More on the Wronskian — Question 1

Question 1 – Solution

Strategy. Differentiate the determinant and substitute the common equation before integrating.

Step 1: Derive Abel’s identity. Differentiation cancels the mixed terms: $$W'=uv''-u''v=u(-pv'-qv)-(-pu'-qu)v=-pW.$$ Multiplication by $\exp(\int_{t_0}^t p(s)\,ds)$ gives a function with zero derivative. Thus, without dividing by $W$, $$\boxed{W(t)=W(t_0)\exp[-\int_{t_0}^t p(s)\,ds].}$$ This includes $W\equiv 0$ when $W(t_0)=0$.

Step 2: Use the nonzero exponential. The exponential factor is positive and finite at every point of $I$. A nonzero initial Wronskian stays nonzero with the same sign. If it is zero, the initial-data columns are dependent; a nontrivial constant combination has zero value and slope at $t_0$. Regular uniqueness forces that combination to vanish on $I$. Therefore, for solutions of this common regular equation, $W\equiv 0$ is equivalent to dependence.

Step 3: Normalize the example. Divide by $1+t^2$, giving $p=2t/(1+t^2)$. Integration gives $$\int_0^t\frac{2s}{1+s^2}\,ds=\ln(1+t^2),\qquad \boxed{W(t)=-\frac 3{1+t^2}.}$$ Thus the Wronskian is determined without finding either solution. The undivided coefficient $2t$ is not the coefficient to use in Abel’s identity.

Step 4: Interpret the limits. We have $W<0$ everywhere, with $W\to 0$ from below as $t\to\pm\infty$. The pair is independent on all of $\mathbb R$, and its data columns are independent at every finite time. A limiting zero outside the interval is not a zero at a point where the initial-data criterion applies.