University:
Massachusetts Institute of Technology
Course:
2.29 | Numerical Marine Hydrodynamics (13.024)
Academic year:

2022
Views:

149

Pages:

30
Author:

Nesbyenvkk

Root Finding Suppose you wish to find the wave number, k, of gravity water waves with a frequency, f, of 0.2 Hz. in water that is 5 meters deep. The circular frequency of 0.2 Hz. waves is w = 2πf = 1.2566 radians/second. The dispersion relation for gravity water waves is: kg tanh kh = w² g is the acceleration of gravity, 9.81 m/s² and h is the water depth, 5 m. This equation can be written as: kg tanh kh - w² 0 If we write an equation: y(k) kg tanh kh - w², The problem at hand is the same as asking: "What is the value of k such that. y(k) 0? The value of a quantity that makes another equal to zero is called a root and the question above is called Root Finding. 4 3 2 > 1 0 -1 -2 0 0.1 0.2 0.3 0.4 0.5 k An mx 1 matrix (m rows, 1 column) is called a column vector or a vector and is written symbolically as: x=x= TI Each represents a numerical value. The standard form for a set of linear equations is: Ax = b where A is an mxn matrix called the coefficient matriz and x is an unknown vector of length n. b is a known vector of length n. Let aj denote the jth column of A. Then the set of equations can be written as: 2181 +2282+I= b The vectors a,, j = 1,2,...,n are linearly dependent if there is a set of numbers 1,2,..., with at least one z; being non zero such that: In this instance, at least one a, is a linear combination of the remaining aj's. The vectors aj, j = 1,2,...,n are linearly independent if they are not linearly dependent. If each of the m linear equations is independent then there is an exact solution if mn. If mn, there are more equations than unknowns and there is no exact solution. Rather, there is an approximate solution for x which is usually chosen to achieve minimum sum of the squared errors from each equation. Example: 3x₁ + x₂ = 5 x₁ + 4x₂ = -3 2x₁ + 3x₂ = -6 [ 3 1 ] [x₁] [ 5 ] [ 1 4 ] [x₂] = [ -3 ] [ 2 3 ] [ -6 ] Approximate Solution is: x₁ = y₁, x₂ = y₂ ; λ = 1 Need To find y. Error Vector is e = [e₁] [e₂] [e₃] e₁ = 5 - 3y₁ - y₂ e₂ = -3 - y₁ - 4y₂ e₃ = -6 + 2y₁ - 3y₂ E = (5 - 3y₁ - y₂)² + (-3 - y₁ - 4y₂)² + (-6 + 2y₁ - 3y₂)² We Seek y such that ∂E/∂y₁ = 0 and ∂E/∂y₂ = 0 which is the same as: 1/2 ∂E/∂y₁ = 0 and 1/2 ∂E/∂y₂ = 0 1/2 ∂E/∂y₁ = -3(5 - 3y₁ - y₂) - 1(-3 - y₁ - 4y₂) + 2(-6 + 2y₁ - 3y₂) = -15 + 9y₁ + 3y₂ + 3 - y₁ - 4y₂ -12 + 4y₁ - 6y₂ = -24 + 14y₁ - 7y₂

Week 4 Notes: Root Finding

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