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I'm looking for a proof that $Ly = (ky')'$ is self-adjoint. My attempt:

\begin{align} \langle Ly(x), w(x)\rangle &= \int_a^b (k(x)y(x)')'w(x) \,\mathrm{d}x \\ &= \int_a^b (k'(x)y(x)' + k(x)y''(x))w(x) \,\mathrm{d}x \\ &=\quad\color{red}{??} \\ &= \int_a^b y(x)(k'(x)w(x)' + k(x)w''(x)) \,\mathrm{d}x \\ &= \langle y(x), Lw(x)\rangle \end{align}

Except I'm not sure how to fill in the center part.

Frank Vel
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  • Tried integration by part? – edm Jan 11 '18 at 09:01
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    The operator $L$ probably acts on functions $u$ such that $$u(a)=u(b)=u'(a)=u'(b)=0$$ hence the identity $$\int u'v=-\int uv'$$ used twice, is all you need. – Did Jan 11 '18 at 09:04
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    $\langle Ly,w\rangle-\langle y,Lw\rangle=k(wy'-w'y)|_{a}^{b}$ – vnd Jan 11 '18 at 09:09

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