![[Graphics:Images/noflownoflux_gr_1.gif]](Images/noflownoflux_gr_1.gif)
Example 1:
![[Graphics:Images/noflownoflux_gr_3.gif]](Images/noflownoflux_gr_3.gif)
![[Graphics:Images/noflownoflux_gr_4.gif]](Images/noflownoflux_gr_4.gif)
Here is a picture of F (the arrows are scaled, to make the picture clearer).
![[Graphics:Images/noflownoflux_gr_5.gif]](Images/noflownoflux_gr_5.gif)
![[Graphics:Images/noflownoflux_gr_6.gif]](Images/noflownoflux_gr_6.gif)
![[Graphics:Images/noflownoflux_gr_7.gif]](Images/noflownoflux_gr_7.gif)
![[Graphics:Images/noflownoflux_gr_8.gif]](Images/noflownoflux_gr_8.gif)
![[Graphics:Images/noflownoflux_gr_9.gif]](Images/noflownoflux_gr_9.gif)
![[Graphics:Images/noflownoflux_gr_10.gif]](Images/noflownoflux_gr_10.gif)
Now look at the level curves of the two components of F1. These components are P1=
, Q1=-2xy.
Their level curves are supposed to be perpendicular to each other. We first show P1, then Q1, then both together.
![[Graphics:Images/noflownoflux_gr_12.gif]](Images/noflownoflux_gr_12.gif)
![[Graphics:Images/noflownoflux_gr_13.gif]](Images/noflownoflux_gr_13.gif)
![[Graphics:Images/noflownoflux_gr_14.gif]](Images/noflownoflux_gr_14.gif)
![[Graphics:Images/noflownoflux_gr_15.gif]](Images/noflownoflux_gr_15.gif)
![[Graphics:Images/noflownoflux_gr_16.gif]](Images/noflownoflux_gr_16.gif)
![[Graphics:Images/noflownoflux_gr_17.gif]](Images/noflownoflux_gr_17.gif)
You can see the curves are perpendicular.
![[Graphics:Images/noflownoflux_gr_18.gif]](Images/noflownoflux_gr_18.gif)
![[Graphics:Images/noflownoflux_gr_19.gif]](Images/noflownoflux_gr_19.gif)
![[Graphics:Images/noflownoflux_gr_20.gif]](Images/noflownoflux_gr_20.gif)
![[Graphics:Images/noflownoflux_gr_21.gif]](Images/noflownoflux_gr_21.gif)
![[Graphics:Images/noflownoflux_gr_22.gif]](Images/noflownoflux_gr_22.gif)
![]()
Example 2: Start with f(x)=Cos[x]. It turns out that Cos[x-iy]=Cos[x]Cosh[y]+iSin[x]Sinh[y],
so P3=Cos[x]Cosh[y], Q3=Sin[x]Sinh[y], and F3=(Cos[x]Cosh[y],Sin[x]Sinh[y]).
Here is a picture of F3:
![[Graphics:Images/noflownoflux_gr_24.gif]](Images/noflownoflux_gr_24.gif)
![[Graphics:Images/noflownoflux_gr_25.gif]](Images/noflownoflux_gr_25.gif)
![[Graphics:Images/noflownoflux_gr_26.gif]](Images/noflownoflux_gr_26.gif)
Now F3=∇(Sin[x]Cosh[y]), and the functions
P4=Sin[x]Cosh[y] Q4=-Cos[x]Sinh[y]
satisfy the Cauchy-Riemann equations, so, as with F1, the level curves of Q4 are the flow lines of F3.
![[Graphics:Images/noflownoflux_gr_27.gif]](Images/noflownoflux_gr_27.gif)
![[Graphics:Images/noflownoflux_gr_28.gif]](Images/noflownoflux_gr_28.gif)
![[Graphics:Images/noflownoflux_gr_29.gif]](Images/noflownoflux_gr_29.gif)
![[Graphics:Images/noflownoflux_gr_30.gif]](Images/noflownoflux_gr_30.gif)
![[Graphics:Images/noflownoflux_gr_31.gif]](Images/noflownoflux_gr_31.gif)
![[Graphics:Images/noflownoflux_gr_32.gif]](Images/noflownoflux_gr_32.gif)