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After doing some research, I've found that most of the literature on operator splitting methods (e.g. Strang Splitting, Fractional Step, etc.) are specifically designed for a standard problem type of the form:

$$u'=(A+B)u.$$

In my application, I have a slightly different form:

$$Mu'=(A+B)u$$

where $M$ is not the identity operator. Can the same methodologies be applied when an operator is applied to the $u'$ term? Are there any other numerical methods to handle this case specific case there $M$ is not the identity operator? In particular, I'm looking at a case where $M$ is singular.

Update

To put my particular problem into greater context, my operator is

$$M=\left[\begin{array}{cc} 0 & 0 \\ I & I\end{array}\right]$$

and

$$u=\left[\begin{array}{c} u_1 \\ u_2 \end{array}\right]$$

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    $\begingroup$ It would help if you gave us an example of what you have in mind. The situation is different if $M$ is zero for a certain block of elements of $u$. This is then often called a "differential-algebraic equation". The time dependent Stokes equations are an example of a DAE, and it leads to all sorts of complications. $\endgroup$ – Wolfgang Bangerth Jun 14 '13 at 8:48
  • $\begingroup$ @WolfgangBangerth: Indeed I am working on the case where $M$ is zero for some blocks. I just updated the question with my particular matrix $M$. $\endgroup$ – Paul Jun 14 '13 at 11:55
  • $\begingroup$ So you can subdivide $u(t)=\{a(t),b(t)\}$ where $a,b$ have the same size? The first set of equations then reads $(A_{11}+B_{11})a(t) + (A_{12}+B_{12})b(t)=0$ which I hope you can solve for either $a(t)$ or $b(t)$ to eliminate one of the two variables from the equations. $\endgroup$ – Wolfgang Bangerth Jun 14 '13 at 20:32
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    $\begingroup$ I should mention that this is the uncommon case. In most DAEs, the number of variables in $a,b$ is different and you can't typically express one of the two in terms of the other because $A_{12}+B_{12}$ is not a square matrix and typically $A_{11}+B_{11}$ is square but singular. $\endgroup$ – Wolfgang Bangerth Jun 14 '13 at 20:34

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