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Tough integral equation

  1. Jan 18, 2012 #1
    Hi, I have an integral equation for which I'd appreciate any tips! The equation is:

    [tex]
    B^2(\vec{r}) = \int_{\vec{r}}^{\infty} \left[ (\vec{B}(\vec{s}) \cdot \nabla) \vec{B}(\vec{s}) \right] \cdot \vec{ds}
    [/tex]

    The path of integration can be chosen arbitrarily, starting at some point r, and ending at infinity (where B = 0). B can be thought of as a dipole magnetic field.

    I've tried expanding the integrand into its components along and normal to the field lines, and choosing ds to be a path always normal to the field lines. This yields the following equation:

    [tex]
    B^2(\vec{r}) = -\int_{\vec{r}}^{\infty} \frac{B^2(\vec{s})}{R_c(\vec{s})} ds
    [/tex]

    where R_c is the radius of curvature of the field line at the point s. But here it seems difficult to continue without numerical assistance. Anyone have any ideas?
     
  2. jcsd
  3. Jan 18, 2012 #2

    Mute

    User Avatar
    Homework Helper

    I would try to solve that numerically before you put any great effort into getting an (approximate) analytical expression. Your integral equation has the trivial solution B = 0, and that might be your only solution. It's worth your time to set up a numerical solution anyways, because there may not be a nice, easy to get analytical expression.

    However, if I were to try to see if I could find a closed-form expression, then because the variable r only appears as part of the limits of the integral, at least in the simplified expression, you might be able to differentiate both sides and derive a differential equation for B^2. (I would do B^2 rather than B because your simplified integral equation is linear in B^2 but non-linear in B). You still might not be able to solve the equation, though, and I'm not sure how nice everything will turn out because it's a multi-dimensional integral.
     
  4. Jan 21, 2012 #3
    I have a solution, but mine is off from your's by a sign and a factor of 2. They key is the vector identity

    [itex] \nabla (\vec{A}\cdot\vec{B}) = \vec{A}\times(\nabla\times\vec{B}) + \vec{B}\times(\nabla\times\vec{A}) + (\vec{A}\cdot\nabla)\vec{B} + (\vec{B}\cdot\nabla)\vec{A} [/itex].

    We also use the fact that [itex] \vec{B} [/itex] is the magnetic field from a dipole (which I assume is not osscilating). This implies that

    [itex] \nabla\times\vec{B} = \displaystyle\frac{\partial \vec{E}}{\partial t} + \vec{J} = 0 [/itex],

    since the fields are constant in time [itex] \left(\displaystyle\frac{\partial\vec{E}}{\partial t} = 0 \right)[/itex] and there is no current [itex] \left( \vec{J}=0 \right) [/itex]. This alows us to rewrite the right hand side of your equation as

    [itex] \displaystyle\int_{\vec{r}}^{\infty} \left[\left(\vec{B}(\vec{s})\cdot\nabla\right)\vec{B} (\vec{s}) \right] \cdot d\vec{s} = \displaystyle\int_{\vec{r}}^{\infty} \left[ \frac{\nabla B^2}{2} - \vec{B}\times(\underbrace{\nabla\times\vec{B}}_0) \right]\cdot d\vec{s} = \frac{B^2(\infty)}{2}-\frac{B^2(\vec{r})}{2} = -\frac{B^2(\vec{r})}{2} [/itex],

    where we have used the Gradient Theorem and the fact that the dipole field goes to zero at infinity.
     
    Last edited: Jan 21, 2012
  5. Jan 23, 2012 #4
    Thanks for your solution, UA, but unfortunately there is a current which is not 0. My original statement was an attempt to solve the integral which comes up in the physical problem which I didn't describe very well. The actual problem is as follows.

    Consider a dipole field [itex]\vec{B}_0[/itex], where plasma is then introduced to the system and allowed to reach an equilibrium state. In this equilibrium state, there will be a diamagnetic current which satisfies
    [tex]
    \nabla P = \vec{J} \times \vec{B}
    [/tex]
    where P = nkT is the plasma pressure, J is the current, and B is the total field:
    [tex]
    \vec{B} = \vec{B}_0 + \vec{B}_d
    [/tex]
    and [itex]\vec{B}_d[/itex] is the field due to the diamagnetic current J. Using
    [tex]
    \vec{J} = \frac{1}{\mu_0} \nabla \times \vec{B}
    [/tex]
    as well as the identity you quoted, the equation becomes
    [tex]
    \nabla \left( P + \frac{B^2}{2 \mu_0} \right) = \frac{1}{\mu_0} \left( \vec{B} \cdot \nabla \right) \vec{B}
    [/tex]
    Dotting both sides by a path length [itex]\vec{ds}[/itex] and then integrating along some path starting at the point of interest [itex]\vec{r}[/itex] and ending at infinity where all the fields go to 0, allows us to use the gradient theorem on the LHS of the equation:
    [tex]
    P(\vec{r}) + \frac{B^2(\vec{r})}{2\mu_0} = -\frac{1}{\mu_0} \int_{\vec{r}}^{\infty} (\vec{B}(\vec{s}) \cdot \nabla) \vec{B}(\vec{s}) \cdot \vec{ds}
    [/tex]
    which was my original equation, except I left out the pressure P term for simplicity.

    I think that this equation cannot in general be solved for [itex]\vec{B}[/itex], but I would be happy if there was a way to use perturbation theory to solve for the scalar correction to the magnitude of B. IE: if
    [tex]
    B = B_0 + b
    [/tex]
    where b is the perturbation/correction to the dipole field magnitude due to the diamagnetic current J. I would be happy with either a solution for [itex]b[/itex] where you can neglect [itex]b^2[/itex] terms, or a full solution for [itex]\vec{B}[/itex]

    If it helps, the following identity is true:
    [tex]
    (\vec{B} \cdot \nabla) \vec{B} = \frac{d}{db} \left( \frac{B^2}{2} \right) \hat{b} - \frac{B^2}{R_c} \hat{n}
    [/tex]
    where [itex]\hat{b}[/itex] is a unit vector along the field line, [itex]\hat{n}[/itex] is normal to the field line and directed anti-radially, and [itex]R_c[/itex] is the radius of curvature of the field line. Because of the derivative in the first term, I thought at first it might be best to integrated along a field line rather than integrate to infinity. This makes it possible to solve the integral, but then you end up with a solution dependent on the second point you choose in the integral which doesn't seem right.
     
    Last edited: Jan 23, 2012
  6. Jan 23, 2012 #5
    Use the "method of moments" aka the method of weighted residuals.
     
  7. Jan 24, 2012 #6
    I dont understand how that will help...?
     
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