Problem source

A damped system with feedback is modelled by the equation 
	\[
	\mathrm{f}'(t)+\mathrm{f}(t)-k\mathrm{f}(t-1)=0,\mbox{ }\tag{$\dagger$}
	\]
	where $k$ is a given non-zero constant. Show that (non-zero) solutions for $\mathrm{f}$ of the form $\mathrm{f}(t)=A\mathrm{e}^{pt},$ where $A$ and $p$ are constants, are possible provided $p$ satisfies
	\[
	p+1=k\mathrm{e}^{-p}.\mbox{ }\tag{*}
	\]
	Show also, by means of a sketch, or otherwise, that equation $(*)$ can have $0,1$ or $2$ real roots, depending on the value of $k$, and find the set of values of $k$ for which such solutions of $(\dagger)$ exist. For what set of values of $k$ do such solutions tend to zero as $t\rightarrow+\infty$?

Solution source

Suppose $f(t) = Ae^{pt}$ is a solution, then

\begin{align*}
&& 0 &= Ape^{pt} + Ae^{pt} - Ake^{p(t-1)} \\
\Leftrightarrow && 0 &= p +1 - ke^{-p}  \\
\Leftrightarrow && p+1 &= ke^{-p}
\end{align*}
\begin{center}
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            plot (\x, {-1/exp(2)*exp(-\x)});
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            plot (\x, {-0.05*exp(-\x)});
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\end{center}


If we sketch $y = x+1$ and $y = ke^{-x}$ we can see that when $k \geq 0$ there will be exactly one solution. If $k < 0$ there can be no solutions (if $k$ is large and negative) and two solutions if $k$ is small and negative. There will be exactly one real root if $y = x+1$ is tangent to $y = ke^{-x}$. The gradient of $y = ke^{-x}$ is $-ke^{-x}$ so to have a gradient $1$ we must have $x+1 = -1 \Rightarrow x = -2$, but also $x = -\ln(-k) \Rightarrow k = -e^{-2}$.

Therefore there will be so solution as long as $k \geq -e^{-2}$.

The solutions will tend to $0$ as $t \to + \infty$ as long as the intersection point is less than zero. For $k \geq 0$ they intersect at $0$ when $k =1$, so we would want $k < 1$. All negative values of k will work since the intersection has to happen for negative $p$. Therefore the range is $-e^{-2} \leq k < 1$