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Champs scalaires et vectoriels

CSV1​

  1. Gradient d'un produit :

    grad→(GK)=∂(GK)∂xe⃗x+∂(GK)∂ye⃗y+∂(GK)∂ze⃗z=(K∂G∂x+G∂K∂x)e⃗x+(K∂G∂y+G∂K∂y)e⃗y+(K∂G∂z+G∂K∂z)e⃗z=K(∂G∂xe⃗x+∂G∂ye⃗y+∂G∂ze⃗z)+G(∂K∂xe⃗x+∂K∂ye⃗y+∂K∂ze⃗z)grad→(GK)=K grad→(G)+G grad→(K)\begin{aligned} \overrightarrow{\text{grad}}(GK) &= \frac{\partial(GK)}{\partial x}\vec{e}_x + \frac{\partial(GK)}{\partial y}\vec{e}_y + \frac{\partial(GK)}{\partial z}\vec{e}_z \\ &= \left(K\frac{\partial G}{\partial x} + G\frac{\partial K}{\partial x}\right)\vec{e}_x + \left(K\frac{\partial G}{\partial y} + G\frac{\partial K}{\partial y}\right)\vec{e}_y + \left(K\frac{\partial G}{\partial z} + G\frac{\partial K}{\partial z}\right)\vec{e}_z \\ &= K\left(\frac{\partial G}{\partial x}\vec{e}_x + \frac{\partial G}{\partial y}\vec{e}_y + \frac{\partial G}{\partial z}\vec{e}_z\right) + G\left(\frac{\partial K}{\partial x}\vec{e}_x + \frac{\partial K}{\partial y}\vec{e}_y + \frac{\partial K}{\partial z}\vec{e}_z\right) \\ \overrightarrow{\text{grad}}(GK) &= K\,\overrightarrow{\text{grad}}(G) + G\,\overrightarrow{\text{grad}}(K) \end{aligned}

  2. Coordonnées sphériques →grad→(Φ(r,θ,φ))=∂Φ∂re⃗r+1r∂Φ∂θe⃗θ+1rsin⁡θ∂Φ∂φe⃗φ\to \overrightarrow{\text{grad}}(\Phi(r, \theta, \varphi)) = \dfrac{\partial\Phi}{\partial r}\vec{e}_r + \dfrac{1}{r}\dfrac{\partial\Phi}{\partial\theta}\vec{e}_\theta + \dfrac{1}{r\sin\theta}\dfrac{\partial\Phi}{\partial\varphi}\vec{e}_\varphi :

    • grad→(r)=e⃗r=r⃗r\overrightarrow{\text{grad}}(r) = \vec{e}_r = \dfrac{\vec{r}}{r} ;
    • grad→(1r)=−1r2e⃗r=−r⃗r3\overrightarrow{\text{grad}}\left(\dfrac{1}{r}\right) = -\dfrac{1}{r^2}\vec{e}_r = -\dfrac{\vec{r}}{r^3} ;
    • grad→(ln⁡r)=1re⃗r=r⃗r2\overrightarrow{\text{grad}}(\ln r) = \dfrac{1}{r}\vec{e}_r = \dfrac{\vec{r}}{r^2} ;
    • grad→(rln⁡r)=(ln⁡r+r1r)e⃗r=(ln⁡(r)+1r)r⃗\overrightarrow{\text{grad}}(r\ln r) = \left(\ln r + r\dfrac{1}{r}\right)\vec{e}_r = \left(\dfrac{\ln(r) + 1}{r}\right)\vec{r}.

    Remarque. grad→(rln⁡r)=ln⁡(r)grad→(r)+r grad→(ln⁡r)=ln⁡(r)r⃗r+rr⃗r2=(ln⁡rr+1r)r⃗=(ln⁡(r)+1r)r⃗\overrightarrow{\text{grad}}(r\ln r) = \ln(r)\overrightarrow{\text{grad}}(r) + r\,\overrightarrow{\text{grad}}(\ln r) = \ln(r)\dfrac{\vec{r}}{r} + r\dfrac{\vec{r}}{r^2} = \left(\dfrac{\ln r}{r} + \dfrac{1}{r}\right)\vec{r} = \left(\dfrac{\ln(r) + 1}{r}\right)\vec{r}.

