Salt Contraction In Connected Tanks With Inflow and Outflow

A system consists of two 50 litre tanks. Salt solution with a concentration of 0.1kg/litre is pumped into tank 1 and pure water is pumped into tank 2. Water is allowed to circulate between the tanks. Water is pumped out of tank 1 to maintain two full tanks. What will be the salt concentration in each tank at  
\[t \gt 0\]
?

salt concentration in two tanks with inflow and outflow

The rate at which the amount of salt in each tank changes is the difference between the inflow and the outflow. Let the amount of salt in tanks 1 and 2 be  
\[x_1, \: x_2\]
  respectively.
For tank 1  
\[\frac{dx_1}{dt}=INFLOW-OUTFLOW=0.1+3\frac{x_2}{50}-2 \frac{x_1}{50}-2 \frac{x_1}{50}=-4 \frac{x_1}{50}+3 \frac{x_2}{50}+0.1\]
.
For tank 2  
\[\frac{dx_2}{dt}=INFLOW-OUTFLOW=2\frac{dx_1}{dt}-3 \frac{dx_2}{dt}\]
.
We can write this in matrix form as
\[\begin{pmatrix} \frac{dx_1}{dt} \\ \frac{dx_2}{dt} \end{pmatrix}= \left( \begin{array}{cc} -4/50 & 3/50 \\ 2/50 & -3/50 \end{array} \right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0.1 \\ 0 \end{pmatrix} \]

We can solve this system by diagonalizing the matrix. First find the eigenvalues. Solve
\[ det \left(\left( \begin{array}{cc} -4/50 & 3/50 \\ 2/50 & -3/50 \end{array} \right) - \lambda \left( \begin{array}{cc} -4/50 & -3/50 \\ 2/50 & -3/50 \end{array} \right) \right)=0\]

\[\begin{equation} \begin{aligned} det \left(\left( \begin{array}{cc} -4/50 - \lambda & 3/50 \\ 2/50 & -3/50 - \lambda \end{array} \right) \right) &= (-4/50 - \lambda )(-3/50-\lambda)- 3/50 \times 2/50 \\ &= \lambda^2+7 \frac{\lambda}{50}+ \frac{6}{2500} \\ &= (\lambda +1/50)(\lambda +6/50) \\ &= 0 \end{aligned} \end{equation}\]

Hence  
\[\lambda_1=-1/50, \lambda_2=-6/50\]
.
Now find the eigenvectors.
\[\lambda_1=-1/50\]
.
\[-1/50 \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}= \left( \begin{array}{cc} -4/50 & 3/50 \\ 2/50 & -3/50 \end{array} \right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \rightarrow \left( \begin{array}{cc} -3/50 & 3/50 \\ 2/50 & -2/50 \end{array} \right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\]

Hence  
\[-3/50x_1+3/50x_2=0, \: 2/50x_1-2/50x_2=0\]
.
We can take  
\[x_1=x_2=1\]
  then the eigenvector corresponding to  
\[\lambda_1=-1/50\]
  is  
\[\begin{pmatrix}1\\1\end{pmatrix}\]
.
\[\lambda_1=-6/50\]
.
\[-6/50 \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}= \left( \begin{array}{cc} -4/50 & 3/50 \\ 2/50 & -3/50 \end{array} \right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \rightarrow \left( \begin{array}{cc} 2/50 & 3/50 \\ 2/50 & 3/50 \end{array} \right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\]

Hence  
\[2/50x_1+3/50x_2=0, \: 2/50x_1+3/50x_2=0\]
.
We can take  
\[x1_-3=x_2=2\]
  then the eigenvector corresponding to  
\[\lambda_1=-6/50\]
  is  
\[\begin{pmatrix}-3\\2\end{pmatrix}\]
.
The matrix of eigenvectors is  
\[P=\left( \begin{array}{cc} 1 & -3 \\ 1 & 2 \end{array} \right)\]
Make the transformation  
\[\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = P \begin{pmatrix} u_1 \\ u_2 \end{pmatrix}\]
  then the system becomes
\[\frac{d}{dt} \left( \begin{array}{cc} 1 & -2 \\ 1 & 3 \end{array} \right) \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \left( \begin{array}{cc} -4/50 & 3/50 \\ 2/50 & -3/50 \end{array} \right) \left( \begin{array}{cc} 1 & -2 \\ 1 & 3 \end{array} \right) \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} + \begin{pmatrix} 0.1 \\ 0 \end{pmatrix} \]

Multiply throughout by  
\[\left( \begin{array}{cc} 1 & -3 \\ 1 & 2 \end{array} \right)^{-1} = 1/5 \left( \begin{array}{cc} 2 & 3 \\ -1 & 1 \end{array} \right)\]
  to get
\[\frac{d}{dt} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \left( \begin{array}{cc} -1/50 & 0 \\ 0 & -6/50 \end{array} \right) \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} + \begin{pmatrix} 0.04 \\ -0.02 \end{pmatrix} \]

this is equivalent to the differential equations
\[\frac{du_1}{dt}=-1/50u_1+0.04\]

\[\frac{du_2}{dt}=-6/50u_1-0.02\]

These have the solutions
\[u_1=2+Ae^{-t/50}, \: u_2= -1/6+Be^{-6t/50}\]

We can write this as  
\[ \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \begin{pmatrix} 2+Ae^{-t/50} \\ -1/6+Be^{-6t/50} \end{pmatrix}\]

Now transform back to  
\[x_1, \: x_2\]
.
\[\begin{equation} \begin{aligned} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} &= P \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} \\ &= \left( \begin{array}{cc} 1 & -3 \\ 1 & 2 \end{array} \right) \begin{pmatrix} 2+Ae^{-t/50} \\ -1/6+Be^{-6t/50} \end{pmatrix} \\ &= \begin{pmatrix} 5/2+Ae^{-t/50}-3Be^{-6t/50} \\ 5/3+Ae^{-t/50}+2Be^{-6t/50} \end{pmatrix} \end{aligned} \end{equation} \]
.
These solutions can be fitted to any initial conditions to find the constants.

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