Number of solutions of a transcendental function

Question

I am studying time-delay differential equations now. When I read "textbook" material, my teacher wrote it, this lemma occurs to me. For a transcendental function, $\lambda*(1-C*e^{-\lambda*t})-A-B*e^{-\lambda*t}=0$, where $A, B,C$ are real entries. If $B \neq 0 $ and $C =0$, then the function above has countably many solutions and there exists a real constant $\alpha$ such that the real part of all $\lambda < \alpha$. Moreover, if the $\lambda_j$ are listed in decreasing order of their real parts, then the real part of $\lambda_j $ approaches to $-\infty$ and $j$ approaches to $\infty$. If $C\neq 0$, then the function has countably many solutions, and there exist real constants $\alpha$ and $\beta$, such that all solutions are contained within the vertical strip $\beta <Re{\lambda_j<\alpha}$ in the complex plane. I wonder why this lemma is true and an relatively easy-understood justification to it, since I didn't know much about transcendental functions. Thank you so much for your help!

Answer 1

I'm assuming $t$ is a given positive number. I'll look at the case $C \ne 0$. Write the equation as $$ e^{-\lambda t} = \dfrac{\lambda - A}{B+ \lambda C} = \dfrac{1}{C} \left(1 - \dfrac{A+B/C}{B/C+\lambda}\right) $$ As $\lambda $ travels on a vertical line $\text{Re}(\lambda) = \sigma$ in the complex plane with $\tau = \text{Im}(\lambda)$ from $-\infty$ to $+\infty$, the left side moves counterclockwise around a circle centred at $0$ with radius $e^{-\sigma t}$, making one revolution each time $\tau$ changes by $2 \pi/t$. The right side also moves along a circle (or in the case $\sigma = -B/C$ a straight line) symmetric about the real axis, but just once: it approaches $1/C$ as $\tau \to \pm \infty$. Moreover the radius of the circle goes to $0$ as $\sigma$ goes to either $-\infty$ or $+\infty$. There is only a certain interval $\alpha < \sigma < \beta$ in which the two circles intersect (that is a necessary condition for the two sides to be equal, though not sufficient).

Now the circle for the left side passes through the point $1/C$ iff $\sigma = \sigma_0$ where $e^{- \sigma_0 t} = 1/|C|$. For $\sigma$ either slightly less or slightly greater than $\sigma_0$ (depending on which side of $1/C$ the circle on the right is on), the two circles will intersect near $1/C$. Let $\tau_1(\sigma)$ (and by symmetry $-\tau_1(\sigma)$ be the $\tau$ values for which the right side is at those intersection points. As $\sigma$ approaches $\sigma_0$, $ \tau_1(\sigma) \to +\infty$ so that the intersection point approaches $1/C$. Meanwhile on the left side of the equation, $e^{-\lambda t} = e^{-(\sigma + i \tau_1(\sigma))\lambda}$ is zipping around its circle, making one revolution each time $\tau_1(\sigma)$ changes by $2\pi/t$. In each of those revolutions it will hit the other circle twice, once on the correct side so that the equation is satisfied.

EDIT:

Here's an animation in the case $t = A = B = 1, C=1/2$. The circle for the left side is in blue, with the point $\exp(-\lambda t)$ in dark blue. The circle for the right side is in pink, with the point $\frac{\lambda - A}{B + \lambda C}$ in red.