# Find the adjoint under the inner product $\langle f, g \rangle = \int_0^1 f(t)g(t)t \ dt$ of $\mathcal{L}(f)(t) = \frac{d^2 f}{dt^2} + f$.

Find the adjoint under the inner product $\langle f, g \rangle = \int_0^1 f(t)g(t)t \ dt$ of $\mathcal{L}(f)(t) = \dfrac{d^2 f}{dt^2} + f$ with $f(0) = 0$ and $f'(1) = 0$.

Note: The weight function is $w(t) = t$.

We need to find the operator $\mathcal{L}^\dagger$ such that $\langle \mathcal{L}f, g \rangle = \langle f, \mathcal{L^\dagger}g \rangle$.

My Solution

$\langle \mathcal{L}f, g \rangle = \int^1_0 (f'' + f)gt = \int_0^1(tg)f'' \ dt + \int_0^1 fgt \ dt$

Using integration by parts twice, we get

$\langle \mathcal{L}f, g \rangle = [(tg)f']^1_0 - \int_0^1 (tg)'f' \ dt + \int_0^1 fgt \ dt$

$= g(1)f'(1) - 0 - [(tg)'f]^1_0 + \int_0^1 (tg)''f \ dt + \int_0^1 fgt \ dt$

Note that $(tg)' = tg' + g$ and $f'(1) = 0$. Therefore, we get

$\langle \mathcal{L}f, g \rangle = 0 - [g'(1) + g(1)]f(1) - 0 + \int_0^1 (tg)'' f \ dt + \int_0^1 fgt \ dt$

Note that $(tg)'' = (tg' + g)' = tg'' + 2g'$. Therefore, we have

$\langle \mathcal{L}f, g \rangle = -[g'(1) + g(1)]f(1) + \int_0^1 (tg'' + 2g' + gt) \ dt$

$\mathcal{L}^\dagger(g)(t) = \dfrac{d^2g}{dt^2} + \dfrac{2}{t} \dfrac{dg}{dt} + g$ with $g'(1) + g(1) = 0$.
However, the provided solution says that the adjoint operator is $\mathcal{L}^\dagger(g)(t) = \dfrac{d^2g}{dt^2} + \dfrac{2}{t} \dfrac{dg}{dt} + \dfrac{1}{t} g$ with $g'(1) + g(1) = 0$.