Question

The system of mixing tanks shown in Figure 8.2.7is a closed system. The tanks A,B,and C initially contain the number of gallons of brine indicated in the figure.

Construct a mathematical model in the form of a linear system of first-order differential equations for the number of pounds of salt x₁(t),x₂(t) and x₃(t) in the tanks A,B and Cat time respectively t. Write the system in matrix form.

(b) Use the eigenvalue method of this section to solve the linear system in part (a) subject to x₁(0) = 30, x₂(0) =20, x₃(0) = 5.

Short answer

(a) The System in Matrix form is $X\text{'}=\left(\begin{array}{ccc}-\frac{1}{20}& 0& \frac{1}{10}\\ \frac{1}{20}& -\frac{1}{20}& 0\\ 0& \frac{1}{20}& -\frac{1}{10}\end{array}\right)X$

(b) The Initial value condition forlocalid="1668092070341" $X\text{'}=\left(\begin{array}{ccc}-\frac{1}{20}& 0& \frac{1}{10}\\ \frac{1}{20}& -\frac{1}{20}& 0\\ 0& \frac{1}{20}& -\frac{1}{10}\end{array}\right)XisX\left(t\right)=11\left(\begin{array}{c}2\\ 2\\ 1\end{array}\right)+2\left(\begin{array}{c}\cos \frac{1}{20}t+\sin \frac{1}{20}t\\ -\cos \frac{1}{20}t\\ -\sin \frac{1}{20t}t\end{array}\right)e^{-\frac{1}{10}t}-6\left(\begin{array}{c}-\cos \frac{1}{20}t+\sin \frac{1}{20}t\\ -\sin \frac{1}{20}t\\ -cos\frac{1}{20t}t\end{array}\right)e^{-\frac{1}{10}t}$

Step-by-step solution

Determine the complex Eigen values

The polynomial of degree n of the variable $\lambda$is the characteristic polynomial of the n x n matrix A.

$p\left(\lambda \right)=det(\lambda I-A)$

Its root is the eigenvalue of A.

Determine the system in matrix form

(a)

For tank A,

$\frac{d{x}_1}{dt}=\text{Input Rate of Salt - Output Rate of Salt}$

$=-\frac{5}{100}{x}_1+\frac{5}{50}{x}_3$

$=-\frac{1}{20}{x}_1+\frac{1}{10}{x}_3$

For tank B,

$\frac{d{x}_2}{dt}=\text{Input Rate of Brine - Output Rate of Brine}$

$=\frac{5}{100}{x}_1-\frac{5}{100}{x}_2$

$=\frac{1}{20}{x}_1-\frac{1}{20}{x}_2$

For tank C,

$\frac{d{x}_2}{dt}=\text{Input Rate of Brine - Output Rate of Brine}$
$=\frac{5}{100}{x}_2-\frac{5}{50}{x}_3$

$=\frac{1}{20}{x}_2-\frac{1}{10}{x}_3$

So now we have

$\left\{\begin{array}{l}\frac{d{x}_1}{dt}=-\frac{1}{20}{x}_1+\frac{1}{10}{x}_3\\ \frac{d{x}_2}{dt}=\frac{1}{20}{x}_1-\frac{1}{20}{x}_2\\ \frac{d{x}_3}{dt}=\frac{1}{20}{x}_2-\frac{1}{10}{x}_3\end{array}\right.$

Which can be written in the matrix form

$X\text{'}=\left(\begin{array}{ccc}-\frac{1}{20}& 0& \frac{1}{10}\\ \frac{1}{20}& -\frac{1}{20}& 0\\ 0& \frac{1}{20}& -\frac{1}{10}\end{array}\right)X$

Where$A=\left(\begin{array}{ccc}-\frac{1}{20}& 0& \frac{1}{10}\\ \frac{1}{20}& -\frac{1}{20}& 0\\ 0& \frac{1}{20}& -\frac{1}{10}\end{array}\right)$

The System inthe matrix form is $A=\left(\begin{array}{ccc}-\frac{1}{20}& 0& \frac{1}{10}\\ \frac{1}{20}& -\frac{1}{20}& 0\\ 0& \frac{1}{20}& -\frac{1}{10}\end{array}\right)$

