I received this homework problem from my professor, but I am unsure of how to get it going. It doesn't correlate exactly with anything in our book.
Consider the heat equation for $u(x,t)$
$$u_t = ku_{xx}, -\infty < x < \infty, t > 0 \tag{1}$$
$$u(x,0) = f(x) \tag{2}$$
assume that a network of sensors is uniformly distributed in space, at a spatial increment of $\Delta x$, and that the smallest absolute amount of change (variation) in $u$ that can be detected by a sensor (sensor's activity) is a constant
$$|\delta u|_{min} = \epsilon > 0 \tag{3}$$
consider the problem detecting the presence of an impulse variation in the initial condition
$$\delta f(x) = a \delta(x - x_{0}) \tag{4}$$
where $a > 0$ is a constant coefficient and the location $x_0$ of the impulse may be unknown.
If $\Delta x = 1$, find the maximum value of $\epsilon$ s.t. any impulse $(4)$ in the initial condition with $a \geq 1$ will be detected by at least one sensor.
Given $\epsilon > 0$ find the maximum value of $\Delta x$ to guarantee that any impulse $(4)$ in the initial condition with $a \geq 1$ will be detected by at least one sensor.
Given the network of parameters $\epsilon > 0$, $\Delta x > 0$, find the minimum value $A > 0$ s.t. any impulse $(4)$ in the initial condition with $a \geq A$ will be detected by at least one sensor.
As the problem is a linear by denoting $u$ the solution with $f$ as initial condition and $u + \delta u$ the solution with $f + \delta f$ as initial condition, you have that $\delta u$ is solution of $$ (\delta u)_t = k(\delta u)_{xx}$$ $$\delta u(x,0)=\delta f(x) a \delta(x-x_0)$$ in this case there is a well known explicit solution, the heat kernel so you obtain $$\delta u(x,t)=\frac{a}{\sqrt{4 \pi t}}e^{-\frac{(x-x_0)^2}{4t}}.$$ When $x$ is fixed the study of the function $t \mapsto \delta u(x,t)$ show that the maximum is $$a \frac{1}{\sqrt{2 \pi e}} \frac{1}{|x-x_0|}$$ so in the worst case the distance betwwen $x_0$ and the closest sensor is $\frac{ \Delta x}{2}$ the initial condition will be detected as long as $$a \sqrt{\frac{2}{ \pi e}} \frac{1}{\Delta x} \geq \epsilon.$$