471 5/1/09
EP -- Engineering Problem Solving II Exercise 20: Parabolic PDEs: Finite Difference Approach
One of the difficulties with the pdepe utility is that we can't see what's going on inside. We can solve this kind of problem using our finite difference representation in much the same way that we solve elliptic equations. To solve this problem numerically, we divide the slab into regions of width h and rewrite the equation in terms of the temperatures at each of these grid points and at each time step. We label each temperature using the notation Ti n , where i denotes the mesh point and n denotes the time step. One way of differencing the time derivative is: T Ti n +1 - Ti n (1) t t where t is the time step. This time derivative is forward-differenced, which is less accurate than a centrally-differenced representation. We'll retain this approximation for now, but correct it later so that the time as well as the spatial differencing is central. The simplest way to solve the equation is to write the spatial derivative term at the current time step, giving: n n 2T Ti +1 - 2Ti n + Ti -1 x2 h2 Substituting these differenced approximations into the partial differential equation, and assuming no heat source, we have: cp k
n n Ti n +1 - Ti n Ti +1 - 2Ti n + Ti -1 = t h2
Since there is only one term in this equation that represents a temperature at the next time step, that future temperature can be written in terms of temperatures at the current time step: n n Ti n +1 = Ti n (1 - 2 ) + (Ti +1 + Ti -1 ) where kt = c ph2 Because temperatures at time step n+1 are not coupled together spatially, we don't have a simultaneous system of equations. We can update each future temperature separately knowing the temperatures at time step n. This is called an explicit method. Unfortunately, there is a drawback to this method; it can be unstable if the time and step mesh spacing are not properly chosen. Specifically, a stable solution requires: 1 2
5/1/09 There is another technique that is unconditionally stable. This is the implicit technique, which is similar to the explicit technique described above except that the spatial derivative is written at the end of the step. That is, cp k
n+ n+ Ti n +1 - Ti n Ti +1 1 - 2Ti n +1 + Ti -1 1 = t h2
Now we have several terms that represent temperatures at time step n+1. Solving for Ti n +1 , we obtain:
n+ n+ Ti n + (Ti +1 1 + Ti -1 1 ) (2) 1 + 2 In this form, an iterative method is required, since there is coupling between spatial temperatures at the future time.
Ti n +1 =
Finally, we have yet to address the forward time-differencing on the previous page. If we combine the explicit and implicit methods we resolve that issue. Here we write the spatial derivative as a simple average of centrally-differenced expressions between the current and next time steps:
n n n+ n+ 2T 1 Ti +1 - 2Ti n + Ti -1 Ti +1 1 - 2Ti n +1 + Ti -1 1 + x 2 2 h2 h2
This provides an estimate for the second spatial derivative that is ce...