On 12/20/18 1:15 AM, miffy....@gmail.com wrote:
> I'm sorry to reply you in a old question mail, I want to know when rhs
> depend on the solution i.e., rhs is a polynomials of the solution, should we
> handle it with test function to get a bilinear form while we construct the
> weak form?
> The description may be confused, for example consider the rhs as f(u,x,t)
> in Step-26 (the heat euqation,the original rhs is f(x,t)), f(u,t) may be
> u^2+u^3+u^4, how to deal with such a problem?
> I have read the toturials to find the answer, but the rhs always only
> depend on x,y,z and Step-15 can't answer my question(maybe) .
No, step-15 is exactly what you need :-) There, the right hand side does
depend on u in some (more complicated) way. The only different is that in
step-15, the iterations we do are Newton steps, whereas what you want to do is
a time iteration; but the principle is still the same.
Can you explain why step-15 doesn't answer your questions?
Best
W
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Wolfgang Bangerth email: bange...@colostate.edu
www: http://www.math.colostate.edu/~bangerth/
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