More on the Wronskian — Question 6

Question 6 – Solution

Strategy. Use the constant sign of the Wronskian first at the endpoints and then in the derivative of a ratio.

Step 1: Exclude repeated and shared zeros. If $u$ and $u'$ both vanished at an endpoint, regular uniqueness would make $u$ identically zero, contrary to independence. Thus both zeros are simple. Since $W$ never vanishes, $W(a)=-u'(a)v(a)$ and $W(b)=-u'(b)v(b)$ force $v(a),v(b)\ne 0$.

Step 2: Compare endpoint signs. Multiply $u$ by $-1$ if necessary so that $u>0$ on $(a,b)$. Simplicity then implies $u'(a)>0$ and $u'(b)<0$. Abel’s identity says $W(a),W(b)$ have the same sign. The two endpoint formulas therefore force $v(a),v(b)$ to have opposite signs. Continuity gives at least one zero of $v$ inside.

Step 3: Prove uniqueness of that zero. On the zero-free interval for $u$, $$\boxed{(v/u)'=\frac{uv'-u'v}{u^2}=\frac{W}{u^2}.}$$ The derivative has a fixed nonzero sign, so the ratio is strictly monotone. Its zeros are precisely the zeros of $v$ there. It can vanish at most once; combined with the preceding existence argument, this proves exactly one zero.

Step 4: Compute the example. Both candidates solve $y''+y=0$, and $$W[\sin t,\cos t+2\sin t]=-1.$$ The endpoint values of $v$ are $1,-1$. In $(0,\pi)$, the equation $\cos t+2\sin t=0$ has its root in the second quadrant, with $$\boxed{t=\pi-\arctan(1/2).}$$ The separation argument proves there are no other roots between these consecutive zeros of $\sin t$.