Problem 8 - A First Course in the Numerical Analysis of Differential Equations

Question

Let \(f\) be analytic. Prove that, for sufficiently small \(h>0\) and an analytic function \(x\), the function $$ \boldsymbol{x}(t+h)-\boldsymbol{x}(t-h)-h \boldsymbol{f}\left(\frac{1}{2}(\boldsymbol{x}(t-h)+\boldsymbol{x}(t+h))\right) $$ can be expanded into power series in odd powers of \(h\). Deduce that the error in the implicit midpoint rule (1.13), when applied to autonomous \(\mathrm{ODEs}\) \(y^{\prime}=f(y)\) also admits an expansion in odd powers of \(h\). [Hint: First try to prove the statement for a scalar function \(f\). Once you solve this problem, \(a\) generalization should present no difficulties.]

Step-by-Step Solution

Answer

The expression can be expanded in odd powers of \(h\), hence, the error in the implicit midpoint rule also admits a series in odd powers of \(h\).

Step 1: Taylor Expansion of Functions

To tackle this problem, recall that both the terms \( \boldsymbol{x}(t+h) \) and \( \boldsymbol{x}(t-h) \) can be expanded using their Taylor series around \( t \). Namely, we have:\[ \boldsymbol{x}(t+h) = \boldsymbol{x}(t) + h \boldsymbol{x}'(t) + \frac{h^2}{2} \boldsymbol{x}''(t) + \frac{h^3}{6} \boldsymbol{x}'''(t) + \ldots \]and,\[ \boldsymbol{x}(t-h) = \boldsymbol{x}(t) - h \boldsymbol{x}'(t) + \frac{h^2}{2} \boldsymbol{x}''(t) - \frac{h^3}{6} \boldsymbol{x}'''(t) + \ldots \]

Step 2: Compute the Difference

Now substitute the Taylor series into the expression \( \boldsymbol{x}(t+h) - \boldsymbol{x}(t-h) \):\[ \begin{align*}\boldsymbol{x}(t+h) - \boldsymbol{x}(t-h) &= \left( \boldsymbol{x}(t) + h \boldsymbol{x}'(t) + \frac{h^2}{2} \boldsymbol{x}''(t) + \ldots \right) - \left( \boldsymbol{x}(t) - h \boldsymbol{x}'(t) + \frac{h^2}{2} \boldsymbol{x}''(t) - \ldots \right) \&= 2h \boldsymbol{x}'(t) + \frac{h^3}{3} \boldsymbol{x}'''(t) + \ldots\end{align*} \]

Step 3: Evaluate the Function Term

We need to compute \( \boldsymbol{f}\left(\frac{1}{2}(\boldsymbol{x}(t-h) + \boldsymbol{x}(t+h))\right) \). Expand this function using Taylor series around \( \boldsymbol{x}(t) \):\[ \frac{1}{2}(\boldsymbol{x}(t-h) + \boldsymbol{x}(t+h)) = \boldsymbol{x}(t) + \frac{h^2}{2} \boldsymbol{x}''(t) + \ldots \]Then,\[ \boldsymbol{f}\left(\frac{1}{2}(\boldsymbol{x}(t-h) + \boldsymbol{x}(t+h))\right) = \boldsymbol{f}(\boldsymbol{x}(t)) + \boldsymbol{f}'(\boldsymbol{x}(t)) \left( \frac{h^2}{2} \boldsymbol{x}''(t) \right) + \ldots \]

Step 4: Assemble and Simplify

Now subtract the function term multiplied by \( h \):\[ h \boldsymbol{f}\left(\frac{1}{2}(\boldsymbol{x}(t-h) + \boldsymbol{x}(t+h))\right) = h \boldsymbol{f}(\boldsymbol{x}(t)) + \ldots \]The main expression simplifies to:\[ \boldsymbol{x}(t+h) - \boldsymbol{x}(t-h) - h \boldsymbol{f}\left(\frac{1}{2}(\boldsymbol{x}(t-h) + \boldsymbol{x}(t+h))\right) = 2h \boldsymbol{x}'(t) - h \boldsymbol{f}(\boldsymbol{x}(t)) + O(h^3) \]Since all higher terms in the expansion are of odd powers starting from \( h^3 \), this expression can be written as a series in odd powers of \( h \).

Step 5: Deduction for Implicit Midpoint Rule Error

This implies that the error in approximations such as the implicit midpoint rule, which is an integral-based rule to approximate solutions to differential equations, also can be expressed in terms of odd powers of \( h \). This is because the leading error term, derived from the expansion, is \( O(h^3) \), which is odd.

Key Concepts

Analytic Functions

Analytic functions are essential in mathematics due to their smooth and infinitely differentiable nature. These functions are expressed in terms of power series, which means that around a point, they can be represented as a sum of their derivatives at that point, each multiplied by a power of the variable. This property not only makes calculations involving analytic functions more straightforward but also aids in their expansion using Taylor or Maclaurin series. Analyticity implies that such functions do not just act locally like polynomials; they also exhibit great flexibility for approximation over larger intervals.
- **Power Series Representation**: An analytic function can be expanded as \[oldsymbol{f}(x) = oldsymbol{f}(a) + oldsymbol{f}'(a)(x-a) + \frac{1}{2!} \boldsymbol{f}''(a)(x-a)^2 + \ldots \] around some point \(a\).
- **Benefits in Applications**: In contexts such as solving differential equations, the ability to expand functions into series allows us to use sophisticated techniques like series solutions, which provide greater accuracy and stability for numerical approximations.

Implicit Midpoint Rule

The implicit midpoint rule is a numerical method primarily used to approximate the solutions of ordinary differential equations (ODEs). Unlike its explicit counterparts, this rule involves solving an equation implicitly, often requiring iterative techniques. Nevertheless, it offers notable advantages, especially concerning stability in stiff equations.
It approximates the function's derivative at the midpoint of the interval, providing a central estimate that helps in achieving better accuracy.
- **Formula**: For a differential equation \( y' = f(y) \), the implicit midpoint rule computes \[ y_{n+1} = y_n + h \cdot f \left( \frac{y_n + y_{n+1}}{2} \right) \], where \( n \) is the time step position.
- **Error Analysis**: The inherent stability of the method is improved through its reliance on an error term expansion in odd powers, typically described as \( O(h^3) \). This slower accumulation of error mitigates the issues found in explicit methods and makes it a preferred choice for long-duration simulations.

Ordinary Differential Equations (ODEs)

Ordinary Differential Equations are equations involving a function and its derivatives, generally providing a relationship between variables and their rates of change. Solving ODEs is critical since they model numerous real-world phenomena, from physics to biology. The presence of derivatives implies that these equations govern the dynamics of systems.
ODEs can range from simple linear equations to more complex non-linear variants. They are often classified based on their order (the highest derivative present) and linearity. Analytical solutions provide exact answers, though they may not always be feasible.
- **Numerical Solutions**: While some simpler ODEs allow for explicit solutions, more complex or nonlinear ones often rely on numerical methods, such as the implicit midpoint rule.
- **Application Examples**:

• Population growth models in ecology.
• Equations of motion in physics.
• Chemical reaction rates in chemistry.

These methods include step-by-step approximations, leveraging techniques like Euler's method, Runge-Kutta methods, and the implicit midpoint rule, each varying in complexity and scope of application.