CSV2​

grad→f=∂∂x(z2x2+y2+z2)e⃗x+∂∂y(z2x2+y2+z2)e⃗y+∂∂z(z2x2+y2+z2)e⃗z\overrightarrow{\text{grad}}f = \frac{\partial}{\partial x}\left(\frac{z^2}{\sqrt{x^2 + y^2 + z^2}}\right)\vec{e}_x + \frac{\partial}{\partial y}\left(\frac{z^2}{\sqrt{x^2 + y^2 + z^2}}\right)\vec{e}_y + \frac{\partial}{\partial z}\left(\frac{z^2}{\sqrt{x^2 + y^2 + z^2}}\right)\vec{e}_z
  • ∂f∂x=∂∂x[z2(x2+y2+z2)−1/2]=z2(−12)(2x)(x2+y2+z2)−3/2=−xz2(x2+y2+z2)−3/2\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x}\left[z^2(x^2 + y^2 + z^2)^{-1/2}\right] = z^2\left(-\dfrac{1}{2}\right)(2x)(x^2 + y^2 + z^2)^{-3/2} = -xz^2(x^2 + y^2 + z^2)^{-3/2} ;
  • ∂f∂y=−yz2(x2+y2+z2)−3/2\dfrac{\partial f}{\partial y} = -yz^2(x^2 + y^2 + z^2)^{-3/2} ;
  • ∂f∂z=2z(x2+y2+z2)−1/2−zz2(x2+y2+z2)−3/2\dfrac{\partial f}{\partial z} = 2z(x^2 + y^2 + z^2)^{-1/2} - zz^2(x^2 + y^2 + z^2)^{-3/2}.
grad→f=−xz2(x2+y2+z2)−3/2e⃗x−yz2(x2+y2+z2)−3/2e⃗y+[2z(x2+y2+z2)−1/2−zz2(x2+y2+z2)−3/2]e⃗zgrad→f=2z(x2+y2+z2)−1/2e⃗z−z2(x2+y2+z2)−3/2(xe⃗x+ye⃗y+ze⃗z)\begin{aligned} \overrightarrow{\text{grad}}f &= -xz^2(x^2 + y^2 + z^2)^{-3/2}\vec{e}_x - yz^2(x^2 + y^2 + z^2)^{-3/2}\vec{e}_y + \left[2z(x^2 + y^2 + z^2)^{-1/2} - zz^2(x^2 + y^2 + z^2)^{-3/2}\right]\vec{e}_z \\ \overrightarrow{\text{grad}}f &= 2z(x^2 + y^2 + z^2)^{-1/2}\vec{e}_z - z^2(x^2 + y^2 + z^2)^{-3/2}(x\vec{e}_x + y\vec{e}_y + z\vec{e}_z) \end{aligned}

Calcul avec grad→f=grad→(GK)\overrightarrow{\text{grad}}f = \overrightarrow{\text{grad}}(GK), f(M)=z2rf(M) = \dfrac{z^2}{r}, r=x2+y2+z2r = \sqrt{x^2 + y^2 + z^2} :

