More on the Wronskian — Question 5

Section 22-7 · Calculus pages 3969–3970

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Question 5

Two independent solutions of a regular equation y″+p(t)y′+q(t)y=0y''+p(t)y'+q(t)y=0 on ℝ\mathbb R have measured ordered Wronskians W(0)=6,W(2)=3,W(5)=12.W(0)=6,\qquad W(2)=3,\qquad W(5)=12. Assume p,qp,q are continuous and the measurements are exact.

Tasks

  1. Find ∫02p(t)dt\int_0^2p(t)\,dt and ∫25p(t)dt\int_2^5p(t)\,dt, and the average value of pp on each interval.

  2. Prove that pp is positive somewhere in (0,2)(0,2), negative somewhere in (2,5)(2,5), and zero somewhere in (0,5)(0,5).

  3. Under the additional assumption that p(t)=a+btp(t)=a+bt, determine a,ba,b and the full Wronskian function. Check all three measurements.

  4. For this affine model, find the unique minimum of WW on [0,5][0,5]. Explain why three Wronskian samples alone do not prove that the true coefficient is affine.

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 ∫02p=ln⁡2,∫25p=−ln⁡4.\int_0^2p=\ln 2,\qquad\int_2^5p=-\ln 4. The average values are (ln⁡2)/2\boxed{(\ln 2)/2} and −(ln⁡4)/3\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 pp at time 22.

Step 3: Fit the affine model. The coefficient equations are 2a+2b=ln⁡22a+2b=\ln 2 and 3a+(21/2)b=−2ln⁡23a+(21/2)b=-2\ln 2. Solving gives p(t)=29−14t30ln⁡2,W(t)=62(7t2−29t)/30.p(t)=\frac{29-14t}{30}\ln 2,\qquad \boxed{W(t)=6\,2^{(7t^2-29t)/30}.} At t=0,2,5t=0,2,5, the exponents are 0,−1,10,-1,1, giving 6,3,126,3,12 as required.

Step 4: Locate and qualify the minimum. Since W′=−pWW'=-pW and W>0W>0, the Wronskian decreases until t=29/14t=29/14 and increases afterward. Its unique minimum on [0,5][0,5] is W(29/14)=62−841/840.\boxed{W(29/14)=6\,2^{-841/840}.} Affineness is an extra modeling assumption. For example, adding any multiple of cos⁡(2πt)\cos(2\pi t) to this pp leaves both measured integrals unchanged, but generally gives a nonaffine coefficient and a different Wronskian between the measurement times.

Diagram shown in the original SVG worksheet.