Linear Homogeneous Differential Equations — Question 10

Question 10 – Solution

Strategy. Compare the root-based behavior with what the conserved expression actually measures.

Step 1: Find growing and decaying oscillatory modes. Since $r^4+4=(r^2-2r+2)(r^2+2r+2)$, the roots are $1\pm i$ and $-1\pm i$. Thus $$\boxed{y=e^x(A\cos x+B\sin x)+e^{-x}(C\cos x+D\sin x).}$$ All roots are simple and the four real modes are independent.

Step 2: Verify the conservation law. The product rule gives exact cancellation: $$E'=y^{(4)}y'+y'''y''-y''y'''+4yy' =y'(y^{(4)}+4y)=0.$$ Hence $E(x)=E(0)$ for every $x$.

Step 3: Test a zero-level growing solution. For $y=e^x\cos x$, the successive derivatives are $$y'=e^x(\cos x-\sin x),\quad y''=-2e^x\sin x,\quad y'''=-2e^x(\sin x+\cos x),\quad y^{(4)}=-4y.$$ At zero its initial vector is $(1,1,0,-2)$, so $E(0)=-2+2=0$. Conservation yields $E\equiv 0$. Yet $y(2\pi n)=e^{2\pi n}\to\infty$, proving unboundedness.

Step 4: Identify the limits of the invariant. On $[0,\infty)$ boundedness requires $A=B=0$: any nonzero growing sinusoid attains its amplitude along an unbounded sequence, while the other term decays. These conditions also suffice, giving a two-dimensional decaying family. Requiring boundedness backward as well forces $C=D=0$ by the same phase argument, so the only entire-line bounded solution is zero.

The expression $E$ is not positive definite in the four initial derivatives: it has a mixed product and a negative square. In particular a nonzero state can have $E=0$. The displayed solution has $|y(0)|^2=1$ and $E(0)=0$, immediately refuting the proposed bound for every positive $C$. Constancy of an indefinite expression does not guarantee bounded motion.