Question 7.3.2: Three brine tanks If V1 = 20, V2 = 40, V3 = 50, r = 10 (gal/...

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

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

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.