Series Solutions — Question 6

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Question 6

The series solution of y‴+y′=0y'''+y'=0 with (y,y′,y″)(0)=(0,1,0)(y,y',y'')(0)=(0,1,0) describes an oscillator derivative pair (y′,y″)(y',y''). A proposed rational approximation is U(x)=1−x2/41+x2/4,V(x)=−x1+x2/4.U(x)=\frac{1-x^2/4}{1+x^2/4},\qquad V(x)=-\frac{x}{1+x^2/4}. Its designer claims that correct initial values and exact conservation of U2+V2U^2+V^2 are enough to make it the true derivative pair.

Tasks

  1. Derive the series recurrence and identify the exact solution. Give its terms through degree seven.

  2. Expand U,VU,V through the first coefficient at which each differs from the true pair. Check their initial values and quadratic invariant.

  3. Test the necessary derivative relation U′=VU'=V. Compute its defect exactly and decide whether any function can have both y′=Uy'=U and y″=Vy''=V on an interval containing zero.

  4. Find the phase τ(x)\tau(x) for which (U,V)=(cos⁡τ,−sin⁡τ)(U,V)=(\cos\tau,-\sin\tau). Explain the actual evolution law and plot the phase lag x−τ(x)x-\tau(x) beside its leading cubic approximation on [0,2][0,2].

Original worksheet page 1: question and worked solution for 7-7-006
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Question 6 – Solution

Strategy. Test derivative compatibility in addition to an invariant and a few Taylor coefficients.

Step 1: Solve the series IVP. Coefficient matching gives an+3=−an+1/[(n+3)(n+2)]a_{n+3}=-a_{n+1}/[(n+3)(n+2)]. The data (a0,a1,a2)=(0,1,0)(a_0,a_1,a_2)=(0,1,0) give the entire series y=sin⁡x=x−x36+x5120−x75040+⋯.\boxed{y=\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\cdots.} The true pair is (cos⁡x,−sin⁡x)(\cos x,-\sin x).

Step 2: Check local accuracy and the invariant. Near zero, U=1−x2/2+x4/8+O(x6),V=−x+x3/4+O(x5).U=1-x^2/2+x^4/8+O(x^6),\qquad V=-x+x^3/4+O(x^5). The correct coefficients are x4/24x^4/24 and x3/6x^3/6, respectively. Nevertheless (U,V)(0)=(1,0)(U,V)(0)=(1,0), and direct squaring gives U2+V2=1U^2+V^2=1 exactly.

Step 3: Expose the derivative defect. Writing d=1+x2/4d=1+x^2/4, we obtain U′−V=x34d2.\boxed{U'-V=\frac{x^3}{4d^2}.} This is nonzero for x≠0x\ne 0. If y′=Uy'=U, differentiation forces y″=U′y''=U', so no such function has both proposed derivatives on a nontrivial interval containing zero. Conservation alone does not enforce the differential equation.

Step 4: Identify the altered clock. The half-angle identities give τ=2arctan⁡(x/2)\tau=2\arctan(x/2) and τ′=1/d\tau'=1/d. Thus U′=V/dU'=V/d, V′=−U/dV'=-U/d: the pair traverses the correct circle with a different speed. The phase lag is x−τ=x3/12+O(x5)x-\tau=x^3/12+O(x^5) and is positive for x>0x>0 because its derivative is x2/(4+x2)>0x^2/(4+x^2)>0.

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Original worksheet page 2: question and worked solution for 7-7-006

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