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Question

Math 6410 • ODE

Spring 2026 • Question 2

Consider the gradient flow defined by the ODE

\[ \dot {x}(t)=-\nabla f(x(t)),\quad \text {with }x(0)=x_0, \]

where \(f(x)=x^{\top }Qx\) with \(Q\in \mathbb {R}^{n\times n}\), \(n\ge 2\) being a symmetric matrix.

  1. Assume \(Q\) is positive semi-definite, show that \[ f(x(t))\le \frac {C\|x_0\|_2^2}{t} \] for some constant \(C\).
  2. Assume \(Q\) is positive definite, show that \[ f(x(t))\le \lambda _{\max }e^{-4\lambda _{\min }\cdot t}\|x_0\|_2^2, \] where \(\lambda _{\min }\) and \(\lambda _{\max }\) are the smallest and largest eigenvalues of \(Q\), respectively.

Raw LaTeX

Consider the gradient flow defined by the ODE

\[
\dot{x}(t)=-\nabla f(x(t)),\quad\text{with }x(0)=x_0,
\]

where \(f(x)=x^{\top}Qx\) with \(Q\in\mathbb{R}^{n\times n}\), \(n\ge 2\) being a symmetric matrix.

\begin{enumerate}

\item
Assume \(Q\) is positive semi-definite, show that
\[
f(x(t))\le \frac{C\|x_0\|_2^2}{t}
\]
for some constant \(C\).



\item
Assume \(Q\) is positive definite, show that
\[
f(x(t))\le \lambda_{\max}e^{-4\lambda_{\min}\cdot t}\|x_0\|_2^2,
\]
where \(\lambda_{\min}\) and \(\lambda_{\max}\) are the smallest and largest eigenvalues of \(Q\), respectively.

\end{enumerate}