More on the Wronskian — Question 6

Question 6

Let $u,v$ be independent real solutions of $y''+p(t)y'+q(t)y=0$ on an open interval $I$, with continuous coefficients. Suppose $a<b$ lie in $I$ and are consecutive zeros of $u$: $u(a)=u(b)=0$ and $u(t)\ne 0$ for $a<t<b$.

Tasks

1. Prove that the zeros at $a,b$ are simple and that $v(a),v(b)$ are nonzero.

2. Use the sign of the Wronskian at the endpoints to prove that $v(a)$ and $v(b)$ have opposite signs.

3. Differentiate $v/u$ on $(a,b)$ and use it to prove that $v$ has exactly one zero there.

4. Apply the result to $u=\sin t$, $v=\cos t+2\sin t$ between $0$ and $\pi$, verifying independence and finding the unique zero explicitly.

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 $\sin t$.