As in the title, I have to find the zeros of a non-linear parametric equation with several parameters as functions of the parameters themselves. I tried using "Solve" in Mathematica and "solve" / "solveset" in Python Sympy. Sympy returns the empty set while even Mathematica crushed. Do you have any suggestion for different packages, functions or software?
$$( A + B)\cdot \frac{- C + e^{-\frac{D-E}{C}}\cdot(C + D - E)}{(D-E)^2} +B\cdot\frac{e^{-\frac{D-E}{C}}}{C}$$
subject to the following constraint: $A + B > 0$; $ C >0$; $E \ge 0$; $ D > E$.
I have to solve for $D$ and the problem is that it is not only in the exponential, therefore symbolic engines cannot solve for it. Any solutions?
That is already a first derivative, but we need also to consider the second derivative so clearly, we have to derivate that again. I tried with simpler functions involving the exponential and our software can provide a solution based on the Lambert W function, but for my case they cannot reach to a solution. Do you think that is just matter of computational power or there is no analyical solution?
$$(A+B)\left(\frac{-C+e^{-\frac{(D-E)}{C}}(C+D-E)}{(D-E)^2}+\frac{Be^{-\frac{(D-E)}{C}}}{C}\right)=0$$ $D\to E-Cx$: $$(A+B)\frac{-C+(C+Cx)e^x}{C^2x^2}+\frac{Be^x}{C^2}=0$$ $$-ACxe^x-BCxe^x+Bx^2e^x+ACe^x+BCe^x-AC-BC=0\tag{1}$$
We see, your equation can be transformed into a polynomial equation of more than one algebraically independent monomials ($x,e^x$), without a univariate factor. We therefore don't know how to rearrange the equation for $D$ by applying only finite numbers of elementary functions (operations) we can read from the equation.
We see, for algebraic $A,B,C$, equation (1) is an irreducible algebraic equation of $x$ and $e^x$ simultaneously. According to the theorems in Lin 1983 and Chow 1999, such kind of equations cannot have solutions except $0$ that are elementary numbers or explicit elementary numbers respectively if Schanuel's conjecture is true.
$$-(A+B)C+((A+B)C-(A+B)Cx+Bx^2)e^x=0$$ $$e^x=\frac{(A+B)C}{(A+B)C-(A+B)Cx+Bx^2}$$ $$\frac{(A+B)C-(A+B)Cx+Bx^2}{(A+B)C}e^x=1$$ $$((A+B)C-(A+B)Cx+Bx^2)e^x=(A+B)C\tag{2}$$
We see, because equation (2) is a polynomial equation of $e^x$, $x$ and $x^2$, the equation is not in a form for applying Lambert W, but Generalized Lambert W.
$$\left(x-\frac{1}{2}\frac{(A+B)C-\sqrt{(A+B)C((A+B)C-4B)C}}{B}\right)\left(x-\frac{1}{2}\frac{(A+B)C+\sqrt{(A+B)C((A+B)C-4B)C}}{B}\right)e^x=(A+B)C$$ $$x=W\left(^{\frac{1}{2}\frac{(A+B)C-\sqrt{(A+B)C((A+B)C-4B)C}}{B},\frac{1}{2}\frac{(A+B)C+\sqrt{(A+B)C((A+B)C-4B)C}}{B}}_{ };(A+B)C\right)$$ $$D=E-C\ W\left(^{\frac{1}{2}\frac{(A+B)C-\sqrt{(A+B)C((A+B)C-4B)C}}{B},\frac{1}{2}\frac{(A+B)C+\sqrt{(A+B)C((A+B)C-4B)C}}{B}}_{ };(A+B)C\right)$$
So we have a closed form for $D$, and the series representations of Generalized Lambert W give some hints for calculating $D$.
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[Mező 2017] Mező, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553
[Mező/Baricz 2017] Mező, I.; Baricz, Á.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)
[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018