# I Differentiation under the integral in retarded potentials

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1. Jun 23, 2017

### 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!

2. Jun 24, 2017

### stevendaryl

Staff Emeritus
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.

3. Jun 24, 2017

### 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!

4. Jul 1, 2017

### 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!

5. Jul 7, 2017

### DavideGenoa

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