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Added J and Jt, untested as of yet.
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@@ -176,6 +176,56 @@ class RichardsProblem(Problem):
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return r, J
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def J(self, m, v, u=None):
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if u is None:
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u = self.field(m)
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Hs = u
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JvC = range(len(Hs)-1) # Cell to hold each row of the long vector.
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# This is done via forward substitution.
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temp, Adiag, B = self.diagsJacobian(Hs[0],Hs[1])
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Adiaginv = Solver(Adiag)
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JvC[0] = Adiaginv.solve(B*v)
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# M = @(x) tril(Adiag)\(diag(Adiag).*(triu(Adiag)\x));
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# JvC{1} = bicgstab(Adiag,(B*v),tolbcg,500,M);
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for ii in range(1,len(Hs)-1):
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Asub, Adiag, B = self.diagsJacobian(Hs[ii],Hs[ii+1])
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Adiaginv = Solver(Adiag)
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JvC[ii] = Adiaginv.solve(B*v - Asub*JvC[ii-1])
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if self.dataType is 'pressureHead':
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Jv = P.Q*np.concatenate(JvC)
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elif self.dataType is 'saturation':
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dT = self.empirical.moistureContentDeriv(np.concatenate(Hs[1:]))
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Jv = P.Q*dT*np.concatenate(JvC)
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def Jt(self, m, v, u=None):
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if u is None:
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u = self.field(m)
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Hs = u
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if self.dataType is 'pressureHead':
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QTz = P.Q.T*z;
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elif self.dataType is 'saturation':
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dT = self.empirical.moistureContentDeriv(np.concatenate(Hs[1:]))
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QTz = dT.T*P.Q.T*z
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# This is done via backward substitution.
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minus = 0
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BJtz = 0
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for ii in range(numel(Hs)-1,-1,-1):
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Asub, Adiag, B = self.diagsJacobian(Hs[ii-1], Hs[ii])
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#select the correct part of z
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# TODO: This might be easier by reshaping the array
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zpart = range((ii-2)*Adiag.shape[0], (ii-1)*Adiag.shape[0])
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AdiaginvT = Solver(Adiag.T)
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JTzC = AdiaginvT.solve(QTz[zpart] - minus)
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minus = Asub.T*JTzC # this is now the super diagonal.
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BJtz = BJtz + B.T*JTzC
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class Haverkamp(object):
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"""docstring for Haverkamp"""
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