grad→(z2r)=1rgrad→(z2)+z2grad→(1r)=1r2ze⃗z+z2(−1r2)e⃗r=2zre⃗z−z2r⃗r3=2zx2+y2+z2e⃗z−z2(xe⃗x+ye⃗y+ze⃗z)(x2+y2+z2)3grad→(z2r)=2z(x2+y2+z2)−1/2e⃗z−z2(x2+y2+z2)−3/2(xe⃗x+ye⃗y+ze⃗z)\begin{aligned} \overrightarrow{\text{grad}}\left(\frac{z^2}{r}\right) &= \frac{1}{r}\overrightarrow{\text{grad}}(z^2) + z^2\overrightarrow{\text{grad}}\left(\frac{1}{r}\right) = \frac{1}{r}2z\vec{e}_z + z^2\left(\frac{-1}{r^2}\right)\vec{e}_r = \frac{2z}{r}\vec{e}_z - z^2\frac{\vec{r}}{r^3} \\ &= \frac{2z}{\sqrt{x^2 + y^2 + z^2}}\vec{e}_z - z^2\frac{(x\vec{e}_x + y\vec{e}_y + z\vec{e}_z)}{\left(\sqrt{x^2 + y^2 + z^2}\right)^3} \\ \overrightarrow{\text{grad}}\left(\frac{z^2}{r}\right) &= 2z(x^2 + y^2 + z^2)^{-1/2}\vec{e}_z - z^2(x^2 + y^2 + z^2)^{-3/2}(x\vec{e}_x + y\vec{e}_y + z\vec{e}_z) \end{aligned}

CSV4​

C(B⃗(M))=∮ΓB⃗.dOM→C(\vec{B}(M)) = \displaystyle\oint_\Gamma \vec{B}.d\overrightarrow{OM}.

  • Calcul de B⃗.dOM→=Bxdx+Bydy+Bzdz\vec{B}.d\overrightarrow{OM} = B_xdx + B_ydy + B_zdz. Or ici, dOM→=dx e⃗x+dy e⃗yd\overrightarrow{OM} = dx\,\vec{e}_x + dy\,\vec{e}_y, donc B⃗.dOM→=Bxdx+Bydy=(2x−y)dx+(x+y)dy\vec{B}.d\overrightarrow{OM} = B_xdx + B_ydy = (2x - y)dx + (x + y)dy. Donc C(B⃗)=∮Γ(2x−y)dx+(x+y)dyC(\vec{B}) = \displaystyle\oint_\Gamma (2x - y)dx + (x + y)dy.

  • Paramétrage de Γ\Gamma : M(Γ){x(p)=Rcos⁡p=3cos⁡py(p)=Rsin⁡p=3sin⁡pz(p)=0M(\Gamma) \begin{cases} x(p) = R\cos p = 3\cos p \\ y(p) = R\sin p = 3\sin p \\ z(p) = 0 \end{cases}, p∈[0,2π]p \in [0, 2\pi] ; dx=−3sin⁡(p)dpdx = -3\sin(p)dp, dy=3cos⁡(p)dpdy = 3\cos(p)dp.

  • On injecte dans l'expression :

    ⇒(2x−y)dx+(x+y)dy=(6cos⁡p−3sin⁡p)(−3sin⁡(p)dp)+(3cos⁡p+3sin⁡p)(3cos⁡(p)dp)=[−18cos⁡psin⁡p+9sin⁡2p+9cos⁡2p+9cos⁡psin⁡p]dp=(−9cos⁡psin⁡p+9)dp=9(1−cos⁡psin⁡p)dp\begin{aligned} \Rightarrow (2x - y)dx + (x + y)dy &= (6\cos p - 3\sin p)(-3\sin(p)dp) + (3\cos p + 3\sin p)(3\cos(p)dp) \\ &= [-18\cos p\sin p + 9\sin^2 p + 9\cos^2 p + 9\cos p\sin p]dp \\ &= (-9\cos p\sin p + 9)dp \\ &= 9(1 - \cos p\sin p)dp \end{aligned}

    d'où

    C(B⃗)=∫02π9(1−cos⁡psin⁡p)dp=9[p−sin⁡2p2]02π=18πC(\vec{B}) = \int_0^{2\pi} 9(1 - \cos p\sin p)dp = 9\left[p - \frac{\sin^2 p}{2}\right]_0^{2\pi} = 18\pi

