# Solving Three-Tank System Modeled by ODEs

• zaboda42
In summary: That's an error.In summary, the three-tank system modeled by the given equations can be solved using eigenvalues and eigenvectors to find a general solution. The solution can then be verified by substituting in the initial conditions. As t→∞, the amounts of grain in each tank will tend to [4, 10, 4]T, representing a stable relationship. However, there may be some errors in the substitution of the initial conditions that need to be rechecked.
zaboda42
Consider a three-tank system modeled by the equations:

$$x_1' = -5x_1+5x_3$$

$$x_2' = 5x_1-2x_2$$

$$x_3' = 2x_2-5x_3$$

(A) Initially there are 10 pounds of grain in each tank. What will the amounts be as $$t \rightarrow \infty$$?

(B) Solve the system and verify your conclusion from (A).

I'm not really sure how I can say anything about (A), so I would appreciate some help there. Here's how I worked out (B), but I'm not really sure how I can finish to "verify my conclusion from (A)":

$${\bf{x}}' = \left[ \begin{array}{cccc} -5&0&5\\5&-2&0\\0&2&-5 \end{array} \right]\left[ \begin{array}{cccc} x_1\\x_2\\x_3 \end{array} \right]$$

$${\bf{A}}-\lambda {\bf{I}} = \left[ \begin{array}{cccc} -5-\lambda&0&5\\5&-2-\lambda&0\\0&2&-5-\lambda \end{array} \right]$$

Which gives $$-\lambda(\lambda^2+12\lambda+45) = 0$$ for solutions $$\lambda = 0$$ and $$\lambda = -6\pm3i$$

In the case of $$\lambda = 0$$:

$$({\bf{A}}-0{\bf{I}}){\bf{v}} \Rightarrow \left[ \begin{array}{cccc} -5&0&5\\5&-2&0\\0&2&-5 \end{array} \right]\left[ \begin{array}{cccc} a\\b\\c \end{array} \right] = \left[ \begin{array}{cccc} 0\\0\\0 \end{array} \right]$$

Which gives the system:

$$-5a+5c = 0$$

$$5a-2b = 0$$

$$2b-5c = 0$$

Thus, $${\bf{x}}_0 = \left[ \begin{array}{cccc} 2\\5\\2 \end{array} \right]$$.

In the case of $$\lambda = -6-3i$$:

$$({\bf{A}}-(-6-3i){\bf{I}}){\bf{v}} \Rightarrow \left[ \begin{array}{cccc} 1+3i&0&5\\5&4+3i&0\\0&2&1+3i \end{array} \right]\left[ \begin{array}{cccc} a\\b\\c \end{array} \right] = \left[ \begin{array}{cccc} 0\\0\\0 \end{array} \right]$$

Which gives $${\bf{x}} = \left[ \begin{array}{cccc} 5\\-4+3i\\-1-3i \end{array} \right]$$

So we have:

$${\bf{x}}(t) = \left[ \begin{array}{cccc} 5\\-4+3i\\-1-ei \end{array} \right]e^{(-6-3i)t}$$

$${\bf{x}}(t) = \left[ \begin{array}{cccc} 5\\-4+3i\\-1-ei \end{array} \right]e^{-6t}(\cos3t-i\sin3t)$$

$${\bf{x}}(t) = e^{-6t}\left[ \begin{array}{cccc} 5\cos3t-5i\sin3t\\3\sin3t-4\cos3t+i(4\sin3t+3\cos3t)\\-3\sin3t-\cos3t+i(\sin3t-3\cos3t) \end{array} \right]$$

And the real and imaginary parts of $${\bf{x}}(t)$$ are real-valued solutions:

$${\bf{x}}_1(t) = e^{-6t}\left[ \begin{array}{cccc} 5\cos3t\\3\sin3t-4\cos3t\\-3\sin3t-\cos3t \end{array} \right]$$

$${\bf{x}}_2(t) = e^{-6t}\left[ \begin{array}{cccc} 5\sin3t\\4\sin3t+4\cos3t\\\sin3t-3\cos3t \end{array} \right]$$