Determine the Eigen values

(b) Obtaining the eigenvalues of the coefficient matrix,

$$\begin{gathered} det(A-\lambda I)=0\to \left|\begin{array}{ccc}-\frac{1}{20}-\lambda & 0& \frac{1}{10}\\ \frac{1}{20}& -\frac{1}{20}-\lambda & 0\\ 0& \frac{1}{20}& -\frac{1}{10}-\lambda \end{array}\right|=0 \\ \left(-\frac{1}{20}-\lambda \right)\left|\begin{array}{cc}-\frac{1}{20}-\lambda & 0\\ \frac{1}{20}& -\frac{1}{10}-\lambda \end{array}\right|+\frac{1}{10}\left|\begin{array}{cc}\frac{1}{20}& -\frac{1}{20}-\lambda \\ 0& \frac{1}{20}\end{array}\right|=0 \\ {\left(-\frac{1}{20}-\lambda \right)}^2\left(-\frac{1}{10}-\lambda \right)+\frac{1}{4000}=0 \end{gathered}$$

$\left(\frac{1}{400}+\frac{1}{10}\lambda +{\lambda }^2\right)\left(-\frac{1}{10}-\lambda \right)+\frac{1}{400}=0$

$-\frac{1}{4000}-\frac{1}{100}\lambda -\frac{1}{10}{\lambda }^2-\frac{1}{400}\lambda -\frac{1}{10}{\lambda }^2-{\lambda }^3+\frac{1}{4000}=0$

$-{\lambda }^3-\frac{1}{5}{\lambda }^2-\frac{1}{80}\lambda =0$

$\lambda \left(-{\lambda }^2-\frac{1}{5}\lambda -\frac{1}{80}\right)=0$

For$-{\lambda }^2-\frac{1}{5}\lambda -\frac{1}{80}=0$, we have

$\lambda =\frac{0.2\pm \sqrt{{(0.2)}^2-4(-1)(-0.0125)}}{2(-1)}$

$=\frac{0.2\pm 0.1i}{-2}$

$=-\frac{1}{10}\pm \frac{i}{20}$

$=-0.1\pm 0.05i$

So our eigenvalues are ${\lambda }_1=0,{\lambda }_2=-0.1+0.05i$and ${\lambda }_3=\overline{{\lambda }_2}=-0.1-0.05i$

Determine the Eigen vector and a corresponding solution vector

For ${\lambda }_1=0$

$$\begin{gathered} \left(A-0I|0\right)=\left(\begin{array}{cccc}-0.05-0& 0& 0.1& 0\\ 0.05& -0.05-0& 0& 0\\ 0& 0.05& -0.1-0& 0\end{array}\right) \\ =\left(\begin{array}{cccc}-0.05& 0& 0.1& 0\\ 0.05& -0.05& 0& 0\\ 0& 0.05& -0.1& 0\end{array}\right) \end{gathered}$$

From the first row we have

$-0.05k_1+0.1k_2=0\to k_1=2k_3$

Choosing k3=1 yields k1=2.

And from the second row we see that

$0.05k_1-0.05k_2=0\to k_1=k_2$

Sok2 =2.

This gives an eigenvector and a corresponding solution vector

$K_1=\left(\begin{array}{c}2\\ 2\\ 1\end{array}\right),X_1=\left(\begin{array}{c}2\\ 2\\ 1\end{array}\right)$

For${\lambda }_{-}2=-0.1+0.05i:$

$(A-(-0.1+0.05i\left)L\right|0)=\left(\begin{array}{cccc}0.05-0.05i& 0& 0.1& 0\\ 0.05& 0.05-0.05i& 0& 0\\ 0& 0.05& -0.05i& 0\end{array}\right)$

From the first row we have

$(0.05-0.05i)k_1+0.1k_3=0\to k_1=(-1-i)k_3$

And from the third row

$0.05k_2-0.05i{k}_3=0\to k_2=i{k}_3$

Choosing k₃=i yields k₁=1-i and k₂=-1.

