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Math Help - Solving a system of differential equations using central difference approximations

  1. #1
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    Solving a system of differential equations using central difference approximations

    Good day,

    I am having a bit a trouble solving this system:

    y^{'} = v

    v^{'} = \lambda y

    Incorporating the following central difference approximations:

    v_{i} =\frac{y_{i+1}-y_{i-1}}{2\Delta x}



    \lambda y_{i} = \frac{v_{i+1}-v_{i-1}}{2\Delta x}

    I've made an attempt to solve this using a method I use when I encounter linear de's rather than a system. I don't know if its valid in a system case...

    First thing, I shifted the indices for v_{i}

    ie. v_{i+1} =\frac{y_{i+2}-y_{i}}{2\Delta x}
    v_{i-1} =\frac{y_{i}-y_{i-2}}{2\Delta x}

    Substituting these into the \lambda y_{i}:

    y_{i+2}+y_{i}\alpha+y_{i-2}=0
    Where, \alpha = -2-4(\Delta x)^{2}\lambda

    From here I let y_{i}=n^{i}, sub it into the equation and basically solve for n.

    I get a solution for y_{i} that contains like 6 constants.

    I use this to solve for v_{i} which results in a long solution with many square roots.

    As I mentioned I'm highly doubtful if this is actually how you do it... Any help would be greatly appreciated.

    Thanks
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  2. #2
    Grand Panjandrum
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    Re: Solving a system of differential equations using central difference approximation

    Quote Originally Posted by MrPhil View Post
    Good day,

    I am having a bit a trouble solving this system:

    y^{'} = v

    v^{'} = \lambda y

    Incorporating the following central difference approximations:

    v_{i} =\frac{y_{i+1}-y_{i-1}}{2\Delta x}



    \lambda y_{i} = \frac{v_{i+1}-v_{i-1}}{2\Delta x}

    I've made an attempt to solve this using a method I use when I encounter linear de's rather than a system. I don't know if its valid in a system case...

    First thing, I shifted the indices for v_{i}

    ie. v_{i+1} =\frac{y_{i+2}-y_{i}}{2\Delta x}
    v_{i-1} =\frac{y_{i}-y_{i-2}}{2\Delta x}

    Substituting these into the \lambda y_{i}:

    y_{i+2}+y_{i}\alpha+y_{i-2}=0
    Where, \alpha = -2-4(\Delta x)^{2}\lambda

    From here I let y_{i}=n^{i}, sub it into the equation and basically solve for n.

    I get a solution for y_{i} that contains like 6 constants.

    I use this to solve for v_{i} which results in a long solution with many square roots.

    As I mentioned I'm highly doubtful if this is actually how you do it... Any help would be greatly appreciated.

    Thanks
    Seems like a lot of work, who said to use central differences?

    CB
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  3. #3
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    Re: Solving a system of differential equations using central difference approximation

    Quote Originally Posted by CaptainBlack View Post
    Seems like a lot of work, who said to use central differences?

    CB
    In a lecture, we had two examples(that weren't systems) and solved them using finite difference methods. We were left to attempt this question along with the two central difference approximations. I would have used Adam's Method to solve something like this.
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  4. #4
    Grand Panjandrum
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    Re: Solving a system of differential equations using central difference approximation

    Quote Originally Posted by MrPhil View Post
    In a lecture, we had two examples(that weren't systems) and solved them using finite difference methods. We were left to attempt this question along with the two central difference approximations. I would have used Adam's Method to solve something like this.
    Write your system as a vector first order ODE:

    {\bf{X}}'=f(t,{\bf{X}})={\bf{AX}}

    where {\bf{X}}=(y,v)^T and {\bf{A}}=\left[\begin{array}{cc}0&1\\\lambda&0 \end{array}\right]

    Now just write your method out in terms of vector \bf{X} and everything should work as in the scalar case.

    But I still always use Euler's method or a (possibly iterated) first order predictor corrector method for these things, but I know that I am in a small minority.

    CB
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