Divergence and curl (rot) of vector fields

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Find lim⁡d(S)→01S∮C(F→,n→) ds\displaystyle \lim_{d(S)\to 0} \dfrac {1}{S}\oint_{C} (\overrightarrow {F},\overrightarrow {n})\, ds, where SS is the area enclosed by the contour CC containing the point (x0,y0)(x_{0},y_{0}), d(S)d(S) is the diameter of the region SS, n→\overrightarrow {n} is the unit vector along the outward normal of the contour CC, and F→(x,y)\overrightarrow {F}(x,y) is a vector that is continuously differentiable on S+CS+C.

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Prove the formula

∇2(uv)=u∇2v+v∇2u+2∇u∇v, \nabla ^{2}(uv) = u\nabla ^{2}v+v\nabla ^{2}u+2\nabla u\nabla v ,

where

∇=i→∂∂x+j→∂∂y+k→∂∂z, \nabla = \overrightarrow {i}\dfrac {\partial }{\partial x}+\overrightarrow {j}\dfrac {\partial }{\partial y}+\overrightarrow {k}\dfrac {\partial }{\partial z} , ∇2=∇∇=∂2∂x2+∂2∂y2+∂2∂z2. \nabla ^{2} = \nabla \nabla = \dfrac {\partial ^{2}}{\partial x^{2}}+\dfrac {\partial ^{2}}{\partial y^{2}}+\dfrac {\partial ^{2}}{\partial z^{2}} .
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Prove by direct computation that the divergence of the vector a→\overrightarrow {a} is independent of the choice of Cartesian coordinate system.

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Prove

div⁡a→(M)=lim⁡d(S)→01V∬Sa→⋅n→ dS, \operatorname {div}\overrightarrow {a}(M) = \lim _{d(S)\to 0} \dfrac {1}{V}\iint _{S} \overrightarrow {a}\cdot \overrightarrow {n}\, dS ,

where SS is a closed surface enclosing the point MM and bounding a volume VV, n→\overrightarrow {n} is the outward normal of the surface SS, and d(S)d(S) is the diameter of the surface SS.

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Find:

div⁡∣i→j→k→∂∂x∂∂y∂∂zωxωyωz∣. \operatorname {div} \begin{vmatrix} \overrightarrow {i} & \overrightarrow {j} & \overrightarrow {k} \\ \dfrac {\partial }{\partial x} & \dfrac {\partial }{\partial y} & \dfrac {\partial }{\partial z} \\ \omega _{x} & \omega _{y} & \omega _{z} \end{vmatrix} .
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Prove:

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(a)

div⁡(a→+b→)=div⁡a→+div⁡b→\operatorname {div}(\overrightarrow {a}+\overrightarrow {b}) = \operatorname {div}\overrightarrow {a}+\operatorname {div}\overrightarrow {b};

(b)

div⁡(uc→)=c→⋅grad⁡u\operatorname {div}(u\overrightarrow {c}) = \overrightarrow {c}\cdot \operatorname {grad} u (where c→\overrightarrow {c} is a constant vector and uu is a scalar);

(c)

div⁡(ua→)=udiv⁡a→+a→⋅grad⁡u\operatorname {div}(u\overrightarrow {a}) = u\operatorname {div}\overrightarrow {a}+\overrightarrow {a}\cdot \operatorname {grad} u.

Prove that for twice differentiable fields uu and vv

Δ(uv)≡∇2(uv)=v∇2u+2(∇u,∇v)+u∇2v \Delta (u v) \equiv \nabla ^{2}(u v) = v \nabla ^{2} u+2(\nabla u, \nabla v)+u \nabla ^{2} v
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Verify the indicated equalities in coordinate form, and also write them and verify them using the symbol ∇\nabla and the rules for operating with it (α,β\alpha , \beta numbers, u,a,bu, \mathbf{a}, \mathbf{b} differentiable scalar and vector fields):

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(1)

div⁡(αa+βb)=αdiv⁡a+βdiv⁡b\operatorname {div}(\alpha \mathbf{a}+\beta \mathbf{b}) = \alpha \operatorname {div} \mathbf{a}+\beta \operatorname {div} \mathbf{b}

(2)

div⁡(ua)=(grad⁡u,a)+udiv⁡a\operatorname {div}(u \mathbf{a}) = (\operatorname {grad} u, \mathbf{a})+u \operatorname {div} \mathbf{a}.