This gives an eigenvector

$K_2=\left(\begin{array}{c}1-i\\ -1\\ i\end{array}\right)=\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)+i\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)$

And column vectors

$B_1=\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right),B_2=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)$

Determine the general solution of the system

Also,

$\lambda =\alpha +\beta i\to \lambda =-\frac{1}{10}+\frac{1}{20}i$

Where$\alpha =-\frac{1}{10}$and$\beta =\frac{1}{20}$

So the corresponding solution vectors are

$$\begin{gathered} X_{\mathbf{2}}=\left[B_1\cos \beta t-B_2\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)\cos \frac{1}{20}t-\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)\sin \frac{1}{20}t\right]e^{-\frac{1}{10}t} \\ =\left(\begin{array}{c}\cos \frac{1}{20}t+\sin \frac{1}{20}t\\ -\cos \frac{1}{20}t\\ -\sin \frac{1}{20}t\end{array}\right)e^{-\frac{1}{10}t} \end{gathered}$$

And,

$$\begin{gathered} X_3=\left[B_2\cos \beta t+B_1\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)\cos \frac{1}{20}t+\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)\sin \frac{1}{20}t\right]e^{-\frac{1}{10}t} \\ =\left(\begin{array}{c}-\cos \frac{1}{20}t+\sin \frac{1}{20}t\\ -\sin \frac{1}{20}t\\ \cos \frac{1}{20}t\end{array}\right)e^{-\frac{1}{10}t} \end{gathered}$$

The general solution of the system is therefore,

$$\begin{gathered} X=c_1{X}_1+c_2{X}_{\mathbf{2}}+c_3{X}_{\mathbf{3}} \\ =c_1\left(\begin{array}{c}2\\ 2\\ 1\end{array}\right)+c_2\left(\begin{array}{c}\cos \frac{{\displaystyle 1}}{{\displaystyle 20}}t+\sin \frac{{\displaystyle 1}}{{\displaystyle 20}}t\\ -\cos \frac{{\displaystyle 1}}{{\displaystyle 20}}t\\ -\sin \frac{{\displaystyle 1}}{{\displaystyle 20}}t\end{array}\right)e^{-\frac{{\displaystyle 1}}{{\displaystyle 10}}t}+c_3\left(\begin{array}{c}-\cos \frac{1}{20}t+\sin \frac{1}{20}t\\ -\sin \frac{1}{20}t\\ \cos \frac{1}{20}t\end{array}\right)e^{-\frac{1}{10}t} \end{gathered}$$

Determine the initial value condition

Applying the initial value condition

$$\begin{gathered} X\left(0\right)=\left(\begin{array}{c}30\\ 20\\ 5\end{array}\right) \\ \left(\begin{array}{c}30\\ 20\\ 5\end{array}\right)=c_1\left(\begin{array}{c}2\\ 2\\ 1\end{array}\right)+c_2\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)+c_3\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right) \end{gathered}$$

Here we have

$$\begin{gathered} 2c_1+c_2-c_3=30.........\left(1\right) \\ 2c_1-c_2=20.......\left(2\right) \\ {c}_1+c_3=5.........\left(3\right) \end{gathered}$$

Rearranging equation (2),

$c_2=2c_1-20$

We substitute this into (1),

$$\begin{gathered} 2c_1+2c_1-20-c_3=30 \\ 4c_1-c_3=50 \end{gathered}$$

Adding this equation to (3) yields

$$\begin{gathered} 5c_1=55 \\ {c}_1=11 \end{gathered}$$

To solve for c₂ we substitute this value of c₁ that we obtained into(2),

$$\begin{gathered} 2\left(11\right)-c_2=20 \\ {c}_2=2 \end{gathered}$$

And to solve for c₃ we substitute the value of c₁ into (3),

$$\begin{gathered} 11+c_3=5 \\ {c}_3=-6 \end{gathered}$$

Therefore, the solution that satisfies the initial condition is

$X\left(t\right)=11\left(\begin{array}{c}2\\ 2\\ 1\end{array}\right)+2\left(\begin{array}{c}\cos \frac{1}{20}t+\sin \frac{1}{20}t\\ -\cos \frac{1}{20}t\\ -\sin \frac{1}{20t}t\end{array}\right)e^{-\frac{1}{10}t}-6\left(\begin{array}{c}-\cos \frac{1}{20}t+\sin \frac{1}{20}t\\ -\sin \frac{1}{20}t\\ -cos\frac{1}{20t}t\end{array}\right)e^{-\frac{1}{10}t}$