Question

Use Newton's method to solve a discretized version of the differential equation $$ y^{\prime \prime}=-\left(y^{\prime}\right)^{2}-y+\ln x, \quad 1 \leq x \leq 2, y(1)=0, y(2)=\ln 2 $$ The discretization on a uniform mesh, with the notation of Example \(9.3\), can be $$ \frac{y_{i+1}-2 y_{i}+y_{i-1}}{h^{2}}+\left(\frac{y_{i+1}-y_{i-1}}{2 h}\right)^{2}+y_{i}=\ln (i h), \quad i=1,2, \ldots, n $$ The actual solution of this problem is \(y(x)=\ln x .\) Compare your numerical results to the solution \(y(x)\) for \(n=8,16,32\), and 64 . Make observations regarding the convergence behavior of Newton's method in terms of the iterations and the mesh size, as well as the solution error.

Short answer

Answer: As the mesh size increases, the solution error decreases, and the convergence behavior of Newton's method improves. The increased accuracy can be attributed to the finer mesh resolution capturing more details of the exact solution.

Step-by-step solution

Form the Discretized Differential Equation

The given discretized version of the differential equation is: $$ \frac{y_{i+1}-2 y_{i}+y_{i-1}}{h^{2}}+\left(\frac{y_{i+1}-y_{i-1}}{2 h}\right)^{2}+y_{i}=\ln(i h), \quad i=1,2, \ldots, n $$

Set up the Uniform Mesh

For n = 8, 16, 32, and 64, set up the uniform mesh with step size h: $$h = \frac{x_{n+1} - x_{n-n}}{n+1}= \frac{2 - 1}{n+1}$$

Apply Newton's Method

Iteratively apply Newton's method for each mesh size and obtain the numerical solution of the discretized equation.

Compare Numerical Results with Actual Solution

For each mesh size (n = 8, 16, 32, and 64), compare the obtained numerical solution with the actual solution y(x) = ln(x) at corresponding mesh points. Calculate the error for each value of n.

Observe Convergence Behavior and Solution Error

Based on the errors calculated in the previous step, observe the convergence behavior of Newton's method in terms of the iterations and mesh size. Make comments on the solution error and the effect of increasing the mesh size on the accuracy of the numerical solution.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are used to describe various phenomena across mathematics, physics, engineering, and many other fields. The primary goal with differential equations is to find a function that satisfies the equation. In the exercise, we have the differential equation \( y^{\prime \prime} = -\left(y^{\prime}\right)^{2} - y + \ln x \), with boundary conditions \( y(1) = 0 \) and \( y(2) = \ln 2 \).To solve this problem, an analytical solution could be complex or impossible, so numerical methods like Newton's Method are applied. This allows us to approximate solutions when an exact solution isn't feasible.

Discretization

Discretization is a process used to transform continuous differential equations into discrete versions. This is achieved by dividing the domain into a finite number of intervals, known as a mesh. By doing so, we convert our differential equation into a difference equation that can be solved using numerical methods.In the exercise, the discretized version of the differential equation is given as:\[ \frac{y_{i+1}-2 y_{i}+y_{i-1}}{h^{2}} + \left(\frac{y_{i+1}-y_{i-1}}{2h}\right)^{2} + y_{i} = \ln (ih), \quad i=1,2, \ldots, n \]Discretization transforms the continuous problem into a form that can be handled with computers, making it possible to solve complex differential equations numerically.

Numerical Solution

Obtaining a numerical solution involves finding approximate solutions to problems that may not have exact solutions. In this context, Newton's Method is used as an iterative approach to solve the discretized differential equations.

Steps of Newton's Method

• Start with an initial guess close to the expected solution.
• Refine this guess iteratively by solving related linear equations.
• Continue the iteration until the solution converges, i.e., the change between successive approximations becomes negligible.

By choosing different mesh sizes (\( n = 8, 16, 32, \) and \( 64 \)), we observe how well the numerical solution approximates the true solution \( y(x) = \ln x \). Newton's Method helps improve the accuracy of such approximations, especially as we refine our mesh size.

Mesh Size

The mesh size, denoted by \( h \), controls the spacing between points in the discretization of a differential equation. It plays a crucial role in determining the accuracy of the numerical solution. The smaller the mesh size, the more points are used to approximate the differential equation, which typically leads to a more accurate solution.In the exercise, for different values of \( n \) (e.g., 8, 16, 32, and 64),\( h \) is computed as:\[ h = \frac{2 - 1}{n+1} = \frac{1}{n+1} \]As \( n \) increases, \( h \) decreases, leading to better approximations. Observing the convergence behavior and error, we can analyze how decreasing \( h \) results in the numerical solutions being closer to the actual solution \( y(x) = \ln x \). However, smaller mesh sizes also require more computations, highlighting a trade-off between accuracy and computational effort.