 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-4/23i/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

(∣ ξ ∣^{4} - ∣ ξ ∣^{2} + 1) u (ξ) = f (ξ) .

m (ξ) = ∣ ξ ∣^{4} - ∣ ξ ∣^{2} + 1 = (∣ ξ ∣^{2} - \frac{1}{2})^{2} + \frac{3}{4}

is everywhere positive. Define

u = f / m

. Since

sup_{ξ \in R^{n}} \frac{( 1 + ∣ ξ ∣ ^{2} ) ^{2}}{m ( ξ )} < \infty,

we have

∥ u ∥_{H^{s + 4}}^{2} = \int (1 + ∣ ξ ∣^{2})^{s + 4} \frac{∣ f ( ξ ) ∣ ^{2}}{m ( ξ ) ^{2}} d ξ \leq C^{2} ∥ f ∥_{H^{s}}^{2} .

Thus

u \in H^{s + 4}

exists and obeys the required estimate. If two solutions existed, their difference would have

m u = 0

; positivity of

m

A classical solution needs continuous derivatives through order four. Applying the Sobolev embedding theorem with

k = 4

u \in H^{s + 4}

shows that this is guaranteed when

s + 4 > 4 + \frac{n}{2},

or equivalently

s > \frac{n}{2} .

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