91Ó°ÊÓ

The identities $∫∇Tâ‹\dots d\mathrm{Ï„}=âˆ\circledR Tâ‹\dots da$, $\underset{}{∫}\text{ â¶Ä‰}\left(∇×v\right)â‹\dots d\mathrm{Ï„}=−\underset{}{∫}\left(v×da\right)\text{ â¶Ä‰}$, $∫T{∇}^{2}U+∇Uâ‹\dots ∇T=∫\left(T∇U\right)\text{ }da\text{ }$,$∫U{∇}^{2}T+∇Tâ‹\dots ∇U=∫\left(U∇T\right)\text{ }da\text{ }$

and $∫∇T×da=−âˆ\circledR Tâ‹\dots dl$have to be proved. Here $T,\text{ }U,\text{ â¶Ä‰}V$are the vector and c is a constant.

According to the Gauss divergence theorem The integral ofdivergenceof a function$f\left(x,y,z\right)$ over an closed surface area is equal to the surfaceintegral of the function $∫\left(∇v\right)â‹\dots \text{ }d\mathrm{Ï„}=\underset{s}{âˆ\circledR }vâ‹\dots \text{ }da$. According to thestokestheorem, the integral of divergence of a function $f\left(x,y,z\right)$over an open surface area is equal to the line integral of the function $∫\left(∇×v\right)â‹\dots \text{ }da=\underset{l}{âˆ\circledR }vâ‹\dots \text{ }dl$.

The divergence theorem is defined as $∫\left(∇v\right)â‹\dots d\mathrm{Ï„}=âˆ\circledR vâ‹\dots da$.Substitute $cT$ for v into $∫\left(∇v\right)â‹\dots d\mathrm{Ï„}=âˆ\circledR vâ‹\dots da$as follows:

$∫\left(∇\left(cT\right)\right)â‹\dots d\mathrm{Ï„}=âˆ\circledR cTâ‹\dots da$ ……….. (1)

Apply the product rule (i) ,$∇\left(fA\right)=f\left(∇â‹\dots A\right)+Aâ‹\dots \left(∇f\right)$ in equation (1),

$\begin{array}{c}∫\left(∇\left(cT\right)\right)â‹\dots d\mathrm{Ï„}=âˆ\circledR cTâ‹\dots da\\ ∫\left(T\left(∇â‹\dots c\right)+câ‹\dots \left(∇T\right)\right)â‹\dots d\mathrm{Ï„}=âˆ\circledR cTâ‹\dots da\end{array}$ ……….. (2)

As c is a constant, so its divergence is 0, that is, $∇â‹\dots c=0$.

Substitute 0 for $∇â‹\dots c$ into equation (2)

$\begin{array}{c}∫\left(T\left(0\right)+câ‹\dots \left(∇T\right)\right)â‹\dots d\mathrm{Ï„}=âˆ\circledR cTâ‹\dots da\\ ∫\left(câ‹\dots \left(∇T\right)\right)â‹\dots d\mathrm{Ï„}=âˆ\circledR cTâ‹\dots da\\ c∫∇Tâ‹\dots d\mathrm{Ï„}=câˆ\circledR Tâ‹\dots da\\ ∫∇Tâ‹\dots d\mathrm{Ï„}=âˆ\circledR Tâ‹\dots da\end{array}$

Thus, $∫∇Tâ‹\dots d\mathrm{Ï„}=âˆ\circledR Tâ‹\dots da$, has been shown.

The gauss divergence theorem states that the volume integral of the divergence of a function v is equal to the surface integral of the function v, that is,$\underset{v}{∫}\text{ }∇vâ‹\dots d\mathrm{Ï„}=\underset{s}{∫}v\text{ }da\text{ }$

Substitute $v×c$for v into $\underset{v}{∫}\text{ }∇vâ‹\dots d\mathrm{Ï„}=\underset{s}{∫}v\text{ }da\text{ }$.

$\underset{v}{∫}\text{ }∇\left(v×c\right)â‹\dots d\mathrm{Ï„}=\underset{s}{∫}\left(v×c\right)\text{ }da\text{ }$…… (3)

Apply the rule $\text{ }∇\left(\text{A}×\text{B}\right)=\text{B}\left(∇×\text{A}\right)−\text{A}\left(∇×\text{B}\right)$ into equation (3)

$\underset{v}{∫}\text{ â¶Ä‰}\left(c\left(∇×v\right)−v\left(∇×c\right)\right)â‹\dots d\mathrm{Ï„}=\underset{s}{∫}\left(v×c\right)\text{ }da\text{ }$

As c is a constant, so it’s curl is 0, that is, $∇×c=0$.

Substitute 0 for $∇×c$ into equation$\underset{v}{∫}\text{ â¶Ä‰}\left(c\left(∇×v\right)−v\left(∇×c\right)\right)â‹\dots d\mathrm{Ï„}=\underset{s}{∫}\left(v×c\right)\text{ }da\text{ }$

$\begin{array}{l}\underset{v}{∫}\text{ â¶Ä‰}\left(c\left(∇×v\right)−v\left(0\right)\right)â‹\dots d\mathrm{Ï„}=\underset{s}{∫}\left(v×c\right)\text{ }da\text{ }\\ \underset{v}{∫}\text{ â¶Ä‰}c\left(∇×v\right)â‹\dots d\mathrm{Ï„}=\underset{s}{∫}c\left(∇×da\right)\text{ â¶Ä‰}\\ c\underset{v}{∫}\text{ â¶Ä‰}\left(∇×v\right)â‹\dots d\mathrm{Ï„}=\underset{s}{∫}c\left(da×v\right)\text{ â¶Ä‰}\\ c\underset{v}{∫}\text{ â¶Ä‰}\left(∇×v\right)â‹\dots d\mathrm{Ï„}=−\underset{s}{∫}c\left(v×da\right)\text{ â¶Ä‰}\end{array}$

