I am trying to solve an optimization problem that is almost separable, but haven't been able to make much progress.
$$ \underset{x_1,x_2, \dots, x_n}{\text{minimize}} \quad \sum_{k=1}^n \left( a_k x_k^2 + b_k |x_k-x_{k-1}| \right) $$
where $x_0=0$ and $a_k >0$ and $b_k$ are real numbers. Taking partial derivatives, it seems that the solution should have the form $$x_k = \pm \frac{b_k}{2a_k}$$ according to the sign of $x_k-x_{k-1}$, though this logic seems too naïve. Does anyone have any thoughts or references on how to approach this problem?
Let us understand a particular case yielding a complete solution (see eq. (3) and (4)) in the case $n=2$, giving a serious track for more general solutions.
I propose a geometrical view of this issue. I will not be fully rigorous, leaving some details to be filled.
Let $C_c$ be the curve with equation :
$2x^2+3y^2 -7 |x-y|=c\tag{1}$
Let us remark that $C_c$ is invariant by central symmetry $x \leftrightarrow -x, y \leftrightarrow -y$.
As can be seen on the following Desmos animation, according to the values of $c$ (move the slider !), curve $C_c$ is:
made of 2 connected arcs of twinned ellipses (through the central symmetry) that can be either the red curves or the blue curves,
split into 2 full small ellipses,
or is void (doesn't exist) when $c<c_0$, here for a $c_0 \approx -10.2$.
The red (resp. violet) curves correspond to the replacement in (1) of $|x-y|$ by $(x-y)$ (resp $(y-x)$).
Let us find $c_0$, the minimal value parameter $c$ can take, in order that the curve exists is nothing else that the global minimum of the left hand side of (1).
Let us consider the conic curve in one of two cases $x>y$ or $x<y$ ; here we will take $x>y$ (red curves) which means that equation (1) becomes
$$2x^2+3y^2-7x+7y-c=0\tag{2}$$
The matrix attached to (2) is
$$A=\pmatrix{2&0&-7/2\\0&3&7/2\\-7/2&7/2&-c}$$
This symmetric matrix defines an ellipse (one of the twined ellipses) iff it is positive definite, i.e., iff its eigenvalues are positive. The value of $c$ where $\det(A)=0$ provides this threshold.
$\det(A)=0$ is a first degree equation :
$$-6c-\frac{245}{4}=0 \ \iff \ c=c_0=-\frac{245}{24}=-10.2083,$$
as expected.
More generaly, for a matrix
$$A=\pmatrix{a_1&0&-b_1/2\\0&a_2&b_1/2\\-b_1/2&b_1/2&-c},$$
The minimum takes the following value :
$$c_0=-\frac{a_1+a_2}{a_1a_2}.\frac{b_1^2}{4}. \tag{3}$$
In the 3D case, we will get a "patchwork of ellipsoids" to which the same kind of treatment can be applied, with $4 \times 4$ determinants, etc.
Edit : Let us now look for the points where the minima takes place. They are plainly the center(s) of the twin ellipse(s).
If the equation of the ellipse is written under the matrix-vector form:
$$E(X)=X^TAX+2B^TX+C=0$$
The center is known to be the point $X$ solution to equation :
$$AX=-B$$
(in short : the derivative of $E, \ 2(AX+B)$ is set to zero).
This gives rise to equation :
$$\pmatrix{a_1&0\\0&a_2}\pmatrix{x\\y}=-\pmatrix{b_1\\-b_1}$$
giving :
$$\begin{cases}x=-b_1/(2a_1)\\y=b_1/(2a_2)\end{cases}\tag{4}$$
(and its symmetrical point with respect to the origin) which was your guess !