Question 7.3.2: Three brine tanks If V1 = 20, V2 = 40, V3 = 50, r = 10 (gal/...
Three brine tanks If V_{1} = 20, V_{2} = 40, V_{3} = 50, r = 10 (gal/min), and the initial amounts of salt in the three brine tanks, in pounds, are
x_{1}(0) = 15, x_{2}(0) = x_{3}(0) = 0,
find the amount of salt in each tank at time t \geqq 0.
Substituting the given numerical values in (14) and (15), we get the initial value problem
x^{′}_{1} = -k_{1} x_{1},
x^{′}_{2} = k_{1} x_{1} – k_{2} x_{2},
x^{′}_{3} = k_{2} x_{2} – k_{3} x_{3}, (14)
k_{i} = \frac{r}{V_{i}} , i = 1, 2, 3. (15)
\textbf{x}^{′}(t) = \left [ \begin{matrix} -0.5 & 0.0 & 0.0 \\ 0.5 & -0.25 & 0.0 \\ 0.0 & 0.25 & -0.2 \end{matrix} \right ] \textbf{x},\textbf{x}(0) = \left [ \begin{matrix} 15 \\ 0 \\ 0 \end{matrix} \right ] (16)
for the vector x(t) = [x_{1}(t) x_{2}(t) x_{3}(t)]^{T} . The simple form of the matrix
A – λI = \left [ \begin{matrix} -0.5-\lambda & 0.0 & 0.0 \\ 0.5 & -0.25-\lambda & 0.0 \\ 0.0 & 0.25 & -0.2-\lambda \end{matrix} \right ] (17)
leads readily to the characteristic equation
|A – λI| = (-0.5 – λ)(-0.25 – λ)(-0.2 – λ) = 0.
Thus, the coefficient matrix A in (16) has the distinct eigenvalues λ_{1} = -0.5, λ_{2} = -0.25, and λ_{3} = -0.2 and therefore has three linearly independent eigenvectors.
CASE 1: λ_{1} = -0.5. Substituting λ = -0.5 in (17), we get the equation
for the associated eigenvector v = [a b c]^{T} . The last two rows, after division by 0.25 and 0:05, respectively, yield the scalar equations
2a + b = 0,
5b + 6c = 0.
The second equation is satisfied by b = -6 and c = 5, and then the first equation gives a = 3. Thus the eigenvector
\textbf{v}_{1} = [3 -6 5]^{T}
is associated with the eigenvalue λ_{1} = -0.5.
CASE 2: λ_{2} = -0.25. Substituting λ = -0.25 in (17), we get the equation
\left [ \begin{matrix} \textbf{A}+ (0.25)\cdot \textbf{I} \end{matrix} \right ]\textbf{v} = \left [ \begin{matrix} -0.25 & 0 & 0 \\ 0.5 & 0 & 0 \\ 0 & 0.25 & 0.05 \end{matrix} \right ] \left [ \begin{matrix} a \\ b \\ c \end{matrix} \right ] =\left [ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right ]for the associated eigenvector v = [a b c]^{T} . Each of the first two rows implies that a = 0, and division of the third row by 0.05 gives the equation
5b + c = 0,
which is satisfied by b = 1, c = -5. Thus the eigenvector
\textbf{v}_{2} = [0 1 -5]^{T}
is associated with the eigenvalue λ_{2} = -0.25.
CASE 3: λ_{3} = -0.2. Substituting λ = -0.2 in (17), we get the equation
for the eigenvector v. The first and third rows imply that a = 0, and b = 0, respectively, but the all-zero third column leaves c arbitrary (but nonzero). Thus
\textbf{v}_{3} = [0 0 1]^{T}
is an eigenvector associated with λ_{3} = -0.2.
The general solution
\textbf{x}(t) = c_{1}\textbf{v}_{1}e^{λ_{1}t} + c_{2} \textbf{v}_{2}e^{λ_{2}t} + c_{3} \textbf{v}_{3}e^{λ_{3}t}
therefore takes the form
\textbf{x}(t) = c_{1} \left [ \begin{matrix} 3 \\ -6 \\ 5 \end{matrix} \right ] e^{(-0.5)t} + c_{2} \left [ \begin{matrix} 0 \\ 1 \\ -5 \end{matrix} \right ] e^{(-0.25)t} +c_{3} \left [ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right ]e^{(-0.2)t} .The resulting scalar equations are
x_{1}(t) = 3c_{1} e^{(-0.5)t} ,
x_{2}(t) = -6c_{1} e^{(-0.5)t} + c_{2}e^{(-0.25)t} ,
x_{3}(t) = 5c_{1}e^{(-0.5)t} – 5c_{2} e^{(-0.25)t} + c_{3} e^{(-0.2)t}.
When we impose the initial conditions x_{1}(0) = 15, x_{2}(0) = x_{3}(0) = 0, we get the equations
3c_{1} = 15,
-6c_{1} + c_{2} = 0,
5c_{1} – 5c_{2} + c_{3} = 0
that are readily solved (in turn) for c_{1} = 5, c_{2} = 30, and c_{3} = 125. Thus, finally, the amounts of salt at time t in the three brine tanks are given by
x_{1}(t) = 15e^{(-0.5)t} ,
x_{2}(t) = -30 e^{(-0.5)t} + 30e^{(-0.25)t} ,
x_{3}(t) = 25e^{(-0.5)t} – 150e^{(-0.25)t} + 125e^{(-0.2)t} .
Figure 7.3.3 shows the graphs of x_{1}(t), x_{2}(t), and x_{3}(t). As we would expect, tank 1 is rapidly “flushed” by the incoming fresh water, and x_{1}(t) → 0 as t → + \infty. The amounts x_{2}(t) and x_{3}(t) of salt in tanks 2 and 3 peak in turn and then approach zero as the whole three-tank system is purged of salt as t → + \infty.
