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Write a MATLAB function that solves tridiagonal systems of equations of size \(n .\) Assume that no pivoting is needed, but do not assume that the tridiagonal matrix \(A\) is symmetric. Your program should expect as input four vectors of size \(n\) (or \(n-1):\) one right-hand-side \(\mathbf{b}\) and the three nonzero diagonals of \(A .\) It should calculate and return \(\mathbf{x}=A^{-1} \mathbf{b}\) using a Gaussian elimination variant that requires \(\mathcal{O}(n)\) flops and consumes no additional space as a function of \(n\) (i.e., in total \(5 n\) storage locations are required). Try your program on the matrix defined by \(n=10, a_{i-1, i}=a_{i+1, i}=-i\), and \(a_{i, i}=3 i\) for all \(i\) such that the relevant indices fall in the range 1 to \(n .\) Invent a right-hand-side vector b.

Short Answer

Expert verified
Short Answer: To solve a tridiagonal system of equations with given input vectors, we can create a MATLAB function that implements the Thomas algorithm, a variant of Gaussian elimination tailored for tridiagonal matrices. The function takes in the right-hand side vector \(b\), and the three nonzero diagonals of matrix \(A\). It performs forward elimination and back substitution, with time complexity O(\(n\)) and 5n storage locations, to efficiently return the solution vector \(x = A^{-1} b\).

Step by step solution

01

Define the MATLAB function header with inputs and outputs

In MATLAB, define the function header with the input vectors b, a_n-1, a_i, and a_n+1. The purpose of this function is to solve a tridiagonal system, so it should return the solution vector x: ```matlab function x = tridiagonalSolver(b, aLower, aDiagonal, aUpper) ```
02

Implement the Thomas algorithm for forward elimination

The Thomas algorithm is a specialized version of Gaussian elimination for tridiagonal matrices. Perform forward elimination on the matrix by iterating through rows 2 through n: ```matlab n = length(aDiagonal); for i = 2:n factor = aLower(i-1) / aDiagonal(i-1); aDiagonal(i) = aDiagonal(i) - factor * aUpper(i-1); b(i) = b(i) - factor * b(i-1); end ```
03

Implement the Thomas algorithm for back substitution

After forward elimination, apply back substitution to obtain the solution vector x by iterating through rows n-1 down to 1: ```matlab x = zeros(n, 1); x(n) = b(n) / aDiagonal(n); for i = n-1:-1:1 x(i) = (b(i) - aUpper(i) * x(i+1)) / aDiagonal(i); end ```
04

Test the function with the given system of equations

Test the function using the given tridiagonal matrix: \(n=10, a_{i-1, i}=a_{i+1, i}=-i\), and \(a_{i, i}=3 i\). Invent a right-hand-side vector b: ```matlab n = 10; aLower = -1 * (1:n-1)'; aDiagonal = 3 * (1:n)'; aUpper = aLower; b = ones(n, 1); % Example right-hand-side vector x = tridiagonalSolver(b, aLower, aDiagonal, aUpper); ``` With this implementation, the MATLAB function `tridiagonalSolver` efficiently solves the given tridiagonal system of equations with time complexity O(\(n\)) and 5n storage locations.

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Key Concepts

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

Gaussian Elimination
The Gaussian elimination method is a fundamental technique in numerical linear algebra to solve systems of linear equations. Typically, it involves two main steps: forward elimination and back substitution. During forward elimination, the matrix is transformed into an upper triangular form. Then, back substitution is used to find the solution to the equation.

For tridiagonal matrices, a specialized version known as the Thomas Algorithm is often employed. This is because a tridiagonal matrix has non-zero elements on only the main diagonal, and the diagonals just above and below it. Gaussian elimination works particularly well here because of its efficiency in reducing computational complexity. It reduces a system to a simpler form, making it easier to solve with fewer operations. This method is prevalent in many fields such as economics, physics, and engineering due to its robustness and reliability.
MATLAB Programming
MATLAB is a powerful computing environment used extensively for numerical analysis thanks to its easy-to-use syntax and extensive built-in functions. In the context of solving tridiagonal systems of equations, understanding how to implement algorithms like the Thomas Algorithm in MATLAB is crucial.

The MATLAB implementation involves creating a function that takes specific inputs and produces output according to the algorithm's steps. By defining the function header, users specify what input variables the function expects. These are typically the vectors that represent the diagonals of the tridiagonal matrix and the right-hand-side vector.

The ease of converting mathematical algorithms into MATLAB code makes it an ideal tool for students and professionals alike. By breaking down the logic into loops and basic arithmetic operations, one can effectively translate complex numerical methods into efficient, executable programs.
Numerical Methods
Numerical methods are techniques used to find approximate solutions to mathematical problems that might be challenging to solve exactly. These methods are vital when dealing with real-world problems where analytical solutions are not possible or practical.

Tridiagonal matrix systems are common in numerical simulations, such as those in computational fluid dynamics or structural analysis. These methods provide strategies to approximate solutions with a high degree of accuracy. Using numerical methods to solve equations allows for flexibility in problem-solving and the ability to handle large and complex systems.