So the general solution can be written as:

$${\bf{x}}(t) = c_1\left[ \begin{array}{cccc} 2\\5\\2 \end{array} \right] + c_2e^{-6t}\left[ \begin{array}{cccc} 5\cos3t\\3\sin3t-4\cos3t\\-3\sin3t-\cos3t \end{array} \right] + c_3e^{-6t}\left[ \begin{array}{cccc} 5\sin3t\\4\sin3t+3\cos3t\\\sin3t-3\cos3t \end{array} \right]$$

With the initial condition, we have a system:

$$2c_1+5c_3 = 10$$

$$5c_1+3c_2+4c_3 = 10$$

$$2c_1 - 3c_2+c_3 = 10$$

So that $$c_1 = 2$$, $$c_2 = -\frac{8}{5}$$, and $$c_3 = \frac{6}{5}$$ so that our solution becomes:

$${\bf{x}}(t) = 2\left[ \begin{array}{cccc} 2\\5\\2 \end{array} \right] -\frac{8}{5}e^{-6t}\left[ \begin{array}{cccc} 5\cos3t\\3\sin3t-4\cos3t\\-3\sin3t-\cos3t \end{array} \right] + \frac{6}{5}e^{-6t}\left[ \begin{array}{cccc} 5\sin3t\\4\sin3t+3\cos3t\\\sin3t-3\cos3t \end{array} \right]$$

I hope I did everything correctly, but what does this tell me about the grain as $$t \rightarrow \infty$$?

As t→∞, two of those terms will tend to zero, no? That leaves you with [4, 10, 4]T.
To check whether that makes sense, substitute those in the ODEs and discover that all three derivatives vanish. So far so good - the limit values represent a stable relationship.
Another check you can do is for an invariant, namely a linear combination of the derivatives which is identically zero. In this case, simply adding the three ODEs gives that:
x1'+x2'+x3' = 0
I.e. x1+x2+x3 should be constant. It starts at 30, but you have it tending to 18. Must be something wrong.
There seem to be multiple errors where you substituted in the initial conditions.

Thanks for the response. I'm not sure where my "multiple errors" are occurring. Can you lend some insight as to where I should be rechecking?

zaboda42 said:
Thanks for the response. I'm not sure where my "multiple errors" are occurring. Can you lend some insight as to where I should be rechecking?
Looks to me like you consistently evaluated cos(0) as 0 and sin(0) as 1.

## 1. What is a three-tank system modeled by ODEs?

A three-tank system modeled by ODEs is a mathematical model used to describe the behavior of a system with three interconnected tanks, where the movement of fluid in each tank is governed by a set of ordinary differential equations (ODEs). This model is commonly used in chemical engineering and fluid dynamics to understand and control the flow of fluids in a complex system.

## 2. Why is it important to solve a three-tank system modeled by ODEs?

Solving a three-tank system modeled by ODEs allows us to predict the behavior of the system and make informed decisions about how to control it. By understanding how changes in one tank affect the levels and flows in the other tanks, we can optimize the system for desired outcomes and avoid potential issues such as overflow or depletion of resources.

## 3. What are some common techniques for solving a three-tank system modeled by ODEs?

Some common techniques for solving a three-tank system modeled by ODEs include numerical methods such as Euler's method, Runge-Kutta methods, and the finite difference method. These methods involve breaking down the system into smaller time intervals and approximating the solution at each interval, eventually converging to a more accurate solution. Additionally, analytical methods such as Laplace transforms and series solutions can be used for simpler systems.

## 4. What are the challenges in solving a three-tank system modeled by ODEs?

One of the main challenges in solving a three-tank system modeled by ODEs is the complexity of the equations and the interdependence of the variables. This can make it difficult to find an analytical solution, and numerical methods may require a significant amount of computation. Additionally, the accuracy of the solution may be affected by errors in initial conditions and parameters.

## 5. How can the solution of a three-tank system modeled by ODEs be used in practical applications?

The solution of a three-tank system modeled by ODEs can be used in practical applications to optimize and control the behavior of the system. For example, in chemical engineering, the solution can be used to determine the optimal flow rates and tank levels to achieve desired reactions or avoid hazardous situations. In fluid dynamics, the solution can be used to predict the behavior of a system and make adjustments to avoid potential issues such as overflow or depletion.

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