Series Solutions — Question 10

Question 10 – Solution

Strategy. Test the forcing’s analyticity before treating its Taylor coefficients as the forcing itself.

Step 1: Compute all center derivatives. The continuous forcing gives a $C^2$ solution; since $f$ is smooth, repeated use of $y''=f-y$ makes $y$ smooth as well. Differentiation gives $$y^{(n+2)}(0)=f^{(n)}(0)-y^{(n)}(0)=-y^{(n)}(0).$$ Starting with $y(0)=y'(0)=0$, induction gives every derivative zero. Its Taylor series is therefore $\boxed{\sum_{n\ge 0}0x^n=0}$, with infinite radius as a series.

Step 2: Recover the actual forced solution. Variation of constants gives, for every real $x$, $$\boxed{y(x)=\int_0^x\sin(x-t)f(t)\,dt.}$$ Here $y'(x)=\int_0^x\cos(x-t)f(t)\,dt$ and $y''(x)=f(x)-y(x)$. At $0$ both integrals vanish. The formula is defined on every finite real segment; any difference between two solutions satisfies $z''+z=0$ with zero initial values, hence is zero. This proves global existence and uniqueness.

Step 3: Prove nonzero values despite zero Taylor data. Because $f$ is even, $y(-x)$ solves the same IVP; uniqueness gives $y(-x)=y(x)$. For $0<x<\pi$ and $0<t<x$, both $\sin(x-t)$ and $f(t)$ are strictly positive, so the integral is positive. Evenness extends this conclusion to $0<|x|<\pi$. The zero Taylor series cannot equal $y$ on any neighborhood of $0$, even though it converges everywhere.

Step 4: Locate the missing analyticity assumption. The leading coefficient is nonzero and the homogeneous coefficients are analytic, but the forcing is not analytic at $0$: its Taylor sum is zero while it is positive nearby. The analytic ordinary-point theorem for a nonhomogeneous equation requires analytic forcing too. Replacing $f$ by its Taylor series changes the equation to $y''+y=0$. The formal zero series solves that changed equation, not the original equation away from $0$.