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Math Help - Differential equations - how to form them?

  1. #1
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    Differential equations - how to form them?

    Forum,

    I seek to gain an understanding of how differential equations are arrived at.

    In most cases, intuition, basic knowledge of physics, or computer simulation can tell us that the unknown variable is dependent on one or more variables. How this translates into an equation is not clear to me.

    I ask because I have solved hundreds of ODE/PDEs, but given a sample real-world situation, how would I even formulate the problem, before solving it, seems to be the challenge.

    Would appreciate advise.

    Best,
    wirefree
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  2. #2
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    Quote Originally Posted by wirefree View Post
    Forum,

    I seek to gain an understanding of how differential equations are arrived at.

    In most cases, intuition, basic knowledge of physics, or computer simulation can tell us that the unknown variable is dependent on one or more variables. How this translates into an equation is not clear to me.

    I ask because I have solved hundreds of ODE/PDEs, but given a sample real-world situation, how would I even formulate the problem, before solving it, seems to be the challenge.

    Would appreciate advise.

    Best,
    wirefree
    Read a book on the subject eg. Amazon.com: Modelling with Differential and Difference Equations (Australian Mathematical Society Lecture Series): Glenn Fulford, Peter Forrester, Arthur Jones: Books
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  3. #3
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    Appreciate the response, mr fantastic.

    I would be much obliged should you be able to provide a little foretaste of the subject. The concept accompanied by a small example would prove beneficial.

    A knack for compression is the mark of brilliance.

    Best regards,
    wirefree
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  4. #4
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    Quote Originally Posted by wirefree View Post
    Appreciate the response, mr fantastic.

    I would be much obliged should you be able to provide a little foretaste of the subject. The concept accompanied by a small example would prove beneficial.

    A knack for compression is the mark of brilliance.

    Best regards,
    wirefree
    A small example from fluid mechanics:

    An irrotational flow is defined a flow having zero vorticity. Vorticity is defined as the curl of the velocity vector (  \vec{V} = (V_x,V_y,V_z)) of a flow:

     \nabla \times \vec{V} = 0

    An identity from vector calculus gives:

     curl(grad \phi) = \nabla \times \nabla \phi = 0

    Hence, for an irrotational flow, there exists a scalar function  \phi whose gradient is equal to the velocity vector of the flow.

     grad \phi =  \vec{V}.

    Or:

     (\frac{d \phi}{dx},\frac{d \phi}{dy},\frac{d \phi}{dz}) = (V_x, V_y, V_z).

    The continuty (mass conservation) equation of an incompressible flow can be written:

     \nabla . \vec{V} = 0

    And hence:

     \nabla . (\nabla \phi) = 0

     => \nabla ^2 \phi = 0

     => \frac{d^2 \phi}{dx^2} + \frac{d^2 \phi}{dy^2}+\frac{d^2 \phi}{dz^2} =0.

    This is Laplace's equation, which is a linear 2nd order homogeneous differential equation. It governs ALL fluid flows which are both irrotational and incompressible.

    The concept: We started with the definitions of an irrotational flow, and an incompressible flow and expressed them in terms of vector calculus using the nabla vector (which is defined as:  \nabla = (\frac{d}{dx},\frac{d}{dy},\frac{d}{dz}) ). Combining these definitions with the identity  curl(grad f) = 0 , we managed to deduce a linear 2nd order homogeneous differential equation which governs the physics behind ALL flows of this type.
    Last edited by Mush; December 26th 2008 at 07:26 PM.
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  5. #5
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    Appreciate the response, Mush. The example you illustrated was a very sophisticated one and very well formulated at that by you.

    As closure, I will be persistent in reducing the thread down to the simplest of ODEs I fished in my course text, right off page 1, and request some thoughts on the nature & complexity of problem it might address and how it might have been formulated:

    dy/dx + 5y = e^x

    Appreciate all advise.

    Best regards,
    wirefree
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