I am self-studying differential equations using MIT's publicly available materials. One of the recitation exercises runs as follows:
Define an inner or dot product on $\mathcal{C}[a,b]$ by \begin{align} \langle u, v \rangle & = \int_a^bu(x)v(x)dx \end{align} Suppose $L[u] = u'' + pu' + qu$ is a differential operator, and $M[u]$ is its adjoint. Show that $\langle L[u], v\rangle = \langle u, M[v]\rangle$ for all $u, v \in \mathcal{C}^2[a,b]$ provided $u(a) = u(b) = v(a) = v(b) = 0$.
In neither the textbook nor the lecture notes has the notion of an adjoint been introduced, so I went to Wikipedia (and various other online sources) for a definition. From what I can tell, if $T$ is a differential operator, then its adjoint $T^*$ is defined as the differential operator such that \begin{align} \langle Tu, v \rangle = \langle u,T^*v\rangle \end{align}
But if this is the case, isn't what the exercise asks for true immediately and trivially? (And, moreover, independent of the fact that $u(a) = u(b) = v(a) = v(b) = 0$?)
No doubt there's something basic I don't know. If so, a gentle hint (rather than a complete solution) would be appreciated.
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$\begingroup$Using the definition of the formal adjoint, we have $M[u] := u'' - pu' + qu$. We verify the adjoint relation as follows:
$$\begin{align}\langle L[u],v\rangle &=\int_a^b(u''(x) + pu'(x) + qu(x))v(x)dx \\ &= \int_a^b u''(x)v(x)dx + p\int_a^bu'(x)v(x)dx + q\int_a^b u(x)v(x)dx \\ &= \int_a^b u(x)v''(x)dx - p\int_a^bu(x)v'(x)dx + q\int_a^b u(x)v(x)dx\\&= \int_a^bu(x)(v''(x)-pv'(x)+qu(v))dx \\&= \int_a^b u(x) M[v](x)dx \\ &= \langle u,M[v]\rangle\end{align}$$
where on the third line we used integration by parts (twice on the first term and once on the second) and the fact that $u$ and $v$ vanish on the boundary.
$\endgroup$ $\begingroup$Have a look at
which gives very clear examples.
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