Question

Consider a linearized version of Example 16.22, given by $$ v^{\prime \prime}+a(t) v=q(t), \quad v(0)=v(1)=0 $$ (a) Converting the linear \(\mathrm{ODE}\) to first order form as in Section \(16.7\), show that $$ \begin{aligned} &A(t)=\left(\begin{array}{cc} 0 & 1 \\ -a(t) & 0 \end{array}\right), \mathbf{q}(t)=\left(\begin{array}{c} 0 \\ q(t) \end{array}\right) \\ &B_{a}=B_{b}=(1 \quad 0), c_{a}=c b=0 \end{aligned} $$ (b) Write down explicitly the linear system of algebraic equations resulting from the application of the midpoint method. Show that the obtained matrix is banded with five diagonals.

Short answer

In summary, we first converted the second-order ODE into a system of first-order ODEs and represented this system using matrices and vectors. Next, we applied the midpoint method to this first-order system, which resulted in a linear system of algebraic equations. Finally, we showed that the resulting matrix of this linear system is banded with five diagonals.

Step-by-step solution

Convert the ODE to first-order form

To rewrite the given second-order ODE in its first order form, let's introduce a new variable: \(u(t) = v'(t)\). So now, we have: $$ u'(t) + a(t)u(t) = q(t) $$ $$ v'(t) = u(t) $$ Now, we have a system of two first-order linear differential equations that we can present them in matrix form:

Represent the first-order form in matrix form

Now, we can create a matrix \(A(t)\) and a vector \(\mathbf{q}(t)\) that describe the first-order system. As we have two equations and two unknowns (\(u(t)\) and \(v(t)\)), our matrix will be of size \(2 \times 2\). The system of first-order ODEs can be represented in the form: $$ \begin{pmatrix} v'(t) \\ u'(t) \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -a(t) & 0 \end{pmatrix} \begin{pmatrix} v(t) \\ u(t) \end{pmatrix} + \begin{pmatrix} 0 \\ q(t) \end{pmatrix} $$ So we have, $$ A(t) = \begin{pmatrix} 0 & 1 \\ -a(t) & 0 \end{pmatrix}, \mathbf{q}(t) = \begin{pmatrix} 0 \\ q(t) \end{pmatrix} $$

Specify boundary conditions in matrix form

The boundary conditions are \(v(0) = v(1) = 0\), the corresponding matrices will be: $$ B_a = B_b = \begin{pmatrix} 1 & 0 \end{pmatrix} $$ and the boundary condition constants will be \(c_a = c_b = 0\).

Apply the midpoint method

The midpoint method formula is : $$ \left(\frac{I}{2} - \frac{\Delta t}{2} A(t)\right)v_{n+1} = \left(\frac{I}{2} + \frac{\Delta t}{2} A(t)\right)v_{n} + \Delta t \mathbf{q}_{n+1/2} $$ Let \(G = \frac{I}{2} - \frac{\Delta t}{2} A(t)\), \(H = \frac{I}{2} + \frac{\Delta t}{2} A(t)\), then $$ G v_{n+1} = H v_{n} + \Delta t \mathbf{q}_{n+1/2} $$ By substituting \(A(t)\) into the expression, we get: $$ \begin{pmatrix} \frac{1}{2} & \frac{\Delta t}{2} \\ \frac{\Delta t a(t)}{2} & \frac{1}{2} \end{pmatrix} v_{n+1} = \begin{pmatrix} \frac{1}{2} & -\frac{\Delta t}{2} \\ -\frac{\Delta t a(t)}{2} & \frac{1}{2} \end{pmatrix} v_{n} + \Delta t \begin{pmatrix} 0 \\ q_{n + \frac{1}{2}} \end{pmatrix} $$ We can rewrite the equation above as a linear system of algebraic equations.

