# Optimization problems on the circle

Consider the optimization problem

$$\min_{x \in \mathbb{R}^2} x^{\top} P x + q^{\top} x$$

subject to:

$$A x = b, \ x \in X, \ x_1^2 + x_2^2 = 1$$

where $X$ is compact and convex.

Then consider the optimization problem

$$\min_{x \in \mathbb{R}^2} x^{\top} P x + q^{\top} x - \lambda ( x_1^2 + x_2^2)$$

subject to:

$$Ax=b, \ x\in X, \ x_1^2 + x_2^2 \leq 1$$

I am wondering if for $\lambda>0$ sufficiently large the optimal solution of the second problem approximates arbitrarily close the optimal solution of the first one. If so, I wonder if there exists a large, finite, $\lambda$ such that the two solutions coincide.

"I am wondering if for $\lambda$ sufficiently large the optimal solution of the second problem approximates arbitrarily close the optimal solution of the first one."
Consider the problem $$\min_{x \in \mathbb{R}^2} x^{\top} P x + q^{\top} x ~~~~{\rm s.t.}~~ A x = b, \ x \in X.$$ If $x^\star$ is the solution and $(x_1^\star)^2 + (x_2^\star)^2 < 1$, then it is impossilbe to "force" $x_1^2 + x_2^2 = 1$ by introducing an Langrange-multiplier $\lambda$. In this case, introducing $\lambda$ would lead to even smaller values for $x_1^2 + x_2^2$.
However, if $(x_1^\star)^2 + (x_2^\star)^2 > 1$, an appropriate Lagrange-multiplier leads to $x_1^2 + x_2^2 = 1$.
• Thanks a lot for the answer. Do you think that if the solution is such that $x_1^2 + x_2^2 < 1$, some $\lambda<0$ may "force" $x_1^2 + x_2^2 = 1$? – user693 Aug 11 '14 at 12:56