Solutions to Systems — Question 9

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Question 9

For x′=y,y′=−x,x'=y,\qquad y'=-x, only the measurements x(0)=1x(0)=1 and x(T)=bx(T)=b are given, where T>0T>0. The second initial component is unknown. You may verify solutions using sine and cosine identities; no phase-plane analysis is required.

Tasks

  1. Find every solution consistent with x(0)=1x(0)=1, using one arbitrary constant, and identify what that constant means.

  2. Classify existence and uniqueness for all T>0T>0 and b∈ℝb\in\mathbb R, including the exceptional times T=kπT=k\pi.

  3. Apply the classification to (T,b)=(π,−1)(T,b)=(\pi,-1) and (π/2,2)(\pi/2,2). Give the full solution set in each case.

  4. When sin⁡T≠0\sin T\ne 0, suppose the reported value of bb has an error δ\delta. Find the exact resulting error in the recovered second initial component. Explain what happens as T→πT\to\pi and why this is not failure of uniqueness for a fully specified IVP.

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Question 9 – Solution

Strategy. Treat the unknown initial velocity as a parameter and ask whether the observations actually determine it.

Step 1: Parametrize all compatible initial states. Since x″=−xx''=-x and x(0)=1x(0)=1, the complete family is x=cos⁡t+csin⁡t,y=−sin⁡t+ccos⁡t.\boxed{x=\cos t+c\sin t,\qquad y=-\sin t+c\cos t.} Direct differentiation verifies both rows. The arbitrary constant is c=y(0)=x′(0)c=y(0)=x'(0), the missing initial component.

Step 2: Test the observation equation. The second measurement requires cos⁡T+csin⁡T=b\cos T+c\sin T=b. If sin⁡T≠0\sin T\ne 0, there is exactly one solution, with c=(b−cos⁡T)/sin⁡T\boxed{c=(b-\cos T)/\sin T}. If T=kπT=k\pi for a positive integer kk, the measurement is independent of cc: there are infinitely many solutions when b=(−1)kb=(-1)^k, and none when b≠(−1)kb\ne(-1)^k. These cases exhaust all positive times.

Step 3: Evaluate the two specified experiments. At (π,−1)(\pi,-1) every real cc is allowed, giving the full family in Step 1. The figure shows three of these indistinguishable endpoint measurements. At (π/2,2)(\pi/2,2), c=2c=2, so the unique solution is X=(cos⁡t+2sin⁡t,−sin⁡t+2cos⁡t)T.\boxed{X=(\cos t+2\sin t,\,-\sin t+2\cos t)^T.}

Step 4: Quantify sensitivity and interpret it. Replacing bb by b+δb+\delta changes cc by exactly Δc=δ/sin⁡T\boxed{\Delta c=\delta/\sin T}. Thus the absolute amplification factor is 1/|sin⁡T|1/|\sin T|, which is unbounded as T→πT\to\pi through nonexceptional times. A fixed nonzero measurement error can then create a large recovered initial-velocity error. At T=πT=\pi, that component is unobservable from these two measurements. Each fully specified initial vector still has a unique solution; the difficulty lies in recovering it from partial observations.

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