'Stiff' Differential Equations - 2

Calculus Level pending

x ˙ = [ − 80.6 119.4 79.6 − 120.4 ] x \dot{x} = \left[\begin{matrix} -80.6&119.4\\79.6&-120.4\end{matrix} \right]x

Where

x = [ x 1 ( t ) x 2 ( t ) ] T x = \left[\begin{matrix} x_1(t)&x_2(t)\end{matrix} \right]^T x ˙ = [ d x 1 d t d x 2 d t ] T \dot{x} = \left[\begin{matrix} \frac{d x_1}{dt}&\frac{d x_2}{dt}\end{matrix} \right]^T x ( 0 ) = [ 1 2 ] T x(0) = \left[\begin{matrix} 1&2\end{matrix} \right]^T

The goal is to solve this system of equations numerically, using the Implicit Euler Integration scheme. Use a time step h = 0.1 s h = 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 x_{k+1} = A x_k

Compute the eigen values of the matrix A A . Let each of them be λ 1 \lambda_1 and λ 2 \lambda_2 .

Enter your answer as ⌊ 100 ( λ 1 + λ 2 ) ⌋ \boxed{\lfloor100 (\lambda_1 +\lambda_2)\rfloor} .

Note: ⌊ . ⌋ \lfloor . \rfloor denotes the Floor function and k k denotes the time series index.

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 h and report your observations with reasoning. How is this method different from the Explicit Euler Integration scheme? Quantitatively compare both methods.


The answer is 95.

Karan Chatrath by Karan Chatrath , 27, Netherlands

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1 solution

Steven Chase
Nov 2, 2020

The one-dimensional version of implicit Euler is:

x k + 1 = x k + x ˙ k + 1 Δ t x_{k+1} = x_k + \dot{x}_{k+1} \Delta t

The difference between this and explicit Euler is that the derivative term comes from the present processing interval, rather than from the previous one. Long story short, the implicit Euler method is more numerically stable than the explicit method, but it is much more computationally expensive to implement for non-linear systems. For linear systems, you still get the stability benefits, but with very little extra processing required. I've always found that to be interesting.

For this problem, the implicit Euler integration looks like:

[ x 1 k + 1 x 2 k + 1 ] = [ x 1 k x 2 k ] + Δ t [ α β γ σ ] [ x 1 k + 1 x 2 k + 1 ] \begin{bmatrix} x1_{k+1} \\ x2_{k+1} \\ \end{bmatrix} = \begin{bmatrix} x1_k \\ x2_k \\ \end{bmatrix} + \Delta t \begin{bmatrix} \alpha & \beta \\ \gamma & \sigma \\ \end{bmatrix} \begin{bmatrix} x1_{k+1} \\ x2_{k+1} \\ \end{bmatrix}

Re-arranging this a bit results in:

[ 1 − α Δ t − β Δ t − γ Δ t 1 − σ Δ t ] [ x 1 k + 1 x 2 k + 1 ] = [ x 1 k x 2 k ] \begin{bmatrix} 1 - \alpha \Delta t & - \beta \Delta t \\ -\gamma \Delta t & 1 - \sigma \Delta t \end{bmatrix} \begin{bmatrix} x1_{k+1} \\ x2_{k+1} \\ \end{bmatrix} = \begin{bmatrix} x1_k \\ x2_k \\ \end{bmatrix}

Pre-multiply both sides by the inverse of this matrix to get it into the requested form. The answer consists of the eigenvalues of the following matrix.

[ 1 − α Δ t − β Δ t − γ Δ t 1 − σ Δ t ] − 1 \begin{bmatrix} 1 - \alpha \Delta t & - \beta \Delta t \\ -\gamma \Delta t & 1 - \sigma \Delta t \end{bmatrix} ^{-1}

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