# I Differentiating a particular integral (retarded potential)

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1. Jul 30, 2017

### DavideGenoa

Hi, friends! Under particular conditions on $\phi:\mathbb{R}^3\times\mathbb{R}\to\mathbb{R}$ - I think, as said here, that it is sufficient that $\phi\in C_c^1(\mathbb{R}^4)$: please correct me if I am wrong - the following equality holds$$\frac{\partial}{\partial r_k}\int_{\mathbb{R}^3} \frac{\phi(\boldsymbol{l},t-c^{-1}\|\boldsymbol{r}-\boldsymbol{l}\|)}{\|\boldsymbol{r}-\boldsymbol{l}\|} d\mu_{\boldsymbol{l}}= \int_{\mathbb{R}^3} \frac{\partial}{\partial r_k}\left(\frac{\phi(\boldsymbol{l})}{\|\boldsymbol{r}-\boldsymbol{l}\|}\right) d\mu_{\boldsymbol{l}}$$where the integral is a Lebesgue integral.
Let $$V(\boldsymbol{x},t):=\frac{1}{4\pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|)}{\|\boldsymbol{x}-\boldsymbol{y}\|} d\mu_{\boldsymbol{y}},$$ which can be interpretated as a Lorenz gauge retarded electric potential in physics if $\varepsilon_0$ is permittivity, and let $\rho$ be under the assumptions stated for $\phi$ above. Then, by differentiating under the integral sign, we get$$\nabla_x V(\boldsymbol{x},t)=\frac{1}{4\pi \varepsilon_0}\int_{\mathbb{R}^3} -\frac{\dot\rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|)}{c} \frac{\boldsymbol{x}-\boldsymbol{y}}{\|\boldsymbol{x}-\boldsymbol{y}\|^2} -\rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|) \frac{\boldsymbol{x}-\boldsymbol{y}}{\|\boldsymbol{x}-\boldsymbol{y}\|^3} d\mu_{\boldsymbol{y}}$$where $\dot\rho$ is the partial derivative taken with respect to the second argument.

I would like to prove to myself that $V$ satisfies the equality $$\nabla_x^2 V(\boldsymbol{x},t)=-\frac{\rho(\boldsymbol{x},t)}{\varepsilon_0}+\frac{1}{c^2}\frac{\partial^2 V(\boldsymbol{x},t)}{\partial t^2}$$which I think to be satisfied by imposing such assumptions on $\rho$, usual in physics.
I think that we cannot differentiate another time under the integral with respect to $x_1$, $x_2$, $x_3$ because informal derivations that I have found of the above equality (as in D.J. Griffiths' Introduction to Electrodynamics) introduce $\delta$ not to get a zero Lebesgue integral. Of course I am trying to get a mathematical proof, a rigourous one, and I know that, if we write a $\delta$, there must be a mathematical justification for that: $\forall\varphi\in C_c^2(\mathbb{R}^3)$ $\int_{\mathbb{R}^3}\frac{\nabla_y^2\varphi(\mathbf{y})}{\| \mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}=-4\pi \varphi(\mathbf{x}),$ while $\int_{\mathbb{R}^3} \varphi(\mathbf{y}) \nabla_x^2 \left( \frac{1}{\| \mathbf{x}-\mathbf{y}\|} \right) d\mu_{\mathbf{y}}=\int_{\mathbb{R}^3} \varphi(\mathbf{y})\cdot 0 \,d\mu_{\mathbf{y}}$$=0$.
Could anybody help me to take the divergence of $\nabla_x V$ to prove that $\nabla_x^2 V=-\frac{\rho}{\varepsilon_0}+\frac{1}{c^2}\frac{\partial^2 V}{\partial t^2}$?
I $\infty$-ly thank you!

Last edited: Jul 30, 2017
2. Aug 4, 2017

### PF_Help_Bot

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3. Aug 6, 2017

### DavideGenoa

I think that another lemma (proved here) useful to prove the desired equality may be: if $\rho\in C_c^k(\mathbb{R}^4)$ and $f:\mathbb{R^3}\mapsto \mathbb{R}$ is locally summable (as $\boldsymbol{z}\mapsto 1/\|\boldsymbol{y}\|$ is), then for any , then we can differentiate under the integral sign in the following way:$$D_{(\boldsymbol{x},t)}^\alpha \int_{\mathbb{R}^3} f(\boldsymbol{x}-\boldsymbol{y}) \rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|)d\mu_{\boldsymbol{y}}= \int_{\mathbb{R}^3}f(\boldsymbol{x}-\boldsymbol{y})D_{(\boldsymbol{y},t)}^\alpha\rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|) d\mu_{\boldsymbol{y}}$$where $D^\alpha$ are the partial derivatives of order $\le k$.
Nevertheless I have not been able to reach the desired result yet...
Could anybody give me help? I heartily thank you all!

4. Aug 6, 2017

### DavideGenoa

Errata corrige: I think that another lemma (proved here) useful to prove the desired equality may be: if $\rho\in C_c^k(\mathbb{R}^4)$ and $f:\mathbb{R^3}\mapsto \mathbb{R}$ is locally summable (as $\boldsymbol{z}\mapsto 1/\|\boldsymbol{y}\|$ is), then for any , then we can differentiate under the integral sign in the following way:$$D_{(\boldsymbol{x},t)}^\alpha \int_{\mathbb{R}^3} f(\boldsymbol{x}-\boldsymbol{y}) \rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|)d\mu_{\boldsymbol{y}}= \int_{\mathbb{R}^3}f(\boldsymbol{x}-\boldsymbol{y})D_{(\boldsymbol{\xi},t)}^\alpha\rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|) d\mu_{\boldsymbol{y}}$$where $D^\alpha$ are the partial derivatives of order $\le k$ and $\boldsymbol{\xi}=(\xi_1,\xi_2,\xi_3)$ are the first three variables of $\rho:(\boldsymbol{\xi},\tau)\mapsto\rho(\boldsymbol{\xi},\tau)$.
Nevertheless I have not been able to reach the desired result yet...
Could anybody give me help? I heartily thank you all![/QUOTE]