Solve further as,

$\begin{array}{c}c\underset{v}{∫}\text{ â¶Ä‰}\left(∇×v\right)â‹\dots d\mathrm{Ï„}=−c\underset{s}{∫}\left(v×da\right)\text{ â¶Ä‰}\\ \underset{}{∫}\text{ â¶Ä‰}\left(∇×v\right)â‹\dots d\mathrm{Ï„}=−\underset{}{∫}\left(v×da\right)\text{ â¶Ä‰}\end{array}$

Thus, the result $\underset{}{∫}\text{ â¶Ä‰}\left(∇×v\right)â‹\dots d\mathrm{Ï„}=−\underset{}{∫}\left(v×da\right)\text{ â¶Ä‰}$has been shown.

Let a function V is defined as, $v=T∇U$and the divergence theorem is defined as $\underset{v}{∫}\text{ }∇vâ‹\dots d\mathrm{Ï„}=\underset{s}{∫}v\text{ }da\text{ }$.

Substitute for into $\underset{v}{∫}\text{ }∇vâ‹\dots d\mathrm{Ï„}=\underset{s}{∫}v\text{ }da\text{ }$.

$\underset{v}{∫}\text{ }∇\left(T∇U\right)â‹\dots d\mathrm{Ï„}=\underset{s}{∫}\left(T∇U\right)\text{ }da\text{ }$…… (4)

Apply the product rule $∇\left(fA\right)=f\left(∇â‹\dots A\right)+Aâ‹\dots \left(∇f\right)$ into equation (4)

$\begin{array}{c}∫\text{ â¶Ä‰}\left(T∇â‹\dots ∇U−∇Uâ‹\dots ∇T\right)â‹\dots d\mathrm{Ï„}=∫\left(T∇U\right)\text{ }da\text{ }\\ ∫T{∇}^{2}U+∇Uâ‹\dots ∇T=∫\left(T∇U\right)\text{ }da\text{ }\end{array}$

Thus, the result $∫T{∇}^{2}U+∇Uâ‹\dots ∇T=∫\left(T∇U\right)\text{ }da\text{ }$has been shown.

Swap the variables $T↔U$in the result of part (c)$∫T{∇}^{2}U+∇Uâ‹\dots ∇T=∫\left(T∇U\right)\text{ }da\text{ }$, as shown below:

$∫U{∇}^{2}T+∇Tâ‹\dots ∇U=∫\left(U∇T\right)\text{ }da\text{ }$

Subtract the resulting equation from $∫T{∇}^{2}U+∇Uâ‹\dots ∇T=∫\left(T∇U\right)\text{ }da\text{ }$,as,

$\begin{array}{c}∫\left(T{∇}^{2}U+∇Uâ‹\dots ∇T\right)−\left(U{∇}^{2}T+∇Tâ‹\dots ∇U\right)d\mathrm{Ï„}=∫\left(T∇U\text{ â¶Ä‰}−U∇T\right)\text{ }da\text{ }\\ ∫\left(T{∇}^{2}U−U{∇}^{2}T\right)d\mathrm{Ï„}=∫\left(T∇U\text{ â¶Ä‰}−U∇T\right)\text{ }da\text{ }\end{array}$

Thus, the result $∫\left(T{∇}^{2}U−U{∇}^{2}T\right)d\mathrm{Ï„}=∫\left(T∇U\text{ â¶Ä‰}−U∇T\right)\text{ }da\text{ }$has been shown.

Sokes theorem is defined as $∫\left(∇×v\right)â‹\dots da=âˆ\circledR vâ‹\dots dl$.Substitute $cT$for v into $∫\left(∇×v\right)â‹\dots da=âˆ\circledR vâ‹\dots dl$as follows:

$∫\left(∇×\left(cT\right)\right)â‹\dots da=âˆ\circledR \left(cT\right)â‹\dots dl$ ……….. (5)

Apply the product rule (ii) ,$∇×\left(fA\right)=f\left(∇×A\right)−A×\left(∇f\right)$in equation (5),

$\begin{array}{c}∫\left(∇×\left(cT\right)\right)â‹\dots da=âˆ\circledR cTâ‹\dots dl\\ ∫\left(T\left(∇â‹\dots ×c\right)−c×\left(∇T\right)\right)â‹\dots da=âˆ\circledR cTâ‹\dots dl\end{array}$……….. (6)

As is a constant, so its curl is 0, that is $∇×c=0$.

Substitute 0 for $∇×c$into equation (6)

$\begin{array}{l}∫\left(T\left(0\right)−c×\left(∇T\right)\right)â‹\dots da=âˆ\circledR cTâ‹\dots dl\\ −∫\left(c×\left(∇T\right)\right)â‹\dots da=âˆ\circledR cTâ‹\dots da\\ c∫\left(∇T×da\right)=−câˆ\circledR Tâ‹\dots dl\\ ∫∇T×da=−âˆ\circledR Tâ‹\dots dl\end{array}$

Thus, $∫∇T×da=−âˆ\circledR Tâ‹\dots dl$has been shown.