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Strategy in solving vector equations involving grad, scalar product operators?

  1. Jul 3, 2011 #1
    What is the general strategy in solving vector equations involving grad and the scalar product?

    In particular, I want to express [itex]\Lambda[/itex] in terms from [itex] \mathbf U \cdot \nabla\Lambda = \Phi[/itex] but it looks impossible, unless there is some vector identity I can use.

    Thanks in advance.
     
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  3. Jul 3, 2011 #2

    I like Serena

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    Hi jason12345! :smile:

    What you have there is a directional derivative.
    See for instance wiki: http://en.wikipedia.org/wiki/Directional_derivative

    If U is for instance a unit vector, then:
    [tex]\vec U \cdot \vec\nabla \Lambda(\vec x) = \Phi(\vec x)[/tex]
    is also written as:
    [tex]\vec\nabla_{\vec U} \Lambda(\vec x) = \Phi(\vec x)[/tex]

    You can find [itex]\Lambda[/itex] with for instance something like:
    [tex]\Lambda(\vec x) = \int \Phi(\vec x + u \vec U) du[/tex]
     
  4. Jul 3, 2011 #3
    That's given me a lot to think about - thanks!

    Maybe I could try for something simpler to start with so:

    How would I find [itex]\nabla\Lambda[/itex] as the other terms with [itex]U[/itex] independent of (x,y,z)?

    Perhaps I could use some vector identity?
     
    Last edited: Jul 3, 2011
  5. Jul 4, 2011 #4

    ideasrule

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    I don't think it's possible to solve. There's an infinite number of solutions, even after we disregard the ones that are off by a constant.

    To illustrate this, suppose we want to construct a Λ such that U⋅∇Λ=Φ is satisfied. The left hand side is a sum of three terms: Ux*dΛ/dx, Uy*dΛ/dy, and Uz*dΛ/dz. For any point, we literally have absolute power to set dΛ/dx to whatever we want. After that, we have absolute power to choose any value for dΛ/dy. There's an infinite number of choices for dΛ/dx, and for each of those choices, there's an infinite number of choices for the other two derivatives.

    This is analogous to a normal dot product, say a*(0,0,1)=0.5. The only thing this equation tells you is that the x component of a is 0.5. The y and z components can be anything, so it's not possible to solve for a in the sense of writing a concise analytical expression that captures all solutions.
     
  6. Jul 5, 2011 #5

    I like Serena

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    In the example a*(1,0,0)=0.5 you can indeed say that the x-component is 0.5.
    So the solution is a=(0.5,y,z) with 2 unknown variables y and z.


    Similarly, if you know for instance that U=(1,0,0), the expression in the OP ([itex]\boldsymbol U \cdot \nabla \Lambda = \Phi[/itex]) reduces to:
    [tex]\frac {\partial \Lambda} {\partial x} = \Phi[/tex]
    So:
    [tex]\Lambda = \int \Phi dx + C(y,z)[/tex]
    And:
    [tex]\nabla \Lambda = \begin{pmatrix} \Phi(x,y,z) \\ \frac {\partial} {\partial y} C(y,z) \\ \frac {\partial} {\partial z} C(y,z) \end{pmatrix}[/tex]

    You can also see here that it is not simpler to solve for [itex]\nabla \Lambda[/itex].
    You need to find [itex]\Lambda[/itex] first.


    When U is not aligned with an axis, you will need to make a coordinate transformation in such a way that U is mapped to for instance (1,0,0).
    Then you can use this same procedure.

    As yet, I don't have a nice complete formula that includes the coordinate transformation though. That will take a bit of mind-benchpressing first. :smile:
     
    Last edited: Jul 5, 2011
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