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0

It turns out that the problem isn't about precision, but about the setup of StopIntegration, where I set to be WhenEvent[z[ρ] == 10^-3, "StopIntegration"], WhenEvent[z[ρ] == 1.1` Sqrt[zt^2 - ρc^2], "StopIntegration"], however, if we have a closer look on the solution $z[ρ]$, we would find it's increasing at first, then decreasing until hitting the ρ-axis. ...


3

To expand a little on MarcoB's answer, if you want the plot to extend to the point {0., -1.99877}, you must tweak the plot options appropriately. Like so: With[{γ = 0.172969}, ParametricPlot[{hR[Tan[α], γ], Areac[Tan[α], γ]}, {α, FBc[γ], FAc[γ] - 0.001}, PlotRange -> {{0, 1.2}, {-10, -1}}, PlotRangePadding -> {{.1, Automatic}, Automatic}, ...


4

The point you want corresponds to the very first lowest value of $\alpha$; playing around with the values of your expression shows that very small changes in the value of $\alpha$ in this region translate to huge changes in the abscissa of the parametric points you want plotted. In other words, it is extremely easy for the adaptive internal plotting routines ...


3

This is not an answer to your question but the solution you get is not what you think it is: When you specify Method -> {"FiniteElement", "MeshOptions" -> {MaxCellMeasure -> 0.005}} You are solving this a 2D spatial problem but you want to solve this as a 1D spatial and time dependent problem. The easiest way to do so is by using: heatdist = ...


4

Decreasing the MaxCellMeasure to $5\times10^{-5}$ improves the quality of the solution considerably while still taking only a few seconds to calculate, so I would suggest that as a viable compromise: heatdist = NDSolve[{HEATIMP == 0, BCTot, ic}, u, {r, 1, 2}, {t, 0.00000, 1}, Method -> {"FiniteElement", "MeshOptions" -> {MaxCellMeasure -> 5*...


3

It seems like Plot always feeds the function a machine number near the left endpoint, and this machine number argument causes the message to be generated: Plot[Exp[-Exp[-Echo@x]], {x, -7, 0}, WorkingPrecision -> 20] -7 -6.99986 General::munfl: Exp[-1096.48] is too small to represent as a normalized machine number; precision may be lost. ...


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