Solution

The principle of minimum potential energy requires that the functional ${\displaystyle \Pi }$ be stationary for the actual displacement field ${\displaystyle w(x_{1},x_{2})}$. Taking the first variation of ${\displaystyle \Pi }$, we get

${\displaystyle \delta \Pi ={\frac {\eta }{2}}\int _{\mathcal {S}}\left[(2w_{,1})\delta w_{,1}+(2w_{,2})\delta w_{,2}\right]~dA-\int _{\mathcal {S}}p~\delta w~dA}$

or,

${\displaystyle \delta \Pi =\eta \int _{\mathcal {S}}\left[w_{,1}~\delta w_{,1}+w_{,2}~\delta w_{,2}\right]~dA-\int _{\mathcal {S}}p~\delta w~dA}$

Now,

${\displaystyle {\begin{aligned}(w_{,1}~\delta w)_{,1}&=w_{,11}~\delta w+w_{,1}~\delta w_{,1}\\(w_{,2}~\delta w)_{,2}&=w_{,22}~\delta w+w_{,2}~\delta w_{,2}\end{aligned}}}$

Therefore,

${\displaystyle w_{,1}~\delta w_{,1}+w_{,2}~\delta w_{,2}=(w_{,1}~\delta w)_{,1}-w_{,11}~\delta w+(w_{,2}~\delta w)_{,2}-w_{,22}~\delta w}$

Plugging into the expression for ${\displaystyle \delta \Pi }$,

${\displaystyle \delta \Pi =\eta \int _{\mathcal {S}}\left[(w_{,1}~\delta w)_{,1}+(w_{,2}~\delta w)_{,2}-(w_{,11}+w_{,22})~\delta w\right]~dA-\int _{\mathcal {S}}p~\delta w~dA}$

or,

${\displaystyle \delta \Pi =\eta \int _{\mathcal {S}}\left[(w_{,1}~\delta w)_{,1}+(w_{,2}~\delta w)_{,2}\right]~dA-\eta \int _{\mathcal {S}}\nabla ^{2}{w}~\delta w~dA-\int _{\mathcal {S}}p~\delta w~dA}$

Now, the Green-Riemann theorem states that

${\displaystyle \int _{\mathcal {S}}(Q_{,1}-P_{,2})~dA=\oint _{\partial {\mathcal {S}}}(P~dx_{1}+Q~dx_{2})}$

Therefore,

${\displaystyle \delta \Pi =\eta \oint _{\partial {\mathcal {S}}}\left[(w_{,1}~\delta w)~dx_{2}-(w_{,2}~\delta w)~dx_{1}\right]-\int _{\mathcal {S}}\left[\eta \nabla ^{2}{w}+p\right]~\delta w~dA}$

or,

${\displaystyle \delta \Pi =\eta \oint _{\partial {\mathcal {S}}}\left[w_{,1}~{\frac {dx_{2}}{ds}}-w_{,2}~{\frac {dx_{1}}{ds}}\right]\delta w~ds-\int _{\mathcal {S}}\left[\eta \nabla ^{2}{w}+p\right]~\delta w~dA}$

where ${\displaystyle s}$ is the arc length around ${\displaystyle \partial {\mathcal {S}}}$.

The potential energy function is rendered stationary if ${\displaystyle \delta \Pi =0}$. Since ${\displaystyle \delta w}$ is arbitrary, the condition of stationarity is satisfied only if the governing differential equation for ${\displaystyle w(x_{1},x_{2})}$ on ${\displaystyle {\mathcal {S}}}$ is

${\displaystyle {\eta \nabla ^{2}{w}+p=0~~~~~\forall ~~(x_{1},x_{2})~\in ~{\mathcal {S}}}}$

The associated boundary conditions are

${\displaystyle {w_{,1}~{\frac {dx_{2}}{ds}}-w_{,2}~{\frac {dx_{1}}{ds}}=0~~~~~\forall ~~(x_{1},x_{2})~\in ~\partial {\mathcal {S}}}}$