0

A problem of solving nonhomogeneous equations：$$\frac{1-e^{-a/x} }{1-e^{-b/x}} =c,$$ where $a, b, c$ are all constants to solve for $x$.

I wonder if there's a good way to solve this equation, I try to approximate it using Newton's method and in some cases I get the answer, but in more cases I can't find the right initial value to solve it

Gary
  • 36,640
yh l
  • 3
  • Solve $\frac{1-y^a}{1-y^b}=c$ and then take $x=-\frac1{\ln y}.$ Not sure what the best numeric approach to the $y$ equation, but it seems like it would be better to write it as $1-c-y^a+cy^b=0$ and maybe apply Newton to that. Do you have any specific information about $a,b,c?$ – Thomas Andrews Jun 21 '23 at 14:23
  • 1
    If $u=y^b$ you can write this as $\frac{1-u^{a/b}}{1-u}=c,$ which the mean value theorem says is $\frac abu_{0}^{a/b-1}=c$ for some $u_0$ between $1$ and $u.$ So $u_0=\left(\frac{bc}a\right)^{b/(a-b)}.$ So you know $u$ must be on the other side from $u_0$ from $1.$ You might want to start Newton from $u_0.$ – Thomas Andrews Jun 21 '23 at 14:32
  • 2
    How is this a differential equation? – Thomas Andrews Jun 21 '23 at 14:35
  • In one case：$a=1.5；b=0.3;c=2063/1836$,in another:$a=1638,b=250,c=1.062$
    Thomas Andrews' thinking helped me，But what bothers me most is how do I choose the initial value of this equation using Newton's method at different values     – yh l Jun 21 '23 at 14:44
  • There are closed form solutions if you substitute $y^b=w$ – Тyma Gaidash Jun 21 '23 at 15:05
  • @ThomasAndrews The linked question, after renaming variables, solves $X^Y=AX-B$ while $w^\frac ab=cw-(c-1)$ is the same with $X=w,Y=\frac ab,A=c,B=c-1$. They are the same form – Тyma Gaidash Jun 21 '23 at 15:36
  • 1
    Yeah, reading comprehension failure on my part. Sorry. @TymaGaidash – Thomas Andrews Jun 21 '23 at 15:57

1 Answers1

1

Using @Thomas Andrews's notation, we look for the zero of function $$f(u)=u^k- c\,u+(c-1) \qquad \text{where} \qquad k=\frac a b$$ As already said, the first derivative cancels at $$u_*=\left(\frac{c}{k}\right)^{\frac{1}{k-1}}$$

Expanding as a series around $u_*$, the first estimate is given by $$u_0=u_*-\sqrt{-2\frac{f(u_*)}{f''(u_*)}}$$ from which we can safely start Newton iterations

For the case where $a=1.5；b=0.3;c=2063/1836$, converted to decimals, this gives $$u_*=0.688516 \qquad \implies \qquad u_0=0.298976$$ Newton iterates are $$\left( \begin{array}{cc} n & u_n \\ 0 & 0.298976 \\ 1 & 0.105273 \\ 2 & 0.110048 \\ \end{array} \right)$$

For the case where $a=1638;b=250;c=1.062$, converted to decimals, this gives $$u_*=0.720550 \qquad \implies \qquad u_0=0.34196$$ Newton iterates are $$\left( \begin{array}{cc} n & u_n \\ 0 & 0.3419602 \\ 1 & 0.0546289 \\ 2 & 0.0583804 \\ \end{array} \right)$$

Claude Leibovici
  • 289,558
  • thanks for @Thomas Andrews and Claude Leibovici.Your method helped me simplify the equation，In particular, the choice of u∗ helped me solve the initial value problem – yh l Jun 22 '23 at 04:52