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PDE Existence/Uniqueness Question

  1. May 1, 2014 #1
    My equation is:

    [tex] \left(\mathbf{\nabla}\sigma\right)\cdot\left(\mathbf{\nabla}V\right) + \sigma\nabla^2V = 0[/tex]

    If I'm given V(r) on the boundary of some volume, and I know σ(r) inside the volume, is there a unique solution V(r) inside that volume for any arbitrary (well-behaved) function σ(r)?

    I suspect the answer is yes, but I've never taken a formal PDE class, so I wanted to double-check.

    Edit: just so it's clear, I don't need to know how to solve for V, I just need to know that it's possible to find V in principle.
     
  2. jcsd
  3. May 2, 2014 #2

    pasmith

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    Homework Helper

    You have [tex]
    \left(\mathbf{\nabla}\sigma\right)\cdot\left(\mathbf{\nabla}V\right) + \sigma\nabla^2V
    = \nabla \cdot ( \sigma \nabla V) = 0
    [/tex] which if [itex]\sigma[/itex] is given is a linear PDE for [itex]V[/itex]. Hence if [itex]V_1[/itex] and [itex]V_2[/itex] satisfy this PDE in a volume [itex]U[/itex] with [itex]V_1 = V_2[/itex] on [itex]\partial U[/itex], then [itex]V = V_1 - V_2[/itex] satisfies the PDE with [itex]V = 0[/itex] on [itex]\partial U[/itex].

    Hence [tex]
    0 = \int_U V \nabla \cdot (\sigma \nabla V)\,dU =
    \int_U \nabla \cdot( V \sigma \nabla V) - \sigma (\nabla V) \cdot (\nabla V)\,dU \\
    = \int_{\partial U} V \sigma \frac{\partial V}{\partial n}\,dS - \int_U \sigma \|\nabla V\|^2\,dU
    [/tex] and as [itex]V[/itex] vanishes on [itex]\partial U[/itex] we have [tex]
    \int_U \sigma \|\nabla V\|^2\,dU = 0.
    [/tex] Hence if [itex]\sigma[/itex] is everywhere strictly positive (or strictly negative) we can conclude that [itex]\|\nabla V\|^2 = 0[/itex] everywhere in [itex]U[/itex], from which it follows that [itex]V[/itex] is constant on [itex]U[/itex], and in view of the boundary conditions [itex]V = 0[/itex]. Hence [itex]V_1 = V_2[/itex].

    It follows that if [itex]\sigma[/itex] is everywhere strictly positive or everywhere strictly negative and a solution exists, then that solution is unique.
     
  4. May 2, 2014 #3
    Perfect, thanks!

    I forgot, I do actually know that [itex]\sigma \geq 0 [/itex], and I could probably restrict it to [itex]\sigma > 0 [/itex].
     
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