Question 7.3.4: Find the amounts x1(t), x2(t), and x3(t) of salt at time t i...

With the given numerical values, (22) takes the form

\frac{dx_{1}}{dt} = -k_{1} x_{1} + k_{3} x_{3} ,

\frac{dx_{2}}{dt} = k_{1} x_{1} – k_{2} x_{2},

\frac{dx_{3}}{dt} = k_{2} x_{2} –  k_{3} x_{3},            (22)

\frac{dx}{dt} = \left [ \begin{matrix} -0.2 & 0 & 0.2 \\ 0.2 & -0.4 & 0 \\ 0 & 0.4 & -0.2 \end{matrix} \right ] x,                        (23)

where x = [x_{1}   x_{2}   x_{3}]^{T} as usual. When we expand the determinant of the matrix

A – \lambda \cdot I = \left [ \begin{matrix} -0.2-\lambda & 0.0 & 0.2 \\ 0.2 & -0.4-\lambda & 0.0 \\ 0.0 & 0.4 & -0.2-\lambda \end{matrix} \right ]        (24)

along its first row, we find that the characteristic equation of A is

(-0.2 – λ)(-0.4 – λ)(-0.2 – λ) + (0.2)(0.2)(0.4)

= – λ³ – (0.8) · λ² – (0.2) · λ

= – λ \left [ \begin{matrix} (\lambda + 0.4)^{2} + (0.2)^{2} \end{matrix} \right ] = 0.

Thus A has the zero eigenvalue λ_{0} = 0 and the complex conjugate eigenvalues λ, \bar{λ}  = -0.4 ± (0.2)i . We anticipate one solution corresponding to the zero eigenvalue and two additional  linearly independent solutions corresponding to the complex conjugate eigenvalues.

CASE 1: λ_{0} = 0. Substitution of λ = 0 in Eq. (24) gives the eigenvector equation

(A – 0\cdot I)v = \left [ \begin{matrix} -0.2 & 0.0 & 0.2 \\ 0.2 & -0.4 & 0.0 \\ 0.0 & 0.4 & -0.2 \end{matrix} \right ] \left [ \begin{matrix} a \\ b \\ c \end{matrix} \right ] =\left [ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right ]           (25)

for v = [a     b     c]^{T} . The first row gives a = c and the second row gives a = 2b, so v_{0} = [2    1    2]^{T} is an eigenvector associated with the eigenvalue λ_{0} = 0. The corresponding solution x_{0}(t) = v_{0} e^{λ_{0}t} of Eq. (23) is the constant solution

x_{0}(t) = \left [ \begin{matrix} 2 \\ 1 \\ 2 \end{matrix} \right ] .                    (26)

CASE 2: λ = – 0.4 – (0.2)i . Substitution of λ = – 0.4 – (0.2)i in Eq. (24) gives the eigenvector equation

The second equation (0.2)a + (0.2)ib = 0 is satisfied by a = 1 and b = i . Then the first equation

[0.2 + (0.2)i ] a + (0.2)c = 0

gives c = -1 – i . Thus, v = [ 1     i     (- 1 – i )]^{T} is a complex eigenvector associated with the complex eigenvalue λ = – 0.4 – (0.2)i . The corresponding complex-valued solution x(t) = v e^{λ_{t}} of (23) is

x(t) = [1     i     -1 – i]^{T}  e^{(-0.4 – 0.2i)t}

= [1      i        – 1 – i]^{T} e^{(-0.4)t }(cos  0.2t  –  i  sin  0.2t )

=  e^{(-0.4)t}  \left [ \begin{matrix} cos 0.2t -i sin 0.2t \\ sin 0.2t + i cos 0.2t\\ -cos 0.2t – sin 0.2t – i cos 0.2t + i sin 0.2t \end{matrix} \right ] .

The real and imaginary parts of x(t) are the real-valued solutions

x_{1}(t) = e^{(-0.4)t} \left [ \begin{matrix} cos 0.2t \\ sin 0.2t\\ -cos 0.2t – sin 0.2t \end{matrix} \right ] .

x_{2}(t) = e^{(-0.4)t} \left [ \begin{matrix} -sin 0.2t \\ cos 0.2t\\ -cos 0.2t +  sin 0.2t \end{matrix} \right ] .                   (27)

The general solution

x(t) = c_{0} x_{0}(t) +  c_{1} x_{1}(t) + c_{2} x_{2}(t)

has scalar components

x_{1}(t) = 2 c_{0} + e^{(-0.4)t}  (c_{1}  cos 0.2t  –  c_{2}   sin 0.2t ),

x_{2}(t) = c_{0} + e^{(-0.4)t} (c_{1}   sin 0.2t + c_{2}  cos 0.2t ),

x_{3}(t) = 2 c_{0} + e^{(-0.4)t} [(-c_{1}  –  c_{2}) cos 0.2t + (- c_{1} + c_{2}) sin 0.2t ]                         (28)

giving the amounts of salt in the three tanks at time t .
Observe that

x_{1}(t) + x_{2}(t) + x_{3}(t) ≡ 5c_{0}.                    (29)

Of course, the total amount of salt in the closed system is constant; the constant c_{0} in (29) is one-fifth the total amount of salt. Because of the factors of e^{(-0.4)t} in (28), we see that

Thus, as t → + \infty the salt in the system approaches a steady-state distribution with 40% of the salt in each of the two 50-gallon tanks and 20% in the 25-gallon tank. So whatever the initial distribution of salt among the three tanks, the limiting distribution is one of uniform concentration throughout the system. Figure 7.3.6 shows the graphs of the three solution functions with c_{0} = 10, c_{1} = 30,  and  c_{2} = – 10, in which case

x_{1}(0) = 50   and    x_{2}(0) = x_{3}(0) = 0.