Variant of Wahba's problem

Question

Wahba's problem is the following:

$$\min_R \sum_{k=1}^K \|v_k - Rw_k\|^2$$ where $v_k$ and $w_k$ are arbitrary $3\times 1$ vectors, and $R$ is a rotation matrix (i.e., orthogonal with $\det(R)=1$). A variant of Wahba's problem is the orthogonal Procrustes problem.

I have a more complicated variant of Wahba's problem.

• I only study $K=3$.
• I have $w_k=R_{u_k}(\alpha_k)v_k$ where $R_{u_k}(\alpha_k)$ is a rotation around the axis $u_k$ with an angle $\alpha_k$.
• The axes $u_k$ are given in advance.
• I can optimize over $\alpha_k$.

Thus, my variant of Wahba's read $$J_{\min}=\min_{R, \alpha_k} \sum_{k=1}^3 \|v_k - RR_{u_k}(\alpha_k)v_k\|^2$$ where $v_k$ and $u_k$ are given. My claim is that there exist non-trivial solutions, i.e., $R\neq I$ and $\alpha_k\neq 0$ so that the minimum value is $J_{\min}=0$. Note that for $R=I, \alpha_k=0$, we trivially obtain $J_{\min}=0$. To be more precise, what I observed in matlab is that if I choose $v_k$ and $u_k$ randomly, I always have multiple solutions achieving $J_{\min}=0$.

Based on the (partial) answer below, I generated the following figure. I found 4 solutions. I plot for a special case here where two circles are parallel, and the third perpendicular to the other two.

Answer 1

This is not a rigorous proof, just some geometrical considerations. You can rephrase your problem to a problem on the unit sphere. The normalized $v_k$ build a triangle on the unit sphere. With the rotations you move the points on a circle on the unit sphere. You can find a minimum value if and only if the new points are congruent to the old ones. Hence your problem is equivalent of asking if we have 3 circles and a triangle with one point on each circle, if there is at least another triangle again with one point on each circle, that is congruent to the first one.

You can choose two nonintersecting circles and the endpoints of the closest connection as two of the starting points. Then you can not choose different points for the two as otherwise the distance would increase. Then the only other congruent triangle with two points fixed is by reflecting the third point along the fixed axis. Hence if the third circle does not contain the reflected point the trivial solution is the unique solution in this case.