Setting r=xi+yj+zk,r=∣r∣\mathbf{r} = x \mathbf{i}+y \mathbf{j}+z \mathbf{k}, r = \left|\mathbf{r}\right|, find div⁡a\operatorname {div} \mathbf{a}, if:

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(1)

a=r\mathbf{a} = \mathbf{r};

(2)

a=rr\mathbf{a} = r \mathbf{r};

(3)

a=r/r\mathbf{a} = \mathbf{r} / r;

(4)

a=(−xi+yj+zk)/x2+y2;\mathbf{a} = (-x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) / \sqrt{x^{2}+y^{2}} ;

(5)

a=(6x2y2−z3+yz−5)i+(4x3+xz+2)j+(xy−3xz2−3)k\mathbf{a} = \left(6 x^{2} y^{2}-z^{3}+y z-5\right) \mathbf{i}+\left(4 x^{3}+x z+2\right) \mathbf{j}+\left(x y-3 x z^{2}-3\right) \mathbf{k}.

Express div⁡grad⁡u\operatorname {div} \operatorname {grad} u in coordinate form.

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Find:

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(1)

div⁡(ugrad⁡u)\operatorname {div}(u \operatorname {grad} u);

(2)

div⁡(ugrad⁡v)\operatorname {div}(u \operatorname {grad} v).

Find (r=xi+yj+zk),r=∣r∣)(\mathbf{r} = x \mathbf{i}+y \mathbf{j}+z \mathbf{k}), r = \left|\mathbf{r}\right|) :

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(1)

div grad⁡r2\operatorname {grad} r^{2};

(2)

div⁡grad⁡(1/r)\operatorname {div} \operatorname {grad}(1 / r);

(3)

div rc,c=r \mathbf{c}, \mathbf{c} = const;

(4)

div⁡(f(r)r)\operatorname {div}(f(r) \mathbf{r});

(5)

div grad⁡f(r)\operatorname {grad} f(r);

(6)

div⁡(f(r)c),c=\operatorname {div}(f(r) \mathbf{c}), \mathbf{c} = const;

(7)

div⁡[c,r],c=\operatorname {div}[\mathbf{c}, \mathbf{r}], \mathbf{c} = const;

(8)

div⁡[r,[c,r]],c=\operatorname {div}[\mathbf{r},[\mathbf{c}, \mathbf{r}]], \mathbf{c} = const.

Solve the equation (r=xi+yj+zk,r=∣r∣)(\mathbf{r} = x \mathbf{i}+y \mathbf{j}+z \mathbf{k}, r = \left|\mathbf{r}\right|) :

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(1)

div⁡(u(r)r)=0\operatorname {div}(u(r) \mathbf{r}) = 0;

(2)

div⁡grad⁡u(r)=0\operatorname {div} \operatorname {grad} u(r) = 0;

(3)

div⁡(u(r)r)=λu(r),λ≠3\operatorname {div}(u(r) \mathbf{r}) = \lambda u(r), \lambda \neq 3.

Find the divergence of the gravitational field of several point masses.

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A medium rotates as a rigid body about an axis with constant angular velocity ω\omega. Find, at a fixed instant of time, the divergence of the field of linear velocities v\mathbf{v} and the field of accelerations w\mathbf{w} of the points of the medium.

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Prove that div a does not depend on the choice of Cartesian coordinate system.

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Find div⁡\operatorname {div} a of the plane field a in polar coordinates.

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Find div a of a three-dimensional field:

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(1)

in cylindrical coordinates;

(2)

in spherical coordinates.

Find (r=x2+y2+z2)\left(r = \sqrt{x^{2}+y^{2}+z^{2}}\right) :

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(1)

div⁡a(r)\operatorname {div} \mathbf{a}(r);

(2)

div⁡(u(r)a(r))\operatorname {div}(u(r) \mathbf{a}(r)).

