Let act on with the single boundary condition

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Linear algebra and vector spaces · Linear algebra and vector spaces

Conditions of Validity, Ch Inner products, orthogonality and the adjoint

Problem

Let Ly=−y′′\mathcal{L}y = -y'' act on [0,1][0,1] with the single boundary condition y(0)=2y′(1)y(0) = 2y'(1), so that D={y∈C2[0,1]:y(0)=2y′(1)}\mathcal{D} = \{y \in C^2[0,1] : y(0) = 2y'(1)\}, and ⟨f,g⟩=∫01f∗g dx\langle f,g\rangle = \int_0^1 f^*g\,\mathrm{d}x. Someone argues: "Integration by parts twice gives, for real-valued f,gf,g, ⟨f,Lg⟩−⟨Lf,g⟩=B(f,g):=[f′g−fg′]01.\langle f,\mathcal{L}g\rangle - \langle \mathcal{L}f,g\rangle = B(f,g) := [f'g - fg']_0^1. Take f=g=yf=g=y for any y∈Dy \in \mathcal{D}. Then B(y,y)=y′(1)y(1)−y(1)y′(1)−(y′(0)y(0)−y(0)y′(0))=0−0=0.B(y,y) = y'(1)y(1) - y(1)y'(1) - \big(y'(0)y(0) - y(0)y'(0)\big) = 0 - 0 = 0. Since the boundary form vanishes for every y∈Dy \in \mathcal{D}, (L,D)(\mathcal{L},\mathcal{D}) is self-adjoint." ((a)) Find the error in this argument. ((b)) Decide whether (L,D)(\mathcal{L},\mathcal{D}) is actually self-adjoint.

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