# Numerical method for this system of ODE's

I'm a total beginner to numerical methods. I just learned about applying Euler's method for first order equations. But now I have the following 2nd order system. The problem stems from 2 masses, which are connected to each other by three springs, and two of the springs are attached to walls. So $$y_1$$ would be the position of the first object and so on. $$y_1''=y_1-(y_1+y_2),$$ $$y_2''=y_2-3(y_1+y_2).$$

with initial conditions $$y_1(0) = a_1, \enspace y_2(0)=a_2,\enspace y_1'(0)=0,\enspace y_2'(0)=0.$$

Where $$a_1,a_2$$ are real numbers. This system can be solved analytically. Let $$z=y_1+y_2$$. The two equations can now be added, and we get $$z'' = z-4z = -3z,$$

which is a homogenic equation for $$z$$ and the solution is $$z=c_1 \cos\sqrt{3t}+c_2\sin\sqrt{3t},$$

where $$c_1,c_2$$ are constants. I then have these two second order equations $$y_1''=y_1 - c_1 \cos\sqrt{3t}-c_2\sin\sqrt{3t},$$ $$y_2''=y_2 -3c_1 \cos\sqrt{3t}-3c_2\sin\sqrt{3t}.$$

I could continue to solve for $$y_1$$ and $$y_2$$, but I'm asked to find a numerical solution. So at this point I have two problems. Whether or not I have developed this problem enough to apply some numerical method, and then what numerical methods should I look into. I'd appreciate any input.

• See math.stackexchange.com/questions/2990229/…, stackoverflow.com/questions/55687304/… for previous attempts at similar problems. Mar 31, 2021 at 13:24
• @Moo Whats wrong with the ICs? They indicate that the initial displacements of the two objects are $a_1$ and $a_2$ respectively. The derivatives indicate we start from rest. As for your second point, yes it is. The masses and spring constants are such that this occurs. Is it a problem? Mar 31, 2021 at 15:21
• Well they are real numbers, I should have mentioned that. But for the displacements, I have not been given explicit numbers. Mar 31, 2021 at 15:26

Recast your system of 2nd order ODEs to a system of 4 first order ones, let's say $$y_1, \quad z_1=y_1', \quad y_2, \quad z_2=y_2'$$

By doing so, you can write $$Y'=AY, Y(0)=Y_0$$

where

$$A= \begin{bmatrix}{} 0&1 &0 & 0\\ 0 &0& -1 & 0 \\ 0&0&0&1\\ -3 & 0 &2 & 0 \end{bmatrix}$$

and $$Y(0)=[y_1(0),y_1'(0),y_2(0),y_2'(0)]^T = [a_1,0,a_2,0]^T$$

The problem is linear and small. You can choose your favourite ODE solver. As you mentioned Forward Euler, here's the recurrence, given $$Y_0$$ and for $$n\geq0$$

$$Y_{n+1} = Y_n + kA Y_n = (1+kA)Y_n$$

Notice that the lack of A-stability requires a bit of care in choosing the step size $$k$$

A good compromise is given by its implicit counterpart, which is A-stable, and reads

$$Y_{n+1} = Y_n + kAY_{n+1}$$ which, solving for $$Y_{n+1}$$ reads $$Y_{n+1}=(I-kA)^{-1} Y_n$$ (DO NOT invert the matrix: solve the associated linear system instead)

If you're also familiar with matrix exponentials and their implementation (which is not straightforward at all), you can some appropriate routine and compute $$Y_{n+1} =e^{kA}Y_n$$ but this probably falls outside the scope of your problem.