 Solution

Take $l>0$ and use the inner product $⟨u,v⟩={\int }_0^l\overline{u}vdx$ on the Hilbert space $L^2(0,l)$. The differential operator has domainwhere $H^2$ is a Sobolev space. Twice applying integration by parts gives the Lagrange identity for a Sturm-Liouville operatorThe Dirichlet boundary conditions make this expression zero for $u,v\in D(\mathcal{L})$, and the formal adjoint is ${\mathcal{L}}^{\ast }=d^2/d{x}^2+1$. To obtain the domain of the adjoint operator, initially impose the boundary conditions only on $u$. The remaining expression is $-[\overline{u^{\prime }}v{]}_0^l$. Since $u^{\prime }(0)$ and $u^{\prime }(l)$ vary independently, it vanishes for all such $u$ exactly when $v(0)=v(l)=0$. The adjoint domain is therefore the original domain: this is full self-adjointness, including the boundary conditions.

Consequently the eigenvalues are real, eigenfunctions for distinct eigenvalues are orthogonal, and regular Sturm-Liouville theory supplies a complete orthonormal eigenbasis. Here these objects can be exhibited explicitly:Indeed $\mathcal{L}{\varphi }_n={\lambda }_n{\varphi }_n$, and the Fourier sine basis is complete in $L^2(0,l)$. There are no additional eigenvalues: for a nonzero eigenfunction $u$, integration by parts givesso $u^{\prime \prime }+(1-\lambda )u=0$ and the endpoint values force $\sqrt{1-\lambda }l=n\pi$. All these eigenvalues are simple. The eigenvectors are precisely the nonzero multiples of these eigenfunctions, regarded as vectors in the Hilbert space.