero for all [asciimath]x[/asciimath] in an open interval, then all its coefficients must be zero. That is
[asciimath]sum_(n=0)^oo\ a_n(x-x_0)^n=0[/asciimath]
Implies [asciimath]a_n=0[/asciimath] for all [asciimath]n[/asciimath].
G. Taylor Series
A Taylor series is a specific type of power series representation of a function based on its derivatives at a specific point, typically at [asciimath]x=x_0[/asciimath]. It is given by
[asciimath]f(x)=sum_(n=0)^oo\ f^(n)(x_0)/(n!)(x-x_0)^n[/asciimath]
Here, [asciimath]f^(n)(x_0)[/asciimath] is the nth derivative of [asciimath]f(x)[/asciimath] evaluated at [asciimath]x=x_0[/asciimath], and [asciimath]n![/asciimath] is the factorial of [asciimath]n[/asciimath].
When [asciimath]x_0=0[/asciimath], the series is often called a Maclaurin series. The Taylor series expansions of a few functions at [asciimath]x=0[/asciimath] (Maclaurin series) are as follows.
[asciimath]e^x=sum_(n=n_0)^oo\ x^n/(n!)=[/asciimath] [asciimath]1+x+x^2/(2!)+x^3/(3!)+...[/asciimath]
[asciimath]sin(x)=sum_(n=n_0)^oo\(-1)^n x^(2n+1)/((2n+1)!)=[/asciimath] [asciimath]x-x^3/(3!)+x^5/(5!)- ...[/asciimath]
[asciimath]cos(x)=sum_(n=n_0)^oo\(-1)^n x^(2n)/((2n)!)=[/asciimath] [asciimath]x-x^2/(2!)+x^4/(4!)- ...[/asciimath]
H. Recursive Relation
A recursive relation for a series provides a way to calculate each term of the series using one or more of the preceding terms. Instead of defining each term independently, a recursive relation relates each term to its predecessors, building the series progressively. This method is particularly useful when the direct calculation of terms is complex or when the relationship between consecutive terms is simpler to express.
Generally, a recursive relation has the following structure.
[asciimath]a_n=f(a_(n-1),a_(n-2),...a_(n-k))[/asciimath] for [asciimath]n>k[/asciimath]
Here, [asciimath]a_n[/asciimath] is the nth term of the series, and [asciimath]f[/asciimath] is a function that defines how to calculate the nth term using the previous [asciimath]k[/asciimath] terms.
The recursive relation allows the calculation of all coefficients in the series from a set of initial conditions or known coefficients. These are usually derived from the initial or boundary conditions of the differential equation.
Suppose the recursive formula for a power series solution is
[asciimath]a_(n+2) = -a_n/((n+1)(n+4))[/asciimath]
Find the second, third, and fourth terms of the series in terms of [asciimath]a_0[/asciimath] and [asciimath]a_1[/asciimath].
Show/Hide Solution
To find the terms we plug [asciimath]n=0,1,2...[/asciimath] into the recursive relation.
[asciimath]n=0\ ->\ a_(0+2)=-a_0/((0+1)(0+4))\ ->\ a_2=-a_0/4[/asciimath]
[asciimath]n=1\ ->\ a_(1+2)=-a_1/((1+1)(1+4))\ ->\ a_3=-a_1/10[/asciimath]
[asciimath]n=2\ ->\ a_(2+2)=-a_2/((2+1)(2+4))\ ->\ a_4=-a_2/18[/asciimath]
Given [asciimath]a_2=-a_0/4[/asciimath] , [asciimath]a_4[/asciimath] can be written in terms of [asciimath]a_0[/asciimath]:
[asciimath]a_4=-1/18 (-a_0/4)=a_0/72[/asciimath]
Try an Example
Section 5.1 Exercises
- Write the following as a single series in terms of [asciimath]x^n[/asciimath].
[asciimath]sum_(n=2)^oo \ n (n-1) a_n x^(n-2)[/asciimath] [asciimath]+xsum_(n=1)^oo\ -2n a_n x^(n-1)[/asciimath] [asciimath]+sum_(n=0)^oo\ -3a_n x^n[/asciimath]
Show/Hide Answer
[asciimath]sum_(n=0)^(oo)[(n+2)(n+1)a_(n+2)-2n a_n-3a_n]x^n[/asciimath]
- Suppose [asciimath]y[/asciimath] can be expressed as a power series [asciimath]y=sum_(n=0)^oo\ a_nx^n[/asciimath] . Express the following as a single series in terms of [asciimath]x^n[/asciimath].
[asciimath]y''-5x y'+2y[/asciimath]
Show/Hide Answer
[asciimath]sum_(n=0)^(oo)[(n+2)(n+1)a_(n+2)-5n a_n+2a_n]x^n[/asciimath]
- Suppose the recursive formula for a power series solution is
[asciimath]a_(n+2) = -a_n/((n+1)(n+2))[/asciimath]
Find the fourth and fifth terms in terms of [asciimath]a_0[/asciimath] and [asciimath]a_1[/asciimath].
Show/Hide Answer
[asciimath]a_4=-a_0/24[/asciimath]
[asciimath]a_5=a_1/120[/asciimath]