Question

Suppose the Runge-Kutta method (12) is applied to the initial value problem \(\mathbf{y}^{\prime}=\) \(A \mathbf{y}, \mathbf{y}(0)=\mathbf{y}_{0}\), where \(A\) is a constant square matrix [thus, \(\left.\mathbf{f}(t, \mathbf{y})=A \mathbf{y}\right] .\) (a) Express each of the vectors \(\mathbf{K}_{j}\) in terms of \(h, A\), and \(\mathbf{y}_{k}, j=1,2,3,4\). (b) Show that the Runge-Kutta method, when applied to this initial value problem, can be unraveled to obtain $$ \mathbf{y}_{k+1}=\left(I+h A+\frac{h^{2}}{2 !} A^{2}+\frac{h^{3}}{3 !} A^{3}+\frac{h^{4}}{4 !} A^{4}\right) \mathbf{y}_{k} $$ (c) Use the differential equation \(\mathbf{y}^{\prime}=A \mathbf{y}\) to express the \(n\)th derivative, \(\mathbf{y}^{(n)}(t)\), in terms of \(A\) and \(\mathbf{y}(t)\). Express the Taylor series expansion $$ \mathbf{y}(t+h)=\sum_{n=0}^{\infty} \mathbf{y}^{(n)}(t) \frac{h^{n}}{n !} $$ in terms of \(h, A\), and \(\mathbf{y}(t)\). Compare the Taylor series with the right-hand side of (15), with \(t=t_{k}\) and \(\mathbf{y}\left(t_{k}\right)=\mathbf{y}_{k}\). How well does (15) replicate the Taylor series?

Short answer

#Answer# (a) \(\mathbf{K}_{1} = A\mathbf{y}_{k}\) \(\mathbf{K}_{2} = A(\mathbf{y}_{k} + \frac{1}{2}hA\mathbf{y}_{k})\) \(\mathbf{K}_{3} = A(\mathbf{y}_{k} + \frac{1}{2}hA(\mathbf{y}_{k} + \frac{1}{2}hA\mathbf{y}_{k}))\) \(\mathbf{K}_{4} = A(\mathbf{y}_{k} + hA(\mathbf{y}_{k} + \frac{1}{2}hA(\mathbf{y}_{k} + \frac{1}{2}hA\mathbf{y}_{k})))\) (b) \(\mathbf{y}_{k+1}=\left(I+h A+\frac{h^{2}}{2 !} A^{2}+\frac{h^{3}}{3 !}A^{3}+\frac{h^{4}}{4 !} A^{4}\right) \mathbf{y}_{k}\) (c) \(\mathbf{y}^{(n)}(t) = A^n \mathbf{y}(t)\), and the Taylor series expansion is \(\mathbf{y}(t+h)=I\mathbf{y}(t) + hA\mathbf{y}(t) + \frac{h^2}{2!}A^2\mathbf{y}(t) + \frac{h^3}{3!}A^3\mathbf{y}(t) + \dots\). The Runge-Kutta method replicates the Taylor series up to the fourth-order term but omits the terms after the \(h^4A^4\) term.

Step-by-step solution

Find the expressions for Kj vectors (j = 1, 2, 3, 4)

Recall the Runge-Kutta method (12) for j = 1 to 4: $$ \mathbf{K}_{j}=\mathbf{f\left(t_{k}+c_{j} h, \mathbf{y}_{k}+h \sum_{i=1}^{j-1} a_{j i} \mathbf{K}_{i}\right)} $$ As given in the problem, \(\mathbf{f}(t, \mathbf{y}) = A\mathbf{y}\). So we can rewrite the equation as: $$ \mathbf{K}_{j}=A\left(\mathbf{y}_{k}+h \sum_{i=1}^{j-1} a_{j i} \mathbf{K}_{i}\right) $$ Now, compute \(\mathbf{K}_{j}\) for each j = 1,2,3, and 4: \(\mathbf{K}_{1} = A\mathbf{y}_{k}\) \(\mathbf{K}_{2} = A(\mathbf{y}_{k} + \frac{1}{2}h\mathbf{K}_{1}) = A(\mathbf{y}_{k} + \frac{1}{2}hA\mathbf{y}_{k})\) \(\mathbf{K}_{3} = A(\mathbf{y}_{k} + \frac{1}{2}h\mathbf{K}_{2}) = A(\mathbf{y}_{k} + \frac{1}{2}hA(\mathbf{y}_{k} + \frac{1}{2}hA\mathbf{y}_{k}))\) \(\mathbf{K}_{4} = A(\mathbf{y}_{k} + h\mathbf{K}_{3}) = A(\mathbf{y}_{k} + hA(\mathbf{y}_{k} + \frac{1}{2}hA(\mathbf{y}_{k} + \frac{1}{2}hA\mathbf{y}_{k})))\) (b)

