# Solving a quadratic equation with perturbation on the exponent.

This is for analysis of chemical rate equations. I have the equation $$A(1-x)^2 = Bx^{2 + \epsilon}$$ which for the trivial case $\epsilon = 0$, and ignoring the negative root, has the solution, $$x = \frac{\sqrt A}{\sqrt A + \sqrt B}$$ I would like to vary $\epsilon$ from say -1 to 0, and have at least an approximate exporession for $x$ in the form $$x = \frac{A^\alpha}{A^\alpha + B^\beta}$$ if that is at all possible (variations for example including cross-terms are acceptable too). With the first equation, I can plug in values for $A$, $B$, and $\epsilon$ and find a numerical solution for x, so I know it can exist, but I cannot come further than this. I can prove the trivial solution is correct but cannot derive a method for a general solution involving the perturbation using this method.

## 1 Answer

Put $u = \sqrt{\dfrac{B}{A}}$, then we have:

$$1-x = u x^{1+\frac{\epsilon}{2}}= u x\exp\left[\frac{\epsilon}{2}\log(x)\right] = u x \left[1 + \frac{\epsilon}{2}\log(x) + \frac{\epsilon^2}{8}\log^2(x)+\cdots\right]$$

We then substitute the formal expansion:

$$x = x_0 + \epsilon x_1 + \epsilon^2 x_2 +\cdots$$

and expand the equation in powers of $\epsilon$ and equate equal powers of $\epsilon$. You then find the unperturbed solution:

$$x_0 = \frac{1}{1+u}$$

The coefficient of $\epsilon$ of the equation yields:

$$(1+u)x_1 + \frac{u}{2} x_0 \log(x_0) = 0$$

therefore:

$$x_1 = \frac{u\log(1+u)}{2(1+u)^2}$$

Extracting the coefficient of $\epsilon^2$ of the equation yields:

$$(1+u)x_2 + \frac{u}{2} x_1 \left[1+\log(x_0)\right] +\frac{u}{8}x_0\log^2(x_0)=0$$

Here we've used the expansion:

$$\log(x) = \log(x_0 + \epsilon x_1 + \cdots) = \log(x_0) + \log(1+ \epsilon \frac{x_1}{x_0}+\cdots) = \log(x_0) + \epsilon \frac{x_1}{x_0}+\cdots$$

So, this way we get an expression for $x_2$ in terms of $u$ and it's easy to proceed in this way to find the higher order terms. A problem you may encounter when proceeding to higher and higher orders is that the perturbation series may not converge for the desired value of $\epsilon$. You can then resort to resummation methods.

• For the log(x) expansion, I cannot see how you get to the last step. I want to invoke something with an exponential expansion (the log has just been dropped), but doesn't this restrict the values of the x coefficients to factorials with $x_0 = 1$? Seems like I am off track here. – Ian Joseph Allen Aug 2 '17 at 13:53
• @IanJosephAllen You take a factor $x_0$ out and then use that $\log(AB) = \log(A) + \log(B)$. Then you expand the term of the form $\log(1+r)$ as $r - r^2/2 + \cdots$ – Count Iblis Aug 2 '17 at 14:13
• In the second term for your extraction of the $\epsilon^2$ coefficient, you have $[1 + log(x_o)]$. I am curious where the 1 came from? I get otherwise the same result for the rest of this line. – Ian Joseph Allen Aug 2 '17 at 15:25
• @IanJosephAllen So, that's the $\epsilon\frac{u}{2}x\log(x)$ term. We then want to extract the coefficient of $\epsilon$ from $x\log(x)$ as that will be multiplied by $\epsilon$ to yield the second order contribution. We have: $$x\log(x) = (x_0 + \epsilon x_1 + \cdots)\log(x_0 + \epsilon x_1 + \cdots)$$ The logarithm can be expanded as: $$\log(x_0) + \epsilon \frac{x_1}{x_0}$$. – Count Iblis Aug 2 '17 at 16:00
• There are then two contributions to order $\epsilon$, we can pick the $\epsilon x_1$ and then we must multiply that by the $\log(x_0)$ term from the expansion of the logarithm, or we take $x_0$ and multiply that by $\epsilon\frac{x_1}{x_0}$. So, we get $$\epsilon x_1\log(x_0) + \epsilon x_1 = \epsilon x_1\left[1+\log(x_0)\right]$$. – Count Iblis Aug 2 '17 at 16:00