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I am looking at a problem arising in finite element analysis. Let $K \subset R^2$ be a triangle, and assume that $\Phi := \{\phi_1, \ldots, \phi_N\}$ span some space $V(K)$. There is some function $v \in V(K)$; expressed in local coefficients, $v = \sum_i v_i \phi_i$.

I have a matrix $A$ (the local stiffness matrix: $A_{ij} = \int_K \nabla \phi_i \cdot \nabla \phi_j$), which is square, symmetric, positive semidefinite, but not invertible. I have a vector $\vec b$ (with $b_i = \int_K \phi_i$).

I am trying to solve the following quadratic programming problem:

Minimize $\frac{1}{2} (\vec x - \vec v)^\top A (\vec x - \vec v)$ for $\vec x$, subject to $\vec x^\top \vec b = \vec v^\top \vec b$.

Looking through quadratic programming texts, $\vec x$ exists and is unique when $A$ is invertible (which I obviously don't have). Can I prove existence and uniqueness some other way?

Thanks!

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  • $\begingroup$ I am going to make a brain-dead remark. If $A=0$, of course you don't have uniqueness. Same goes with the matrix $A=\begin{bmatrix} 0&0\\ 0&1 \end{bmatrix}$ on the constraint $\vec x \cdot (0,1)=0$. Why do you expect uniqueness in your case? $\endgroup$ – Giuseppe Negro Apr 6 '17 at 14:41
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Let $y = x-v$, your original problem becomes \begin{align*} \min\quad& \frac12y'Ay \\ \text{s.t}\quad & b'y = 0 \end{align*} If $y^*$ is a solution, then $y^*+z$ is also a solution as long as $z$ satisfies \begin{align} Az & = 0 \\ b'z & = 0 \end{align} That means $z$ is a eigenvector of $A$ with zero eigenvalue, and also $z$ is orthogonal to $b$. Only in the case $A$ has only one zero eigenvalue and $b$ is not orthogonal to the corresponding eigenvector, such $z$ doesn't exist.

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