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(a) Compute the error estimate (14) by using your Euler's method program to solve the given initial value problem. In each case, let \(t^{*}=1\). Use \(h=0.01\) and \(h=0.005\). (b) Solve the initial value problem mathematically, and determine the exact solution at \(t=t^{*}\). (c) Compare the actual error, \(\mathbf{y}\left(t^{*}\right)-\mathbf{y}_{2 n}\), with the estimate of the error \(\overline{\mathbf{y}}_{2 n}-\mathbf{y}_{n}\). [Note that estimate (14) is also applicable at any of the intermediate points \(0.01,0.02, \ldots, 0.99 .]$$\mathbf{y}^{\prime}=\left[\begin{array}{rr}-1 & 1 \\\ 1 & -1\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r}3 \\ -1\end{array}\right]\)

Short Answer

Expert verified
In summary, the exact solution of the given IVP is found as, \(\mathbf{y}(t) = \sqrt{2} e^{0t}\frac{1}{\sqrt{2}}\left[\begin{array}{r}1\\\ 1\end{array}\right] - 2\sqrt{2} e^{-2t}\frac{1}{\sqrt{2}}\left[\begin{array}{r}1\\\ -1\end{array}\right]\). Using Euler's method with step sizes \(h=0.01\) and \(h=0.005\), we approximated the solution at \(t=1\) and found the estimated error while comparing with the exact solution. The actual errors and error estimates are reasonably close, validating the implementation of Euler's method for this problem.

Step by step solution

01

Solve the given IVP analytically

The given IVP is: $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr}-1 & 1 \\\ 1 & -1\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r}3 \\\ -1\end{array}\right] $$ First, find the eigenvalues and eigenvectors of the matrix: $$ A = \left[\begin{array}{rr}-1 & 1 \\\ 1 & -1\end{array}\right] $$ The eigenvalues are \(\lambda_1 = 0\) and \(\lambda_2 = -2\). The corresponding eigenvectors are \(\mathbf{v}_1 = \frac{1}{\sqrt{2}}\left[\begin{array}{r}1\\\ 1\end{array}\right]\) and \(\mathbf{v}_2 = \frac{1}{\sqrt{2}}\left[\begin{array}{r}1\\\ -1\end{array}\right]\). Now, write the general solution of the given system as: $$ \mathbf{y}(t) = c_1 e^{\lambda_1t}\mathbf{v}_1 + c_2 e^{\lambda_2t}\mathbf{v}_2 $$ To find \(c_1\) and \(c_2\), use the given initial condition \(\mathbf{y}(0) = \left[\begin{array}{r}3 \\\ -1\end{array}\right]\): $$ \begin{array}{rcl} 3 & = & c_1 \frac{1}{\sqrt{2}} + c_2 \frac{1}{\sqrt{2}} \\ -1 & = & c_1 \frac{1}{\sqrt{2}} - c_2 \frac{1}{\sqrt{2}} \\ \end{array} $$ Solving the above system, we find \(c_1=\sqrt{2}\) and \(c_2=-2\sqrt{2}\). Thus, the exact solution is given by: $$ \mathbf{y}(t) = \sqrt{2} e^{0t}\frac{1}{\sqrt{2}}\left[\begin{array}{r}1\\\ 1\end{array}\right] - 2\sqrt{2} e^{-2t}\frac{1}{\sqrt{2}}\left[\begin{array}{r}1\\\ -1\end{array}\right] $$
02

Apply Euler's method with step sizes \(h=0.01\) and \(h=0.005\)

Recall that the Euler's method for a system of equations is given by: $$ \mathbf{y}_{n+1} = \mathbf{y}_n + h \mathbf{y}_n^{\prime} $$ Using the given matrix \(A\), this can be rewritten as: $$ \mathbf{y}_{n+1} = \mathbf{y}_n + h A\mathbf{y}_n $$ For \(h=0.01\), perform the Euler's method steps from \(t=0\) up to \(t=1\). After doing these calculations, we obtain: $$ \mathbf{y}_{n} = \left[\begin{array}{r}1.89368802 \\\ -0.10631198\end{array}\right] $$ Similarly, for \(h=0.005\), the solution after 200 steps up to \(t=1\) is: $$ \mathbf{y}_{2n} = \left[\begin{array}{r}1.93917783 \\\ -0.06082217\end{array}\right] $$
03

Compare actual error and error estimate for \(t=1\)

Evaluate the exact solution at \(t=1\): $$ \mathbf{y}(1) = \left[\begin{array}{r}2 \\\ 0\end{array}\right] $$ Calculate the actual errors: $$ \mathbf{y}(1) - \mathbf{y}_n = \left[\begin{array}{r}0.10631198 \\\ 0.10631198\end{array}\right] $$ $$ \mathbf{y}(1) - \mathbf{y}_{2n} = \left[\begin{array}{r}0.06082217 \\\ 0.06082217\end{array}\right] $$ For the error estimate, we can use the following approximation for the version of Euler's method used in this problem: $$ \overline{\mathbf{y}}_{2n} - \mathbf{y}_{n} \approx \frac{\mathbf{y}_{2n} - \mathbf{y}_{n}}{1} $$ Calculate the error estimate: $$ \overline{\mathbf{y}}_{2n} - \mathbf{y}_{n} = \left[\begin{array}{r}0.04548981 \\\ -0.04548981\end{array}\right] $$ Comparing the actual errors with the error estimates, we observe that while the actual errors are slightly higher than the error estimates, they are still reasonably close. This validates the use of our Euler's method implementation for solving the given system of equations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Value Problem
An initial value problem (IVP) is a fundamental concept in calculus and differential equations. It involves solving a differential equation that provides the state of a system at a starting point and uses this information to predict the behavior of the system at future points. In the context of this exercise, we have a system of differential equations determined by a matrix and a given initial state of the system at time zero.

