Question 5.3.4: Suppose that two brine storage tanks are connected with two ...
Suppose that two brine storage tanks are connected with two pipes used to exchange solutions between them. The first pipe allows water from tank 1 to enter tank 2 at a rate of 5 gal/min. The second pipe reverses the process allowing water to flow from tank 2 to tank 1, also at a rate of 5 gal/min. Initially, the first tank contains a well-mixed solution of 8 lb of salt in 50 gal of water, while the second tank contains 100 gal of pure water.
a. Find the linear system of differential equations to describe the amount of salt in each tank at time t .
b. Solve the system of equations by reducing it to an uncoupled system.
c. Determine the amount of salt in each tank as t increases to infinity and explain the result
a. Let y_{1}(t) \text{ and } y_{2}(t) be the amount of salt (in pounds) in each tank after t min. Thus, y^{\prime }_{1}(t) \text{ and } y^{\prime }_{2} (t) are, respectively, the rates of change for the amount of salt in tank 1 and tank 2. To develop a system of equations, note that for each tank
Rate of change of salt = rate in − rate out
Since the volume of brine in each tank remains constant, for tank 1, the rate in is \frac{5}{100} y_{2}(t) while the rate out is \frac{5}{50} y_{1}(t). For tank 2, the rate in is \frac{5}{50} y_{1}(t) while the rate out is \frac{5}{100} y_{2} (t). The system of differential equations is then given by
Since the initial amounts of salt in tank 1 and tank 2 are 8 and 0 lb, respectively, the initial conditions on the system are y_{1}(0) = 8 \text{ and } y_{2}(0) = 0.
b. The system of equations in matrix form is given by
y′ = \begin{bmatrix} -\frac{1}{10} &\frac{1}{20} \\ \frac{1}{10} &-\frac{1}{20} \end{bmatrix} y \quad \text{with} \quad y(0) = \begin{bmatrix} 8 \\ 0 \end{bmatrix}
The eigenvalues of the matrix are \lambda _{1} = − \frac{3}{20} \text{ and } \lambda _{2} = 0 with corresponding eigenvectors \begin{bmatrix} -1\\ 1 \end{bmatrix} and \begin{bmatrix} -1\\ 2 \end{bmatrix} Thus, the matrix that uncouples the system is
P =\begin{bmatrix} -1&1\\ 1&2 \end{bmatrix} with P^{-1}= \begin{bmatrix} -\frac{2}{3}& \frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} \end{bmatrix}
The uncoupled system is then given by
w′ = \begin{bmatrix} -\frac{2}{3}& \frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} \end{bmatrix} \begin{bmatrix} -\frac{1}{10} &\frac{1}{20} \\ \frac{1}{10} &-\frac{1}{20} \end{bmatrix} \begin{bmatrix} -1 &1\\ 1&2 \end{bmatrix} w
= \begin{bmatrix} -\frac{3}{20} &0 \\ 0&0 \end{bmatrix}w
The solution to the uncoupled system is
w(t) = \begin{bmatrix} e^{-\frac{3}{20} t} &0\\ 0&1 \end{bmatrix} w(0)
Hence, the solution to the original system is given by
y(t) = \begin{bmatrix} -1 &1\\ 1&2 \end{bmatrix} \begin{bmatrix} e^{-\frac{3}{20} t} &0\\ 0&1 \end{bmatrix} \begin{bmatrix} -\frac{2}{3}& \frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} \end{bmatrix} y(0)
= \frac{1}{3} \begin{bmatrix} 2 e^{-\frac{3}{20} t} + 1 &- e^{-\frac{3}{20} t} + 1 \\ -2 e^{-\frac{3}{20} t} + 2 & e^{-\frac{3}{20} t} + 2 \end{bmatrix}\begin{bmatrix} 8\\ 0 \end{bmatrix}
= \frac{8}{3} \begin{bmatrix} 2 e^{-\frac{3}{20} t} + 1 \\ -2 e^{-\frac{3}{20} t} + 2 \end{bmatrix}
c. The solution to the system in equation form is given by
y_{1}(t) = \frac{8}{3}( 2 e^{-\frac{3}{20} t} + 1 ) \quad and \quad y_{2}(t) = \frac{8}{3}( -2 e^{-\frac{3}{20} t} + 2 )To find the amount of salt in each tank as t goes to infinity, we compute the limits
\underset{t\rightarrow \infty }{\lim }\frac{8}{3}( 2 e^{-\frac{3}{20} t} + 1 )= \frac{8}{3} (0 + 1) = \frac{8}{3}and
\underset{t\rightarrow \infty }{\lim }\frac{8}{3}( -2 e^{-\frac{3}{20}} + 2 )= \frac{8}{3} (0 + 2) = \frac{16}{3}These values make sense intuitively as we expect that the 8 lb of salt should eventually be thoroughly mixed, and divided proportionally between the two tanks in a ratio of 1 : 2.