Question

For the given initial value problem, an exact solution in terms of familiar functions is not available for comparison. If necessary, rewrite the problem as an initial value problem for a first order system. Implement one step of the fourth order Runge-Kutta method (14), using a step size \(h=0.1\), to obtain a numerical approximation of the exact solution at \(t=0.1\). \(y^{\prime \prime \prime}-t y=0, \quad y(0)=1, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=-1\)

Short answer

Based on the given initial value problem involving a third-order ordinary differential equation, we implemented the fourth-order Runge-Kutta method to find a numerical approximation of the exact solution at t=0.1. After performing the calculations, we found that y(0.1) is approximately equal to 0.995.

Step-by-step solution

Rewrite the problem as a first-order system

To rewrite the IVP as a first-order system, we introduce new variables for each derivative: $$ x_1 = y \\ x_2 = y^{\prime} \\ x_3 = y^{\prime \prime} $$ Then, the system becomes: $$ x_1^{\prime} = x_2 \\ x_2^{\prime} = x_3 \\ x_3^{\prime} = ty $$ The given initial conditions are: $$ x_1(0) = 1 \\ x_2(0) = 0 \\ x_3(0) = -1 $$

Set up the fourth-order Runge-Kutta method

To implement the fourth-order Runge-Kutta method for the given system, we need to compute the increments k1, k2, k3, k4 for each variable \(x_i\). The general formulae for the Runge-Kutta method are: $$ k_1 = hF(t, x) \\ k_2 = hF(t + \frac{h}{2}, x + \frac{k_1}{2}) \\ k_3 = hF(t + \frac{h}{2}, x + \frac{k_2}{2}) \\ k_4 = hF(t + h, x + k_3) \\ x(t + h) = x(t) + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) $$ In our case, the function F(t, x) is defined as follows: $$ F(t, x) =\begin{pmatrix} x_2 \\ x_3 \\ t x_1 \end{pmatrix} $$ Now, we can implement the Runge-Kutta method for one step, using a step size of \(h=0.1\).

Implement the fourth-order Runge-Kutta method

First, we need to define the initial values \(t_0 = 0\) and \(\begin{pmatrix}x_{1}(0) \\ x_{2}(0) \\ x_{3}(0) \end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ -1 \end{pmatrix}\). Next, we will calculate the increments k1, k2, k3, k4 for the given step size (\(h=0.1\)): $$ k_{1} = 0.1F(0,\begin{pmatrix}1 \\ 0 \\ -1 \end{pmatrix})=0.1\begin{pmatrix}0 \\ -1 \\ 0 \end{pmatrix}=\begin{pmatrix}0 \\ -0.1 \\ 0 \end{pmatrix} \\ k_{2} = 0.1F(0.05,\begin{pmatrix}1 \\ -0.05 \\ -0.5 \end{pmatrix})=0.1\begin{pmatrix}-0.05 \\ -0.5 \\ 0.025 \end{pmatrix}=\begin{pmatrix}-0.005 \\ -0.05 \\ 0.0025 \end{pmatrix} \\ k_{3} = 0.1F(0.05,\begin{pmatrix}1 \\ -0.025 \\ -0.75 \end{pmatrix})=0.1\begin{pmatrix}-0.025 \\ -0.75 \\ 0.0125 \end{pmatrix}=\begin{pmatrix}-0.0025 \\ -0.075 \\ 0.00125 \end{pmatrix} \\ k_{4} = 0.1F(0.1,\begin{pmatrix}1 \\ 0 \\ -0.875 \end{pmatrix})=0.1\begin{pmatrix}0 \\ -0.875 \\ 0 \end{pmatrix}=\begin{pmatrix}0 \\ -0.0875 \\ 0 \end{pmatrix} $$ Finally, the numerical approximation for the solution at \(t=0.1\) is given by the updated variables \(x_{1}(0.1), x_{2}(0.1), x_{3}(0.1)\): $$ \begin{pmatrix}x_{1}(0.1) \\ x_{2}(0.1) \\ x_{3}(0.1) \end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ -1 \end{pmatrix} + \frac{1}{6}(\begin{pmatrix}0 \\ -0.1 \\ 0 \end{pmatrix} + 2(\begin{pmatrix}-0.005 \\ -0.05 \\ 0.0025 \end{pmatrix}) + 2(\begin{pmatrix}-0.0025 \\ -0.075 \\ 0.00125 \end{pmatrix})+\begin{pmatrix}0 \\ -0.0875 \\ 0 \end{pmatrix}) $$ After calculations, we find: $$ \begin{pmatrix}x_{1}(0.1) \\ x_{2}(0.1) \\ x_{3}(0.1) \end{pmatrix} \approx \begin{pmatrix}0.995 \\ -0.05416 \\ -1.00833 \end{pmatrix} $$ Thus, the numerical approximation of the exact solution at \(t=0.1\) is \(y(0.1) \approx x_{1}(0.1) \approx 0.995\).

Key concepts

Numerical Approximation

Numerical approximation is a process used in math to find an estimated solution for mathematical problems that might be difficult or impossible to solve exactly. It's like finding a close answer when the exact one is too tricky. When working with differential equations, especially ones that don't have simple solutions, numerical methods become incredibly helpful. They allow us to compute approximate values for functions or their derivatives at various points. Runge-Kutta methods, like the one mentioned in the exercise, are popular approaches for these tasks. They involve calculating new estimates through intermediate points within a step, thus providing a more accurate approximation of the solution. The typical goal is to minimize errors and gain insights into the behavior of the solution over time.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation that comes with a set of initial conditions, telling you the state of the system at the start. Basically, it's like starting a journey with a known starting point and trying to figure out where you'll end up. IVPs are crucial in modeling real-world phenomena where the initial conditions can be controlled or measured. The differential equation will dictate how the system evolves over time, while the initial conditions specify the system's starting state. The challenge often faced with IVPs is finding the solution at a future point in time, especially when no exact formula exists.

First-order System

In mathematical terms, rewriting a problem as a first-order system means expressing higher-order differential equations as a set of first-order equations. Imagine translating a complex idea into smaller, more speakable parts. This allows for easier manipulation and solution using numerical methods like the Runge-Kutta. To achieve this, each derivative of a function up to the order of the original equation is treated as an independent variable. So, a third-order differential equation, as seen in the exercise, transforms into three interrelated first-order equations. Not only does this simplification accommodate easier computation, but it also aligns with how many computer algorithms handle differential equations.

Higher-order Differential Equations

Higher-order differential equations involve derivatives of an unknown function higher than the first derivative. They express relationships involving rates of change of rates of change, like how the acceleration of a car implies a change in its speed. Though their complexity is greater than first-order equations, they are pivotal in accurately describing physical and engineering systems. The exercise's highlighted equation, an example of a third-order differential equation, needs rewriting to a first-order system to use numerical methods effectively. Tackling higher-order equations often involves breaking down the problem into simpler parts, calculating step-by-step results that compound to reach accurate approximations.