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What advantages does Fortran Calculus provide for PDEs?
1. Computer size requirement is lowered to a small (near PC) size. Supercomputers are -not- necessary.
2. Solution methods are meshless! Faster since there is No looping as done in todays standard solver methods for PDEs.
3. Improved accuracy with Automatic Differentiation.
4. Partial variables are available to print; e.g. Ux, Uy, Uz, etc. This allows one to plot isolines of their partials.
Oakland, CA Hills fire of 1991
Example PDE problem
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This fire covered 3 miles along a ridge line and 1 mile inland. The fire lasted several days. A group [ 1 ] at Lawrence Livermore National Laboratory (LLNL) built a PDE model and solved their model to simulate the fire. The simulation required several computers running in parallel for several hours. The mesh size was a big problem. They could only show/calculate a point every 5 to 10 feet along the fire.
Using Fortran Calculus this fire could be calculated every micrometer. The execution time is less relative to LLNL's but would increase if step size was reduced a thousand times otherwise no other problem.
Solution of either LLNL or FC could be shown on isoline plots. We suggest trying to see if isolines of partials would help one understand where the fire was heading. How about the 2nd order partials? Would their plots be helpful? If you need a partial say Uyy it can be calculated using Uyy = partial( Uy, y).
FC provides savings in programming time, computer size & run time for executions, and accuracy improvements.
[Note: At present, PDEs must be converted to a system of ODEs in order for Fortran Calculus to solve them. When there is enough interest in solving PDEs, we will look into other ways of doing it.]
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