3
$\begingroup$

My question is the following:

What is the minimum regularity for a continuous loop $\gamma: S^1 \rightarrow M$ in a Riemannian manifold $M$ to have short-time existence for the harmonic map flow in a suitably sense?

It is well-known that if you start with a $C^1$ loop $\gamma$, then the harmonic map flow $$ \frac{\mathrm{d}}{\mathrm{d} t} \gamma_t = \frac{\nabla}{\partial \theta} \frac{\partial}{\partial \theta}{\gamma}_t(\theta), ~~~~~ \gamma_0(\theta) = \gamma(\theta)$$ exists in the classical sense, and $\gamma_t$ is smooth for $t \rightarrow 0$.

It is also clear that if the target is $\mathbb{R}^n$, then the differential equation is linear, so it makes sense to say when a solution solves the equation in the distributional sense, and the result is that solutions to any distributional initial value exist and become smooth immediately. In the general case that $M$ is curved, the equation will be non-linear, and it is harder to say what weak solutions are.

So precisely: Is there a good notion of a weak solution of the harmonic map flow for, say, $C^\alpha$ functions $\alpha > 0$? If so, do solutions exist for short time and become immediately smooth? And do solutions depend continuously on initial data?

$\endgroup$
  • 2
    $\begingroup$ The obvious thing to try is to solve the equation for a family of smooth initial data, converging to $\gamma$ and prove a priori estimates that imply that there exists a subsequence, which converges to a solution with initial data $\gamma$. The a priori estimates will probably imply uniqueness, too. $\endgroup$ – Deane Yang Nov 20 '17 at 15:14
  • 1
    $\begingroup$ It also appears to me that a contraction mapping argument using the solution operator for the linearized equation should work, too. $\endgroup$ – Deane Yang Nov 20 '17 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.