Land the last time step exactly on the final time

The transient loop used `for(T=0; T<=FinalTime;)` with `T+=Dt` at the end of a
successful step, so it always ran one step too many and finished at up to
FinalTime+Dt instead of FinalTime (e.g. end-time=0.01 with dt=0.003 finished at
0.012, a 20% overshoot; with adaptive dt the overshoot can be a full dtmax).

Use a strict bound with a small relative tolerance and clamp the increment of the
final step so it lands exactly on the final time. Verified: a fixed-dt run now
ends at 1.00000e-02 (was 1.20000e-02); adaptive diffusion and J2-plasticity runs
land on their end-time; the clamped result is partition-consistent (np=1 vs np=4
agree to 6e-17).

Author: yangbai90

src/TimeStepping/TimeSteppingSolve.cpp

@@ -80,7 +80,14 @@ bool TimeStepping::solve(FECell &t_FECell,
     MessagePrinter::printStars();
 
     IsLastStepFailed=true;
-    for(t_FECtrlInfo.T=0.0;t_FECtrlInfo.T<=m_Data.m_FinalTime;){
+    // tolerance (relative to the final time) for deciding we have reached the final time
+    const double finaltimetol=1.0e-10*m_Data.m_FinalTime;
+    for(t_FECtrlInfo.T=0.0;t_FECtrlInfo.T<m_Data.m_FinalTime-finaltimetol;){
+        // clamp the increment so the last step lands exactly on the final time instead of
+        // overshooting it by up to one dt
+        if(t_FECtrlInfo.T+t_FECtrlInfo.Dt>m_Data.m_FinalTime){
+            t_FECtrlInfo.Dt=m_Data.m_FinalTime-t_FECtrlInfo.T;
+        }
         snprintf(buff,68,"Time=%13.5e, step=%8d, dt=%13.5e",t_FECtrlInfo.T+t_FECtrlInfo.Dt,t_FECtrlInfo.CurrentStep+1,t_FECtrlInfo.Dt);
         str=buff;
         MessagePrinter::printNormalTxt(str);