1-dimensional quadratic placement with distances in [1], Hall shows solution to the following graph optimization problem for a vector of node positions X, given a cost matrix C (normalizing the solution with the constraint $X^TX=1$
$$\text{argmin}_X \sum_i \sum_j (x_i - x_j)^2 \cdot c_{ij}$$
by reducing it to $\text{argmin}_X X^T B X$ where $B$ is a positive semi-definite matrix of rank $n - 1$ (where $n$ is the size of $X$), obtained from $C$ (not important how).
Then he finds the solution as the first but minimal eigenvector of $B$.
I have a similar problem, where instead of forcing the adjacent vectors as close as possible to each other, I'd like to force them to be in the desired distance $d_{ij}$ where $D$ is a skew-symmetric matrix.
$$\text{argmin}_X \sum_i \sum_j (x_i - x_j + d_{ij})^2 \cdot c_{ij}$$
I managed, similarly as in [1] (and hopefully correctly), to reduce the problem into
$$\text{argmin}_X (X^TB + U)X$$
where $B$ is the same as in the original problem and $U$ is a horizontal vector, obtained by adding up columns of a skew-symmetric matrix (for what it's worth).
I wonder if this problem has been investigated before or whether the solution can be elegantly found using eigenvectors, similarly as for the first problem. By the way, I'm not keen on the quadratic metric in the problem definition, an absolute value or something similar is also good.
[1] Hall, Kenneth M. "An r-dimensional quadratic placement algorithm." Management science 17.3 (1970): 219-229.
 A: Got it. Guided by the differentiation in this thread and then doing some experimentation, I've found a solution that looks very simple and the results look OK.
To clean up the notation, without loss of generality, I will replace the question's $U$ with $-U^T$ (using $^T$ to show that it's a horizontal vector) and the question's $B$ with $\frac{1}{2}B$. To sum up, we will minimize
$$f(X) = \frac{1}{2}(X^TB - U^T)X = \frac{1}{2}X^TBX - U^TX$$
where X and U are vectors and B is a symmetric matrix (for how it was obtained, see the PS section).
First, I've found the minimum of $f(X)$ by deriving by $X$ and doing some magic with transpositions (we can remove some of them because they are applied on scalars, other ones due to $B$'s symmetry):
$$\partial f(X) = \frac{1}{2}(\partial X)^TBX + \frac{1}{2}X^TB\partial X - U^T\partial X = \partial X(BX - U)$$
We could solve $BX = U$ using $X = B^{-1}U$. Issue here is that $B$, as defined in [1], is singular, which could be expected because the problem is invariant to translation of $X$. To get a solution, I've added parabolic cost on the position of $x_0$ to $f(X)$:
$$f(x) = \frac{1}{2}X^TBX - U^TX + X^TE_0X$$
where $E_0$ is a matrix with all zeroes except the 1 in the left most corner. This way I've actually added $x_0^2$ to $f(x)$ and thus glued the position of $x_0$ to 0. Using the same technique as before, I've got to an equation $(B + E_0)X = U$. The result $X = (B + E_0)^{-1}U$ is now defined.
PS: To map the solution to the original problem
$$\text{argmin}_X \sum_i \sum_j (x_i - x_j + d_{ij})^2 \cdot c_{ij}$$
the matrix $B$ has sums of rows of $C$ on the diagonal and values of $-C$ elsewhere. The vector $U$ is obtained by summing up rows of the pointwise product of $C$ and $-D$.
