vector-field-F-R-3-R-3-F-h-C-and-Let-s-define-as-A-F-can-we-find-vector-field-F-Curl-and-Divergence-inverse-operator-dose-exist-1-A-

Question Number 222828 by wewji12 last updated on 09/Jul/25

vector field F^→ ;R^3 →R^3 , F_h ∈C^ω and Let′s define as A^→ =▽^→ ×F^→ can we find vector field F^→ .....??? Curl and Divergence inverse operator dose exist?? (▽_ ^→ ×)^(−1) A^→ , (▽_ ^→ ∗)^(−1) A^→ ex. ( ((d )/dx))^(−1) =∫

$$\mathrm{vector}\:\mathrm{field}\:\:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}};\mathbb{R}^{\mathrm{3}} \rightarrow\mathbb{R}^{\mathrm{3}} \:,\:{F}_{{h}} \in\mathcal{C}^{\omega} \\ $$$$\mathrm{and}\:\mathrm{Let}'\mathrm{s}\:\mathrm{define}\:\mathrm{as}\:\overset{\rightarrow} {\boldsymbol{\mathrm{A}}}=\overset{\rightarrow} {\bigtriangledown}×\overset{\rightarrow} {\boldsymbol{\mathrm{F}}} \\ $$$$\mathrm{can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{vector}\:\mathrm{field}\:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}…..??? \\ $$$$\mathrm{Curl}\:\mathrm{and}\:\mathrm{Divergence}\:\:\mathrm{inverse}\:\mathrm{operator}\:\mathrm{dose}\:\mathrm{exist}?? \\ $$$$\left(\overset{\rightarrow} {\bigtriangledown}_{\:} ×\right)^{−\mathrm{1}} \overset{\rightarrow} {\boldsymbol{\mathrm{A}}}\:,\:\left(\overset{\rightarrow} {\bigtriangledown}_{\:} \ast\right)^{−\mathrm{1}} \overset{\rightarrow} {\boldsymbol{\mathrm{A}}} \\ $$$$\mathrm{ex}.\:\left(\:\frac{\mathrm{d}\:\:\:}{\mathrm{d}{x}}\right)^{−\mathrm{1}} =\int\: \\ $$

Answered by MrGaster last updated on 09/Jul/25

(▽^→ ×)^(−1) A^→ =−(1/(4π))∫∫∫_R^3 ((A^→ (r^→ ′)×(r^→ ′−r^→ ′))/(∣r^→ −r^→ ′∣^3 ))dr′ (▽^→ ∙)^(−1) D=(1/(4π))▽^→ ∫∫∫_R^3 ((D(r^→ ′))/(∣r^→ −r^→ ′∣))dr′

$$\left(\overset{\rightarrow} {\bigtriangledown}×\right)^{−\mathrm{1}} \overset{\rightarrow} {\boldsymbol{{A}}}=−\frac{\mathrm{1}}{\mathrm{4}\pi}\int\int\int_{\mathbb{R}^{\mathrm{3}} } \frac{\overset{\rightarrow} {\boldsymbol{{A}}}\left(\overset{\rightarrow} {\boldsymbol{{r}}}'\right)×\left(\overset{\rightarrow} {\boldsymbol{{r}}}'−\overset{\rightarrow} {\boldsymbol{{r}}}'\right)}{\mid\overset{\rightarrow} {\boldsymbol{{r}}}−\overset{\rightarrow} {\boldsymbol{{r}}}'\mid^{\mathrm{3}} }{d}\boldsymbol{{r}}' \\ $$$$\left(\overset{\rightarrow} {\bigtriangledown}\centerdot\right)^{−\mathrm{1}} {D}=\frac{\mathrm{1}}{\mathrm{4}\pi}\overset{\rightarrow} {\bigtriangledown}\int\int\int_{\mathbb{R}^{\mathrm{3}} } \frac{{D}\left(\overset{\rightarrow} {\boldsymbol{{r}}}'\right)}{\mid\overset{\rightarrow} {\boldsymbol{{r}}}−\overset{\rightarrow} {\boldsymbol{{r}}}'\mid}{d}\boldsymbol{{r}}' \\ $$

Commented by wewji12 last updated on 09/Jul/25

can u explain why (▽_ ^→ ×)^(−1) A=−(1/(4π)) ∫_R^3 d^3 r ((A(r′)×(r−r′))/(∣r−r′∣^3 )) and (▽_ ^→ ∗)^(−1) D=(1/(4π)) ▽^→ ∫_R^3 d^3 r ((D(r′))/(∣r−r′∣))

$$\mathrm{can}\:\mathrm{u}\:\mathrm{explain}\:\mathrm{why} \\ $$$$\left(\overset{\rightarrow} {\bigtriangledown}_{\:} ×\right)^{−\mathrm{1}} \boldsymbol{\mathrm{A}}=−\frac{\mathrm{1}}{\mathrm{4}\pi}\:\int_{\mathbb{R}^{\mathrm{3}} } \:\mathrm{d}^{\mathrm{3}} \boldsymbol{\mathrm{r}}\:\frac{\boldsymbol{\mathrm{A}}\left(\boldsymbol{\mathrm{r}}'\right)×\left(\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\right)}{\mid\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\mid^{\mathrm{3}} }\:\:\mathrm{and} \\ $$$$\left(\overset{\rightarrow} {\bigtriangledown}_{\:} \ast\right)^{−\mathrm{1}} \mathcal{D}=\frac{\mathrm{1}}{\mathrm{4}\pi}\:\overset{\rightarrow} {\bigtriangledown}\int_{\mathbb{R}^{\mathrm{3}} } \:\mathrm{d}^{\mathrm{3}} \boldsymbol{\mathrm{r}}\:\:\frac{\mathcal{D}\left(\boldsymbol{\mathrm{r}}'\right)}{\mid\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\mid} \\ $$

Commented by MrGaster last updated on 09/Jul/25