Question

A more elegant proof of the second uniqueness theorem uses Green's

identity (Prob. 1.61c), with $\mathit{T}\mathbf{=}\mathit{U}\mathbf{=}{\mathit{V}}_{\mathbf{3}}$. Supply the details.

Short answer

Answer

It is proved as second uniqueness of theorem by using greens identity.

Step-by-step solution

Define function

Write the Greens identity.

$\underset{\mathbf{~}\mathbf{N}}{\int }\left[\left(\mathbf{T}{\mathbf{~}\mathbf{N}}^{\mathbf{2}}\mathbf{U}\right)\mathbf{+}\mathbf{~}\mathbf{NU}\mathbf{\times }\mathbf{~}\mathbf{NT}\right]\mathbf{d\tau }\mathbf{=}\oint \left(\mathbf{T}\mathbf{~}\mathbf{NU}\right)\mathbf{\times }\mathbf{da}$…… (1)

Given that $T=U=V_3$,

Therefore, the greens identity is changes as,

$\underset{\mathbf{~}\mathbf{N}}{\int }\left[\left({\mathbf{V}}_{\mathbf{3}}{\mathbf{~}\mathbf{N}}^{\mathbf{2}}{\mathbf{V}}_{\mathbf{3}}\right)\mathbf{+}\mathbf{~}{\mathbf{NV}}_{\mathbf{3}}\mathbf{\times }\mathbf{~}{\mathbf{NV}}_{\mathbf{3}}\right]\mathbf{d\tau }\mathbf{=}\oint {\mathbf{V}}_{\mathbf{3}}\left(\mathbf{~}{\mathbf{NV}}_{\mathbf{3}}\right)\mathbf{\times }\mathbf{da}$ …… (2)

Determine proof of Greens identity

Since

$$\begin{gathered} {\mathbf{\nabla }}^{\mathbf{2}}{\mathit{V}}_{\mathbf{3}}\mathbf{=}{\mathbf{\nabla }}^{\mathbf{2}}{\mathit{V}}_{\mathbf{1}}\mathbf{-}{\mathbf{\nabla }}^{\mathbf{2}}{\mathit{V}}_{\mathbf{2}} \\ {\mathbf{\nabla }}^{\mathbf{2}}{\mathit{V}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{-}\mathbf{\rho }}{{\mathbf{\epsilon }}_{\mathbf{0}}} \\ {\mathbf{\nabla }}^{\mathbf{2}}{\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{-}\mathbf{\rho }}{{\mathbf{\epsilon }}_{\mathbf{0}}} \end{gathered}$$

Then,

$$\begin{gathered} {\mathbf{\nabla }}^{\mathbf{2}}{\mathit{V}}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{-}\mathbf{\rho }}{{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{+}\frac{\mathbf{\rho }}{{\mathbf{\epsilon }}_{\mathbf{0}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \end{gathered}$$

As known to us,

$\mathbf{\nabla }{\mathit{V}}_{\mathbf{3}}\mathbf{=}{\mathit{E}}_{\mathbf{3}}$

${\mathit{E}}_{\mathbf{3}}\mathbf{=}\mathbf{\nabla }{\mathit{V}}_{\mathbf{3}}$as per derivation.

Determine proof of Greens identity

Substitute the above values in equation (1)

$\begin{array}{rcl}\underset{\nabla }{\int }\left[V_3\left(0\right)+E_3^2\right]d\tau & =& -\oint V_3{E}_3·da\\ \underset{\nabla }{\int }{E}_3^{2}d\tau & =& -\oint V_3{E}_3·da\end{array}$ …… (3)

As,

${\mathit{E}}_{\mathbf{3}}\mathbf{=}{\mathit{E}}_{\mathbf{1}}\mathbf{-}{\mathit{E}}_{\mathbf{2}}$ …… (4)

If V is specified as ${\mathit{V}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$or ${\mathit{E}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$then,

The equation (4) will be,

$$\begin{gathered} \mathbf{0}\mathbf{=}{\mathit{E}}_{\mathbf{1}}\mathbf{-}{\mathit{E}}_{\mathbf{2}} \\ {\mathit{E}}_{\mathbf{1}}\mathbf{=}{\mathit{E}}_{\mathbf{2}} \end{gathered}$$

Thus, filed is uniquely determined.

And it is proved as this is a second uniqueness theorem.