The first heuristic

My GSoC work had been on and off, and I’m nowhere near the required 40 hours for the past one week, due to various reasons. However this is a report of what I had accomplished during the past week.

The lie group solver for solving Ordinary Differential Equations of the first order, requires as one of the steps, (the major actually, in fact the first six weeks of my project) to solve the following Partial Differential Equation \frac{\partial \eta}{\partial x} + (\frac{\partial \eta}{\partial y} - \frac{\partial \xi}{\partial x}) * h(x, y) - \frac{\partial \xi}{\partial y}*h(x,y)^{2} - \xi*\frac{\partial h}{\partial x} - \eta*\frac{\partial h}{\partial y} for \xi(x, y) and \eta(x, y)where \frac{dy}{dx} = h(x, y). It is pretty strange (atleast I find it) that solving a first order ODE requires solving a much complicated Partial Differential Equation , having two functions in two variables. However that is how things are, and as for how this PDE helps in solving a first order ODE, my blog posts would slowly lead up to that, much like how seemingly unrelated events lead to an awesome climax in a Hollywood thriller flick.

For now the focus remains on solving the huge PDE that I had mentioned. It looks almost impossible to do so at first sight. However an intelligent method to solve it, would be to guess solutions for \xi(x, y) and \eta(x, y) and substitute it back in the PDE and check how much it simplifies (almost like how I used to ‘intelligently’ guess solutions to my Organic Chemistry objective examinations).

The first heuristic and the easiest (according to Maple) involves assuming one of \xi(x,y) and \eta(x,y) to be zero and the other to be a function of x or y. This gives us four possibilities

1.  \xi(x,y)=0 and \eta(x, y)=f(x)
2.  \xi(x,y)=0 and \eta(x, y)=f(y)
3.  \eta(x,y)=0 and \xi(x, y)=f(x)
4. \eta(x,y)=0 and \xi(x, y)=f(y)

I would explain how the first case works and the remaining would be self-inferred. Putting \xi as 0 and \eta as f(x) , the PDE reduces to \frac{df}{dx} - f*\frac{\partial h}{\partial y} . Clearly to solve for f easily, \frac{\partial h}{\partial y} should be a function of x for the ODE to be solved easily , since y is dependent on x. The same logic applies to the other four assumptions. I also wrote a small function called checkinfosol that checks if the infinitesimals actually satisfy the PDE. Works fine for now!. On a side note, I just began working on the second heuristic and it seems that they are getting more challenging and I cannot afford to slack any more.

For those who are interested in the source-code, .

P.S : Sorry, my posts are getting really TL;DR. Too much of this and I’ll consider renaming my blog TL;DR


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