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发表于 2010-9-22 16:54:32
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来自 美国
看了tonnyw的回复,顿时豁然开朗。不知能否按如下方法解答:
先取v的初始值,带到第一个方程,解出u;
然后带到第二个方程,解出v;
再带到第一个方程,解出u,如此反复,直到前后两次u和v的差值向量范数小于某个 ...
panzichao 发表于 2010-9-22 10:32
No. That's not what I mean. First, assume the [c] matrix is not a function of (u, v), you can put
| k1 0 |
[c] =| k2|
You can get the finite element solution for this, which will make sure that your temporal discretization and spatial discretization is okay.
Then you make [c] matrix the function of (u,v) to sort out the iteration solver.
What I suggest is that you tackle this problem step by step. |
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