Differential Equations Solutions 138

Differential Equations Solutions 138 - Then the Jacobian of...

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148 Chapter 24. Solutions: Nonlinear Systems It is silly to refactor the matrix B each time, when it is just a rank-1 update of the previous matrix. Instead, update a decomposition or (less desirable) update the inverse using the techniques of Chapter 7. CHALLENGE 24.3. For some vector r , we need to form ( A ZV T ) 1 r = A 1 r + A 1 Z ( I V T A 1 Z ) 1 V T A 1 r , where A = B ( k ) , Z = y B ( k ) s , and V = s / ( s T s ). Thus we need to form t = A 1 r and u = A 1 Z , at a cost of 2 p multiplications, form α = 1 V T u at a cost of n multiplications, form w = (( V T t ) ) u at a cost of 2 n multiplications and 1 division add t and w . The total number of multiplications is 2 p + 3 n + 1. Again, updating a matrix decomposition has many advantages over this approach! CHALLENGE 24.4. (Partial Solution.) (a) We compute the partial of ρ a ( λ, x ) with respect to λ : s = x 2 y 3 + xy 2 ( x a 1 ) 2 xy 2 + x 2 y + xy ( y
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Unformatted text preview: . Then the Jacobian of ρ a is the 2 × 3 matrix b J ( λ, x ) = £ s , (1 − λ ) I + λ J ( x ) ¤ , where J ( x ) is the matrix from Problem 1. (b) In order for the function to be transversal to zero, the matrix b J ( λ, x ) must be full rank (i.e., rank-2) at every point λ ∈ [0 , 1), x, y ∈ ( −∞ , ∞ ). The matrix J ( x ) has two eigenvalues – call them α 1 and α 2 . The matrix K = (1 − λ ) I + λ J ( x ) has eigenvalues (1 − λ )+ λα i , so it is singular only if λ = 1 / (1 − α 1 ) or λ = 1 / (1 − α 2 ). Even if that happens, it is likely that the vector s will point in a di±erent direction, making the rank of J ( λ, x ) equal to 2....
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