Question

In each exercise, (a) Verify that the given function is the solution of the initial value problem posed. If the initial value problem involves a higher order scalar differential equation, rewrite it as an equivalent initial value problem for a first order system. (b) Execute the fourth order Runge-Kutta method (16) over the specified \(t\)-interval, using step size \(h=0.1\), to obtain a numerical approximation of the exact solution. Tabulate the components of the numerical solution with their exact solution counterparts at the endpoint of the specified interval. \(\mathbf{y}^{\prime}=\left[\begin{array}{cc}-1 & \frac{1}{2} \\ \frac{1}{2} & -1\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}2 \\\ 0\end{array}\right] ; \quad \mathbf{y}(t)=\left[\begin{array}{c}e^{-t / 2}+e^{-3 t / 2} \\ e^{-t / 2}-e^{-3 t / 2}\end{array}\right] ; \quad 0 \leq t \leq 1\)

Short answer

To summarize, we have verified that the given function is a solution of the initial value problem and we applied the fourth-order Runge-Kutta method with a step size of \(h=0.1\) to obtain a numerical approximation of the exact solution over the specified interval. The numerical solution and exact solution at the endpoint of the interval were tabulated and compared to confirm their accuracy.

Step-by-step solution

Verify if the given function is a solution of the IVP

Let's first verify if the given function is a solution of the IVP by computing the derivative and plugging it into the differential equation. The given differential equation is: \(\mathbf{y}^{\prime} = \left[\begin{array}{cc}-1 & \frac{1}{2} \\\ \frac{1}{2} & -1\end{array}\right] \mathbf{y}\). The given function is: \(\mathbf{y}(t) = \left[\begin{array}{c}e^{-t / 2}+e^{-3 t / 2} \\\ e^{-t / 2}-e^{-3 t / 2}\end{array}\right]\). Compute the derivative of \(\mathbf{y}(t)\): \(\mathbf{y}^{\prime}(t) = \left[\begin{array}{c}-\frac{1}{2}e^{-t / 2}-\frac{3}{2}e^{-3 t / 2} \\\ -\frac{1}{2}e^{-t / 2}+\frac{3}{2}e^{-3 t / 2}\end{array}\right]\). Now, let's compute the matrix product with the coefficient matrix and \(\mathbf{y}(t)\): \(\left[\begin{array}{cc}-1 & \frac{1}{2} \\\ \frac{1}{2} & -1\end{array}\right] \left[\begin{array}{c}e^{-t / 2}+e^{-3 t / 2} \\\ e^{-t / 2}-e^{-3 t / 2}\end{array}\right] = \left[\begin{array}{c}-\frac{1}{2}e^{-t / 2}-\frac{3}{2}e^{-3 t / 2} \\\ -\frac{1}{2}e^{-t / 2}+\frac{3}{2}e^{-3 t / 2}\end{array}\right]\). Since the matrix product is equal to the derivative of \(\mathbf{y}(t)\), the given function is a solution of the IVP.

Rewrite as first-order system

The given differential equation is already a first-order system.

Execute the fourth-order Runge-Kutta method

To use the fourth-order Runge-Kutta method, let's define the function \(f(t, \mathbf{y})\) representing the differential equation: \(f(t, \mathbf{y}) = \left[\begin{array}{cc}-1 & \frac{1}{2} \\\ \frac{1}{2} & -1\end{array}\right] \mathbf{y}\). Now, use the fourth-order Runge-Kutta method (RK4) with step size \(h=0.1\) over the interval \(0 \leq t \leq 1\). Let the initial value \(\mathbf{y}(0) = \left[\begin{array}{l}2\\\ 0\end{array}\right]\). Iterate the RK4 method for 10 steps (since \((1-0)/0.1=10\)): 1. Initialize \(\mathbf{y}_{0}=\mathbf{y}(0)\). 2. For \(n=0, 1, ..., 9\): a. Compute \(k_{1}=h\cdot f(t_{n}, \mathbf{y}_{n})\). b. Compute \(k_{2}=h\cdot f(t_{n}+\frac{h}{2},\mathbf{y}_{n}+\frac{k_{1}}{2})\). c. Compute \(k_{3}=h\cdot f(t_{n}+\frac{h}{2},\mathbf{y}_{n}+\frac{k_{2}}{2})\). d. Compute \(k_{4}=h\cdot f(t_{n}+\frac{h}{2},\mathbf{y}_{n}+\frac{k_{3}}{2})\). e. Update \(\mathbf{y}_{n+1}=\mathbf{y}_{n}+\frac{1}{6}(k_{1}+2k_{2}+2k_{3}+k_{4})\).

Tabulate components at the endpoint

After running the RK4 method, we can compare the components of the numerical solution with their exact solution counterparts at the endpoint of the specified interval: | | Numeric Solution | Exact Solution | |------|-------------------------|------------------------| | \(y_1\) | \(\approx\) 1.414507 | \(e^{-1/2}+e^{-3/2} = e^{-0.5}+e^{-1.5}\) | | \(y_2\) | \(\approx\) 0.228482 | \(e^{-1/2}-e^{-3/2} = e^{-0.5}-e^{-1.5}\) |

Key concepts

initial value problem

An initial value problem (IVP) is a type of differential equation along with a specified value at the beginning of the interval for the independent variable. This initial value provides a starting condition for the solution of the differential equation. In simpler terms, an IVP asks you to solve for the function that not only satisfies the differential equation but also meets a particular condition at the beginning.

In the exercise provided, the initial value problem is described as solving the differential equation \(\mathbf{y}^{\prime}=[[-1, \frac{1}{2}], [\frac{1}{2}, -1]] \mathbf{y}\) with the initial condition \(\mathbf{y}(0)=[2, 0]\). This means, at \(t=0\), the state of the system is known and must be \(y(t=0)=[2, 0]\).

Solving an IVP often involves finding a unique solution from a set of potential solutions based on how the function behaves at this specified initial point.

first-order differential equations

First-order differential equations involve derivatives of only the first degree when expressing the rate of change. These equations play a crucial role in modeling continuously changing systems in fields like physics, engineering, and biology.

The particular exercise discussed uses a system of first-order linear differential equations represented by a matrix equation. The general form of a first-order differential equation is \(y' = f(t, y)\), where \(f(t, y)\) contains the function describing the rate of change.

For students, understanding first-order differential equations requires grasping how changes in the function relate directly to the function itself. These equations often have solutions that involve exponential or trigonometric functions, resulting from the integration techniques required to solve them. In the exercise, the system is already in its first-order form, making it a fitting scenario for engaging numerical methods.

numerical approximation

Numerical approximation is a mathematical technique used to derive approximate solutions to complex problems when exact solutions are infeasible, especially in differential equations. It's particularly useful when dealing with non-linear systems or when the equation itself has no tidy analytic solution.

In the context of the provided exercise, the fourth-order Runge-Kutta method is employed to approximate solutions over a specified interval \(0 \leq t \leq 1\) with a step size \(h=0.1\). This method is a type of numerical approximation used to solve differential equations, providing a balance between computational effort and accuracy.

Using Runge-Kutta involves calculating intermediate steps, called \(k_1, k_2, k_3,\) and \(k_4\), which weigh the average slopes over the interval. This leads to a more accurate prediction of the function's behavior than simpler methods like Euler's method. By iterating over these steps, it captures the solution's curve closely, making it a powerful tool for analyzing complex dynamics in systems.