Let $g_{\lambda}$ be a one parameter family of Riemannian metrics, which are complete and with bounded curvature, on the unit disk, depending smoothly on the parameter $\lambda$. Let $\Delta_{\lambda}$ be the corresponding family of Laplacian operators. Consider the following family of non linear parabolic equations $$\frac{\partial u}{\partial t}=e^{u}\Delta_{\lambda}ue^{u}R_{\lambda}+r,$$ where $R_{\lambda}$ is the curvature of $g_{\lambda}$ and $r$ is a constant. If we have a solution $u_{\lambda}(x,t)$ with $u_{\lambda}(\cdot,0)\equiv0$, is it true that $u_{\lambda}$ depends smoothly on $\lambda$?
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$\begingroup$ Most likely yes. Calculate the parabolic PDE satisfied by the difference quotient with respect to $\lambda$. That has a solution for some time interval with zero initial data. Use a priori estimates ($L^2$ energy estimates) to show that the norm of the difference quotient is bounded, so the limit exists and therefore the derivative of $u$ with respect to $\lambda$ exists and satisfies a parabolic PDE. Repeat to get estimates for higher derivatives. $\endgroup$ – Deane Yang Nov 30 '17 at 20:37

$\begingroup$ The argument is essentially the same as for ODEs. $\endgroup$ – Deane Yang Nov 30 '17 at 20:38

$\begingroup$ Hi Deane. Thank you for your answer, it was helpful (more or less what I was thinking about). Do you know any reference for the continuous dependence of the solution on the parameter $\lambda$? I don't know how to prove that, and I can't find any proof in the literature. $\endgroup$ – Richard Muniz Jun 28 '18 at 13:38

$\begingroup$ It's hard to find general theorems about PDEs, because usually in each case, you want specific estimates beyond what a general theorem would say. For example, usually you don't just want to know that a solution depends smoothly on a parameter, but you also want to have some explicit estimates on the dependence. In your case, the coefficients of the PDE depend on the metric and its dervatives (the Christoffel symbols). So the difference between the two solutions will depend on the difference between the two metrics and the between the corresponding Christoffel symbols, which is a tensor. $\endgroup$ – Deane Yang Jun 28 '18 at 15:36