Example 1: Let us consider an initial value problem (IVP) for a homogeneous linear differential equation:

First, we create a differentail operator corresponding to the linear differential equation and set up the initial conditions: We set up our Taylor series as a symbolic expansion using derivatives of x evaluated at the origin. I use an order of 15 but that is something one would probably make as an argument to a function, if automating all this. Next apply the differential operator and add the initial conditions. Then find a solution that makes all powers of x vanish. Let's look at the Taylor polynomial. To assess how accurate it might be I will compare with NDSolve. This is a good start. Power series solutions, though, are frequently used to obtain recursion equations for the coefficients (of any solution that might be analytic within a neighborhood of the point of expansion). It would be nice, then, to have a function that outputs these equations (given a differential operator as input), rather than just obtaining an approximate solution with a limited radius of accuracy. In order to analyze singular points, it would also be useful to consider slightly more general series of the Frobenius form or non-integral α.

Mathematica has a special command that can provide a series solution even for non-linear differential equations.

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Example 2: Consider the differential equation that we met previously in section

Using the ansatz about solution form we substitutte this series into the given differential equation. This yields We check our answer with Mathematica:    ■

Example 3: Consider the differential equation

The corresponding coefficients are holomorphic functions in the circle |x - 2| < 2. Therefore, we expect that our differential equation has a solution that can be represented by a convergent in the same circle series: Substituting this series into the given differential equation, we obtain Equating like power coefficients, we get For the first two coefficients, we have So two first values are \( c_2 = -3\,c_0 /8 , \quad c_3 = 0. \) For other coefficients, we get we check our answer with Mathematica: So we claim that the general solution becomes This is a general solution composed by two linearly independent solutions, where c₀, c₁ are arbitrary constants.

Use of the nearest singular point is of some worth since having a lower bound is stronger than not knowing anything about the radius of convergence. However, for particular types of differential equations, there is a method that has been proven to give an exact value.    ■

Example 4: Let us consider a second order differential equation

We are going to determine a lower bound on the radius of convergence for the series solution about x₀ = 0 and x₀ = 3.

The point x₀ = 0 is an ordinary point since the coefficients of the given differential equation \( \displaystyle y'' + p(x)\, y' + q(x)\, y = 0 \) are holomorphic functions:

The radius of convergence of the series solution will be at least as large as the minimum of the radius of convergence of the series for p(x) = (x² and q(x) = (x³ about x₀ = 0.

Since p(x) and q(x) are already expanded in power series, and these series are not infinite, the radius of convergence for them is ρ = ∞. Therefore, the series solution about x₀ = 0 must have a radius of convergence that is at least as large as ρ = ∞, which of course means it must be ρ = ∞.

A similar argument holds for x₀ = 3.    ■

Example 5: Let us consider a second order differential equation

We are going to determine a lower bound on the radius of convergence for the series solution about x₀ = 0 and x₀ = 5.

The point x₀ = 0 is an ordinary point because the coefficients

are both analytic about x₀ = 0.

Let’s determine the radius of convergence of p and q without working out the Taylor series for them. The complex poles of p and p all occur when x² −x + 3 = 0, which means x = 1 and x = −3. These roots are at distance 1 and 3 from the origin. Since the nearest complex pole is 1, we conclude that the radius of convergence of the series for p(x) and q(x) is ρ = 1. Therefore, the minimum radius of convergence for the series solution about x₀ = 0 to the differential equation is ρ = 1.

The distance from x₀ = 5 to the nearest complex pole is ρ = 4. The minimum radius of convergence for the series solution about x₀ = 5 to the differential equation is ρ = 4.    ■

Example 6: The conservative form of the Burgers' equation can be reduce to the following form

where ν is a positive constant. Substituting the proposed series solution into Burgers' equation, we obtain Matching coefficients with the equal exponents, we derive the sequence of recursive relations:

1. n = 0:      \( a_0 a_1 = 2\nu\,a_2 \)
2. n = 1:      \( a_1^2 + 2\,a_0 a_2 = 6\nu\,a_3 \)
3. n = 2:      \( a_1 a_2 + a_0 a_3 = 4\nu\,a_4 \)
4. n = 3:      \( a_1 a_3 + 2\,a_2^2 + a_0 a_5 = 5\nu\,a_5 \)

Solving the algebraic system of 9 equations, for the 9 unknown variables [a₂, a₃, ... ,a₉] as functions of two known initial values a₀ = v(0) and a₁ = v'(0) and the physical parameterν, we express coefficients explicitly:    ■

Example 7: Consider the initial value problem for undamped Duffing equation:

We can rewrite the Duffing equation in the operator form: Physically, the resilience of the oscillator is directly proportional to N[y]. When k ≥ 0 and ε < 0, there is one unique equilibrium point y = 0 where the nonlinear term vanishes. However, when k ≤ 0 and ε > 0, there are two equilibrium points: y = 0 and \( y = \sqrt{|k/\varepsilon |} . \) From the physical points of view, the sum of the kinetic and potential energy of the oscillator keeps the same, therefore it is clear that the oscillation motion is periodic, no matter k is positive or negative. Thus, from physical points of view, it is easy to know that y(t) is periodic, even if we do not directly solve the Duffing equation.    ■

Example 8:

Example: Move to chapter 4, ndsolve Consider the second-order nonlinear differential equation with the product of the derivative and a sinusoidal nonlinearity of the solution

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Example 9:

Example: Consider the pendulum like equation

This equation arises from the general sine-Gordon equation which is a model for the nonlinear meson fields with periodic properties for the unified description of mesons and their particle sources. A simple solution of the sine-Gordon equation may be found by representing u as a function of \( \xi = \left( x - vt \right) /\sqrt{1-v^2} . \) In this case, the sine-Gordon equation is reduced to the pendulum like equation \( {\text d}^2 u/{\text d} \xi^2 = \sin u . \) For a real physical velocity v < 1, there are implicit solutions given by with total energy \( \displaystyle E = \frac{2}{\pi} \,\frac{1}{\sqrt{1-v^2}} . \) These solutions may be interpreted as the fields associated with a particle of mass 2/π centered at ξ = ξ₀ and moving with velocity v.

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Exampl 10: The system of partial differential equations, called KdV equations, can be reduced by appropriate substitution to the following ordinary differential equation

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In all problems, y' denotes the derivative of function y(x) with respect to independent variable x and \( \dot{y} = {\text d} y/{\text d}t \) denotes the derivative with respect to time variable.

1. Using ADM, solve the initial value problem: \( e^x \,y'' + x\,y = 0 , \quad y(0) =A, \quad y' (0) = B . \)
2. Using ADM, solve the initial value problem: \( e^x \,y'' + x\,y = 0 , \quad y(0) =A, \quad y' (0) = B . \)

1. Dettman, J.W., Power Series Solutions of Ordinary Differential Equations, The American Mathematical Monthly, 1967, Vol. 74, No. 3, pp. 428--430.
2. Grigorieva, E., Methods of Solving Sequence and Series Problems, Birkhäuser; 1st ed. 2016.