Series Solutions — Question 6

Question 6

The series solution of $y'''+y'=0$ with $(y,y',y'')(0)=(0,1,0)$ describes an oscillator derivative pair $(y',y'')$. A proposed rational approximation is $$U(x)=\frac{1-x^2/4}{1+x^2/4},\qquad V(x)=-\frac{x}{1+x^2/4}.$$ Its designer claims that correct initial values and exact conservation of $U^2+V^2$ 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 $U,V$ 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 $U'=V$. Compute its defect exactly and decide whether any function can have both $y'=U$ and $y''=V$ on an interval containing zero.

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

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 $a_{n+3}=-a_{n+1}/[(n+3)(n+2)]$. The data $(a_0,a_1,a_2)=(0,1,0)$ give the entire series $$\boxed{y=\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\cdots.}$$ The true pair is $(\cos x,-\sin x)$.

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

Step 3: Expose the derivative defect. Writing $d=1+x^2/4$, we obtain $$\boxed{U'-V=\frac{x^3}{4d^2}.}$$ This is nonzero for $x\ne 0$. If $y'=U$, differentiation forces $y''=U'$, 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 $\tau=2\arctan(x/2)$ and $\tau'=1/d$. Thus $U'=V/d$, $V'=-U/d$: the pair traverses the correct circle with a different speed. The phase lag is $x-\tau=x^3/12+O(x^5)$ and is positive for $x>0$ because its derivative is $x^2/(4+x^2)>0$.