CSV5​

  • A⃗.dOM→=3xy dx−5z dy+10x dz\vec{A}.d\overrightarrow{OM} = 3xy\,dx - 5z\,dy + 10x\,dz.
  • M(Γ){x(t)=1+t2y(t)=2t2z(t)=t3M(\Gamma) \begin{cases} x(t) = 1 + t^2 \\ y(t) = 2t^2 \\ z(t) = t^3 \end{cases}, t∈[0,1]t \in [0, 1] ; {dx=2t dtdy=4t dtdz=3t2dt\begin{cases} dx = 2t\,dt \\ dy = 4t\,dt \\ dz = 3t^2dt \end{cases}
  • 3xy dx−5z dy+10x dz=(12t5+10t4+12t3+30t2)dt3xy\,dx - 5z\,dy + 10x\,dz = (12t^5 + 10t^4 + 12t^3 + 30t^2)dt.
C(A⃗)=∫ΓA⃗.dOM→=∫01(12t5+10t4+12t3+30t2)dt=[2t6+2t5+3t4+10t3]01=17C(\vec{A}) = \int_\Gamma \vec{A}.d\overrightarrow{OM} = \int_0^1 (12t^5 + 10t^4 + 12t^3 + 30t^2)dt = [2t^6 + 2t^5 + 3t^4 + 10t^3]_0^1 = 17

CSV6​

  1. F⃗(M)=xy e⃗x−z2e⃗y−x2e⃗z\vec{F}(M) = xy\,\vec{e}_x - z^2\vec{e}_y - x^2\vec{e}_z.

    • WOABP(F⃗)=∫OABPF⃗.dOM→=∫OAF⃗.dOM→+∫ABF⃗.dOM→+∫BPF⃗.dOM→W_{OABP}(\vec{F}) = \displaystyle\int_{OABP} \vec{F}.d\overrightarrow{OM} = \int_{OA} \vec{F}.d\overrightarrow{OM} + \int_{AB} \vec{F}.d\overrightarrow{OM} + \int_{BP} \vec{F}.d\overrightarrow{OM} :

      • ∫OAF⃗.dOM→\displaystyle\int_{OA} \vec{F}.d\overrightarrow{OM} →\to dOM→=dx e⃗x⇒F⃗.dOM→=xy dxd\overrightarrow{OM} = dx\,\vec{e}_x \Rightarrow \vec{F}.d\overrightarrow{OM} = xy\,dx ; M∈[OA]{x∈[0,x0]y=0z=0⇒∫OAF⃗.dOM→=0M \in [OA] \begin{cases} x \in [0, x_0] \\ y = 0 \\ z = 0 \end{cases} \Rightarrow \displaystyle\int_{OA} \vec{F}.d\overrightarrow{OM} = 0 ;
      • ∫ABF⃗.dOM→\displaystyle\int_{AB} \vec{F}.d\overrightarrow{OM} →\to dOM→=dy e⃗y⇒F⃗.dOM→=−z2dyd\overrightarrow{OM} = dy\,\vec{e}_y \Rightarrow \vec{F}.d\overrightarrow{OM} = -z^2dy ; M∈[AB]{x0y∈[0,y0]z=0⇒∫ABF⃗.dOM→=0M \in [AB] \begin{cases} x_0 \\ y \in [0, y_0] \\ z = 0 \end{cases} \Rightarrow \displaystyle\int_{AB} \vec{F}.d\overrightarrow{OM} = 0 ;
      • ∫BPF⃗.dOM→\displaystyle\int_{BP} \vec{F}.d\overrightarrow{OM} →\to dOM→=dz e⃗z⇒F⃗.dOM→=−x2dzd\overrightarrow{OM} = dz\,\vec{e}_z \Rightarrow \vec{F}.d\overrightarrow{OM} = -x^2dz ; M∈[BP]{x0y0z∈[0,z0]⇒∫BPF⃗.dOM→=∫0z0−x02dz=−x02z0M \in [BP] \begin{cases} x_0 \\ y_0 \\ z \in [0, z_0] \end{cases} \Rightarrow \displaystyle\int_{BP} \vec{F}.d\overrightarrow{OM} = \int_0^{z_0} -x_0^2dz = -x_0^2z_0 ;

      ⇒WOABP(F⃗)=−x02z0\Rightarrow W_{OABP}(\vec{F}) = -x_0^2z_0.

    • WOCDP(F⃗)=∫OCF⃗.dOM→+∫CDF⃗.dOM→+∫DPF⃗.dOM→=0+0+x02y02=x02y02W_{OCDP}(\vec{F}) = \displaystyle\int_{OC} \vec{F}.d\overrightarrow{OM} + \int_{CD} \vec{F}.d\overrightarrow{OM} + \int_{DP} \vec{F}.d\overrightarrow{OM} = 0 + 0 + \frac{x_0^2y_0}{2} = \frac{x_0^2y_0}{2}.

    • WOCBP(F⃗)=∫OCF⃗.dOM→+∫CBF⃗.dOM→+∫BPF⃗.dOM→=0+x02y02−x02z0=x02y02−x02z0W_{OCBP}(\vec{F}) = \displaystyle\int_{OC} \vec{F}.d\overrightarrow{OM} + \int_{CB} \vec{F}.d\overrightarrow{OM} + \int_{BP} \vec{F}.d\overrightarrow{OM} = 0 + \frac{x_0^2y_0}{2} - x_0^2z_0 = \frac{x_0^2y_0}{2} - x_0^2z_0.

  2. WOABP≠WOCDP≠WOCBP⇒∄f(M)/F⃗=grad→f⇔rot→(F⃗)W_{OABP} \neq W_{OCDP} \neq W_{OCBP} \Rightarrow \nexists f(M) / \vec{F} = \overrightarrow{\text{grad}}f \Leftrightarrow \overrightarrow{\text{rot}}(\vec{F}) n'est pas =0⃗ ∀(x,y,z)= \vec{0}\ \forall (x, y, z).

    rot→(F⃗)=2z e⃗x+2x e⃗y−x e⃗z≠0⃗∀(x,y,z)\overrightarrow{\text{rot}}(\vec{F}) = 2z\,\vec{e}_x + 2x\,\vec{e}_y - x\,\vec{e}_z \neq \vec{0} \quad \forall (x, y, z)

CSV8​

rot→B⃗=∇⃗∧B⃗=(0−0)e⃗x+(0−0)e⃗y+[2x(x+y)+x2+y2−(2y(x−y)−(x2+y2))]e⃗z=[3x2+2xy+y2−2xy+3y2+x2]e⃗z=(4x2+4y2)e⃗z≠0⃗\begin{aligned} \overrightarrow{\text{rot}}\vec{B} = \vec{\nabla} \wedge \vec{B} &= (0 - 0)\vec{e}_x + (0 - 0)\vec{e}_y + [2x(x + y) + x^2 + y^2 - (2y(x - y) - (x^2 + y^2))]\vec{e}_z \\ &= [3x^2 + 2xy + y^2 - 2xy + 3y^2 + x^2]\vec{e}_z \\ &= (4x^2 + 4y^2)\vec{e}_z \neq \vec{0} \end{aligned}

CSV9​

  1. rot→C⃗=0⃗ ∀M⇒∃f/grad→f=C⃗\overrightarrow{\text{rot}}\vec{C} = \vec{0}\ \forall M \Rightarrow \exists f / \overrightarrow{\text{grad}}f = \vec{C}.

  2. Recherche de HH :

    rot→C⃗=0⃗⇒(∂H∂y−y)e⃗x+(2x−∂H∂x)e⃗y+(0−0)e⃗z\overrightarrow{\text{rot}}\vec{C} = \vec{0} \Rightarrow \left(\frac{\partial H}{\partial y} - y\right)\vec{e}_x + \left(2x - \frac{\partial H}{\partial x}\right)\vec{e}_y + (0 - 0)\vec{e}_z
    {∂H∂y−y=02x−∂H∂x=0⇒{∂H∂y=y(1)∂H∂x=2x(2)\begin{cases} \dfrac{\partial H}{\partial y} - y = 0 \\ 2x - \dfrac{\partial H}{\partial x} = 0 \end{cases} \Rightarrow \begin{cases} \dfrac{\partial H}{\partial y} = y & (1) \\ \dfrac{\partial H}{\partial x} = 2x & (2) \end{cases}

    D'après (1), H(x,y)=y22+g(x)⇒∂H∂x=dgdx=2xH(x, y) = \dfrac{y^2}{2} + g(x) \Rightarrow \dfrac{\partial H}{\partial x} = \dfrac{dg}{dx} = 2x, d'après (2). Donc g(x)=x2+cte⇒H(x,y)=y22+x2+cteg(x) = x^2 + cte \Rightarrow H(x, y) = \dfrac{y^2}{2} + x^2 + cte.

    Or C⃗(0)=0⃗⇒cte=0\vec{C}(0) = \vec{0} \Rightarrow cte = 0. Donc H(x,y)=y22+x2H(x, y) = \dfrac{y^2}{2} + x^2.

  3. Potentiel :

    grad→f=C⃗⇔{2xz=∂f∂x(1)yz=∂f∂y(2)y22+x2=∂f∂z(3)\overrightarrow{\text{grad}}f = \vec{C} \Leftrightarrow \begin{cases} 2xz = \dfrac{\partial f}{\partial x} & (1) \\ yz = \dfrac{\partial f}{\partial y} & (2) \\ \dfrac{y^2}{2} + x^2 = \dfrac{\partial f}{\partial z} & (3) \end{cases}

    D'après (1), f(x,y,z)=x2z+g(y,z)⇒∂f∂y=∂g∂y=yzf(x, y, z) = x^2z + g(y, z) \Rightarrow \dfrac{\partial f}{\partial y} = \dfrac{\partial g}{\partial y} = yz, d'après (2) ⇒g(y,z)=y22z+h(z)\Rightarrow g(y, z) = \dfrac{y^2}{2}z + h(z).

    Donc f(x,y,z)=x2z+y22z+h(z)⇒∂f∂z=x2+y22+dhdz=x2+y22f(x, y, z) = x^2z + \dfrac{y^2}{2}z + h(z) \Rightarrow \dfrac{\partial f}{\partial z} = x^2 + \dfrac{y^2}{2} + \dfrac{dh}{dz} = x^2 + \dfrac{y^2}{2}, d'après (3) ⇒h(z)=cte\Rightarrow h(z) = cte.

    Donc f(x,y,z)=x2z+y22z+ctef(x, y, z) = x^2z + \dfrac{y^2}{2}z + cte. Or f(x,y,z)=0f(x, y, z) = 0 dans le plan xOyxOy, c'est-à-dire si z=0⇒cte=0z = 0 \Rightarrow cte = 0. Donc f(x,y,z)=(x2+y22)zf(x, y, z) = \left(x^2 + \dfrac{y^2}{2}\right)z.

CSV11​

  1. div A⃗=∂Ax∂x+∂Ay∂y=1+1=2\text{div}\,\vec{A} = \dfrac{\partial A_x}{\partial x} + \dfrac{\partial A_y}{\partial y} = 1 + 1 = 2 ; rot→A⃗=(0−0)e⃗z=0⃗\overrightarrow{\text{rot}}\vec{A} = (0 - 0)\vec{e}_z = \vec{0}.
  2. div B⃗=∂(−y)∂x+∂x∂y=0\text{div}\,\vec{B} = \dfrac{\partial(-y)}{\partial x} + \dfrac{\partial x}{\partial y} = 0 ; rot→B⃗=2e⃗z\overrightarrow{\text{rot}}\vec{B} = 2\vec{e}_z.