Verify the indicated equalities in coordinate form, and also write and verify them using the symbol ∇\nabla and the rules for operating with it (α,β\alpha , \beta numbers, u,a,bu, \mathbf{a}, \mathbf{b} differentiable scalar and vector fields, c a constant vector):

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(1)

rot⁡(αa+βb)=αrot⁡a+βrot⁡b\operatorname {rot}(\alpha \mathbf{a}+\beta \mathbf{b}) = \alpha \operatorname {rot} \mathbf{a}+\beta \operatorname {rot} \mathbf{b};

(2)

rot⁡(uc)=[grad⁡u,c]\operatorname {rot}(u \mathbf{c}) = [\operatorname {grad} u, \mathbf{c}];

(3)

rot⁡(ua)=urot⁡a+[grad⁡u,a]\operatorname {rot}(u \mathbf{a}) = u \operatorname {rot} \mathbf{a}+[\operatorname {grad} u, \mathbf{a}];

(4)

rot⁡[c,a]=cdiv⁡a−(c,∇)a\operatorname {rot}[\mathbf{c}, \mathbf{a}] = \mathbf{c} \operatorname {div} \mathbf{a}-(\mathbf{c}, \nabla ) \mathbf{a};

(5)

rot⁡[a,b]=adiv⁡b−bdiv⁡a+(b,∇)a−(a,∇)b\operatorname {rot}[\mathbf{a}, \mathbf{b}] = \mathbf{a} \operatorname {div} \mathbf{b}-\mathbf{b} \operatorname {div} \mathbf{a}+(\mathbf{b}, \nabla ) \mathbf{a}-(\mathbf{a}, \nabla ) \mathbf{b};

(6)

div⁡[a,b]=(b,rot⁡a)−(a,rot⁡b)\operatorname {div}[\mathbf{a}, \mathbf{b}] = (\mathbf{b}, \operatorname {rot} \mathbf{a})-(\mathbf{a}, \operatorname {rot} \mathbf{b}).

Find (r=xi+yj+zk,r=∣r∣(\mathbf{r} = x \mathbf{i}+y \mathbf{j}+z \mathbf{k}, r = \left|\mathbf{r}\right|, a\mathbf{a} and b\mathbf{b} constant vectors, u(r)u(r) a differentiable field):

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(1)

rot⁡r\operatorname {rot} \mathbf{r};

(2)

rot⁡(ra)\operatorname {rot}(r \mathbf{a});

(3)

rot⁡((r,a)b)\operatorname {rot}((\mathbf{r}, \mathbf{a}) \mathbf{b});

(4)

rot⁡(u(r)a)\operatorname {rot}(u(r) \mathbf{a});

(5)

rot⁡(u(r)r)\operatorname {rot}(u(r) \mathbf{r}).

Calculate rot⁡a\operatorname {rot} \mathbf{a} at the point M0M_{0}, if:

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(1)

a=xyzi+(2x+3y−z)j+(x2+z2)k;M0(1;3;2)\mathbf{a} = x y z \mathbf{i}+(2 x+3 y-z) \mathbf{j}+\left(x^{2}+z^{2}\right) \mathbf{k} ; M_{0}(1 ; 3 ; 2);

(2)

a=yzi+zxj+xyk;M0(1;2;−2)\mathbf{a} = \frac{y}{z} \mathbf{i}+\frac{z}{x} \mathbf{j}+\frac{x}{y} \mathbf{k} ; M_{0}(1 ; 2 ;-2).

For any vector p\mathbf{p} the vectors [p,a][\mathbf{p}, \mathbf{a}] and a are perpendicular (if they are nonzero). Is this true for the vectors [∇,a][\nabla , \mathbf{a}] and a\mathbf{a}?

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Find the angle between rot⁡a(M1)\operatorname {rot} \mathbf{a}\left(M_{1}\right) and rot⁡a(M2)\operatorname {rot} \mathbf{a}\left(M_{2}\right), if:

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(1)

a=(x2+y2)i+(y2+z2)j+(z2+x2)kM1(1;2;3),M2(1;1;−1)\mathbf{a} = \left(x^{2}+y^{2}\right) \mathbf{i}+\left(y^{2}+z^{2}\right) \mathbf{j}+\left(z^{2}+x^{2}\right) \mathbf{k} M_{1}(1 ; 2 ; 3), M_{2}(1 ; 1 ;-1)

(2)

a=z3i+(x3+y3)j+xyzk;M1(1;2;0),M2(1;12;4)\mathbf{a} = z^{3} \mathbf{i}+\left(x^{3}+y^{3}\right) \mathbf{j}+x y z \mathbf{k} ; M_{1}(1 ; 2 ; 0), M_{2}(1 ; 12 ; 4).