Unravel the Runge-Kutta method to find y(k+1)

Using the expression for \(\mathbf{K}_{j}\) from part (a), unravel the Runge-Kutta method: $$ \mathbf{y}_{k+1} = \mathbf{y}_k + \frac{1}{6}h(\mathbf{K}_{1} + 2\mathbf{K}_{2} + 2\mathbf{K}_{3} + \mathbf{K}_{4}) $$ Plugging in the values of \(\mathbf{K}_{j}\) and simplifying, we get: $$ \mathbf{y}_{k+1}=\left(I+h A+\frac{h^{2}}{2 !} A^{2}+\frac{h^{3}}{3 !}A^{3}+\frac{h^{4}}{4 !} A^{4}\right) \mathbf{y}_{k} $$ (c)

Find the nth derivative of y(t) and express the Taylor series expansion in terms of given variables

To find the nth derivative, use the given differential equation \(\mathbf{y'} = A\mathbf{y}\): $$ \mathbf{y}^{(n)}(t) = A^n \mathbf{y}(t) $$ Now, express the Taylor series expansion in terms of h, A, and \(\mathbf{y}(t)\): $$ \mathbf{y}(t+h)=\sum_{n=0}^{\infty} \mathbf{y}^{(n)}(t) \frac{h^{n}}{n !} = I\mathbf{y}(t) + hA\mathbf{y}(t) + \frac{h^2}{2!}A^2\mathbf{y}(t) + \frac{h^3}{3!}A^3\mathbf{y}(t) + \dots $$ Comparing the Taylor series with the right-hand side of (15), with \(t=t_{k}\) and \(\mathbf{y}\left(t_{k}\right)=\mathbf{y}_{k}\), we can observe that the Runge-Kutta method equation in (b) replicates the Taylor series up to the fourth-order term, providing a good approximation for the Taylor series. However, it omits the terms that come after the \(h^4A^4\) term.

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation with a given initial condition. Essentially, it means you have a differential equation, which is like an equation involving derivatives, and also know the function's value at a certain point. Consider it like a roadmap where you not only know the path but also the starting point.

The problem in the exercise is \\(\mathbf{y}^{\prime} = A \mathbf{y}, \mathbf{y}(0) = \mathbf{y}_{0}\), with \(A\) as a constant matrix. Here, \(\mathbf{y}(0) = \mathbf{y}_{0}\) is your initial condition at \(t = 0\). This means that at the start, the function \(\mathbf{y}(t)\) has a specific value \(\mathbf{y}_{0}\).

Initial value problems are fundamental because they allow us to predict the behavior of dynamical systems over time, once we know how they start. Using methods like the Runge-Kutta, we can approximate solutions to these problems even if finding an exact solution is difficult.

Differential Equations

Differential equations involve rates of change and are used to describe many natural phenomena. They are equations that relate a function with its derivatives, allowing us to model physical situations by expressing how one quantity changes in relation to another.

In this exercise, our differential equation is \(\mathbf{y}^{\prime} = A \mathbf{y}\), where \(\mathbf{y}^{\prime}\) is the derivative of \(\mathbf{y}\), and it shows how \(\mathbf{y}\) changes with respect to time. Here, \(A \mathbf{y}\) suggests that the rate of change of \(\mathbf{y}\) is directly proportional to the current state \(\mathbf{y}\).
This form is a linear differential equation because \(A\) is a constant matrix, and these equations often have straightforward solutions. Such equations are prevalent in physics and engineering, modeling anything from population growth to electrical circuits.

Taylor Series Expansion

The Taylor series expansion is a way to express functions as infinite sums of terms calculated from the values of its derivatives at a particular point. It’s a useful tool when you need to approximate functions that are otherwise difficult to express.

The exercise asks us to match the Taylor series with \\(\mathbf{y}(t+h) = \sum_{n=0}^{\infty} \mathbf{y}^{(n)}(t) \frac{h^{n}}{n !}\). This means you can think of \(\mathbf{y}(t+h)\), the function at a slight future time, in terms of its current value, its rate of change, and higher derivatives—each term providing more precision.

The terms \(I\mathbf{y}(t) + hA\mathbf{y}(t) + \frac{h^2}{2!}A^2\mathbf{y}(t)\) etc., are analogous to the Taylor series terms. The Runge-Kutta method effectively captures the first few terms of this series providing an accurate solution by approximating future values. The closer these terms match, the closer the approximation is to the real behavior of the system.