Solution of Quasi Polynomial $x=-a-\frac{a}{2}e^{-3x}-ae^{-2x}-\frac{3a}{2}e^{-x}$

Question

I am trying to solve a multiple time delay differential equation analytically. The characteristic polynomial is, $$x=-a-\frac{a}{2}e^{-3x}-ae^{-2x}-\frac{3a}{2}e^{-x}$$ where $a$ is an arbitrary positive real value.

I tried to plot the solution of $x$ with smallest real part numerically for different value of $a$ (dominant pole). The shape of the plot is similar to the Lambert-W function. And there is critical value $a_c$, that there is no real solutions for $a>a_c$.

Is the characteristic equation have the analytical solution? Any method to study the critical value $a_c$ analytically?

Answer 1

As @IV_ explained, you cannot obtain an explicit expression for $a_c$. However, you can generate more and more rational approximations of the corresponding $x_c$ which is the maximum value of the function $$f(x)=-\frac{2 x e^{3 x}}{3 e^{2 x}+2 e^{3 x}+2 e^x+1}$$ The first derivative cancel at the zero of $$g(x)=1+3 x+2 e^x (2 x+1)+3 e^{2 x} (x+1)+2 e^{3 x}$$ If we use Newton method with $x_0=0$, we shall have an underestimate of the solution because $$g(0)\times g''(0)= 8\times 52=416 >0$$ (This is by Darboux theorem). But we can expand as a Taylor series $$g(x)=8+24x+\sum_{n=2}^\infty\frac {\left(3\times 2^{n-1}+4\right) n+3\times 2^n+2\times 3^n+2 }{ n! }\,x^n$$

Truncating to some order and using series reversion $$x=\sum_{n=1}^\infty a_n \, t^n \qquad \text{where} \qquad t=\frac{g(x)-8}{24} $$ Since we look for $g(x)=0$, this generates for $x_c$ the sequence $$\left\{-\frac{1}{3},-\frac{12}{23},-\frac{69}{122},-\frac{244}{435},- \frac{2175}{3898},-\frac{93552}{167611},\cdots\right\}$$ which are closer and closer to the "exact" solution given by Newton method.

Thanks to the $ISC$, a good approximation could be $$x_c \sim -\frac{13 \sqrt{21}-\sqrt{14}}{100} $$ (absolute error : $1.72 \times 10^{-7}$).

Using this approximation $a_c=0.0597423$ while the "exact" solution is $\cdots$ the same.

Approximate solution for $x$

For an approximation of the first negative root of $$a=-\frac{2 x e^{3 x}}{3 e^{2 x}+2 e^{3 x}+2 e^x+1}$$ building $P_n$ (the $[n,n]$ Padé approximant of the rhs), makes the problem simple

For example $$P_2=-\frac{23 x^2+30 x}{35 x^2-58 x+120}$$ gives $$x_{(2)}\sim \frac{29a-15+\sqrt{-3359 a^2-3630 a+225}}{35 a+23}$$

Using this approximation and expanding as a series around $a=0$ gives $$f(x_{(2)})=a+\frac{1504 }{45}a^5+O\left(a^6\right)$$

A few values

$$\left( \begin{array}{ccc} a & \text{estimate} & \text{solution} \\ 0.005 & -0.020524 & -0.020524 \\ 0.010 & -0.042202 & -0.042202 \\ 0.015 & -0.065228 & -0.065228 \\ 0.020 & -0.089852 & -0.089851 \\ 0.025 & -0.116411 & -0.116408 \\ 0.030 & -0.145374 & -0.145362 \\ 0.035 & -0.177426 & -0.177392 \\ 0.040 & -0.213646 & -0.213551 \\ 0.045 & -0.255913 & -0.255642 \\ 0.050 & -0.308091 & -0.307259 \\ 0.055 & -0.381452 & -0.378013 \\ \end{array} \right)$$ Using the next approximation $$P_3=\frac{3 x \left(32 x^2+101 x+94\right)}{46 x^3-279 x^2+198 x-1128}$$ would require solving the cubic $$(96-46 a) x^3+(279 a+303) x^2+(282-198 a) x+1128 a=0$$ would give $x=-0.377979$ for $a=0.055$

Answer 2

It's not known if your equation can be solved in closed form in the general case.

$$x=-a-\frac{a}{2}e^{-3x}-ae^{-2x}-\frac{3a}{2}e^{-x}$$

$$x=-a-\frac{a}{2}(e^{-x})^3-a(e^{-x})^2-\frac{3a}{2}e^{-x}$$

Your equation is an exponential polynomial equation.
For $a=0$, the solution is $x=0$.
For $a\neq 0$, your equation is a polynomial equation of $e^{-x}$ of degree $3$.
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a)

Because your equation is an algebraic equation in dependence of two algebraically independent monomials ($x,e^x$), it cannot be rearranged for $x$ by only elementary functions/operations that we can read from the equation. But it's not known if elementary partial inverse functions exist for your equation.
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b)

Because the coefficients of $e^{-x}$, $(e^{-x})^2$ and $(e^{-x})^3$ are not zero, the general equation cannot be solved by Lambert W.

$x\to -t$:

$$-t=-a-\frac{a}{2}(e^t)^3-a(e^t)^2-\frac{3a}{2}e^t$$

Let's solve for $e^t$:

$$e^t=R(t),$$

where $R(t)$ is an $a$-containing root expression in dependence of $t$ that comes from Cardano's solution formula for the cubic equation.

Because $R(t)$ is, for the general equation, a non-rational expression of $t$, the general equation cannot be solved by Generalized Lambert W of Mezö et al.
Because $R(t)$ is somewhat complicated, the general equation cannot be solved by Generalized Lambert W of [Castle 2018] (first formula on page 2) with only rational exponents $p_i$.

I don't know if the general equation can be solved by Generalized Lambert W of [Castle 2018] with some non-rational exponents $p_i$.
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c)

Let's solve for $a$:

$$a=-\frac{2x}{2+e^{-3x}+2e^{-2x}+3e^{-x}}.$$

We see - because $a$ is equal to an elementary expression of $x$: if a solution $x$ is a closed-form number (e.g. an elementary number), $a$ is a closed-form number (an elementary number). This means in the reverse conclusion: if $a$ is not a closed-form number (or not an elementary number respectively), $x$ is not a closed-form number (or not an elementary number respectively).

Let $x_0\in\mathbb{C}$. If $a=-\frac{2x_0}{2+e^{-3x_0}+2e^{-2x_0}+3e^{-x_0}}$, $x=x_0$ is a solution.
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d)

If $a=cx$ ($c$ a complex constant), we get

$$1=-\frac{1}{2}c\left(2+(e^{-x})^3+2(e^{-x})^2+3e^{-x}\right),$$

and the equation has an elementary solution in dependence of $c$ therefore.
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[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: 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