Question

Solve Using Dirac Deltla/discontinuous forcing

Consider the following scenario. A salt tank that initially contains 50 gallons of pure water. A brine solution containing 1/5 lb/gal of salt flows in to the tank at a rate of 5 gal/min. Brine flows out of the tank at the same rate. At time t = 15 minutes the mechanism regulating salt flow in to the tank breaks and 20 pounds of salt is dumped instantaneously in to the tank. Although water continues to flow in to the tank at the original 5 gal/min, there is no salt in the water.

1. Write an IVP describing the amount of salt in the tank at time t.

2. Solve the IVP from Problem 1.

3. Plot the solution found in Problem 2 and explain the results.

Answer 1

Let $A(t)$ denote the amount of salt in the tank at time $t$. We're told that $A(0)=0$.

For $0\le t\le15$, the salt flows in at a rate of (1/5 lb/gal)*(5 gal/min) = 1 lb/min. When the regulating mechanism fails, 20 lbs of salt is dumped and no more salt flows for $t>15$. We can capture this in terms of the unit step function $u(t)$ and Dirac delta function $\delta(t)$ as

$\text{rate in}=u(t)-u(t-15)+20\delta(t-15)$

(in lb/min)

The salt from the mixed solution flows out at a rate of

$\text{rate out}=\left(\dfrac{A(t)\,\mathrm{lb}}{50+(5-5)t\,\mathrm{gal}}\right)\left(5\dfrac{\rm gal}{\rm min}\right)=\dfrac A{10}\dfrac{\rm lb}{\rm min}$

Then the amount of salt in the tank at time $t$ changes according to

$\dfrac{\mathrm dA}{\mathrm dt}=u(t)-u(t-15)+20\delta(t-15)-\dfrac A{10}$

Let $\hat A(s)$ denote the Laplace transform of $A(t)$, $\hat A(s)=\mathcal L_s\{A(t)\}$. Take the transform of both sides to get

$s\hat A(s)-A(0)=\dfrac1s-\dfrac{e^{-15s}}s+20e^{-15s}-\dfrac1{10}\hat A(s)$

Solve for $\hat A(s)$, then take the inverse of both sides.

$\hat A(s)=\dfrac{\frac{10-10e^{-15s}}{s^2}+\frac{200e^{-15s}}s}{10s+1}$

$\implies\boxed{A(t)=10-10e^{-t/10}+\left(30e^{3/2-t/10}-10\right)u(t-15)}$