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The backward heat solution map is unbounded

Statement

Assume Countable Choice. Let T>0 and consider the heat equation ut=uxx on (0,π)×(0,T] with homogeneous Dirichlet boundary data u(0,t)=u(π,t)=0.

Well-definedness of the terminal-to-initial map. If a C4,2 function w on [0,π]×[0,T] solves this problem with w(x,T)=0 for all x, then w≡0 (Backward uniqueness for the heat equation on a bounded interval). Hence on the class of C4,2 solutions the terminal data determine the solution, and the map S sending a terminal datum u(⋅,T) to the initial state u(⋅,0) is well defined.

Unboundedness. For every integer k≥1 the function uk(x,t):=ek2(T−t)sin⁡(kx) is a classical solution of ut=uxx on (0,π)×(0,T] with uk(0,t)=uk(π,t)=0, initial state uk(⋅,0)=ek2Tsin⁡(k⋅) and terminal data uk(⋅,T)=sin⁡(k⋅). Consequently there is no constant C<∞ with ∥u(⋅,0)∥L2(0,π)≤C ∥u(⋅,T)∥L2(0,π) for all C4,2 solutions u of this problem: the solutions u~k(x,t)=e−k2tsin⁡(kx) have terminal data gk=e−k2Tsin⁡(k⋅) with ∥gk∥2=π/2 e−k2T→0 while ∥u~k(⋅,0)∥2=π/2 for every k. Since Sgk=sin⁡(k⋅)=ek2Tgk, the terminal-to-initial map multiplies the k-th sine mode by ek2T and is unbounded with respect to the L2(0,π) norms on its domain and range: the backward heat problem on a bounded interval has no norm-stable solution operator, and the amplification factor of the k-th Dirichlet mode is exactly ek2T.

Facts & Assumptions

Given: Countable Choice, T>0, the interval (0,π), and an integer k≥1.

[A1]

Countable Choice is the ambient hypothesis, inherited through the backward-uniqueness and L2 suppliers (The Axiom of Countable Choice (ACω)).

[F1]

Backward uniqueness: a C4,2 function on [0,π]×[0,T] satisfying wt=wxx, w(0,t)=w(π,t)=0 and w(x,T)=0 vanishes identically (Backward uniqueness for the heat equation on a bounded interval).

[F3]

The L2(0,π) inner product is ⟨f,g⟩=∫0πfg on the quotient space of The space Lp(μ) as the quotient by null functions, with ∥f∥22=∫0πf2 (L2 with the integral pairing is a Hilbert space), and ∥sin⁡(k⋅)∥2=π/2 (L2 normalisation of the sine modes on an interval).

[F4]

e−k2T→0 as k→∞, faster than every polynomial (The exponential dominates every fixed nonnegative integer power at +∞).

Proof

Given: Countable Choice, T>0 and k≥1.

1.1A1F1given

If u1,u2 are C4,2 solutions with ui(0,t)=ui(π,t)=0 and the same terminal data u1(⋅,T)=u2(⋅,T), then w:=u1−u2 is C4,2 and satisfies wt=wxx, w(0,t)=w(π,t)=0 and w(x,T)=0, so [F1] gives w≡0; hence the terminal-to-initial map S is well defined on the terminal data of the class.

1.2F2given

For every k≥1 the function uk(x,t)=ek2(T−t)sin⁡(kx) has ∂tuk=−k2uk and, by [F2], ∂x2uk=−k2sin⁡(kx)ek2(T−t)=−k2uk, with boundary values uk(0,t)=sin⁡0=0 and uk(π,t)=sin⁡(kπ)=0; the same computation with e−k2t shows that u~k(x,t)=e−k2tsin⁡(kx) is a solution with u~k(⋅,0)=sin⁡(k⋅) and terminal data gk=e−k2Tsin⁡(k⋅).

2.1step 1.2F3given

By [F3] and scaling by the positive factor e−k2T, ∥gk∥2=e−k2T∥sin⁡(k⋅)∥2=π/2 e−k2T, while ∥u~k(⋅,0)∥2=∥sin⁡(k⋅)∥2=π/2.

2.2step 1.2F4given

By [F4] the sequence e−k2T tends to 0; consequently the terminal data gk of step 1.2 satisfy ∥gk∥2=π/2 e−k2T→0 while ∥u~k(⋅,0)∥2=π/2 stays constant.

3.1step 1.1step 1.2step 2.2F3given∎

Since gk are admissible terminal data with ∥gk∥2→0 but Sgk=sin⁡(k⋅) and ∥sin⁡(k⋅)∥2=π/2 for every k, no finite constant C satisfies ∥u(⋅,0)∥2≤C∥u(⋅,T)∥2 on the class, so S is unbounded; explicitly S multiplies the k-th sine mode by ek2T, which is the amplification factor asserted.

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