12.520 Lecture Notes 25
The Stream Function
For continuum mechanics in general and fluid mechanics specifically, a number of
“laws” are expressed in terms of differential equations. For example,
1) Newton’s second law (F = ma)
∂σ ji
∂x j
+ ρ fi = ρ
(general)
Dvi
Dt
2) Rheology (constitutive equation)
(Newtonian fluid)
σ ij = − pδ ij + 2ηε ij
3) Definition of strain rate
1 ⎛ ∂v
(general)
∂v ⎞
ε ij = ⎜⎜ i + j ⎟⎟
2 ⎝ ∂x j ∂xi ⎠
4) Continuity (conservation of mass)
(incompressible)
∂vi
=0
∂xi
These 4 coupled first order differential equations, plus boundary conditions, can be
solved to determine fluid flow for a variety of interesting applications.
Alternatively, they can be combined to form a single fourth order differential equation.
For fluids, this fourth order equation often involves the stream function.
Consider a 2-D flow with velocities v1, v3 in the x1, x3 plane (v2 = 0)
If v1 = −
∂Ψ
∂x3
∂vi
∂2Ψ
∂2Ψ
∂Ψ
v3 =
⇒ ∇⋅v =
=−
+
=0
∂xi
∂x1∂x3 ∂x3∂x1
∂x1
Incompressibility is automatically satisfied!
[In general, if v = ∇ × Ψ , ∇ ⋅ v = 0. Here Ψ = (0, Ψ, 0)]
Substituting into the (steady) Navier-Stokes equation
⎛ ∂3 Ψ
∂p
∂3 Ψ ⎞
− η⎜ 2
+
+ ρ f1 = 0
3⎟
∂x1
⎝ ∂x1 ∂x3 ∂x3 ⎠
⎛ ∂3 Ψ
∂3 Ψ ⎞
∂p
+ η⎜ 3 +
−
+ ρ f3 = 0
2⎟
∂x3
⎝ ∂x1 ∂x1∂x3 ⎠
−
Now take
∂
∂
of first,
of second
∂x3
∂x1
⎛ ∂4 Ψ
∂2 p
∂4 Ψ ⎞
∂f
− η⎜ 2 2 +
+ρ 1 =0
4 ⎟
∂x1∂x3
∂x3
⎝ ∂x1 ∂x3 ∂x3 ⎠
⎛ ∂4 Ψ
∂4 Ψ ⎞
∂2 p
∂f
+ η⎜ 4 + 2 2 ⎟ + ρ 3 = 0
−
∂x1∂x3
∂x1
⎝ ∂x1 ∂x1 ∂x3 ⎠
−
Subtract:
⎛ ∂4 Ψ
η⎜
⎝ ∂x1
4
+2
⎛ ∂f3 ∂f1 ⎞
∂4 Ψ
∂4 Ψ ⎞
+
+
ρ
−
⎟
⎜
⎟=0
∂x12∂x32 ∂x3 4 ⎠
⎝ ∂x1 ∂x3 ⎠
∂4 Ψ
∂4 Ψ
∂4 Ψ
+2 2 2 +
= ∇ 2 (∇ 2 Ψ) = ∇ 4 Ψ
4
4
∂x1
∂x1 ∂x3 ∂x3
∇ 4 is called biharmonic operator.
For uniform or no f :
∇4 Ψ = 0
Advantages of using the biharmonic operator are
1. only one equation
2. efficient solution
Disadvantage: Loss of “physical insight”.
Physical Interpretation of Stream Function
Consider triangle APB.
x3
B
δs
A
δx3
δx1
v1
P
v3
x1
For incompressible fluid,
fluxAP + fluxBP + fluxAB = 0
−v3δ x1 + v1δ x3 + flux AB = 0
flux AB = v3δ x1 − v1δ x3 =
B
or
∫ dΨ = Ψ
B
∂Ψ
∂Ψ
δ x1 +
δ x3 = δΨ
∂x1
∂x3
− ΨA
A
Difference in Ψ represents the flux crossing the curve.
