Lie Group hint for SymPy

Hi, this week started with two of my previous PR’s getting merged (finally).

1.  The heuristics PR –
2. The variable coefficient PDE –

I started working on the integration of the infinitesimals to the present dsolve architecture. Before telling about the issues I ran into (a number of them actually), let me explain the algorithm in a few lines.

As you know (If you have been following my blog),the past few weeks my focus was on solving this monster PDE.
\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}
Why? Well, without going too much into detail, the solution to this PDE, \xi and \eta, give the infinitesimals of the ODE \frac{dy}{dx} = h(x, y). After getting the infinitesimals, this method is adopted.
1] One has to solve the Partial Differential Equations (again?)
a] \xi\frac{\partial r}{\partial x} + \eta\frac{\partial r}{\partial y} = 0
b] \xi\frac{\partial r}{\partial x} + \eta\frac{\partial s}{\partial y} = 1
2] Now one knows r and s in terms of x and y, doing \frac{ds}{dr} = \frac{\frac{\partial s}{\partial x} + \frac{\partial s}{\partial y}*h}{\frac{\partial s}{\partial x} + \frac{\partial s}{\partial y}*h} and converting the R.H.S in terms of r and s, reduces into a quadrature, which can be solved quite easily with the ode_separable hint
3] After solving the ODE, it can be converted back into the original coordinates.

On a scale of optimism to pessimism, I am somewhere in between a realist and a pessimist, and I have to admit I was slightly disappointed with the effectiveness of the hint, since I was running into various issues, with a few ODEs that I had tested. These were some of them that I had identified.

1] Problem with Integral: I was testing an ODE in which, r = x, and I had to integrate a huge expression, which couldn’t be integrated, something like \frac{1}{\sqrt{a0 + a1*r + a2*r^{2} + a3^r{3}}}, it gave me an output of the form Integral, and when I substitued r as x, it gave me a definite value, this is because doing subs on an Integral object, doesn’t substitute for the variable with respect to which you are integrating to. I was pointed to this issue by Aaron, which I haven’t looked at yet.

2] Assumptions: When I was applying to SymPy for Google Summer of Code, I saw this awesome proposal by Tom Bachmann, which seemed Greek and Latin to me then (Some parts of it does still, but never mind). The bottom line, is \frac{dr}{ds} simplifies to a quadrature, sometimes only when, there are certain assumptions on r and s. Lets take a random example, suppose \frac{dr}{ds} = log{r^{s}} - s*log{r} reduces to zero only when r > 0, Since the input variable is x, giving assumptions on x, doesn’t seem to affect the assumptions on r.

3] There are some cases when the final expression, cannot be solved explicitly for y like this . I think the best way would be to return it as it is.

4] Recursion: Take the case of this wonderful ODE, x^{2}*(-f(x)^{2} + \frac{df}{dx})- a*x^{2}*f(x) + 2 - a*x, calculating the infinitesimals, give me \xi = x^{2} and \eta = a*x^{2}*f(x) - 2 + a*x + x^{2}*(-f(x)^{2}). Since the first step in solving the PDE, \frac{dy}{dx} = \frac{\eta}{\xi} , it gives the same ODE again.

Apart from this I believe rest of the code is good.

P.S: First ten days of college is over. There have been huge disappointments, but apart from that I have enjoyed either roaming outside, and working on my SymPy project, and I have done nothing other than that.


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