I am interested in minimizing the Dirichlet energy
$$E[f]=\iint_{D_1(\mathbf{0})} \|\nabla f\|^2 \mathrm{d}A ,$$
over the unit disk $D_1(\mathbf{0})=\{(x_1,x_2) \in \mathbf{R}^2:x_1^2+x_2^2 \leq 1\}$ given equality constraints of the form $\{f(\mathbf{x}_n)=y_n:1 \leq n \leq N \}$ with $\|\mathbf{x}_n\|=1$ and $y_n \in \mathbf{R}$.
I know that specifying $f$ on the boundary in full, in terms of a continuous function, determines the solution entirely as the harmonic function given by Poisson's formula
$$f(r \cos \theta, r \sin \theta)=\frac{1}{2\pi} \int_{0}^{2\pi} \frac{1-r^2}{1-2r\cos(\theta-t)+r^2}f(\cos t,\sin t) \mathrm{d}t. $$
However, I'm not sure what happens when only a select few boundary values are provided. In summary, I want to solve the following optimization problem: $$\operatorname{argmin} _f\iint_{x_1^2+x_2^2\leq 1} \|\nabla f(x_1,x_2)\|^2 \mathrm{d}x_1 \mathrm{d}x_2\\ s.t. f(\cos(t_n),\sin(t_n))=y_n, \quad 1 \leq n \leq N. $$
My attempt:
Obviously if there is only one constraint $f(\mathbf{x}_1)=y_1$ the solution is the constant function $f(x_1,x_2) \equiv y_1$.
I've also managed to express the Dirichlet energy in smaller disks as a quadratic form of the boundary values:
$$E_R[f]=\iint_{D_R(\mathbf{0})} \|\nabla f\|^2 \mathrm{d}A = \frac{1}{4\pi^2} \int_0^{2\pi} \int_0^{2\pi} \frac{4 \pi R^2 \left( (1+R^4) \cos(t_1-t_2)-2R^2 \right)}{\left(1-2R^2 \cos(t_1-t_2)+R^4\right)^2} f(\cos t_1, \sin t_1) f(\cos t_2, \sin t_2)\mathrm{d}t_1 \mathrm{d}t_2. $$ Here $0 \leq R < 1$, and unfortunately a straightforward limit $R \to 1$ appears to make the integral divergent.
Letting $$K(t_1-t_2):=\frac{1}{4 \pi^2} \frac{4 \pi R^2 \left( (1+R^4) \cos(t_1-t_2)-2R^2 \right)}{\left(1-2R^2 \cos(t_1-t_2)+R^4\right)^2},$$ I computed the variational derivative with respect to $f$ and found the condition $$\int_0^{2\pi} K(\tau-t) f(\cos t, \sin t) \equiv 0, \quad \text{ for all } \tau.$$
I am stuck here as I don't know how to use the constraints in this variational approach.
Any help on solving the problem with more than 1 constraint would be greatly appreciated. I would love to have the solution of the most general problem, but if that's too hard, special cases such as boundary values given on the vertices of a regular polygon would also be informative. Thanks.
I believe the variational problem does not have a solution unless $y_1=\ldots = y_N$, in which case the solution is constant. The reason is that functions with singularities exist in $W^{1,2}$, meaning that we can satisfy the conditions at the boundary points, with arbitrarily little cost in terms of the $L^2$-norm of the gradient.
Let me sketch the idea for $N=2$, and $y_1=1$ and $y_2=0$ (and $\mathbf{x}_1,\mathbf{x}_2$ arbitrary). The general case can be done similarly, although slightly more involved. We consider for $\epsilon>0$ the continuous function $$f_{\epsilon}:\mathbb{R}^2 \to \mathbb{R}, \quad f_{\epsilon}(x) = \begin{cases} 1-c_\epsilon+\log\left(\log\left(1+\frac{1}{|\mathbf{x}-\mathbf{x}_1|+\epsilon}\right)\right) & \text{when $|\mathbf{x}-\mathbf{x}_1|\leq \frac{1}{\exp(\exp(c_\epsilon-1))-1}-\epsilon$},\\ 0 & \text{else,} \end{cases}$$ where $c_\epsilon = \log(\log(1+\frac{1}{\epsilon}))$. Note that $c_\epsilon \to \infty$ as $\epsilon \to 0$, and hence, we have that $f_{\epsilon}(\mathbf{x}_1)=1=y_1$ and $f_{\epsilon}(\mathbf{x}_2)=0=y_2$ for $\epsilon$ small enough. Since $u(x) = \log(\log(1+\frac{1}{|x|})) \in W^{1,2}_{loc}$, we deduce that $$\|\nabla f_{\epsilon}\|_{L^2(D_1(\mathbf{0}))}\leq \|\nabla u\|_{L^2(D_{r_\epsilon}(\mathbf{0}))} \to 0 \quad \text{as $\epsilon \to 0$},$$ with $r_{\epsilon}=\frac{1}{\exp(\exp(c_\epsilon-1))-1}-\epsilon \to 0$ as $\epsilon \to 0$. This implies that the infimum of the variational problem is zero, which is not attained since constant functions do not satisfy the boundary condition.