Analyze the matrix resulting from the midpoint method

In order to see that the obtained matrix has five diagonals, let's represent the linear system as $$ E_n V_n = F_n V_n + \Delta t \mathbf{Q}_{n+1/2}, $$ where \(E_n\) and \(F_n\) are block-diagonal matrices constructed from matrices \(G\) and \(H\) respectively at each time step \(t_n\). Further, let \(V_n\) represent the vector of u and v at each time step. Considering the structure of matrices \(G\) and \(H\), we can see that \(E_n\) and \(F_n\) will be banded matrices with three diagonals each, where one diagonal is a sequence of \(\frac{1}{2}\), while the other two contain only non-zero elements from \(A(t)\) (i.e., \(\frac{\Delta t}{2}\) and \(\frac{\Delta t a(t)}{2}\)) and \(-\frac{\Delta t}{2} A(t)\) (i.e., \(-\frac{\Delta t}{2}\) and \(-\frac{\Delta t a(t)}{2}\)). In the multiplication of two banded matrices with three diagonals, the resulting matrix will be banded with five diagonals. Hence, the matrix resulting from the midpoint method is banded with five diagonals.

Key concepts

First-order Differential Equations

First-order differential equations are equations involving the first derivative of a function. They often describe rates of change and are fundamental in understanding many physical phenomena. For example, Newton’s law of cooling and exponential growth or decay models are based on first-order differential equations.

In the exercise, a second-order differential equation for the function \( v(t) \) is given. To convert it into a first-order system, a new variable \( u(t) = v'(t) \) is introduced. This transformation converts the original problem into a system involving the variables \( v \) and \( u \), each described by a first-order differential equation.

• Equation for \( u \): \( u'(t) + a(t)u(t) = q(t) \)
• Equation for \( v \): \( v'(t) = u(t) \)

These equations can be written in matrix form, facilitating numerical solutions using methods like the Midpoint Method.

Midpoint Method

The Midpoint Method is a numerical technique used to approximate solutions to differential equations. It is a specific type of Runge-Kutta method, designed to be more accurate than simple methods like Euler's. This method calculates the slope (or derivative) at the midpoint between known points and uses it to estimate the function's next value.

In the context of the exercise, the Midpoint Method is applied to the first-order equations developed from the original problem. It involves using matrices to handle the system of equations. The midpoint of the interval \( (t_n, t_{n+1}) \) is leveraged for better approximation.

• Consider interval steps of size \( \Delta t \).
• Adjust the system variables using the midpoint slope: \( v_{n+1} = v_{n} + f(\frac{t_{n} + t_{n+1}}{2}, \frac{v_{n} + v_{n+1}}{2}) \Delta t \).

Hence, using the method, one obtains a discrete system that approximates the continuous problem.

Boundary Conditions

Boundary conditions specify the values that a solution to a differential equation must satisfy at the boundaries of the domain. They are essential in defining a well-posed problem. Without these conditions, solutions to differential equations can be indeterminate.

In the exercise, the boundary conditions are \( v(0) = v(1) = 0 \). These conditions anchor the solutions, ensuring they are unique and calculated relative to specified values at the endpoints of the interval under consideration.

• The boundary conditions are represented in matrix form in numerical calculations.
• The matrices \( B_a \) and \( B_b \) ensure the conditions are met across the solution process. These are paired with constants \( c_a = c_b = 0 \) holding the values of the solution at the boundaries.

They are integral when translating continuous problems into a discrete framework like the Midpoint Method.

Algebraic Systems

An algebraic system is a set of equations that can be written in matrix form to solve for unknown variables. In numerical methods involving differential equations, continuous differential equations are often approximated by algebraic systems when using discretization techniques.

In this task, using the Midpoint Method results in a linear system of algebraic equations. These equations incorporate matrices capturing the necessary transformations from the differential equations using the matrices \( G \) and \( H \).

• Represents the dynamics of the whole system in a compact form.
• Often results in a banded matrix, as seen in the exercise, where the structure simplifies computationally efficient solutions.

The solved algebraic system provides approximations to solutions of the original differential equations, rendered in a discrete step-by-step format.