'Stiff' Differential Equations
x ˙ = [ − 8 0 . 6 7 9 . 6 1 1 9 . 4 − 1 2 0 . 4 ] x
Where
x = [ x 1 ( t ) x 2 ( t ) ] T x ˙ = [ d t d x 1 d t d x 2 ] T x ( 0 ) = [ 1 2 ] T
The goal is to solve this system of equations numerically, using the Explicit Euler Integration scheme. Use a time step h = 0 . 0 1 s . By Applying the method, the above set of differential equations can be converted into a set of difference equations of the form:
x k + 1 = A x k
Compute the eigen values of the matrix A . Let each of them be λ 1 and λ 2 .
Enter your answer as ∣ λ 1 ∣ + ∣ λ 2 ∣ , rounded to the nearest integer.
Bonus: Plot the solutions and comment on why you get such a result, after comparing it with the closed form solution of the differential equations. Vary the value of the time step h and report your observations with reasoning.
The answer is 2.
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The Euler approximation to the differential equation x ˙ = C x is x ( k + 1 ) − x ( k ) = h C x ( k ) , hence x ( k + 1 ) = ( I + h C ) x ( k )
Thus our matrix A = I + h C . We are given that C = [ − 8 0 . 6 7 9 . 6 1 1 9 . 4 − 1 2 0 . 4 ] , and h = 0 . 0 1 . Therefore,
A = [ 1 0 0 1 ] + [ − 0 . 8 0 6 0 . 7 9 6 1 . 1 9 4 − 1 . 2 0 4 ] = [ 0 . 1 9 4 0 . 7 9 6 1 . 1 9 4 − 0 . 2 0 4 ]
The eigenvalues of A are the roots of the characteristic equation ∣ λ I − A ∣ = 0 . This translates into ( λ − 0 . 1 9 4 ) ( λ + 0 . 2 0 4 ) − ( 1 . 1 9 4 ) ( 0 . 7 9 6 ) = 0 . Simplifying, this becomes λ 2 + 0 . 0 1 λ − 0 . 9 9 = 0 . Factoring, it becomes ( λ + 1 ) ( λ − 0 . 9 9 ) = 0 , so that λ 1 = − 1 and λ 2 = 0 . 9 9 , from which ∣ λ 1 ∣ + ∣ λ 2 ∣ = 1 . 9 9 ≈ 2 .