Systems of Differential Equations — Question 5

Question 5 – Solution

Strategy. Compare the derivative-state, scalar-elimination and normal-mode descriptions of the same four-dimensional motion.

Step 1: Convert and transfer the initial state. For $X=(x,x',z,z')^T$, $$X'=\begin{pmatrix}0&1&0&0\\-2&0&1&0\\0&0&0&1\\1&0&-2&0\end{pmatrix}X.$$ Since $z=x''+2x$, substitution into the second equation gives $$\boxed{x^{(4)}+4x''+3x=0,\qquad(x,x',x'',x''')(0)=(1,0,-2,0).}$$

Step 2: Solve the normal modes. Adding and subtracting give $u''+u=0$, $v''+3v=0$, with initial displacements one and velocities zero. Thus $$\boxed{x=\tfrac 12(\cos t+\cos(\sqrt 3t)),\qquad z=\tfrac 12(\cos t-\cos(\sqrt 3t)).}$$

Step 3: Prove reversibility. For any scalar solution, define $z=x''+2x$. The first coupled equation is then an identity, and the second has residual $x^{(4)}+4x''+3x=0$. This formula also proves uniqueness of $z$. The displayed cosines satisfy the scalar factors $(D^2+1)(D^2+3)$ and give $z(0)=z'(0)=0$ as required.

Step 4: Check energy and boundedness. Multiplying the equations by $x',z'$ and adding proves conservation of $$E=\tfrac 12(x'^2+z'^2)+x^2+z^2-xz =\tfrac 14(u'^2+u^2+v'^2+3v^2)=1.$$ Since $x^2+z^2-xz\geq(x^2+z^2)/2$, both displacements are bounded; the explicit formulas sharpen this to $|x|,|z|\leq 1$. Each displacement mixes the frequencies $1$ and $\sqrt 3$.