Euler Integration Comparison (Part 2)

Calculus Level pending

Consider the time-domain differential equation:

d d t x ( t ) = x 2 ( t ) \frac{d}{dt} x(t) = x^2(t)

Let us discretize the function and time itself. Function x 1 x_1 is the explicit Euler approximation, and x 2 x_2 is the implicit Euler approximation. Let the time step Δ t = 0.001 \Delta t = 0.001

x 1 k = x 1 k − 1 + x ˙ 1 k − 1 Δ t = x 1 k − 1 + x 1 k − 1 2 Δ t x 2 k = x 2 k − 1 + x ˙ 2 k Δ t = x 2 k − 1 + x 2 k 2 Δ t t k = t k − 1 + Δ t x_{1 k} = x_{1 \, k-1} + \dot{x}_{1 \, k-1} \Delta t \\ = x_{1 \, k-1} + x^2_{1 \, k-1} \Delta t \\ \,\\ x_{2 k} = x_{2 \, k-1} + \dot{x}_{2 \, k} \Delta t \\ = x_{2 \, k-1} + x^2_{2 \, k} \Delta t \\ \, \\ t_k = t_{k - 1} + \Delta t

In the above equations, the k k subscript denotes the present processing step, and the k − 1 k-1 subscript denotes the previous processing step. The simulation is initialized as follows:

t 0 = 0 x 1 0 = 1 x 2 0 = 1 t_0 = 0 \\ x_{1 \, 0} = 1 \\ x_{2 \, 0} = 1 \\

When t = 0.9 t = 0.9 , what is x 2 − x 1 x_2 - x_1 , rounded to the nearest integer multiple of 0.05 0.05 ?

Note: For the implicit case, there are two conceivable ways to solve. One of them gives nonsensical results, so use the other.


The answer is 0.45.

Steven Chase by Steven Chase , 37, USA

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1 solution

Karan Chatrath
Mar 16, 2021

For the implicit case, we obtain x k + 1 x_{k+1} by solving the following quadratic equation:

x k + 1 = x k + Δ t x k + 1 2 x_{k+1} = x_{k} + \Delta t x_{k+1}^2

In this case, The roots of the quadratic equation lead to the iteration scheme:

x k + 1 = 1 ± 1 − 4 Δ t x k 2 Δ t x_{k+1} = \frac{1 \pm \sqrt{1 - 4 \Delta t x_k}}{2\Delta t}

If: x k + 1 = 1 + 1 − 4 Δ t x k 2 Δ t x_{k+1} = \frac{1 + \sqrt{1 - 4 \Delta t x_k}}{2\Delta t}

Then the limit as Δ t → 0 \Delta t \to 0 does not lead to x k + 1 = x k x_{k+1}=x_{k} . However, using the second root and evaluating the same limit results in x k + 1 = x k x_{k+1}=x_{k} which is expected. Essentially, as the time step goes to zero, solutions in two successive time steps become the same.

x k + 1 = 1 − 1 − 4 Δ t x k 2 Δ t x_{k+1} = \frac{1 - \sqrt{1 - 4 \Delta t x_k}}{2\Delta t}

Hence, the above root is the appropriate choice. The following script allows us to evaluate the required result. 

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clear all
clc

dt      = 1e-3;                                      % Time step
t       = 0:dt:0.9;                                  % Time array

x1(1)   = 1;                                         % Initial cond: Explicit soln.             
x2(1)   = 1;                                         % Initial cond: Implicit soln.

for k = 1:length(t)-1

  x1(k+1)   = x1(k) + dt*x1(k)^2;                    % Explicit Euler
  x2(k+1)   = (1 - sqrt(1 - 4*dt*x2(k)))/(2*dt);     % Implicit Euler
end

ANSWER  =  x2(end) - x1(end)                         
% ANSWER = 0.4612

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