What is the orthogonal complement of $H^1_0$ in $H^1$?

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Let $\Omega$ be a closed domain with smooth boundary in $\mathbb{R}^n$. Let $H^1_0(\Omega)$ be the closure of compactly supported smooth functions under the norm $\|u\|_1 = \int_\Omega u^2 + |\nabla u|^2\ dx$ and let $H^1(\Omega)$ be the closure of smooth, continuous functions under the same norm.

Any $H^1$ function which has nonvanishing trace cannot be approximated by any sequence of functions in $H^1_0$. So $H^1_0$ is a closed subspace of the Hilbert space $(H^1, \|\cdot\|_1)$, hence has an orthogonal complement.

What is a generating set of the orthogonal complement of $H^1_0$ in $H^1$?

Motivation is to get my hands on some concrete examples, rather than to just appeal to theorems that establish the existence of a right inverse to a trace operator.

Of course if anyone has references, I'm happy to follow them up. I've skimmed through Gilbarg-Trudinger and Evans and found nothing, but maybe I'm looking in the wrong place.

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As reuns has stated in the comments, the answer depends on the inner product you choose in $H^1(\Omega)$. Let us fix the most common choice $$ (u,v)_{H^1(\Omega)} = \int_\Omega \nabla u \cdot \nabla v + u \, v \, \mathrm{d}x.$$

In this case, the orthogonal complement of $H_0^1(\Omega)$ consists precisely of the (weak) solutions $u \in H^1(\Omega)$ of $$ -\Delta u + u = 0$$ (without B.C.). Indeed, the weak formulation of this PDE is $$(u,v)_{H^1(\Omega)} = 0 \quad\forall v \in H_0^1(\Omega).$$

For different inner products, you get different PDEs.