 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-13/3/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

u_{1} = T v_{1}

and

u_{2} = T v_{2}

, linearity of the weak equation makes

α u_{1} + β u_{2}

solve the equation with forcing

α v_{1} + β v_{2}

. Uniqueness therefore gives

T (α v_{1} + β v_{2}) = α T v_{1} + βT v_{2}

u = T v

with

u

∥ D u ∥_{2}^{2} = - \int vu \leq ∥ v ∥_{2} ∥ u ∥_{2} \leq C_{P} ∥ v ∥_{2} ∥ D u ∥_{2} .

If the gradient norm is nonzero, divide by it; if it is zero, Poincare gives

u = 0

and the estimate is automatic. In either case

∥ T v ∥_{W^{1, 2}} \leq C_{P} 1 + C_{P}^{2} ∥ v ∥_{2} .

Hence

T : L^{2} (Ω) \to W_{0}^{1, 2} (Ω)

is a bounded linear operator. Notice that the energy identity has a minus sign, so

⟨ T v, v ⟩ = - ∥ D (T v) ∥_{2}^{2} \leq 0

. Testing the equations for

T v

and

T z

against each other gives

⟨ T v, z ⟩ = - \int D (T v) \cdot D (T z) = ⟨ v, T z ⟩

. Thus the bounded Dirichlet inverse is a negative self-adjoint operator on

L^{2}

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