In essence, these methods prioritize efficiency and accuracy and make use of algorithms like those used in Gaussian elimination or the Thomas Algorithm. They are designed to minimize computational costs while still providing reliable results. Understanding these methods is essential for any field that requires computational precision and modeling.
Thomas Algorithm
The Thomas Algorithm is a streamlined method for solving tridiagonal systems of equations. It is a variant of Gaussian elimination that is specifically optimized for tridiagonal matrices. The algorithm is both powerful and efficient, requiring only linear time complexity, which is \( O(n) \), in the number of equations.

The process involves two main phases:
  • Forward Elimination: Here, elements are manipulated to simplify the system, preparing it for back substitution by creating zeros below the main diagonal. This step modifies the main diagonal and right-hand side vector to remove the influence of the lower diagonal.
  • Back Substitution: After forward elimination, this phase calculates the solution vector by starting from the last row and moving upwards, using the simplified upper triangular form to find the solutions successively.
The Thomas Algorithm's memory efficiency also makes it attractive for computational tasks, as it operates using minimal storage beyond the initial input size. It is exceedingly applicable in fields requiring repetitive solutions to tridiagonal systems, showcasing its utility in disciplines like mathematics and computer science.

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Most popular questions from this chapter

Apply your program from Exercise 17 to the problem described in Example \(4.17\) using the second set of boundary conditions, \(v(0)=v^{\prime}(1)=0\), for \(g(t)=\left(\frac{\pi}{2}\right)^{2} \sin \left(\frac{\pi}{2} t\right)\) and \(N=100\). Compare the results to the vector \(\mathbf{u}\) composed of \(u(i h)=\sin \left(\frac{\pi}{2} i h\right), i=1, \ldots, N\), by recording \(\|\mathbf{v}-\mathbf{u}\|_{\infty}\)

Let \(A\) and \(T\) be two nonsingular, \(n \times n\) real matrices. Furthermore, suppose we are given two matrices \(L\) and \(U\) such that \(L\) is unit lower triangular, \(U\) is upper triangular, and $$ T A=L U $$ Write an algorithm that will solve the problem $$ A \mathbf{x}=\mathbf{b} $$ for any given vector \(\mathbf{b}\) in \(\mathcal{O}\left(n^{2}\right)\) complexity. First explain briefly yet clearly why your algorithm requires only \(\mathcal{O}\left(n^{2}\right)\) flops (you may assume without proof that solving an upper trian- gular or a lower triangular system requires only \(\mathcal{O}\left(n^{2}\right)\) flops). Then specify your algorithm in detail (including the details for lower and upper triangular systems) using pseudocode or a MATLAB script.

Apply the modified solver obtained in Exercise 10 to the following systems. In each case, check the difference between the computed solution \(\mathbf{x}\) and the result of MATLAB's built-in solver \(A \backslash \mathbf{b}\). (a) $$ \begin{aligned} x_{1}+x_{2}+x_{4} &=2 \\ 2 x_{1}+x_{2}-x_{3}+x_{4} &=1 \\ 4 x_{1}-x_{2}-2 x_{3}+2 x_{4} &=0 \\ 3 x_{1}-x_{2}-x_{3}+x_{4} &=-3 . \end{aligned} $$ (b) Same as the previous system, but with the coefficient of \(x_{4}\) in the last equation set to \(a_{4,4}=2\) (c) $$ \begin{aligned} x_{1}+x_{2}+x_{3} &=1, \\ x_{1}+\left(1+10^{-15}\right) x_{2}+2 x_{3} &=2, \\ x_{1}+2 x_{2}+2 x_{3} &=1 . \end{aligned} $$ (d) Same as the previous system, but with the second equation multiplied by \(10^{20}\). (e) $$ A=\left(\begin{array}{lll} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{array}\right), \mathbf{b}=\left(\begin{array}{l} 1 \\ 2 \\ 1 \end{array}\right) $$ (f) $$ A=\left(\begin{array}{lll} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 3 & 2 & 1 \end{array}\right) $$ with the same right-hand side as before.

Consider the LU decomposition of an upper Hessenberg (no, it's not a place in Germany) matrix, defined on the facing page, assuming that no pivoting is needed: \(A=L U\). (a) Provide an efficient algorithm for this LU decomposition (do not worry about questions of memory access and vectorization). (b) What is the sparsity structure of the resulting matrix \(L\) (i.e., where are its nonzeros)? (c) How many operations (to a leading order) does it take to solve a linear system \(A \mathbf{x}=\mathbf{b}\), where \(A\) is upper Hessenberg? (d) Suppose now that partial pivoting is applied. What are the sparsity patterns of the factors of \(A ?\)

Run the MATLAB script of Example \(5.22\), plugging in different values for \(d\). In particular, try \(d=.25, d=.85\), and \(d=-.75 .\) What do you observe? What will happen as \(d \downarrow(-1) ?\)

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