Solution of biharmonic
v0 x32
Polynomials (e.g., for Conette flow, Ψ = −
)
2h
Separation of variables:
Ψ = X ( x )Z ( z )
∇ 4 Ψ = 0 ⇒ X '''' Z + 2 X '' Z ''+ XZ '''' = 0
X ''''
X '' Z '' Z ''''
+2
+
=0
X
X Z
Z
Harmonic Ψ = sin
2π x
λ
Z(z)
Solution: Ψ = [(A + Bz)exp(
2π z
λ
Physical boundary conditions:
) + (C + Dz)exp(−
Tn = 0
λ
)]sin(
2π x
λ
)
Tτ = 0
x3 '
x3
ξ
2π z
α
x1 '
x1
In x1 ', x3 ' coordinates, at x3 = ξ (x1 ) :
σ 3' 3' = 0
σ 3'1' = σ 1' 3' = 0
Have solution to biharmonic in terms of x1 , x3 -- easily applied at x3 = 0.
Need to take physical ( x1 ', x3 ' ) boundary conditions and
1. rotate to x1 , x3 space
2. Taylor’s series expansion
3. subtract out hydrostatic reference state
Result (to first order in ξ / λ )
⎛? ?
0 ⎞
⎜
⎟
σ = ⎜? ? 0 ⎟
⎜
⎟
⎝ 0 0 ρ gξ ⎠
4. solve biharmonic.
Postglacial Rebound
Decay of Boundary Undulations (1/2 space, uniform η )
k=
τ = τ0 cos (kx1)
X3
Figure 25.1
Figure by MIT OCW.
•
Assume uniform η
•
Subtract out lithostatic pressure P = p − ρ gx3
•
Assume ρ g uniform
•
Use stream function Ψ
v1 = −
∂Ψ
∂x3
v3 =
∂Ψ
∂x1
⇒ ∇4 Ψ = 0
Solution: Ψ = ⎡⎣(A + Bkx3 )exp (−kx3 ) + (C + Dkx3 )exp (kx3 )⎤⎦ ⋅ sin kx1
Boundary conditions:
at x3 = 0 (mathematical, not physical)
σ 33 = ρ gζ
⎛ ∂v1
σ 13 = 0 = η ⎜
⎝ ∂x3
+
∂v3 ⎞
⎟
∂x1 ⎠
at x3 → ∞ , must be bounded
2π
λ
X1
⇒C = D = 0
In order that σ 13 = 0 at x 3 = 0 ,
∂2 Ψ ∂2 Ψ
− 2 + 2 =0
∂x3
∂x1
⇒B=A
or Ψ = A(1+ kx3 )exp (−kx 3 ) ⋅ sin kx1
Then
v1 = Ak 2 x3 exp (−kx3 ) ⋅ sin kx1
v3 = Ak(1+ kx3 )exp (−kx3 ) ⋅ cos kx1
at x3 = 0 v3 = ζ = Ak cos( kx1 )
Now
σ 33 = − p + 2ηε 33
ε 33 = 0 at x3 = 0
To get p, use −
∂p
∂2 vi
+η
+ ρ x1 = 0
∂xi
∂x j ∂x j
for i = 1
⇒ −
∂p
∂2 v ∂ 2 v
+ η( 21 + 21 ) = 0
∂x1
∂x1 ∂x3
Substitute for v1 and integrating ⇒ p x
But p = −ρ gζ ⇒ A = −
Or ζ 0 = −
3 =0
ρ gζ 0
2 k 2η
ρ gζ 0
ρ gλζ 0
=−
2kη
4 πη
Or ζ 0 = ζ 0
where τ =
t =0
exp(−
ρ gt
t
) = ζ 0 t =0 exp(− )
2kη
τ
2kη 4 πη
=
ρ g ρ gλ
Solving for η : η =
For curves shown,
ρ gλτ
4π
= 2ηk 2 A cos kx1
τ : 5000 yr ⎫
⎬ ⇒ η : 10 21 Pa
λ : 3000 km ⎭
Note: stream function ∼ exp(−kx3 ) = exp(−
Falls off to ∼ 1 / e at x3 :
2 π x3
λ
)
λ
2π
Senses to fairly great depth
⇒ postglacial rebound doesn’t reveal the details of mantle viscosity structure,
but only the gross structure.
Note: Behavior at Hudson Bay and Boston different:
Hudson Bay
t
Continuous uplift
Boston
t
Subsidence, then uplift
Is this consistent with uniform 1/2 space?
τ=
4 πη
ρ gλ
Decompose into Fourier components
⇒
?
Details depend on geometry of ice load and elastic support of load.
Suppose we require faster relaxation for short λ than for long λ.
elastic “lithosphere”
low viscosity “asthenosphere”
depth
higher viscosity “mesosphere”
How to get solution? What are the boundary conditions?