1
$\begingroup$

Since the leaky bucket problem follows Torricelli's law in physics, which says that the rate at which water drains from the bucket is proportional to the square root of the height of the water remaining inside the bucket, we have

$$h'(t) = -k\sqrt{h(t)},\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:[1]$$ $$h(0) = h_0 , t\geq0$$

where $h(t)$ is the height of the water in the bucket at time $t$, and $h_0$ is the initial height at time $0$, some positive constant $k$ and the negative sign is needed because $h(t)$ is a decreasing function.

Now solving for the nonzero solutions of [1], we solve

$$\int \frac{1}{\sqrt{h}} = \int -kdt$$

and obtain $$h(t) = \frac{k^2}{4} \left[t-\frac{2\sqrt{h_0}}{k}\right]^2, \:\:\:\: 0 \leq t \leq \frac{2\sqrt{h_0}}{k}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:[2]$$

where the initial condition $h(0) = h_0$ is used. The formula $[2]$ indicates the height of the water becomes zero or empty when $t = \frac{2\sqrt{h_0}}{k}$. Thus, for things to be meaningful in physics, the bucket will remain empty or the height will remain zero after $t = \frac{2\sqrt{h_0}}{k}$. Therefore we define

$$x(t,0,h_0)= \begin{cases}\frac{k^2}{4} \left[t-\frac{2\sqrt{h_0}}{k}\right]^2, & 0 \leq t \leq \frac{2\sqrt{h_0}{k}} \ \\ 0, & t>\frac{2\sqrt{h_0}}{k} \ \end{cases} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\: [3]$$

Then for $x(t,0,h_0)$ defined in $[3]$, we can verify that

$$x'(t)= \begin{cases}\frac{k^2}{2} \left[t-\frac{2\sqrt{h_0}}{k}\right]=-k\sqrt{x(t)}, & 0 \leq t \leq \frac{2\sqrt{h_0}{k}} \ \\ 0 = -k\sqrt{x(t)}, & t>\frac{2\sqrt{h_0}}{k} \ \end{cases} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\: [4]$$

$\textbf{That is, for any}$ $h_0 \geq 0$ $\textbf{fixed,}$ $x(t,0,h_0)$ $\textbf{defined in}$ $[3]$ $\textbf{gives the unique solution of}$ $[1]$ $\textbf{for}$ $t \geq 0,$ $\textbf{which uniquely determines the height of the water at any time}$ $t \geq 0$.

Well, I really can't understand the bold part, do I have to differentiate $[4]$ to arrive at $[1]$ to show uniqueness?

(Note: Problem is derived from J.H Liu "A First Course in the Qualitative Theory of Differential Equations")

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.