I am not a user of mathematica (It is kind of new for me) so I am not sure if what I want to do is possible.
Imagine I have the following cauchy problem:
\frac{dw}{dt} = F(t, w(t)) = f1(w(t), t) + f2(w(t), t)
where I have split F into f1 and f2.
I now would like to solve the problem in a sequential way either:
case a:
\begin{align} \frac{dw^{*}}{dt} &= f_{1}( w^{*}(t),t) \quad t[t_{n}, t_{n+1}], \quad w^{*}(t_{n}) = w0 \end{align}
\begin{align} \frac{dw^{**}}{dt} &= f_{2}( w^{**}(t),t) \quad t[t_{n}, t_{n+1}], \quad w^{**}(t_{n}) = w^{*}(t_{n+1}) \end{align}
Namely, for a time step between t_n and t_n+1, I solved first w*’ =f1 using the initial boundary condition, and then I use the solution of z1 as boundary condtion to solve w**'2 = f2.
case b:
\begin{align} \frac{dw^{*}}{dt} &= f_{1}( w^{*}(t),t) \quad t[t_{n}, t_{n+1/2}], \quad w^{*}(t_{n}) = w0 \end{align}
\begin{align} \frac{dw^{**}}{dt} &= f_{2}( w^{**}(t),t) \quad t[t_{n}, t_{n+1}], \quad w^{**}(t_{n}) = w^{*}(t_{n+1/2}) \end{align}
\begin{align} \frac{dwb^{*}}{dt} &= f_{1}( wb^{*}(t),t) \quad t[t_{n+1/2}, t_{n+1}], \quad wb^{*}(t_{n+1/2}) = w^{**}(t_{n+1}) \end{align}
w* and wb* are actually the same process.
So my question is how can I do the Taylor series of the solution of case a and case b (i.e. taylor of w**(w*) and taylor of wb*(w**(w*))) in Mathematica if possible.
For case a, The taylor series at t_{n} with dt = t_{n+1} - t_{n} of w** is:
\begin{align} w^{**} = w0 + dt(f_{1}+f_{2})+\frac{dt^{2}}{2}&[\frac{\partial f1}{\partial w{*}}f_{1} + \frac{\partial f_{1}}{\partial t}+\frac{\partial f_{2}}{\partial w{**}}f_{2}+\frac{\partial f_{2}}{\partial t}]+dt^{2}\frac{\partial f_{2}}{\partial w{**}}f_{1} + O(dt^{3}) \end{align}
Everthing is evaluated at t_{n}, w*(t_{n}), the solution of case a can be found here:
http://www.sciencedirect.com/science/article/pii/S0898122105003317
Thanks