Find:

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(1)

rot⁡[c,r],c=\operatorname {rot}[\mathbf{c}, \mathbf{r}], \mathbf{c} = const;

(2)

rot⁡[r,[c,r]]\operatorname {rot}[\mathbf{r},[\mathbf{c}, \mathbf{r}]], c=\mathbf{c} = const.

Verify in coordinate form:

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(1)

formula (14): rot⁡grad⁡u=[∇,∇u]=[∇,∇]u=0\operatorname {rot} \operatorname {grad} u = [\nabla , \nabla u] = [\nabla , \nabla ] u = \mathbf{0};

(2)

formula (15): div⁡rot⁡a=(∇,[∇,a])=([∇,∇],a)=0\operatorname {div} \operatorname {rot} \mathbf{a} = (\nabla , [\nabla , \mathbf{a}]) = ([\nabla , \nabla ], \mathbf{a}) = 0.

Verify the equality rot⁡rot⁡a=grad⁡div⁡a−Δa\operatorname {rot} \operatorname {rot} \mathbf{a} = \operatorname {grad} \operatorname {div} \mathbf{a}-\Delta \mathbf{a} in coordinate form, and also write and derive it using the symbol ∇\nabla and the rules for operating with it.

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Find rot⁡grad⁡(1/r)\operatorname {rot} \operatorname {grad}(1 / r).

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Derive the formulas:

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(1)

∇(∇,uc)=(c,∇)∇u,c=\nabla (\nabla , u \mathbf{c}) = (\mathbf{c}, \nabla ) \nabla u, \mathbf{c} = const;

(2)

∇(∇,ua)=u∇(∇,a)+(∇,a)∇u+[∇u,[∇,a]]+(∇u,∇)a+(a,∇)∇u\nabla (\nabla , u \mathbf{a}) = u \nabla (\nabla , \mathbf{a})+(\nabla , \mathbf{a}) \nabla u+[\nabla u,[\nabla , \mathbf{a}]]+(\nabla u, \nabla ) \mathbf{a} + (\mathbf{a}, \nabla ) \nabla u

(3)

[∇,[∇u,c]]=(c,∇)∇u−cΔu[\nabla ,[\nabla u, \mathbf{c}]] = (\mathbf{c}, \nabla ) \nabla u-\mathbf{c} \Delta u.

Show that:

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(1)

div⁡[∇u,∇v]=0\operatorname {div}[\nabla u, \nabla v] = 0

(2)

the vectors a=ugrad⁡v\mathbf{a} = u \operatorname {grad} v and rot⁡a\operatorname {rot} \mathbf{a} are perpendicular.

Find the components of rot⁡a\operatorname {rot} \mathbf{a} of the plane field a in polar coordinates.

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Find the components of rot⁡a\operatorname {rot} \mathbf{a} of a three-dimensional field a:

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(1)

in cylindrical coordinates;

(2)

in spherical coordinates.

Find rot⁡(u(r)a(r)),r=x2+y2+z2\operatorname {rot}(u(r) \mathbf{a}(r)), r = \sqrt{x^{2}+y^{2}+z^{2}}.

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Write Δu=div⁡grad⁡u\Delta u = \operatorname {div} \operatorname {grad} u :

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(1)

in cylindrical coordinates;

(2)

in spherical coordinates.

A medium rotates as a rigid body about an axis with constant angular velocity ω\omega. Let v\mathbf{v} be the field of linear velocities of points at a fixed instant of time. Find rot⁡v\operatorname {rot} \mathbf{v} (use cylindrical coordinates).

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In the simplest case, the system of Maxwell's equations for the electromagnetic field has the form

εc∂E∂t=[∇,H],−μc∂H∂t=[∇,E],(∇,E)=0,(∇,H)=0. \frac{\varepsilon }{c} \frac{\partial \mathbf{E}}{\partial t} = [\nabla , \mathbf{H}], \quad -\frac{\mu }{c} \frac{\partial \mathbf{H}}{\partial t} = [\nabla , \mathbf{E}], \quad (\nabla , \mathbf{E}) = 0, \quad (\nabla , \mathbf{H}) = 0.

Here E\mathbf{E} and H\mathbf{H} are the vector fields of electric and magnetic intensity, ε,μ,c=\varepsilon , \mu , c = const >0> 0. Assuming all functions sufficiently smooth, prove that E\mathbf{E} and H\mathbf{H} satisfy the wave equation

∂2E∂t2=c2εμΔE,∂2H∂t2=c2εμΔH \frac{\partial ^{2} \mathbf{E}}{\partial t^{2}} = \frac{c^{2}}{\varepsilon \mu } \Delta \mathbf{E}, \quad \frac{\partial ^{2} \mathbf{H}}{\partial t^{2}} = \frac{c^{2}}{\varepsilon \mu } \Delta \mathbf{H}
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Let an orthogonal system of curvilinear coordinates (ξ;η;ζ)(\xi ; \eta ; \zeta ) be introduced in the region Ω\Omega :

x=x(ξ;η;ζ),y=y(ξ;η;ζ),z=z(ξ;η;ζ) x = x(\xi ; \eta ; \zeta ), \quad y = y(\xi ; \eta ; \zeta ), \quad z = z(\xi ; \eta ; \zeta )

where the right-hand sides are continuously differentiable functions. Let eξ,eη,eζ\mathbf{e}_{\xi }, \mathbf{e}_{\eta }, \mathbf{e}_{\zeta } be the unit vectors of this system (vectors tangent to the coordinate lines and directed toward increasing coordinates, eξ⊥eη,eη⊥eζ,eζ⊥eξ\mathbf{e}_{\xi } \perp \mathbf{e}_{\eta }, \mathbf{e}_{\eta } \perp \mathbf{e}_{\zeta }, \mathbf{e}_{\zeta } \perp \mathbf{e}_{\xi })∗^{*}. Let Hξ,Hη,HζH_{\xi }, H_{\eta }, H_{\zeta } be the Lame coefficients, i.e. Hξ=(∂x∂ξ)2+(∂y∂ξ)2+(∂z∂ξ)2H_{\xi } = \sqrt{\left(\frac{\partial x}{\partial \xi }\right)^{2}+\left(\frac{\partial y}{\partial \xi }\right)^{2}+\left(\frac{\partial z}{\partial \xi }\right)^{2}} etc. Prove that:

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(1)

grad⁡u=1Hξ∂u∂ξeξ+1Hη∂u∂ηeη+1Hζ∂u∂ζeζ\operatorname {grad} u = \frac{1}{H_{\xi }} \frac{\partial u}{\partial \xi } \mathbf{e}_{\xi }+\frac{1}{H_{\eta }} \frac{\partial u}{\partial \eta } \mathbf{e}_{\eta }+\frac{1}{H_{\zeta }} \frac{\partial u}{\partial \zeta } \mathbf{e}_{\zeta };

(2)

div⁡a=1HξHηHζ(∂(HηHζaξ)∂ξ+∂(HζHξaη)∂η+∂(HξHηaζ)∂ζ)\operatorname {div} \mathbf{a} = \dfrac {1}{H_{\xi } H_{\eta } H_{\zeta }}\left(\dfrac {\partial \left(H_{\eta } H_{\zeta } a_{\xi }\right)}{\partial \xi }+\dfrac {\partial \left(H_{\zeta } H_{\xi } a_{\eta }\right)}{\partial \eta }+\dfrac {\partial \left(H_{\xi } H_{\eta } a_{\zeta }\right)}{\partial \zeta }\right);

(3)

rot⁡a=1HξHηHζ∣HξeξHηeηHζeζ∂∂ξ∂∂η∂∂ζHξaξHηaηHζaζ∣\operatorname {rot} \mathbf{a} = \dfrac {1}{H_{\xi } H_{\eta } H_{\zeta }} \begin{vmatrix} H_{\xi } \mathbf{e}_{\xi } & H_{\eta } \mathbf{e}_{\eta } & H_{\zeta } \mathbf{e}_{\zeta } \\ \frac{\partial }{\partial \xi } & \frac{\partial }{\partial \eta } & \frac{\partial }{\partial \zeta } \\ H_{\xi } a_{\xi } & H_{\eta } a_{\eta } & H_{\zeta } a_{\zeta }\end{vmatrix}

Here, for orthogonal curvilinear coordinates (ξ;η;ζ)(\xi ; \eta ; \zeta ) with unit vectors eξ,eη,eζ\mathbf{e}_{\xi }, \mathbf{e}_{\eta }, \mathbf{e}_{\zeta } and Lame coefficients Hξ,Hη,HζH_{\xi }, H_{\eta }, H_{\zeta } (see problem 66), formulas (39)-(41) are

grad⁡u=1Hξ∂u∂ξeξ+1Hη∂u∂ηeη+1Hζ∂u∂ζeζ,(39) \operatorname {grad} u = \frac{1}{H_{\xi }} \frac{\partial u}{\partial \xi } \mathbf{e}_{\xi }+\frac{1}{H_{\eta }} \frac{\partial u}{\partial \eta } \mathbf{e}_{\eta }+\frac{1}{H_{\zeta }} \frac{\partial u}{\partial \zeta } \mathbf{e}_{\zeta }, \quad (39) div⁡a=1HξHηHζ(∂(HηHζaξ)∂ξ+∂(HζHξaη)∂η+∂(HξHηaζ)∂ζ),(40) \operatorname {div} \mathbf{a} = \dfrac {1}{H_{\xi } H_{\eta } H_{\zeta }}\left(\dfrac {\partial \left(H_{\eta } H_{\zeta } a_{\xi }\right)}{\partial \xi }+\dfrac {\partial \left(H_{\zeta } H_{\xi } a_{\eta }\right)}{\partial \eta }+\dfrac {\partial \left(H_{\xi } H_{\eta } a_{\zeta }\right)}{\partial \zeta }\right), \quad (40) rot⁡a=1HξHηHζ∣HξeξHηeηHζeζ∂∂ξ∂∂η∂∂ζHξaξHηaηHζaζ∣.(41) \operatorname {rot} \mathbf{a} = \dfrac {1}{H_{\xi } H_{\eta } H_{\zeta }} \begin{vmatrix} H_{\xi } \mathbf{e}_{\xi } & H_{\eta } \mathbf{e}_{\eta } & H_{\zeta } \mathbf{e}_{\zeta } \\ \frac{\partial }{\partial \xi } & \frac{\partial }{\partial \eta } & \frac{\partial }{\partial \zeta } \\ H_{\xi } a_{\xi } & H_{\eta } a_{\eta } & H_{\zeta } a_{\zeta }\end{vmatrix}. \quad (41)

Using formulas (39)-(41), derive expressions for grad⁡u,div⁡a,rot⁡a\operatorname {grad} u, \operatorname {div} \mathbf{a}, \operatorname {rot} \mathbf{a} :

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(1)

in cylindrical coordinates;

(2)

in spherical coordinates.

Prove formula (24).

Formula (24): let a\mathbf{a} be a continuously differentiable vector field in a region Ω\Omega, M∈ΩM \in \Omega, and consider the regions G⊂ΩG \subset \Omega containing MM for which the Gauss--Ostrogradsky formula (23)

∬∂G(a,n)dS=∭Gdiv⁡adV=∭G(∇,a)dV \iint _{\partial G} (\mathbf{a}, \mathbf{n}) d S = \iiint _{G} \operatorname {div} \mathbf{a} d V = \iiint _{G} (\nabla , \mathbf{a}) d V

holds (GG is a bounded region, Gˉ⊂Ω\bar{G} \subset \Omega, with piecewise smooth boundary ∂G\partial G oriented by the outward normal n\mathbf{n}). Let d(G)d(G) be the diameter and μG\mu G the volume of GG. Then

div⁡a(M)=lim⁡d(G)→01μG∬∂G(a,n)dS. \operatorname {div} \mathbf{a}(M) = \lim _{d(G) \rightarrow 0} \frac{1}{\mu G} \iint _{\partial G} (\mathbf{a}, \mathbf{n}) d S.
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Let the field a be continuously differentiable in Ω,G\Omega , G an arbitrary region with piecewise smooth boundary, Gˉ⊂Ω\bar{G} \subset \Omega. Prove that the flux of rot⁡a\operatorname {rot} \mathbf{a} through ∂G\partial G equals zero.