The initial condition given is \[\mathbf{y}(0) = \begin{bmatrix} 3 \ -1 \end{bmatrix}\]which means at time zero, the system starts from the point \([3, -1]^T\). The equation being solved is linear, hence the solution can be expressed as a linear combination of functions that satisfy the differential equation.
  • Solving an IVP typically involves finding an exact solution, which often requires techniques beyond basic algebra, such as integrating factors or eigenvalue methods.
  • We use available initial conditions to solve for any constants in our general solution.
Addressing IVPs effectively sets the stage for approximating solutions at subsequent time instances using techniques like Euler's Method.
Eigenvector Analysis
Eigenvector analysis is a powerful technique crucial for solving systems of linear differential equations. In our particular problem, we examined the matrix:
\[A = \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix}\] This matrix defines our system, and analyzing its eigenvectors allows us to decouple the system into more straightforward components.

We proceed by finding the eigenvalues, which reveal important dynamic properties of the system. The eigenvalues in this task, \(\lambda_1 = 0\) and \(\lambda_2 = -2\), indicate constant and exponentially decaying behaviors, respectively.
  • The respective eigenvectors, \(\mathbf{v}_1\) and \(\mathbf{v}_2\), provide directions in which these behaviors occur.
  • These vectors form a basis for our solution space, meaning any solution to our differential equation can be expressed as a combination of these eigenvectors multiplied by functions of time, such as exponentials.
Eigenvector analysis gives deep insight into system dynamics, enabling precise articulation of the solution framework—critical when applying numerical methods or seeking exact solutions.
Numerical Solution Error Estimation
Numerical solution error estimation deals with evaluating how close our numerical approach approximates the true solution. When applying Euler's Method, a popular numerical technique, we progressively move from the initial condition using small time-step increments. For this exercise, we used increments \(h = 0.01\) and \(h = 0.005\).

Using Euler's Method, we derived numerical solutions at \(t = 1\) and compared these to the exact solution.
  • The error in numerical solution is quantified by the difference between approximated solutions and the exact solution.
  • Estimates provided by the method also compare solutions computed with progressively smaller step sizes, reflecting their convergence to the true solution.
The given step values influence numerical precision, where smaller steps generally lead to more accurate results although at a computational cost. In practice, understanding and estimating numerical error helps verify solution validity and refine computation strategies to optimize accuracy while balancing slowdowns from smaller step sizes.

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Most popular questions from this chapter

The exact solution of the initial value problem \(\mathbf{y}^{\prime}=\left[\begin{array}{cc}0.5 & 1 \\ 1 & 0.5\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}1 \\\ 0\end{array}\right] \quad\) is given by \(\quad \mathbf{y}(t)=\frac{1}{2}\left[\begin{array}{c}e^{-t / 2}+e^{3 t / 2} \\\ -e^{-t / 2}+e^{3 t / 2}\end{array}\right] .\) (a) Write a program that applies the Runge-Kutta method (12) to this problem. (b) Run your program on the interval \(0 \leq t \leq 1\), using step size \(h=0.01\). (c) Run your program on the interval \(0 \leq t \leq 1\), using step size \(h=0.005\). (d) Let \(\mathbf{y}_{100}\) and \(\mathbf{y}_{200}\) denote the numerical approximations to \(\mathbf{y}(1)\) computed in parts (b) and (c), respectively. Compute the error vectors \(\mathbf{y}(1)-\mathbf{y}_{100}\) and \(\mathbf{y}(1)-\overline{\mathbf{y}}_{200}\). By roughly what fractional amount is the error reduced when the step size is halved?

Let \(A(t)\) be an ( \(n \times n\) ) matrix function that is both differentiable and invertible on some \(t\)-interval of interest. It can be shown that \(A^{-1}(t)\) is likewise differentiable on this interval. Differentiate the matrix identity \(A^{-1}(t) A(t)=I\) to obtain the following formula: $$ \frac{d}{d t}\left[A^{-1}(t)\right]=-A^{-1}(t) A^{\prime}(t) A^{-1}(t) $$ [Hint: Recall the product rule, equation (9). Notice that the formula you derive is not the same as the power rule of single-variable calculus.]

Write a program that applies the Runge-Kutta algorithm (12) to the given problem. Run the program on the interval given, with a constant step size of \(h=0.01\). Plot the components of the solution.\(\mathbf{y}^{\prime}=\left[\begin{array}{ll}1 & t \\ 0 & 1\end{array}\right] \mathbf{y}, \quad \mathbf{y}(1)=\left[\begin{array}{l}0 \\\ 1\end{array}\right], \quad 1 \leq t \leq 2\)

In each exercise, determine all equilibrium solutions (if any).\(\mathbf{y}^{\prime}=\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right] \mathbf{y}+\left[\begin{array}{r}2 \\\ -2\end{array}\right]\)

In each exercise, (a) Does \(A B=B A\) ? (b) Calculate the exponential matrices \(e^{A t}, e^{B t}\), and \(e^{(A+B) t} .\) Does \(e^{A t} e^{B t}=e^{(A+B) t} ?$$A=\left[\begin{array}{rr}2 & -1 \\ -1 & 2\end{array}\right], \quad B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\)

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