Solving ODEs with Forward Euler & Sampled Data Systems

[adoprea]

Homework Statement

Sampled Data system
Using the forward Euler integration algorithm, convert these differential equations to a set of difference equations. Use a stept size of deltaT = 0.1s.

Homework Equations

x₁(dot) = x₂
x₂(dot) = x₃
x₃(dot) = -2x₁-3x₂-4x₃
y = 7x₁-5x₂

The Attempt at a Solution

Nothing relevant! I have scribbled around and played with stuff from wikipedia, mainly with the form y_n = y_n-1+deltaT*f(t_n-1, y_n-1), but to no avail. Do I start with:
dy/dt = 7x1(dot)-5x2(dot) and so on? I end up in a loop as soon as iI get to the derivation of x3. IIt doesn't feel like I am on the right path.

What I am looking for is the starting point (so far). I learned about this stuff in school about 5 years ago, but I haven't touched it ever since. Now I wanted to do some reading on system modeling and I thought I could give a go to the exercises in the book as well. First attempt, first failure. Help!

 

[BvU]

With forward euler you shouldn't end up in a loop: you use $\vec x_n$ to calculate $\dot {\vec x}_n$ which you use to calculate $\vec x_{n+1}$.

Y is just a calculated variable $y(t) =y (\vec x(t))$. The differential equation is in $\vec x$

note: to actually solve, you need initial conditions (so it's funny you should ask where to start because without them the problem statement is incomplete)

Can't resist:

Ever tried. Ever failed.
No matter. Try Again.
Fail again. Fail better.

Samuel Beckett​

 

[adoprea]

Haha! yes, "fail better" seems to apply here.

There are no initial conditions. The book asks for a path to the solution, not actually solving the equations. But your post gave me the starting point I was looking for.

The frustration I experience when I look at something that I was able to solve some years ago is infinite. I hope I am not the only one in this situation.

 

[BvU]

There are no initial conditions. The book asks for a path to the solution, not actually solving the equations.

The better !

The frustration I experience when I look at something that I was able to solve some years ago is infinite. I hope I am not the only one in this situation.

This is a third order system. I don't think many people can solve that offhand; I sure can't. At first I thought I saw a glimpse of the equations for a harmonic oscillator, but that didn't last long.

Frustration is a self-imposed handicap. Unlike most other handicaps, you can throw it off: it doesn't help at all. And often, once you throw it aside, you find that it wasn't necessary in the first place (you know, like breaking in a door that isn't locked...)

--

 

[HallsofIvy]

There are two rather different ways to solve a system like this. The simpler (at least the method I learned first) is to differentiate the last equation again:
x_3''= -2x_1'- 3x_2'- 4x_3'. Replace the derivatives of x_1 and x_2 from the first two equations: x_3''= -2x_2- 3x_3- 4x_3'. Differentiate again: x_3'''= -2x_2'- 3x_3'- 4x_3'' which is the same as the "third order linear differential equation with constant coefficients", x_3''+ 4x_3''+ 3x_3'+ 2x_3 which has characteristic equation r^3+ 4r^2+ 3r+ 2= 0

Second, more "sophisticated" method: write this linear system as a matrix equation. We can write
\begin{bmatrix}x_1' \\ x_2' \\ x_3'\end{bmatrix}= \begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -3 & -4 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2\\ x_3\end{bmatrix}
Simplify that by finding the eigenvalues and corresponding eigenvectors for that three by three matrix.

 

[Ray Vickson]

Haha! yes, "fail better" seems to apply here.

There are no initial conditions. The book asks for a path to the solution, not actually solving the equations. But your post gave me the starting point I was looking for.

The frustration I experience when I look at something that I was able to solve some years ago is infinite. I hope I am not the only one in this situation.

Do you remember how you used to do it? I bet you started with the assumptions that $x_1 = A e^{rt}, x_2 = B e^{rt}, x_3 = C e^{rt}$, then substituted these into the DE's and turned the crank. Let's do that:
\begin{array}{ccl} \dot{x_1} = x_2 &amp;\Rightarrow&amp; r A = B \\<br /> \dot{x_2}= x_3 &amp; \Rightarrow &amp; r B = C \\<br /> \dot{x_3} = -2x_1-3x_2-4x_3 &amp;\Rightarrow &amp; r C = -2A - 3B - 4C<br /> \end{array}<br />
This implies
\pmatrix{r &amp; -1 &amp; 0 \\<br /> 0 &amp; r &amp; -1 \\<br /> 2 &amp; 3 &amp; r+4} \pmatrix{A\\B\\C\\} = \pmatrix{0\\0\\0}
So, you need the determinant of the matrix to = 0 in order to have a non-vanishing solution, and that produces a cubic equation in $r$, Generally, there will be three roots $r_1, r_2, r_3$, so you will actually have solutions of the form $x_1 = A_1 e^{r_1 t} + A_2 e^{r_2 t} + A_3 e^{r_3 t}$, etc.

 

[BvU]

Not sure ado is really interested in the actual solution (#1, #3). Could well be a step-up for a numerical solution exercise.
But still interesting !