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Prove that

∭G(∇u,[∇,a])dV=∬∂G(a,∇u,n)dS. \iiint _{G}(\nabla u,[\nabla , \mathbf{a}]) d V = \iint _{\partial G}(\mathbf{a}, \nabla u, \mathbf{n}) d S.
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Let uu and a be continuously differentiable fields in Ω\Omega, GG a region of Ω,Gˉ⊂Ω,∂G\Omega , \bar{G} \subset \Omega , \partial G a piecewise smooth surface, oriented by the outward normal. Prove that

∬∂G(ua,n)dS=∭G(u(∇,a)+(a,∇u))dV \iint _{\partial G}(u \mathbf{a}, \mathbf{n}) d S = \iiint _{G}(u(\nabla , \mathbf{a})+(\mathbf{a}, \nabla u)) d V
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Let uu and a\mathbf{a} be continuously differentiable fields in Ω\Omega, M∈ΩM \in \Omega. Consider the collection of regions G⊂ΩG \subset \Omega containing MM for which formula (23) holds.

Formula (23), the Gauss--Ostrogradsky formula: if GG is a bounded region, Gˉ⊂Ω\bar{G} \subset \Omega, with piecewise smooth boundary ∂G\partial G oriented by the outward normal n\mathbf{n}, then ∬∂G(a,n)dS=∭Gdiv⁡adV=∭G(∇,a)dV\iint_{\partial G} (\mathbf{a}, \mathbf{n}) d S = \iiint_{G} \operatorname {div} \mathbf{a} d V = \iiint_{G} (\nabla , \mathbf{a}) d V.

Let d(G)d(G) be the diameter, μ(G)\mu (G) the volume of GG. Prove that:

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(1)

grad⁡u(M)=lim⁡d(G)→01μ(G)∬∂GundS;\operatorname {grad} u(M) = \lim_{d(G) \rightarrow 0} \frac{1}{\mu (G)} \iint_{\partial G} u \mathbf{n} d S ;

(2)

rot⁡a(M)=lim⁡d(G)→01μ(G)∬∂G[n,a]dS\operatorname {rot} \mathbf{a}(M) = \lim_{d(G) \rightarrow 0} \frac{1}{\mu (G)} \iint_{\partial G}[\mathbf{n}, \mathbf{a}] d S.

Prove that the potential uu of a continuously differentiable field a satisfies the equation Δu=div⁡a\Delta u = \operatorname {div} \mathbf{a}.

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Let uu be a field twice continuously differentiable in Ω\Omega, a\mathbf{a} and b\mathbf{b} differentiable fields in Ω,a=b+grad⁡u\Omega , \mathbf{a} = \mathbf{b}+\operatorname {grad} u. Prove that, for the field b\mathbf{b} to be solenoidal, it is necessary and sufficient that the field uu satisfy the equation Δu=div⁡a\Delta u = \operatorname {div} \mathbf{a}.

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From the equations of electrostatics

(∇,E)=ρ/ε0,[∇,E]=0 (\nabla , \mathbf{E}) = \rho / \varepsilon _{0}, \quad [\nabla , \mathbf{E}] = \mathbf{0}

where E\mathbf{E} is the electric field intensity, ρ\rho the charge density, ε0=\varepsilon_{0} = const >0> 0, derive Gauss's law

∬∂G(n,E)dS=Qε0 \iint _{\partial G}(\mathbf{n}, \mathbf{E}) d S = \frac{Q}{\varepsilon _{0}}

on the proportionality of the flux of the intensity through the boundary of the region GG (with outward normal n\mathbf{n}) and the total charge QQ located in this region.

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References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Функции многих переменных. 2-е изд., ФИЗМАТЛИТ, 2003.