# numerical solution to pde on an ellipse

Looking for advice on discretization (preferably finite difference) schemes for pdes on curves in general, but in this case it is an ellipse (so given by $$(a\cos(r), b\sin(r)$$). The problem is the advection equation

$$\frac{ \partial u_s}{\partial t}+ \frac{ \partial u_s}{\partial s}=0,$$ with $$s$$ corresponding to arc length and the initial condition $$u(s,0)=\cos^2(2\pi s/L)$$. The solution should be $$u(s,t)=\cos^2(2\pi(s-t)/L)$$. I don't want to do this using the closest point method, I want to keep the curve parameterized and just discretize on the curve but I'm not sure how to proceed.

• Why is the curve shape important at all? Isn't it the same as solving on a straight 1D periodic domain? Sep 9 '20 at 22:48
• I agree with @MaximUmansky. I think that it's the same as a 1d periodic domain for which you can write down the analytic solution (which I believe you've actually already done, though with a typo -- the time is missing on the lhs). Sep 10 '20 at 2:45
• Thank you, I see now that the solution is just the same as substituting say $s=x$ for a normal pde. I guess my main problem is the overall formatting - I originally had the general advection equation on a manifold, which from my understanding for a curve simplifies to the derivative (or 2nd derivative for the laplace beltrami operator) with respect to arc length. I'm wondering generally how I can relate the solution to the curve x(r), y(r), so that I can visualize the solution on the curve rather than the solution just being dependent on arc length. I'm also Sep 10 '20 at 12:54
• Assuming you have the solution as a function of the arc length $s$, and you have the parametric representation of the curve $x(s), y(s)$, you have all what's needed to plot the solution as a function of the spatial coordinates. Sep 10 '20 at 14:54
• Would you be able to give an more insight as to how I would do this given x(s), y(s)? I have the solution u(s,t), and I want to plot the solution in cartesian coordinates along with the curve. Sep 10 '20 at 17:33