Differentiation under the integral in retarded potentials

[DavideGenoa]

Hello, friends! I know, thanks to @Hawkeye18 who proved this identity to me, that, if $\phi:V\to\mathbb{R}$ is a bounded measurable function defined on the bounded measurable domain $V\subset\mathbb{R}^3$, then, for any $k\in\{1,2,3\}$,
$$\frac{\partial}{\partial r_k}\int_V \frac{\phi(\boldsymbol{r})}{\|\boldsymbol{r}-\boldsymbol{r}\|} d\mu_{\boldsymbol{r}}= \int_V \frac{\partial}{\partial r_k}\left(\frac{\phi(\boldsymbol{r})}{\|\boldsymbol{r}-\boldsymbol{r}\|}\right) d\mu_{\boldsymbol{r}}.$$
An important application of this result is the proof that the magnetostatic vector potential $\frac{\mu_0}{4\pi}\int_\mathbb{R}^3 \frac{\boldsymbol{J}(\boldsymbol{r})}{\|\boldsymbol{r}-\boldsymbol{r}\|} d\mu_{\boldsymbol{r}}$ satisfies Ampère's law.
Since I am, fuitlessly until now, trying to prove to myself that the Lorenz gauge magnetodynamic retarded vector potential satisfies Maxwell's equations, I am wondering whether, given a constant $c$, it is also true that $$\frac{\partial}{\partial r_k}\int_V \frac{\phi(\boldsymbol{r},t-c^{-1}\|\boldsymbol{r}-\boldsymbol{r}\|)}{\|\boldsymbol{r}-\boldsymbol{r}\|} d\mu_{\boldsymbol{r}}= \int_V \frac{\partial}{\partial r_k}\left(\frac{\phi(\boldsymbol{r},t-c^{-1}\|\boldsymbol{r}-\boldsymbol{r}\|)}{\|\boldsymbol{r}-\boldsymbol{r}\|}\right) d\mu_{\boldsymbol{r}}$$for, say $\phi\in C^2(\mathbb{R}^4)$ and $\phi(-,t)$ compactly supported as a function of the first vector variable.
If it is true, how could we prove it?
I $\infty$-ly thank any answerer!

 

[stevendaryl]

Since I am, fuitlessly until now, trying to prove to myself that the Lorenz gauge magnetodynamic retarded vector potential satisfies Maxwell's equations, I am wondering whether, given a constant $c$, it is also true that $$\frac{\partial}{\partial r_k}\int_V \frac{\phi(\boldsymbol{r},t-c^{-1}\|\boldsymbol{r}-\boldsymbol{r}\|)}{\|\boldsymbol{r}-\boldsymbol{r}\|} d\mu_{\boldsymbol{r}}= \int_V \frac{\partial}{\partial r_k}\left(\frac{\phi(\boldsymbol{r},t-c^{-1}\|\boldsymbol{r}-\boldsymbol{r}\|)}{\|\boldsymbol{r}-\boldsymbol{r}\|}\right) d\mu_{\boldsymbol{r}}$$for, say $\phi\in C^2(\mathbb{R}^4)$ and $\phi(-,t)$ compactly supported as a function of the first vector variable.
If it is true, how could we prove it?
I $\infty$-ly thank any answerer!

I assume that you mean \| \mathbf{r} - \mathbf{r'} \|, and that you mean \frac{\partial}{\partial r'_k}?

I have not worked out the extent to which \phi being compactly supported affects whether you can pull the partial derivative inside the integral. However, your modified problem is obtained from the original by replacing \phi(\mathbf{r}) by \tilde{\phi}(\mathbf{r}) \equiv \phi(\mathbf{r}, t - \frac{1}{c} \| \mathbf{r} - \mathbf{r'} \|). But knowing that \phi(\mathbf{r}, t) is compactly supported in the first argument doesn't necessarily imply that \tilde{\phi} is compactly supported.

Here's a analogous problem:

Let f(x,y) be a function that is zero unless |x| < 2|y|. So that's compactly supported in the first argument. But if we define \tilde{f}(x,y) = f(x, y-x), then \tilde{f} is not compactly supported in the first argument.

