Systems of Differential Equations — Question 2

Question 2 – Solution

Strategy. Apply Newton’s law to each mass and use energy to check the coupling signs independently.

Step 1: Account for the spring forces. The wall spring exerts $-2u$ on mass $1$. The coupling spring exerts $3(v-u)$ on mass $1$ and its opposite on mass $2$. Therefore $$\boxed{u''=-5u+3v,\qquad 2v''=3u-3v+f(t).}$$ Dividing the second equation by its mass is essential.

Step 2: Include positions and velocities. Writing $p=u'$, $q=v'$ gives $$X'=\begin{pmatrix}0&1&0&0\\-5&0&3&0\\0&0&0&1\\3/2&0&-3/2&0\end{pmatrix}X +\begin{pmatrix}0\\0\\0\\f(t)/2\end{pmatrix},\quad X(0)=(1,0,0,0)^T.$$ Equal positions can have different velocities, so positions alone do not determine the first derivatives. The four coordinates supply the complete state.

Step 3: Verify the work-energy balance. Differentiate the stated energy before substituting the equations: $$E'=p(-5u+3v)+q(3u-3v+f)+2up+3(v-u)(q-p).$$ Every spring term cancels, leaving $\boxed{E'=f(t)v'}$. Thus $f=0$ conserves $E$; for the stated initial data, $E(0)=5/2$. The two kinetic terms reflect the different masses. The remaining terms are the nonnegative spring potential energies.

Step 4: Balance a constant force. An equilibrium has $p=q=0$, $-5u+3v=0$, and $3u-3v+F=0$. Consequently $u=F/2$, $v=5F/6$, and $$\boxed{X_*=(F/2,0,5F/6,0)^T.}$$ This is the unique equilibrium. Zero acceleration at one instant would not make a state constant if either velocity were nonzero.