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Second derivatives of magnetic potential

  1. Feb 23, 2016 #1
    Hello, friends! I have been told that, if ##\mathbf{J}## is of class ##C^2## and ##V\subset \mathbb{R}^3## is a ##\mu##-measurable and bounded set, where ##\mu## is the ordinary Lebesgue measure on ##\mathbb{R}^3##, then, for all ##\mathbf{x}\in\mathbb{R}^3##, $$\mathbf{A}(\mathbf{x}):=\frac{\mu_0}{4\pi}\int_V\frac{\mathbf{J}(\mathbf{y})}{\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}$$where ##\mu_0## is a constant defines a function of class ##C^2(\mathbb{R})##. If ##\mathbf{J}## is the density of current, the physical interpretation of ##\mathbf{A}## is that of the magnetic potential.
    By using this argument I have not found it difficult to prove, by deriving under the integral sign, that ##\mathbf{A}\in C^1(\mathbb{R}^3)##.

    How can we prove that the second derivatives of ##\mathbf{A}## belong to ## C(\mathbb{R}^3)##?

    I have thought that, if ##\mathbf{J}\in C^2(\mathbb{R}^3)## and its support is contained in ##V##, we might write $$\mathbf{A}(\mathbf{x})=\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{l}-\mathbf{x}\|}d\mu_{\mathbf{l}}=\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{y}+\mathbf{x})}{\|\mathbf{y}\|}d\mu_{\mathbf{y}}$$and try to derive under the integral sign by using the corollary to the dominated convergence theorem that I quote here, but I cannot find a ##\varphi## (notation of the linked post) to use in the desired inequality...
    I ##\infty##-ly thank you!
     
  2. jcsd
  3. Feb 28, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Feb 29, 2016 #3
    By using this result, with ##f:\mathbf{y}\mapsto \|\mathbf{y}\|^{-1}## and ##g## as each component of ##\frac{\mu_0}{4\pi}\mathbf{J}##, we can see that, if ##\mathbf{J}\in C^k(\mathbb{R}^3)## is compactly supported, then, if ##k_1+k_2+k_3=k##, $$\frac{\partial^k\mathbf{A}(\mathbf{x})}{\partial x_1^{k_1}\partial x_2^{k_2}\partial x_3^{k_3}}=\frac{\mu_0}{2\pi}\int_{\mathbb{R}^3}\frac{1}{\|\mathbf{x}-\mathbf{y}\|} \frac{\partial^k\mathbf{J}(\mathbf{y})}{\partial y_1^{k_1}\partial y_2^{k_2}\partial y_3^{k_3}}\, d\mu_{\mathbf{y}}$$and ##\mathbf{A}\in C^k(\mathbb{R}^3)##. If I am not wrong...
     
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