6.4 Green’s theorem  (Page 5/12)

Note that the domain of F is all of ${ℝ}^2,$ which is simply connected. Therefore, to show that F is source free, we can show any of items 1 through 4 from the previous list to be true. In this example, we show that item 4 is true. Let $P\left(x,y\right)=y$ and $Q\left(x,y\right)=\text{−}x.$ Then $P_x+0=Q_y,$ and therefore $P_x+Q_y=0.$ Thus, F is source free.

To find a stream function for F , proceed in the same manner as finding a potential function for a conservative field. Let g be a stream function for F . Then $g_y=y,$ which implies that

Since $\text{−}{g}_x=Q=\text{−}x,$ we have $h\text{′}\left(x\right)=x.$ Therefore,

Letting $C=0$ gives stream function

To confirm that g is a stream function for F , note that $g_y=y=P$ and $\text{−}{g}_x=\text{−}x=Q.$

Notice that source-free rotation vector field $\text{F}(x,y)=⟨y,\text{−}x⟩$ is perpendicular to conservative radial vector field $\text{∇}g=⟨x,y⟩$ ( [link] ).

Let $P\left(x,y\right)=e^x\text{sin}y$ and $Q\left(x,y\right)=e^x\text{cos}y.$ Notice that the domain of F is all of two-space, which is simply connected. Therefore, we can check the cross-partials of F to determine whether F is conservative. Note that $P_y=e^x\text{cos}y=Q_x,$ so F is conservative. Since $P_x=e^x\text{sin}y$ and $Q_y=e^x\text{sin}y,P_x+Q_y=0$ and the field is source free.

To find a potential function for F , let $f$ be a potential function. Then, $\text{∇}f=\text{F},$ so $f_x=e^x\text{sin}y.$ Integrating this equation with respect to x gives $f\left(x,y\right)=e^x\text{sin}y+h\left(y\right).$ Since $f_y=e^x\text{cos}y,$ differentiating $f$ with respect to y gives $e^x\text{cos}y=e^x\text{cos}y+h\text{′}\left(y\right).$ Therefore, we can take $h\left(y\right)=0,$ and $f\left(x,y\right)=e^x\text{sin}y$ is a potential function for $f.$

To verify that $f$ is a harmonic function, note that $f_{xx}=\frac{\partial }{\partial x}\left(e^x\text{sin}y\right)=e^x\text{sin}y$ and

$f_{yy}=\frac{\partial }{\partial x}\left(e^x\text{cos}y\right)=\text{−}{e}^x\text{sin}y.$ Therefore, $f_{xx}+f_{yy}=0,$ and $f$ satisfies Laplace’s equation.