 

[DavideGenoa]

Thank you very much! Yes, @stevendaryl, I meant:

Hello, friends! I know, thanks to @Hawkeye18 who proved this identity to me, that, if $\phi:V\to\mathbb{R}$ is a bounded measurable function defined on the bounded measurable domain $V\subset\mathbb{R}^3$, then, for any $k\in\{1,2,3\}$,
$$\frac{\partial}{\partial r_k}\int_V \frac{\phi(\boldsymbol{l})}{\|\boldsymbol{r}-\boldsymbol{l}\|} d\mu_{\boldsymbol{l}}= \int_V \frac{\partial}{\partial r_k}\left(\frac{\phi(\boldsymbol{l})}{\|\boldsymbol{r}-\boldsymbol{l}\|}\right) d\mu_{\boldsymbol{l}}.$$
An important application of this result is the proof that the magnetostatic vector potential $\frac{\mu_0}{4\pi}\int_\mathbb{R}^3 \frac{\boldsymbol{J}(\boldsymbol{l})}{\|\boldsymbol{r}-\boldsymbol{l}\|} d\mu_{\boldsymbol{l}}$ satisfies Ampère's law.
Since I am, fuitlessly until now, trying to prove to myself that the Lorenz gauge magnetodynamic retarded vector potential satisfies Maxwell's equations, I am wondering whether, given a constant $c$, it is also true that $$\frac{\partial}{\partial r_k}\int_V \frac{\phi(\boldsymbol{l},t-c^{-1}\|\boldsymbol{r}-\boldsymbol{l}\|)}{\|\boldsymbol{r}-\boldsymbol{l}\|} d\mu_{\boldsymbol{l}}= \int_V \frac{\partial}{\partial r_k}\left(\frac{\phi(\boldsymbol{l},t-c^{-1}\|\boldsymbol{r}-\boldsymbol{l}\|)}{\|\boldsymbol{r}-\boldsymbol{l}\|}\right) d\mu_{\boldsymbol{l}}$$for, say $\phi\in C^2(\mathbb{R}^4)$ and $\phi(-,t)$ compactly supported as a function of the first vector variable.
If it is true, how could we prove it?
I $\infty$-ly thank any answerer!

 

[DavideGenoa]

Dear @stevendaryl , I have tried to adapt the proof by Hawkeye18 to this problem, and, as far as I understand, it can used in an almost identical way, with $\frac{\partial }{\partial r_k}\left(\frac{\phi(\boldsymbol{l},t-c^1\|\boldsymbol{r}-\boldsymbol{l}\|)}{\|\boldsymbol{r}-\boldsymbol{l}\|} \right)$ instead of $\frac{\partial }{\partial r_k}\left(\frac{\phi(\boldsymbol{l})}{\|\boldsymbol{r}-\boldsymbol{l}\|} \right)$, provided for example that $\phi\in C^2(\mathbb{R}^4)$ and that the two functions $$\boldsymbol{l}\mapsto \phi(\boldsymbol{l},t-c^1\|\boldsymbol{r}-\boldsymbol{l}\|)$$$$\boldsymbol{l}\mapsto \dot\phi(\boldsymbol{l},t-c^1\|\boldsymbol{r}-\boldsymbol{l}\|)$$ are compactly supported. This last condition is met if $\phi\in C_2^2(\mathbb{R}^4)$.
By following Hawkeye18's reasoning we also see that the derivatives of $V:=\frac{1}{4\pi\varepsilon}\int_{\mathbb{R}^3} \frac{\phi(\boldsymbol{l},t-c^1\|\boldsymbol{r}-\boldsymbol{l}\|)}{\|\boldsymbol{r}-\boldsymbol{l}\|} d\mu_{\boldsymbol{l}}$ are continuous and also that $V\in C^1(\mathbb{R}^3)$.
Am I right? I $\infty$-ly thank you!

 

[DavideGenoa]

@stevendaryl and anyone else reading: Are my conclusions in the preceding post correct?