# Thread: Help with a long applied DE problem.

1. ## Help with a long applied DE problem.

Consider two tanks, A and B, each holding 200 litres of water. A pipe pumps water from tank A to tank B at a rate of 5l/min. At the same time another pipe pumps liquid from tank B to tank A at the same rate. At time t=0, $\displaystyle x_0$kg of a chemical X is dissolved into tank A, and tank B has $\displaystyle y_0$kg of the same chemical X dissolved into it.
i). Write down the system of differential equations satisfied by x(t) and y(t), the quantity of the chemical X in tanks A and B respectively.
I put this:

$\displaystyle x(t)=\frac{-5x_0}{200}+\frac{5y_0}{200}=\frac{1}{40}(-x_0+y_0)$
$\displaystyle y(t)=\frac{5x_0}{200}-\frac{5y_0}{200}=\frac{1}{40}(x_0-y_0)$

ii). Find the eigenvalues and the eigenvectors of the resulting matrix form.
$\displaystyle \frac{1}{40}\begin{pmatrix} {-1}&{1}\\ {1}&{-1} \end{pmatrix}\begin{pmatrix} {x_0}\\ {y_0} \end{pmatrix}=\begin{pmatrix} {x(t)}\\ {y(t)} \end{pmatrix}$

$\displaystyle \frac{1}{40}\begin{vmatrix} {-1-\lambda}&{1}\\ {1}&{-1-\lambda} \end{vmatrix}=\frac{1}{40}((1+\lambda)^2-1)=0$

$\displaystyle (1+\lambda^2)-1=0$
$\displaystyle \lambda^2+2\lambda=0 \Rightarrow \lambda(\lambda+2)=0$

The eigenvalues are 0 and -2.

$\displaystyle \lambda=0$, Eigenvector:$\displaystyle \begin{pmatrix} {-1}\\ {1} \end{pmatrix}$

$\displaystyle \lambda=-2$, Eigenvector: $\displaystyle \begin{pmatrix} {1}\\ {1} \end{pmatrix}$

iii). Show that the amount of the chemical X in either tank approaches $\displaystyle \frac{1}{2}(x_0+y_0)$ as t approaches infinity.
$\displaystyle X(t)=A\begin{pmatrix} {-1}\\ {1} \end{pmatrix}+Be^{-2t}\begin{pmatrix} {1}\\ {1} \end{pmatrix}$

At t=0:

$\displaystyle x(0)=x_0$
$\displaystyle y(0)=y_0$

$\displaystyle x_0=B-A$
$\displaystyle y_0=A+B$

Working this out gives:

$\displaystyle B=\frac{1}{2}(x_0+y_0)$
$\displaystyle A=\frac{1}{2}(y_0-x_0)$

$\displaystyle X(t)=\frac{1}{2}(y_0-x_0)\begin{pmatrix} {-1}\\ {1} \end{pmatrix}+\frac{1}{2}(x_0+y_0)e^{-2t}\begin{pmatrix} {1}\\ {1} \end{pmatrix}$

as $\displaystyle t \rightarrow \infty$:

$\displaystyle X(t) \rightarrow \begin{pmatrix} {\frac{1}{2}(x_0+y_0)}\\ {\frac{1}{2}(y_0-x_0} \end{pmatrix}$

I get really far but this doesn't work out. It suggests that my inital formulae are wrong but I can't see where my mistake is. If I reversed my A and B my formulae would work, but I can't see how I can do this.

Help would be appreciated greatly!

2. Hello,

Just a remark (or a correction) :
I put this:

$\displaystyle x(t)=\frac{-5x_0}{200}+\frac{5y_0}{200}=\frac{1}{40}(-x_0+y_0)$
$\displaystyle y(t)=\frac{5x_0}{200}-\frac{5y_0}{200}=\frac{1}{40}(x_0-y_0)$
How are these differential equations ?
There is not even t in these expressions

This is how I see the problem.
Let's care about tank A first.
There are initially $\displaystyle x_0$ kg of X. So the ratio chemical/litre is $\displaystyle \frac{x_0}{200}$
Hence if you remove a certain amount of litres, $\displaystyle \alpha$, there will be $\displaystyle \alpha \cdot \frac{x_0}{200}$ that will be removed.

Now, assuming that t is in minutes... We know that 5 litres are removed within 1 minutes.
So during t minutes, there are $\displaystyle 5t$ litres that are removed.

Can you see that if we consider the amount of chemical X removed during t, $\displaystyle \alpha=5t$ ?

So for tank A, there are $\displaystyle \frac{x_0}{40} t$ kg of X that will be removed and go into tank B.

Similarly, there are $\displaystyle \frac{y_0}{40}t$ kg of X that will be removed from tank B and that will go into tank A.

(don't forget that there is initially a certain amount of chemical !)

$\displaystyle x(t)=x_0-\frac{x_0}{40}t+\frac{y_0}{40} t=x_0-t \left(\frac{y_0}{40}-\frac{x_0}{40}\right)$

and $\displaystyle y(t)=y_0-\frac{y_0}{40}t+\frac{x_0}{40} t=y_0-t \left(\frac{x_0}{40}-\frac{y_0}{40}\right)$

do tell me if it looks incorrect to you

3. Okay, I understand where that came from. It does make more sense than what I put, I forgot to incude a time factor.

I might be doing something wrong, but with those equations in matrix form I get:

$\displaystyle \begin{pmatrix} {1+\frac{t}{40}}&{\frac{-t}{40}}\\ {\frac{-t}{40}}&{1+\frac{t}{40}} \end{pmatrix}\begin{pmatrix} {x_0}\\ {y_0} \end{pmatrix}=\begin{pmatrix} {x(t)}\\ {y(t)} \end{pmatrix}$

When I find the determinant I get:

$\displaystyle \lambda^2-2(1+\frac{t}{40}) \lambda+(1+\frac{t}{40})^2-\frac{t^2}{40^2}=0$

This doesn't look right :s

I was thinking that I have to remove t so I just have constants in the 2x2 matrix. However, I don't kno how to do this.

4. Originally Posted by Moo
(don't forget that there is initially a certain amount of chemical !)

$\displaystyle x(t)=x_0-\frac{x_0}{40}t+\frac{y_0}{40} t=x_0-t \left(\frac{y_0}{40}-\frac{x_0}{40}\right)$

and $\displaystyle y(t)=y_0-\frac{y_0}{40}t+\frac{x_0}{40} t=y_0-t \left(\frac{x_0}{40}-\frac{y_0}{40}\right)$

do tell me if it looks incorrect to you
Looks perfect to me...however, there is a slight problem: they're not differential equations!!!!

Differentiating the two equations will give you the system you're looking for: \displaystyle \left\{\begin{aligned} \dot{x}(t)&=\tfrac{1}{40}\left(x_0-y_0\right)\\ \dot{y}(t)&=\tfrac{1}{40}\left(y_0-x_0\right) \end{aligned}\right.

...which happens to be the system [although there's a little something wrong with his system notation] that Showcase_22 used!!

5. I think you wrote those the wrong way round (did you? or am I missing something?? =S)

So if my initial DEs are correct then the mistake must be later on.

I know i'm close because I get $\displaystyle \frac{1}{2}(x_0+y_0)$ appearing towards the end. I just can't see where the hole in my logic lies.