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Math Help - Help with a long applied DE problem.

  1. #1
    Super Member Showcase_22's Avatar
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    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, x_0kg of a chemical X is dissolved into tank A, and tank B has y_0kg 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:

    x(t)=\frac{-5x_0}{200}+\frac{5y_0}{200}=\frac{1}{40}(-x_0+y_0)
    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.
    \frac{1}{40}\begin{pmatrix}<br />
{-1}&{1}\\ <br />
{1}&{-1}<br />
\end{pmatrix}\begin{pmatrix}<br />
{x_0}\\ <br />
{y_0}<br />
\end{pmatrix}=\begin{pmatrix}<br />
{x(t)}\\ <br />
{y(t)}<br />
\end{pmatrix}

    \frac{1}{40}\begin{vmatrix}<br />
{-1-\lambda}&{1}\\ <br />
{1}&{-1-\lambda}<br />
\end{vmatrix}=\frac{1}{40}((1+\lambda)^2-1)=0

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

    The eigenvalues are 0 and -2.

    \lambda=0, Eigenvector: \begin{pmatrix}<br />
{-1}\\ <br />
{1}<br />
\end{pmatrix}

    \lambda=-2, Eigenvector: \begin{pmatrix}<br />
{1}\\ <br />
{1}<br />
\end{pmatrix}

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

    At t=0:

    x(0)=x_0
    y(0)=y_0

    x_0=B-A
    y_0=A+B

    Working this out gives:

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

    X(t)=\frac{1}{2}(y_0-x_0)\begin{pmatrix}<br />
{-1}\\ <br />
{1}<br />
\end{pmatrix}+\frac{1}{2}(x_0+y_0)e^{-2t}\begin{pmatrix}<br />
{1}\\ <br />
{1}<br />
\end{pmatrix}

    as t \rightarrow \infty:

    X(t) \rightarrow \begin{pmatrix}<br />
{\frac{1}{2}(x_0+y_0)}\\ <br />
{\frac{1}{2}(y_0-x_0}<br />
\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!
    Last edited by Showcase_22; November 30th 2008 at 01:30 PM.
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  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
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    Hello,

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

    x(t)=\frac{-5x_0}{200}+\frac{5y_0}{200}=\frac{1}{40}(-x_0+y_0)
    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 x_0 kg of X. So the ratio chemical/litre is \frac{x_0}{200}
    Hence if you remove a certain amount of litres, \alpha, there will be \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 5t litres that are removed.

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

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



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



    So your equations are :
    (don't forget that there is initially a certain amount of chemical !)

    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 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
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  3. #3
    Super Member Showcase_22's Avatar
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    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:

    \begin{pmatrix}<br />
{1+\frac{t}{40}}&{\frac{-t}{40}}\\ <br />
{\frac{-t}{40}}&{1+\frac{t}{40}}<br />
\end{pmatrix}\begin{pmatrix}<br />
{x_0}\\ <br />
{y_0}<br />
\end{pmatrix}=\begin{pmatrix}<br />
{x(t)}\\ <br />
{y(t)}<br />
\end{pmatrix}

    When I find the determinant I get:

    \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.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Moo View Post
    So your equations are :
    (don't forget that there is initially a certain amount of chemical !)

    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 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: \left\{\begin{aligned}<br />
\dot{x}(t)&=\tfrac{1}{40}\left(x_0-y_0\right)\\<br />
\dot{y}(t)&=\tfrac{1}{40}\left(y_0-x_0\right)<br />
\end{aligned}\right.<br />

    ...which happens to be the system [although there's a little something wrong with his system notation] that Showcase_22 used!!
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  5. #5
    Super Member Showcase_22's Avatar
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    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 \frac{1}{2}(x_0+y_0) appearing towards the end. I just can't see where the hole in my logic lies.
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