More on the Wronskian — Question 8

Question 8 – Solution

Strategy. Periodicity forces the Wronskian to return to its starting value, while Abel’s identity specifies its multiplicative change.

Step 1: Compare one full period. Differentiating $y(t+T)=y(t)$ shows that the slope is periodic too. Thus two periodic solutions have $W(T)=W(0)$. Independence gives $W(0)\ne 0$, so Abel’s identity implies $$\exp[-\int_0^T p(t)\,dt]=1.$$ The integral is real; consequently $$\boxed{\int_0^T p(t)\,dt=0.}$$

Step 2: Exclude nontrivial nonnegative damping. By continuity and periodicity, if $p\ge 0$ and is not identically zero, it is positive on some interval within a period. Hence its integral over a period is strictly positive. The necessary condition fails, so two independent $T$-periodic solutions cannot exist.

Step 3: Disprove sufficiency. For $y''-y=0$, $p=0$ satisfies the integral condition for every $T$. Every solution is $Ae^t+Be^{-t}$. A continuous periodic function is bounded on $\mathbb R$; boundedness as $t\to\infty$ forces $A=0$, and boundedness as $t\to-\infty$ forces $B=0$. Thus there are no nonzero periodic solutions, much less two independent ones.

Step 4: Test a single periodic solution. In $y''+y'=0$, the constant solution $u=1$ is nonzero and periodic for every $T>0$, while $p=1$ has integral $T\ne 0$. A companion is $v=e^{-t}$, with $W[u,v]=-e^{-t}\ne 0$; it is not periodic. One periodic solution does not make a nonzero pair Wronskian periodic and imposes no such zero-integral requirement.