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Green's Theorem

Usagi

Member
Jun 26, 2012
46


For the above expression, I was told that it can be proven using Green's Theorem on the line integral on the RHS, however I can't seem the prove the equality.

Note that $G$, $H$, $f$ are functions of $x_1$ and $x_2$.

So I apply Green's Theorem:

$\displaystyle{\oint_C f\left(Gdx_1 - H dx_2\right) = \oint_C fG dx_1 + \left(-fH\right)dx_2 = \iint_D -\frac{\partial \left(fH\right)}{\partial x_1} - \frac{\partial \left(fG\right)}{\partial x_2} dA}$

But then what? I can't seem to get the RHS to equal the LHS.

Any help would be appreciated :)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703


For the above expression, I was told that it can be proven using Green's Theorem on the line integral on the RHS, however I can't seem the prove the equality.

Note that $G$, $H$, $f$ are functions of $x_1$ and $x_2$.

So I apply Green's Theorem:

$\displaystyle{\oint_C f\left(Gdx_1 - H dx_2\right) = \oint_C fG dx_1 + \left(-fH\right)dx_2 = \iint_D -\frac{\partial \left(fH\right)}{\partial x_1} - \frac{\partial \left(fG\right)}{\partial x_2} dA}$

But then what? I can't seem to get the RHS to equal the LHS.

Any help would be appreciated :)
This is just the product rule: $\dfrac{\partial(fH)}{\partial x_1} = \dfrac{\partial f}{\partial x_1}H + f\dfrac{\partial H}{\partial x_1}$ and $\dfrac{\partial (fG)}{\partial x_2} = \dfrac{\partial f}{\partial x_2}G + f\dfrac{\partial G}{\partial x_2}.$
 

Usagi

Member
Jun 26, 2012
46
This is just the product rule: $\dfrac{\partial(fH)}{\partial x_1} = \dfrac{\partial f}{\partial x_1}H + f\dfrac{\partial H}{\partial x_1}$ and $\dfrac{\partial (fG)}{\partial x_2} = \dfrac{\partial f}{\partial x_2}G + f\dfrac{\partial G}{\partial x_2}.$
Thanks, Yup I did that however how does it simplify down the RHS to equal the LHS?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Thanks, Yup I did that however how does it simplify down the RHS to equal the LHS?
$$\begin{aligned}\oint_C f(Gdx_1 - H dx_2) &= \iint_D \Bigl(-\frac{\partial (fH)}{\partial x_1} - \frac{\partial (fG)}{\partial x_2}\Bigr)\, dA \\ &= \iint_D \Bigl(-\dfrac{\partial f}{\partial x_1}H - f\dfrac{\partial H}{\partial x_1} - \dfrac{\partial f}{\partial x_2}G - f\dfrac{\partial G}{\partial x_2} \Bigr)\,dA \\ &= \iint_D \Bigl(-f\Bigl[\dfrac{\partial H}{\partial x_1} + \dfrac{\partial G}{\partial x_2}\Bigr] -\Bigl[G\dfrac{\partial f}{\partial x_2} + H\dfrac{\partial f}{\partial x_1}\Bigr]\Bigr)\,dA, \end{aligned}$$ from which it appears that $$\iint_D\Bigl[G\dfrac{\partial f}{\partial x_2} + H\dfrac{\partial f}{\partial x_1}\Bigr]\,dA = -\iint_D f\Bigl[\dfrac{\partial H}{\partial x_1} + \dfrac{\partial G}{\partial x_2}\Bigr]\,dA -\oint_C f(Gdx_1 - H dx_2). $$ Hmm, it looks as though the sign of that last term is wrong – not sure where that happened (or maybe the original problem had the wrong sign).
 

Usagi

Member
Jun 26, 2012
46
Awesome, thanks Opalg, I had a feeling the initial question had a mistake in it :)