Power Series Method Of Solving Differential Equations

The power series method is a technique used to find solutions to differential equations by expressing them as power series. This method is particularly useful for solving linear differential equations with variable coefficients, where traditional methods may not be applicable. Let's delve deeper into understanding and applying this method.

Power Series: An Introduction

A power series is an infinite series of the form:

$ \sum_{n=0}^{\infty} a_n (x - c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + a_3(x-c)^3 + \dots $

where:

• (x) is a variable,
• (a_n) are the coefficients,
• (c) is a constant representing the center of the series.

The key idea behind using power series to solve differential equations is to assume that the solution to the differential equation can be expressed as a power series. By substituting the power series into the differential equation and solving for the coefficients (a_n), we can find the solution.

Why Use Power Series?

• Applicability to Variable Coefficients: Power series are effective for solving differential equations with variable coefficients, which are common in many physical applications.
• Approximation of Solutions: When an exact solution is hard to find, a power series can provide an approximation that is accurate within its radius of convergence.
• Foundation for Advanced Techniques: Understanding power series is crucial for more advanced techniques in differential equations, such as Frobenius method.

Steps for Solving Differential Equations Using Power Series

Here's a step-by-step guide on how to apply the power series method:

1. Assume a Power Series Solution: Start by assuming that the solution (y(x)) to the differential equation can be expressed as a power series centered at a point (c). Typically, (c = 0) is chosen for simplicity unless otherwise specified by the problem.

   $ y(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1x + a_2x^2 + a_3x^3 + \dots $

2. Compute Derivatives: Calculate the necessary derivatives of the power series. If the differential equation involves (y'(x)) and (y''(x)), compute these derivatives by differentiating the power series term by term.

   $ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} = a_1 + 2a_2x + 3a_3x^2 + \dots $

   $ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = 2a_2 + 6a_3x + 12a_4x^2 + \dots $

3. Substitute into the Differential Equation: Substitute the power series and its derivatives into the given differential equation.

4. Manipulate the Series: Adjust the indices of summation so that all series have the same power of (x). This often involves shifting indices by replacing (n) with (n+k) for some integer (k). The goal is to combine all series into a single series.

5. Equate Coefficients: Set the coefficient of each power of (x) equal to zero. This is because the only way for a power series to be identically zero is if all of its coefficients are zero.

6. Find a Recurrence Relation: Derive a recurrence relation that expresses (a_n) in terms of previous coefficients. This relation will allow you to determine all coefficients in terms of a few initial coefficients, such as (a_0) and (a_1).

7. Determine the Coefficients: Use the recurrence relation to find the coefficients (a_n). The initial coefficients (a_0) and (a_1) will typically be arbitrary constants, reflecting the general solution to a second-order linear differential equation.

8. Write the General Solution: Substitute the coefficients back into the power series to obtain the general solution. The solution will typically be expressed as a linear combination of two linearly independent power series, each multiplied by an arbitrary constant.

9. Determine the Radius of Convergence (Optional): Find the radius of convergence (R) of the power series solution. This indicates the interval ((-R, R)) in which the series converges and represents a valid solution.

Detailed Examples

Example 1: Solving (y' - y = 0)

Let's solve the differential equation (y' - y = 0) using the power series method.

1. Assume a Power Series Solution:

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

2. Compute Derivatives:

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

3. Substitute into the Differential Equation:

   $ \sum_{n=1}^{\infty} n a_n x^{n-1} - \sum_{n=0}^{\infty} a_n x^n = 0 $

4. Manipulate the Series: To combine the series, we shift the index of the first series. Let (m = n - 1), so (n = m + 1). When (n = 1), (m = 0). Thus,

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

   Replacing (m) with (n) for consistency, we get

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

   Combining the series:

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

5. Equate Coefficients:

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

6. Find a Recurrence Relation:

   $ a_{n+1} = \frac{a_n}{n+1} $

7. Determine the Coefficients: Let's find the first few coefficients:

   • (a_1 = \frac{a_0}{1})
   • (a_2 = \frac{a_1}{2} = \frac{a_0}{2!})
   • (a_3 = \frac{a_2}{3} = \frac{a_0}{3!})
   • (a_n = \frac{a_0}{n!})

8. Write the General Solution:

   $ y(x) = \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} \frac{a_0}{n!} x^n = a_0 \sum_{n=0}^{\infty} \frac{x^n}{n!} = a_0 e^x $

   The general solution is (y(x) = a_0 e^x), where (a_0) is an arbitrary constant.

Example 2: Solving (y'' + y = 0)

Let's solve the differential equation (y'' + y = 0) using the power series method.

1. Assume a Power Series Solution:

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

2. Compute Derivatives:

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

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

3. Substitute into the Differential Equation:

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

4. Manipulate the Series: Shift the index of the first series. Let (m = n - 2), so (n = m + 2). When (n = 2), (m = 0). Thus,

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

   Replacing (m) with (n) for consistency, we get

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

   Combining the series:

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

5. Equate Coefficients:

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

6. Find a Recurrence Relation:

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

7. Determine the Coefficients: Let's find the first few coefficients:

   • For even terms:
     • (a_2 = -\frac{a_0}{2 \cdot 1} = -\frac{a_0}{2!})
     • (a_4 = -\frac{a_2}{4 \cdot 3} = \frac{a_0}{4!})
     • (a_{2k} = (-1)^k \frac{a_0}{(2k)!})
   • For odd terms:
     • (a_3 = -\frac{a_1}{3 \cdot 2} = -\frac{a_1}{3!})
     • (a_5 = -\frac{a_3}{5 \cdot 4} = \frac{a_1}{5!})
     • (a_{2k+1} = (-1)^k \frac{a_1}{(2k+1)!})

8. Write the General Solution:

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

   $ y(x) = a_0 \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!} $

   The general solution is (y(x) = a_0 \cos(x) + a_1 \sin(x)), where (a_0) and (a_1) are arbitrary constants.

