More on the Wronskian — Question 3

Question 3 – Solution

Strategy. Check the counterexample directly, then identify the hypothesis missing from the student’s assertion.

Step 1: Check regularity. For $t\ne 0$, $v'=3t|t|$ and $v''=6|t|$. At zero, $$v'(0)=\lim_{h\to 0}|h|^3/h=0,\qquad v''(0)=\lim_{h\to 0}3h|h|/h=0.$$ These agree continuously with the formulas away from zero, so $v$ is $C^2$; $u$ is a polynomial and is smooth.

Step 2: Compare the Wronskian and independence. On $t>0$, $v=u$; on $t<0$, $v=-u$. Thus $W=0$ on each side, and at zero both data columns are zero, so $\boxed{W\equiv 0}$. Yet $Au+Bv=0$ on all of $\mathbb R$ forces $A+B=0$ on the positive side and $A-B=0$ on the negative side. Hence $A=B=0$: the functions are independent on $\mathbb R$.

Step 3: Restore the missing hypothesis. The zero-Wronskian implication requires solutions of one common regular homogeneous linear equation on a connected interval, not merely two $C^2$ functions. The nonzero solution $u$ would have zero value and slope at zero, contradicting regular uniqueness. Thus no normalized equation with continuous coefficients on $\mathbb R$ can have both as solutions.

Step 4: Verify a singular equation. For $u$, the residual is $t^2(6t)-2t(3t^2)=0$. On each side $v=\pm u$, so its residual also vanishes; at zero both terms vanish. The leading coefficient is zero there. The ratio $v/u$ is $1$ on $t>0$ and $-1$ on $t<0$, and is undefined at the shared zero. Local constant ratios need not join into one global constant.