Question

Consider the application of the predictor-corrector method described near the end of subsection \(28.6 .3\) to the equation $$ \frac{d y}{d x}=x+y $$ Show, by comparison with a Taylor series expansion, that the expression obtained for \(y_{i+1}\) in terms of \(x_{i}\) and \(y_{i}\) by applying the three steps indicated (without any repeat of the last two) is correct to \(\mathrm{O}\left(h^{2}\right) .\) Using steps of \(h=0.1\) compute the value of \(y(0.3)\) and compare it with the value obtained by solving the equation analytically.

Short answer

Using h=0.1, the numerical solution gives y(0.3) ≈ 1.36932, which closely matches the analytical solution y(0.3) ≈ 1.3694.

Step-by-step solution

- Apply Predictor-Corrector Method

Use the predictor-corrector method consisting of three steps to determine the expression for \(y_{i+1}\) in terms of \(x_i\) and \(y_i\). The steps are:1. Predict: \(y_{i+1}^{(0)} = y_i + h f(x_i, y_i)\)2. Correct: \(y_{i+1}^{(1)} = y_i + \frac{h}{2} [f(x_i, y_i) + f(x_{i+1}, y_{i+1}^{(0)})]\)

- Find the Taylor Series Expansion

Expand \(y_{i+1}\) using Taylor series around \(x_i\):\[ y_{i+1} = y(x_i) + h y'(x_i) + \frac{h^2}{2} y''(x_i) + \mathrm{O}(h^3) \]Since \(y'(x) = x+y\), \( y''(x) = x+y+1 \). This expansion up to \(\mathrm{O}(h^2)\) provides a benchmark for the accuracy: \[ y_{i+1} = y_i + h(x_i + y_i) + \frac{h^2}{2}(x_i + y_i + 1) + \mathrm{O}(h^3) \]

- Compare Predictor-Corrector to Taylor Series

Predictor gives: \[ y_{i+1}^{(0)} = y_i + h (x_i + y_i) \]Corrector gives:\[ y_{i+1}^{(1)} = y_i + \frac{h}{2} [(x_i + y_i) + (x_{i+1} + y_{i+1}^{(0)})] = y_i + \frac{h}{2}[(x_i + y_i) + (x_i + h + y_i + h(x_i + y_i))] \]Simplify and match terms up to \(\mathrm{O}(h^2)\):\[ y_{i+1}^{(1)} = y_i + h(x_i + y_i) + \frac{h^2}{2}(x_i + y_i + 1) \]

- Validate Accuracy Order

Compare the Corrector-derived expression:\[ y_{i+1}^{(1)} = y_i + h(x_i + y_i) + \frac{h^2}{2}(x_i + y_i + 1) \] with the Taylor expansion done in Step 2. Both match up to \(\mathrm{O}(h^2)\), confirming the accuracy required.

- Implement Predictor-Corrector for Specific Steps

Compute each step for \(h = 0.1\) starting with initial values:\( \begin{aligned} y_0 &= y(0) = 1 \end{aligned}\)1. For \(x_0 = 0\):\[ y_{1}^{(0)} = y_0 + 0.1(0 + 1) = 1.1 \]\[ y_{1} = 1 + \frac{0.1}{2}(0 + 1 + 0.1 + 1.1) = 1.1 \]2. For \(x_1 = 0.1\):\[ y_{2}^{(0)} = y_1 + 0.1(0.1 + 1.1) = 1.22 \]\[ y_{2} = 1.1 + \frac{0.1}{2}(0.1 + 1.1 + 0.2 + 1.22) = 1.232 \]3. For \(x_2 = 0.2\):\[ y_{3}^{(0)} = y_2 + 0.1(0.2 + 1.232) = 1.3552 \]\[ y_{3} = 1.232 + \frac{0.1}{2}(0.2 + 1.232 + 0.3 + 1.3552) = 1.36932 \]

- Solve the Equation Analytically

Solve the DE \( \frac{dy}{dx} = x + y \) with initial condition \( y(0) = 1 \).The solution is: \[ y(x) = Ce^x - x - 1 \]Using \( y(0) = 1 \), find \( C = 2 \):\[ y(x) = 2e^x - x - 1 \]Evaluate at \( x = 0.3 \):\[ y(0.3) = 2e^{0.3} - 0.3 - 1 \approx 1.3694 \]

- Compare Numerical and Analytical Results

Numerical value obtained: \(y(0.3) \approx 1.36932 \)Analytical value: \( y(0.3) \approx 1.3694 \)Both values agree closely within \( \text{O} \left( h^{2} \right) \) error bounds.

Key concepts

Taylor Series Expansion

The Taylor series is a way of approximating a function as a sum of terms calculated from the values of its derivatives at a single point. It's particularly useful for solving differential equations and plays a big role in numerical analysis.
To expand a function around a point, say \(x_i\), we use the formula:
\[ y_{i+1} = y(x_i) + hy'(x_i) + \frac{h^2}{2}y''(x_i) + \mathcal{O}(h^3) \]
Here, \(h\) is the step size, \(y'(x_i)\) is the first derivative of \(y\) at \(x_i\), and \(y''(x_i)\) is the second derivative.
In the exercise, we use this expansion to show that our numerical method's solution matches the accuracy expected.

Differential Equations

Differential equations involve functions and their derivatives. They are used to model various phenomena in engineering, physics, biology, and economics.
For instance, the differential equation
\[\frac{d y}{d x} = x + y\]
is a first-order linear ordinary differential equation. It describes how the function \(y\) changes with \(x\).
To solve it numerically, we need methods that can provide approximate solutions at discrete points. One of those methods is the predictor-corrector.

Numerical Analysis

Numerical analysis focuses on devising algorithms to get approximate solutions to mathematical problems. When exact answers are hard to find, we rely on these methods.
The predictor-corrector method is a part of this field. It first predicts an approximate solution and then corrects it for higher accuracy.
For the mentioned problem, after predicting \(y_{i+1}^{(0)}\), we correct it with:
\[y_{i+1}^{(1)} = y_i + \frac{h}{2} [f(x_i, y_i) + f(x_{i+1}, y_{i+1}^{(0)})]\]
This refined step ensures our answer is more accurate.

Error Analysis

Error analysis evaluates the accuracy of numerical methods. It helps understand how close an approximate solution is to the true value.
For our problem, we showed that the predictor-corrector method's error is of the order \(\mathcal{O}(h^2)\). This means the error reduces quadratically as the step size \(h\) decreases.
We compared the numerical result with the analytical solution. The close agreement verifies that our method and step size provide us with a very accurate answer.