Example 3: Solving ((1-x^2)y'' - xy' + 4y = 0)

Let's solve the differential equation ((1-x^2)y'' - xy' + 4y = 0) using the power series method. This is a more complex example involving variable coefficients.

1. Assume a Power Series Solution:

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

2. Compute Derivatives:

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

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

3. Substitute into the Differential Equation:

   $ (1-x^2) \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - x \sum_{n=1}^{\infty} n a_n x^{n-1} + 4 \sum_{n=0}^{\infty} a_n x^n = 0 $

   Expanding, we get

   $ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - \sum_{n=2}^{\infty} n(n-1) a_n x^{n} - \sum_{n=1}^{\infty} n a_n x^{n} + 4 \sum_{n=0}^{\infty} a_n x^n = 0 $

4. Manipulate the Series: We need to shift indices to have all series in terms of (x^n).

   • For the first series, let (m = n - 2), so (n = m + 2). When (n = 2), (m = 0). Thus,

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

   Now we have:

   $ \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^{n} - \sum_{n=2}^{\infty} n(n-1) a_n x^{n} - \sum_{n=1}^{\infty} n a_n x^{n} + 4 \sum_{n=0}^{\infty} a_n x^n = 0 $

   We need to separate out terms from the series to align the starting indices:

   $ [2 \cdot 1 \cdot a_2 + 4a_0] + [3 \cdot 2 \cdot a_3 - a_1 + 4a_1]x + \sum_{n=2}^{\infty} [(n+2)(n+1) a_{n+2} - n(n-1) a_n - n a_n + 4 a_n] x^n = 0 $

   Simplify:

   $ [2 a_2 + 4a_0] + [6 a_3 + 3a_1]x + \sum_{n=2}^{\infty} [(n+2)(n+1) a_{n+2} - (n^2 - 4) a_n] x^n = 0 $

5. Equate Coefficients:

   • (2 a_2 + 4a_0 = 0)
   • (6 a_3 + 3a_1 = 0)
   • ((n+2)(n+1) a_{n+2} - (n^2 - 4) a_n = 0)

6. Find a Recurrence Relation:

   • (a_2 = -2 a_0)
   • (a_3 = -\frac{1}{2} a_1)
   • (a_{n+2} = \frac{(n^2 - 4)}{(n+2)(n+1)} a_n) for (n \geq 2)

7. Determine the Coefficients:

   • (a_2 = -2 a_0)
   • (a_3 = -\frac{1}{2} a_1)
   • (a_4 = \frac{(2^2 - 4)}{(4)(3)} a_2 = 0)
   • (a_5 = \frac{(3^2 - 4)}{(5)(4)} a_3 = \frac{5}{20} \left(-\frac{1}{2} a_1\right) = -\frac{1}{8} a_1)
   • (a_6 = \frac{(4^2 - 4)}{(6)(5)} a_4 = 0)
   • (a_7 = \frac{(5^2 - 4)}{(7)(6)} a_5 = \frac{21}{42} \left(-\frac{1}{8} a_1\right) = -\frac{1}{16} a_1)

   Notice that all even terms beyond (a_2) are zero because (a_4 = 0), which implies (a_6 = a_8 = \dots = 0).

8. Write the General Solution:

   $ y(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5 + \dots $

   $ y(x) = a_0 + a_1 x - 2 a_0 x^2 - \frac{1}{2} a_1 x^3 + 0 x^4 - \frac{1}{8} a_1 x^5 + \dots $

   $ y(x) = a_0 (1 - 2x^2) + a_1 \left(x - \frac{1}{2} x^3 - \frac{1}{8} x^5 - \frac{1}{16} x^7 - \dots\right) $

   The general solution is (y(x) = a_0 (1 - 2x^2) + a_1 \left(x - \frac{1}{2} x^3 - \frac{1}{8} x^5 - \dots\right)), where (a_0) and (a_1) are arbitrary constants.

Convergence of Power Series Solutions

An important aspect of power series solutions is their convergence. The radius of convergence (R) determines the interval ((-R, R)) within which the series converges. The convergence of the power series can be determined using various tests, such as the ratio test.

For a power series (\sum_{n=0}^{\infty} a_n (x - c)^n), the ratio test gives:

$ \lim_{n \to \infty} \left| \frac{a_{n+1}(x - c)^{n+1}}{a_n(x - c)^n} \right| = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| |x - c| $

For the series to converge, this limit must be less than 1:

$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| |x - c| < 1 $

From this, the radius of convergence (R) can be found as:

$ R = \frac{1}{\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|} $

If the limit is infinite, (R = 0), and if the limit is zero, (R = \infty).

Advantages and Limitations

Advantages:

• General Solutions: Provides a systematic way to find general solutions to linear differential equations.
• Variable Coefficients: Effective for equations with variable coefficients, where other methods may fail.
• Approximations: Offers a way to approximate solutions when exact solutions are not obtainable.

Limitations:

• Convergence: The power series solution may only converge within a certain radius, limiting its applicability.
• Complexity: Can be algebraically intensive, especially for higher-order equations or equations with complex coefficients.
• Not Always Practical: In some cases, the recurrence relation may be too complex to find a closed-form expression for the coefficients.

Conclusion

The power series method is a versatile and powerful technique for solving differential equations, particularly those with variable coefficients. By expressing solutions as power series, we can find general solutions or approximations when other methods are not applicable. While it has its limitations, understanding and mastering the power series method provides valuable insights into the behavior of differential equations and their solutions. This method is an essential tool in the arsenal of any mathematician, physicist, or engineer dealing with differential equations.