CSV12​

  1. div r⃗=∂x∂x+∂y∂y+∂z∂z=3\text{div}\,\vec{r} = \dfrac{\partial x}{\partial x} + \dfrac{\partial y}{\partial y} + \dfrac{\partial z}{\partial z} = 3.

    div u⃗=div(e⃗r)=div(r⃗r)\text{div}\,\vec{u} = \text{div}(\vec{e}_r) = \text{div}\left(\dfrac{\vec{r}}{r}\right) avec

    r⃗r=xx2+y2+z2e⃗x+yx2+y2+z2e⃗y+zx2+y2+z2e⃗z=x(x2+y2+z2)−1/2e⃗x+y(… )e⃗y+z(… )e⃗z\frac{\vec{r}}{r} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}\vec{e}_x + \frac{y}{\sqrt{x^2 + y^2 + z^2}}\vec{e}_y + \frac{z}{\sqrt{x^2 + y^2 + z^2}}\vec{e}_z = x(x^2 + y^2 + z^2)^{-1/2}\vec{e}_x + y(\dots)\vec{e}_y + z(\dots)\vec{e}_z

    Méthode directe :

    div(r⃗r)=(x2+y2+z2)−1/2+x(−12)(2x)(x2+y2+z2)−3/2+(… ) en y+(… ) en z=(… )−1/2−x2(… )−3/2+(… )−1/2−y2(… )−3/2+(… )−1/2−z2(… )−3/2=3x2+y2+z2−x2+y2+z2(x2+y2+z2)3/2=3x2+y2+z2−1x2+y2+z2=2x2+y2+z2\begin{aligned} \text{div}\left(\frac{\vec{r}}{r}\right) &= (x^2 + y^2 + z^2)^{-1/2} + x\left(-\frac{1}{2}\right)(2x)(x^2 + y^2 + z^2)^{-3/2} + (\dots) \text{ en } y + (\dots) \text{ en } z \\ &= (\dots)^{-1/2} - x^2(\dots)^{-3/2} + (\dots)^{-1/2} - y^2(\dots)^{-3/2} + (\dots)^{-1/2} - z^2(\dots)^{-3/2} \\ &= \frac{3}{\sqrt{x^2 + y^2 + z^2}} - \frac{x^2 + y^2 + z^2}{(x^2 + y^2 + z^2)^{3/2}} = \frac{3}{\sqrt{x^2 + y^2 + z^2}} - \frac{1}{\sqrt{x^2 + y^2 + z^2}} \\ &= \frac{2}{\sqrt{x^2 + y^2 + z^2}} \end{aligned}

    Méthode avec la relation du CSV10, div(GA⃗)=A⃗.grad→ G+G div A⃗\text{div}(G\vec{A}) = \vec{A}.\overrightarrow{\text{grad}}\,G + G\,\text{div}\,\vec{A} :

    div(r⃗r)=r⃗.grad→1r+1rdiv r⃗=r⃗.[∂∂r(1r)e⃗r]+1r3=−1r2e⃗r.r⃗+3r=−1r2r⃗r.r⃗+3r=3r−1r\text{div}\left(\frac{\vec{r}}{r}\right) = \vec{r}.\overrightarrow{\text{grad}}\frac{1}{r} + \frac{1}{r}\text{div}\,\vec{r} = \vec{r}.\left[\frac{\partial}{\partial r}\left(\frac{1}{r}\right)\vec{e}_r\right] + \frac{1}{r}3 = -\frac{1}{r^2}\vec{e}_r.\vec{r} + \frac{3}{r} = -\frac{1}{r^2}\frac{\vec{r}}{r}.\vec{r} + \frac{3}{r} = \frac{3}{r} - \frac{1}{r}
    div(r⃗r)=2r=2x2+y2+z2\text{div}\left(\frac{\vec{r}}{r}\right) = \frac{2}{r} = \frac{2}{\sqrt{x^2 + y^2 + z^2}}
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