2. Determine Green's Identity for the Given Differential Operator

To construct in a systematic way the solution to the differential equation specified by (a) and (b) above, one introduces the adjoint differential equation and its solution. This philosophy is an extension of the approach already used to solve ordinary differential equations of second order.

Central to this approach was the consideration of the linear operator

$$L\psi=\frac{d^2\psi}{dx^2}+\beta \frac{d\psi}{dx}+\gamma\psi$$

and its adjoint

$$L^*\phi=\frac{d^2\phi}{dx^2}- \frac{\beta d\phi}{dx}+\gamma\phi$$

which was determined by the compatibility condition, the Lagrange identity Eq.(3.15)

$$\phi L\psi -\psi L^*\phi =\frac{d}{dx}P(\phi,\psi)~.$$ (618)

Here the right hand side was the derivative of

$$P(\phi,\psi)=\phi\frac{d\psi}{dx}-\psi\frac{d\phi}{dx} +\beta\phi\psi~,$$

the ``bilinear concomitant'' introduced with Eq.(4.4) on page for a one-dimensional domain.

The extension of Lagrange's identity to a two-dimensional domain is straight forward. Given the differential operator, Eq.(6.17), one seeks the corresponding adjoint, $L^*\phi$ , which is determined by the compatibility condition

$$\boxed{ \phi L\psi -\psi L^*\phi =\frac{\partial P_u}{\partial u}+\frac{\partial P_v}{\partial v} }$$ (619)

Here the right hand side is the divergence of a two-component vectorial concomitant. It replaces the total derivative of the scalar concomitant, Eq.(6.18).

What are $L^*\phi$ and $(P_u,P_v)$ ? Comparing the sought-after expressions in Eq.(6.19), with the known result, Eq.(6.18), one finds with a little scrutiny that

$$L^*\phi= \frac{\partial^2\phi}{\partial u\partial v} -\frac{\partial D\phi}{\partial u} -\frac{\partial E\phi}{\partial v} +F\phi$$ (620)

and

$$P_u = \frac{1}{2}\left( \phi\frac{\partial\psi}{\partial v}-\psi\frac{\partial\phi}{\partial v} \right) +D\phi\psi$$ (621)

$$P_v = \frac{1}{2}\left( \phi\frac{\partial\psi}{\partial u}-\psi\frac{\partial\phi}{\partial u} \right) +E\phi\psi$$ (622)

The boxed Eq.(6.19) is the key to success. By integrating it over a triangular region, say $RSQ$ in Figure 6.2, one obtains Green's identity adapted to the given differential operator, Eq.(6.17). Indeed applying Stokes' theorem to the right hand side, one obtains

$$\int_{RSQ}\int (\phi L\psi -\psi L^*\phi)du\,dv=\int_{S\rightarrow Q\rightarrow R\rightarrow S} P_udv-P_vdu$$ (623)