More on the Wronskian — Question 5

Question 5 – Solution

Strategy. Ratios of Wronskian measurements give integrals of the coefficient, rather than its pointwise values.

Step 1: Extract the averages. Abel’s identity gives $$\int_0^2p=\ln 2,\qquad\int_2^5p=-\ln 4.$$ The average values are $\boxed{(\ln 2)/2}$ and $\boxed{-(\ln 4)/3}$ respectively.

Step 2: Force a sign change. A continuous function with positive integral must be positive at an interior point; likewise the negative integral requires an interior negative value. Choosing one point from each interval, the intermediate value theorem supplies a zero between them. This conclusion does not specify the value of $p$ at time $2$.

Step 3: Fit the affine model. The coefficient equations are $2a+2b=\ln 2$ and $3a+(21/2)b=-2\ln 2$. Solving gives $$p(t)=\frac{29-14t}{30}\ln 2,\qquad \boxed{W(t)=6\,2^{(7t^2-29t)/30}.}$$ At $t=0,2,5$, the exponents are $0,-1,1$, giving $6,3,12$ as required.

Step 4: Locate and qualify the minimum. Since $W'=-pW$ and $W>0$, the Wronskian decreases until $t=29/14$ and increases afterward. Its unique minimum on $[0,5]$ is $$\boxed{W(29/14)=6\,2^{-841/840}.}$$ Affineness is an extra modeling assumption. For example, adding any multiple of $\cos(2\pi t)$ to this $p$ leaves both measured integrals unchanged, but generally gives a nonaffine coefficient and a different Wronskian between the measurement times.