Second derivatives of magnetic potential

[math771]

Hi there! It looks like you are trying to prove that the second derivatives of the magnetic potential function $\mathbf{A}$ belong to the class $C(\mathbb{R}^3)$. This is a great question and involves some advanced mathematical techniques.

One approach you can take is to use the dominated convergence theorem, as you mentioned. However, in order to do this, you will need to find a function $\varphi$ that satisfies the conditions of the theorem. One possible choice for $\varphi$ is the function $\frac{1}{\|\mathbf{y}\|^2}$. This function is positive and integrable on $\mathbb{R}^3$, and it dominates the integrand $\frac{\mathbf{J}(\mathbf{y}+\mathbf{x})}{\|\mathbf{y}\|}$, which is also positive.

Using this function as your $\varphi$, you can then apply the dominated convergence theorem and derive under the integral sign to show that the second derivatives of $\mathbf{A}$ exist and are continuous on $\mathbb{R}^3$. This will then prove that $\mathbf{A}\in C^2(\mathbb{R}^3)$, and therefore the second derivatives belong to the class $C(\mathbb{R}^3)$.

I hope this helps! Let me know if you have any further questions.

 

[DavideGenoa]

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!

 

[DavideGenoa]

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...