Let $k(x, y): \mathbb{R}^2 \to \mathbb{R}$ be a kernel and $T: H^1(a, b) \to L_2(c, d)$ $$ T u(x) = \int\limits_{a}^{b} k(x, s) u(s) ds$$ Find the adjoint operator $T^*$.
It is easy to see the if $T: L_2(a, b) \to L_2(c, d)$, then $T^*$ can be expressed as $$ T^*v(s) = \int\limits_{c}^{d} k(x, s) v(x) dx$$
However, here we need $$ (Tu, v)_{L_2(c, d)} = (u, T^*v)_{H^1(a, b)} = (u, T^*v)_{L_2(a, b)} + (u', (T^*v)')_{L_2(a, b)}$$
I have tried to use the identity $$ \int\limits_{a}^{b} uv dx = \int\limits_{a}^{b} u'v dx + \int\limits_{a}^{b} uv' dx $$
Then $$ \int\limits_{c}^{d} \int\limits_{a}^{b} k(x, s) u(s) v(x) ds dx = \int\limits_{a}^{b} u(s) \left[ \int\limits_{c}^{d} k'_s(x, s) v(x) dx \right] ds + \int\limits_{a}^{b} u'(s) \left[ \int\limits_{c}^{d} k(x, s) v(x) dx \right] ds $$
And it does not look like a form of adjoint operator...
Thank you in advance for any help!
The line $$ (Tu, v)_{L_2(c, d)} = (u, T^*v)_{H^1(a, b)} = (u, T^*v)_{L_2(a, b)} + (u', (T^*v)')_{L_2(a, b)}$$ contains the answer. Define $z:=T^*v$. Then $z$ has to solve the equation $$ (u, z)_{L_2(a, b)} + (u', z')_{L_2(a, b)}= (Tu, v)_{L_2(c, d)} $$ for all $u\in H^1(a,b)$, or equivalently, $$ \int_a^b z'(s)u'(s)+z(s)u(s) \ ds = \int_a^b \int_a^b k(x,s) u(s)v(x) \ dx \ ds . $$ This is the weak formulation in $H^1$ of the boundary value problem $$ -z''(x) + z(x) = \int_a^b k(x,s) u(s)\ ds $$ with boundary conditions $z'(a)=z'(b)=0$.
Given $v$, one has to solve the integral equation or the boundary value problem to obtain $z:=T^*v$. Without further information about the problem this cannot be simplified. Note that the right-hand side of the boundary-value problem is equal to the expression of the adjoint operator of $T$ when considered as mapping from $L^2$ to $L^2$.