Instructor's solution for the windbreak problem (transect along z/H=0.4).
Instructor's solutions for lab 2:
For gridlengths Δ=(0.25,0.05), the numerical solution should give temperature peaks (at the origin) of respectively T(0,0) ≈(0.375,0.636)o. The analytic solution gives T(0,0)=0.6357o. As the number of terms included in the analytical solution increases to infinity, ever shorter waves (higher wavenumbers) contribute and the analytic solution captures the "peakiness" of the delta function. In the numeric solution, the equivalent progression towards the true solution is made by refining the resolution (reducing the gridlength Δ), so that the heat source, ideally a delta function (if seen in cross section through y=0), is represented by an increasingly narrow (but higher) triangle (always preserving unit area).