Question

使用FiPy求解圆柱几何中扩散方程的稳态解与从其他软件（例如，Mathematica）获得的解有很大不同。

等式是：

$ 0 = \ frac {1} {r} \ frac {d} {dr} \ left（\ frac {r} {T ^ {1/2}} \ frac {dT} {dr} \ right）+ cte * T ^ {3/2} $

这意味着，通过使用CylindricalGrid1D网格，我们可以将公式写为：

mesh = CylindricalGrid1D(nr=100, dr=0.01, origin=0.0)
T = CellVariable(name='temperature', mesh=mesh, hasOld=True)
r = mesh.cellCenters()

#BC's
T.constrain(0., mesh.facesRight)
T.faceGrad.constrain(0.,mesh.facesLeft)

#initial temperature profile
T.setValue( 1-r**2)

eq = 0 == DiffusionTerm( coeff=T**(-1/2), var=T) + 20*ImplicitSourceTerm(coeff=T**(1/2), var=T)

viewer = Viewer(vars=T)
eq.solve()

viewer.plot()
raw_input(" Press <enter> to proceed...")

在这里，我设置cte = 20，但是无论此值是多少，问题仍然存在。我在左边得到解决方案，而Mathematica提供的解决方案在右边：

plots

然后，我尝试按照建议对此类非线性方程进行扫描。因此，我没有eq.solve()，而是这样做了：

current_residual = 1.0e100
desired_residual = 1e-5

while current_residual > desired_residual:

    current_residual = eq.sweep()
    T.updateOld()

但是我得到了错误：

/home/antonio/.local/lib/python2.7/site-packages/fipy/solvers/scipy/linearLUSolver.py:66: RuntimeWarning: overflow encountered in square
  error0 = numerix.sqrt(numerix.sum((L * x - b)**2))
/home/antonio/.local/lib/python2.7/site-packages/fipy/solvers/scipy/linearLUSolver.py:71: RuntimeWarning: overflow encountered in square
  if (numerix.sqrt(numerix.sum(errorVector**2)) / error0)  <= self.tolerance:
/home/antonio/.local/lib/python2.7/site-packages/fipy/solvers/scipy/linearLUSolver.py:71: RuntimeWarning: invalid value encountered in double_scalars
  if (numerix.sqrt(numerix.sum(errorVector**2)) / error0)  <= self.tolerance:
/home/antonio/.local/lib/python2.7/site-packages/fipy/tools/numerix.py:966: RuntimeWarning: overflow encountered in square
  return sqrt(add.reduce(arr**2))
/home/antonio/.local/lib/python2.7/site-packages/fipy/solvers/scipy/linearLUSolver.py:58: RuntimeWarning: overflow encountered in multiply
  b = b * (1 / maxdiag)
Traceback (most recent call last):
  File "stack.py", line 26, in <module>
    current_residual = eq.sweep()
  File "/home/antonio/.local/lib/python2.7/site-packages/fipy/terms/term.py", line 254, in sweep
    solver._solve()
  File "/home/antonio/.local/lib/python2.7/site-packages/fipy/solvers/scipy/scipySolver.py", line 61, in _solve
    self.var[:] = numerix.reshape(self._solve_(self.matrix, self.var.ravel(), numerix.array(self.RHSvector)), self.var.shape)   
  File "/home/antonio/.local/lib/python2.7/site-packages/fipy/solvers/scipy/linearLUSolver.py", line 64, in _solve_
    permc_spec=3)
  File "/usr/lib/python2.7/dist-packages/scipy/sparse/linalg/dsolve/linsolve.py", line 257, in splu
    ilu=False, options=_options)
RuntimeError: Factor is exactly singular

最后，我通过使用变量V = T ^ {1/2}以等价形式重写了初始方程。很容易看到，随着V方程变为

$ 0 = \ frac {1} {r} \ frac {d} {dr} \ left（r \ frac {dV} {dr} \ right）+ \ frac {cte} {2} V ^ 3 $ < / p>

所以我然后使用了代码：

mesh = CylindricalGrid1D(nr=100, dr=0.01, origin=0.0)
V = CellVariable(name='V', mesh=mesh, hasOld = True)
r = mesh.cellCenters()

#BC's
V.constrain(0., mesh.facesRight)
V.faceGrad.constrain(0.,mesh.facesLeft)

#initial V profile
V.setValue( 1-r**2)

eqV = 0 == DiffusionTerm( coeff=1., var=V) + 20*0.5*ImplicitSourceTerm(coeff=V*V, var=V)

T = V*V
viewer = Viewer(vars=T)

eqV.solve()

viewer.plot()
raw_input(" Press <enter> to proceed...")

和获得的轮廓非常相似，但是第一个FiPy解决方案或Mathematica的y轴值都不相同！扫描产生的错误与以前相同。

Answer 1

我不认为这个问题除了T = 0之外没有其他解决方案。此外，对于初始条件和/或cte的不同值，该解决方案似乎不稳定。考虑到T形式的方程在T = 0时具有无限扩散性，因此这种不稳定性并不完全令人惊讶。

我怀疑Mathematica在执行第一组图形中的工作大致与FiPy所做的一样，这是显示此非线性问题的第一步。这不是答案；只是第一个猜测。但是，无论是解析还是数字，我对使用Mma解决PDE一无所知。

V解决方案经过一扫后的图看起来不一样，顺便说一句，因为您没有调整初始条件。应该是：

V.setValue( numerix.sqrt(1-r**2))

使用FiPy求解圆柱坐标中的扩散方程不正确

1 个答案: