Chapter 9 Appendices
9.1 Transformation of Euler’s equation to an ac- celerating frame of reference
The Euler equation may be written
∂u
∂t +u· ∇u+f ∧u=−1 ρ∇p and
u· ∇u=ω∧u+∇(1 2u2) Therefore
∂u
∂t + (ω+f)∧u=−1 ρ∇
µ p+ 1
2u2
¶
Now transform the coordinate system (x, t) to (X, T), where x=X+xc(t), t =T
and dxc
dt =c(t) = (c1(t), c2(t)).
Then ∂
∂x = ∂
∂X
∂X
|{z}∂x
= 1
+ ∂
∂Y
∂Y
|{z}∂x
= 0
+ ∂
∂T
∂T
|{z}∂x
= 0
∂
∂y = ∂
∂X
∂X
|{z}∂y
= 0
+ ∂
∂Y
∂Y
|{z}∂y
= 1
+ ∂
∂T
∂T
|{z}∂y
= 0
∂
∂t = ∂
∂X
∂X
|{z}∂t
=−c1
+ ∂
∂Y
∂Y
|{z}∂t
=−c2
+ ∂
∂T
∂T
|{z}∂t
= 1
118
CHAPTER 9. APPENDICES 119
Figure 9.1:
∂
∂x
∂
∂y
∂
∂t
=
1 0 0
0 1 0
−c1 −c2 1
∂
∂X
∂
∂Y
∂
∂T
,
or ∂
∂t = ∂
∂T −c· ∇X, ∇x=∇X
Note that although
T =t, ∂/∂T 6=∂/∂t.
LetUbe the velocity in the moving frame, i.e.,U=u−c. Then the Euler equation transforms to
µ ∂
∂T −c· ∇
¶
(U+c) + (ω+f)∧(U+c) = 1
ρ∇p− ∇ µ1
2(U+c)2
¶ , where
ω=∇x∧u=∇X∧U and U∧c= 0.
Further reduction gives
∂U
∂T + (ω+f)∧U=−1
ρ∇p− ∇(12U2)− ∂c
∂t +c· ∇U−(ω+f)∧c− ∇(U·c), using the fact that ∇c=0 because c=c(t). Now
∇(U·c) =U· ∇c|{z}
= 0
+c· ∇U+U∧(∇ ∧c)
| {z }
= 0
+c(∇ ∧U)
| {z }
=ω
∴ ∂U
∂T + (ω+f)∧U=−1
ρ∇(p+ 12ρU2)−f ∧c− dc
dt. (9.1)
CHAPTER 9. APPENDICES 120 The vorticity equation takes the form
∂ω
∂T +U· ∇(ω+f) = (ω+f)· ∇U−c· ∇f (9.2) using
∇ ∧(f ∧c) =f(∇ ·c)
| {z }
= 0
− (∇ ·f)c
| {z }
f=(0,0,f)=0 and∂f /∂z=0.
+c· ∇f − f|{z}· ∇
= 0
c
