Symbolic entropy maximization in SymPy - sympy

Question

A simple problem of entropy maximization in statistical mechanics in Physics, is formulated as follows.
The goal is to maximize an entropy function (LaTeX is still missing in stackexchange?):
H = - sum_x P_x ln P_x
subject to the following constraints: the normalization constraint
1 = sum_x P_x
and the constraint of average energy
U = sum_i E_x P_x
where the index i runs over x=1,2,...,n. E_x represents the energy of the system when it is in microscopic state x and P_x is the probability for the system to be in the microscopic state x.
The solution to such a problem can be obtained by the method of Lagrange multipliers. In this context, it works as follows...
Firstly, the Lagrangian is defined as
L = H + a( 1 - sum_i P_x ) + b( U - sum_i P_x E_x )
Here, a and b are the Lagrange multipliers. The Lagrangian L is a function of a, b and the probabilities P_x for x=1,2,...,n. The term a( 1 - sum_x P_x ) correspond to the normalization constraint and the term b( E - sum_x P_x E_x ) to the average energy constraint.
Secondly, the partial derivatives of L with respect to a, b and the P_x for the different x=1,2,...,n are calculated. These result in
dL/da = 1 - sum_x P_x
dL/db = E - sum_x E_x P_x
dL/P_x = dH/P_x - a - b E_x
= - ln P_x - 1 - a - b E_x
Thirdly, we find the solution by equating these derivatives to zero. This makes sense since there are 2+n equations and we have 2+n unknowns: the P_x, a and b. The solution from these equations read
P_x = exp( - b E_x ) / Z
where
Z = sum_x exp( - b E_x )
and b is implicitly determined by the relation
E = sum_x P_x E_x = ( 1 / Z ) sum_x exp( -b E_x ) E_x
Now that the "mathematical" problem is defined, lets state the "computational" problem (which is the one I want to ask here).
The computational problem is the following: I would like to reproduce the previous derivation in Sympy.
The idea is to automatize the process so, eventually, I can attack similar but more complicate problems.
I already made certain progress. Still, I think I haven't used the best approach. This is my solution.
# Lets attempt to derive these analytical result using SymPy.
import sympy as sy
import sympy.tensor as syt
# Here, n is introduced to specify an abstract range for x and y.
n = sy.symbols( 'n' , integer = True )
a , b = sy.symbols( 'a b' ) # The Lagrange-multipliers.
x = syt.Idx( 'x' , n ) # Index x for P_x
y = syt.Idx( 'y' , n ) # Index y for P_y; this is required to take derivatives according to SymPy rules.
>>> P = syt.IndexedBase( 'P' ) # The unknowns P_x.
>>> E = syt.IndexedBase( 'E' ) # The knowns E_x; each being the energy of state x.
>>> U = sy.Symbol( 'U' ) # Mean energy.
>>>
>>> # Entropy
>>> H = sy.Sum( - P[x] * sy.log( P[x] ) , x )
>>>
>>> # Lagrangian
>>> L = H + a * ( 1 - sy.Sum( P[x] , x ) ) + b * ( U - sy.Sum( E[x] * P[x] , x ) )
>>> # Lets compute the derivatives
>>> dLda = sy.diff( L , a )
>>> dLdb = sy.diff( L , b )
>>> dLdPy = sy.diff( L , P[y] )
>>> # These look like
>>>
>>> print dLda
-Sum(P[x], (x, 0, n - 1)) + 1
>>>
>>> print dLdb
U - Sum(E[x]*P[x], (x, 0, n - 1))
>>>
>>> print dLdPy
-a*Sum(KroneckerDelta(x, y), (x, 0, n - 1)) - b*Sum(KroneckerDelta(x, y)*E[x], (x, 0, n - 1)) + Sum(-log(P[x])*KroneckerDelta(x, y) - KroneckerDelta(x, y), (x, 0, n - 1))
>>> # The following approach does not work
>>>
>>> tmp = dLdPy.doit()
>>> print tmp
-a*Piecewise((1, 0 <= y), (0, True)) - b*Piecewise((E[y], 0 <= y), (0, True)) + Piecewise((-log(P[y]) - 1, 0 <= y), (0, True))
>>>
>>> sy.solve( tmp , P[y] )
[]
>>> # Hence, we try an ad-hoc procedure
>>> Px = sy.Symbol( 'Px' )
>>> Ex = sy.Symbol( 'Ex' )
>>> tmp2 = dLdPy.doit().subs( P[y] , Px ).subs( E[y] , Ex ).subs( y , 0 )
>>> print tmp2
-Ex*b - a - log(Px) - 1
>>> Px = sy.solve( tmp2 , Px )
>>> print Px
[exp(-Ex*b - a - 1)]
Is there a "better" way to proceed? In particular, I don't like the idea of substituting P[y] with Px and E[y] with Ex in order to solve the equations. Why the equations cannot be solved in terms of P[y]?