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

5.4
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a. Let y1(t) and y2(t)y_{1}(t) \text{ and } y_{2}(t) be the amount of salt (in pounds) in each tank after t min. Thus, y1(t) and  y2(t)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 5100 y2(t)\frac{5}{100}  y_{2}(t) while the rate out is 550 y1(t)\frac{5}{50}  y_{1}(t). For tank 2, the rate in is  550 y1(t)\frac{5}{50}  y_{1}(t) while the rate out is 5100 y2(t) \frac{5}{100}  y_{2} (t). The system of differential equations is then given by

{y1(t)=5100 y2(t) – 550 y1(t)y2(t)=550 y1(t) – 5100 y2(t)  that is ,{y1(t)= –110 y1(t)+120 y2(t)y2(t)=110 y1(t) – 120 y2(t)\begin{cases} y^{\prime }_{1}(t) = \frac{5}{100}  y_{2}(t)  –  \frac{5}{50}  y_{1}(t) \\y^{\prime }_{2}(t) = \frac{5}{50}  y_{1}(t)  –  \frac{5}{100}  y_{2}(t)\end{cases}    \text{that is}  , \quad \begin{cases} y^{\prime }_{1}(t) =  – \frac{1}{10}  y_{1}(t) + \frac{1}{20}  y_{2}(t) \\y^{\prime }_{2}(t) = \frac{1}{10}  y_{1}(t)  –  \frac{1}{20}  y_{2}(t)\end{cases}

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 y1(0) = 8   and y2(0)=0.y_{1}(0)  =  8   \text{ and } y_{2}(0) = 0.
b. The system of equations in matrix form is given by

y′ =  [110120110120 ] ywith y(0)=[80]\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 λ1=320  and λ2\lambda _{1} = − \frac{3}{20}  \text{ and } \lambda _{2} = 0 with corresponding eigenvectors [11  ]and[12  ]\begin{bmatrix} -1\\ 1   \end{bmatrix} and \begin{bmatrix} -1\\ 2   \end{bmatrix} Thus, the matrix that uncouples the system is

P =[1112  ]  with  P1= [2313  1313]\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′ =  [2313  1313][110120110120 ][1112 ]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

= [320000]w\begin{bmatrix} -\frac{3}{20} &0 \\ 0&0 \end{bmatrix}w

The solution to the uncoupled system is

w(t) = [e320t001]\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) = [1112 ][e320t001][2313  1313]\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)

= 13[2e320t+1e320t+12e320t+2 e320t+2 ][80 ]\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}

= 83[2e320t+1 2e320t+2]\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

y1(t)=83(2e320t+1)  and  y2(t)=83(2e320t+2) 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

limt83(2e320t+1)=83(0+1)=83\underset{t\rightarrow \infty }{\lim }\frac{8}{3}( 2 e^{-\frac{3}{20} t} + 1 )= \frac{8}{3} (0 + 1) = \frac{8}{3}

and

limt83(2e320+2)=83(0+2)=163\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.

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