Simultaneous exponential equations

Question

I don't know exactly where exactly the equation applies. I don't think it is that important for solving the problem. It's about the following equation:

$$ E(t) = E_0 (1- e^{-\frac{t}{\tau}}) $$

We are looking for $\tau$ and $E_0$ under the condition that we have given two points $A$ and $B$ on the graph.

Is there a unique analytic solution to the above problem? No approximate solution should be determined using methods such as regression or Lagrange interpolation.

My approach:

Given the two points $A(x_1|y_1)$ and $B(x_2|y_2)$, we obtain the following system

\begin{align*} E_0 (1- e^{-\frac{x_1}{\tau}}) &= y_1 \\ E_0 (1- e^{-\frac{x_2}{\tau}}) &= y_2 \end{align*}

From here I don't know exactly how to solve the system of equations. Is there a solution at all? If a solution exists, under what conditions is the solution unique?

Answer 1

This is a typical exponential decay problem which will be treated as such (that is to say assuming that $(x_1,y_1,c_2,y_2)$ are all positive

To make the problem better conditioned, let $\tau=\frac 1k$

\begin{align*} E_0 (1- e^{-k x_1}) &= y_1 \\ E_0 (1- e^{-k x_2}) &= y_2 \end{align*}

Making the ratio $$\frac {y_2}{y_1}=\frac{1- e^{-k x_2} } {1- e^{-k x_1} }$$ which will require a numerical method. To make the solver more happy, rewrite the above and consider that you look for the zero of function $$F(k)=\log \left(y_2+y_1 \,e^{-k x_2}\right)-\log \left(y_1+y_2\, e^{-k x_1}\right)$$

For an example, using $A(100|45)$ and $B(150|60)$. Since $F(0)=0$, compute a few values $F(n \Delta)$ until you have $$F(n \Delta)<0 \qquad \text{and} \qquad F((n+1) \Delta)>0 $$ For this example, using $\Delta=0.002$ the solution is between $k=0.004$ and $k=0.006$.

Start Newton iterations at the midpoint of the range and the iterates will be $$\left( \begin{array}{cc} n & k_n \\ 0 & 0.005000000000 \\ 1 & 0.005304125986 \\ 2 & 0.005289971732 \\ 3 & 0.005289941886 \\ \end{array} \right)$$

Using one of the equations, then $E_0=109.542$ and $\tau=189.038$.

Edit (to be handled with care, may be)

Suppose that $k$ is small. Perform a series expansion around $k=0$ to obtain as a possible estimate $$k_0=\frac{2 (y_1+y_2) (x_1 y_2-x_2 y_1)}{\left(x_1^2-x_2^2\right) y_1 y_2}$$

For the working example, this gives $$k_0=\frac{7}{1500}=0.00466667 \qquad \qquad\qquad \color{red}{\huge !!}$$

Edit

Consider now the function $$G(k)= \left(y_2+y_1 \,e^{-k x_2}\right)-\left(y_1+y_2\, e^{-k x_1}\right)$$

Its first derivative cancels at $$k_0=\frac 1 {x_2-x_1}\,\log \left(\frac{x_2\, y_1}{x_1 \,y_2}\right)$$ as already wriiten by @Griboullis.

To approximate the solution, expand $G(k)$ as a series around $k_0$ to generate $$k_1=k_0+\sqrt{-2\,\frac{G(k_0) }{G''(k_0) }}$$ For the worked example, this gives $k_1=0.00495774$.

Now, $k_1$ being the starting guess, $\color{red}{\text{one single iteration}}$ of Newton-like method of order $n$ gives an $\color{red}{\text{explicit formula}}$ for $k_2^{(n)}$.

For the worked example, this gives

$$\left( \begin{array}{ccc} n & k_2^{(n)} &\text{method} \\ 2 & 0.00530379935784 & \text{Newton} \\ 3 & 0.00528890969515 & \text{Halley} \\ 4 & 0.00529001046153 & \text{Householder} \\ 5 & 0.00528993729050 & \text{no name} \\ 6 & 0.00528994219427 & \text{no name} \\ 7 & 0.00528994186568 & \text{no name} \\ 8 & 0.00528994188770 & \text{no name} \\ \cdots & \cdots \\ \infty & 0.00528994188633 & \text{solution} \\ \end{array} \right)$$

Answer 2

Here is how the number of solutions can be determined. To simplify things, we suppose that $y_2 \not = y_1$. This implies $E_0\not = 0$ and we can write the equation in the unknown $k = \frac{1}{\tau}\not = 0$ \begin{equation} g(k):= \frac{y_2}{y_2 - y_1} e^{-k x_1} - \frac{y_1}{y_2 - y_1}e^{-k x_2} = 1 \end{equation} Clearly $g(0)=1$ but $k=0$ is not a valid solution. One has \begin{equation} g'(k)= - \frac{x_1 y_2}{y_2 - y_1} e^{-k x_1} + \frac{x_2 y_1}{y_2 - y_1}e^{-k x_2} \end{equation} Again we simplify things by assuming that $x_1 x_2 y_1 y_2 \not = 0$ and $x_1\not = x_2$.

If the quantities $x_1 y_2$ and $x_2 y_1$ have opposite signs, it implies that the sign of $g'(k)$ is constant, hence $g$ is strictly monotonic. In that case the equation cannot have a solution $k\not = 0$.

When $x_1 y_2$ and $x_2 y_1$ have the same signs, we have $g'(k_0) = 0$ where \begin{equation} k_0 = \frac{1}{x_2-x_1}\log \frac{x_2 y_1}{x_1 y_2} \end{equation} It is now easy to see that the equation $g(k) = 1$ can have a valid solution in two cases

• $k_0 > 0$ and $1$ belongs to the interval which bounds are $g(k_0)$ and $\displaystyle\lim_{k\to\infty}g(k)$. This solution satisfies $\frac{1}{\tau}\in (k_0,+\infty)$
• $k_0 < 0$ and $1$ belongs to the interval which bounds are $g(k_0)$ and $\displaystyle\lim_{k\to-\infty}g(k)$. This solution satisfies $\frac{1}{\tau}\in (-\infty, k_0)$.

In both cases, the solution is unique.