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NUMERICAL METHODS FOR DIFFERENTIAL EQNS
NAVIER - STOKES
∂t (u(y) v) + ∇ · (u(y) v ⊗ v) + ∇p - ∇ · ϵ(v) + F = 0
- go to WEAK form, performing integrals piecewise
∫∫ Ω ∇· (u(y) v) q = 0
∫∫ Ω ϵ(v) : ∇ q = ∫∫ ∂Ω φ q = 0; this is already used to ensure div(u) = 0
(u(y) v · ∇) v = nonlinear term remains as it was
- Non-linear term ∫Ω (u · ∇v) vT is called a trilinear form
- Usually we add a condition to Q : Q = L2(α) s.t. ∫ Q = 0
- Algebraic counterpart of 1st eqn no time derivative
- Iterative schemes for c(vn, vn, vn) = F(v)∀ v ∈ Hp
- Idea is to use at differential level a scheme like: initial guess for velocity u0; having discarded time it's only an iteration for nonlinearity
- Convergence strongly depends on choice of initial guess
[ A + C(v)][U] = [F] [ B ][ P ]
- Convergence strongly depends on properties of the flow
Recall of Newton method
E(X) = 0 → fi(x1, ..., xn) = 0
x(0)
to line to F = 0, in intersection with x-axis
f' (x(0)) (x - x(0)) + f (x(0)) = 0
f' (x(0)) (x(1) - x(0)) + f (x(0)) = 0
f' (x(0)) (x(1)) + f (x(0)) = 0
to be generalized to our case
DF (x(i))
(x(i+1) = x(i) + δx)
DF (x(i)) = [∂fi/∂xj … ∂fi/∂xn](i)
J (x(i))
we'll have to invert Jacobian matrix
J (x(i)) δx = -F (x(i))
δx is the solution of the linear system
Stopping Criteria
- Residue: |P (x(M))| must be as close as possible to zero
- f (x(M)) = x(M), choose a-priori tolerance ε
If |P (x(M))|2 l2 ε , stop
F (x(M)) = x(M)
Increments (difference between 2 successive iterates: small enough)
In most cases both criterias are checked, to impose stronger requirements
In practical cases for our problems, residue is the most commonly used
What is DF (x) = J (x) in our case?
Consider a generic non-linear operator N (u)
DN (u+)
active d.o.f., it could be (ℓ(M+1-M))
h direction
DN (u+) h = limε→0 N (u+ + εB) - N (u+) / ε
(definition of directional derivative)
N (u ) is LINEAR: DN (u+) h = limε→0 N (u+) -εEN (B) - N (u+)/ ε = N (B)1
N (u+) = 0
(DN (u+) hb = N (B)b = ∂h)
N (u+) is NONLINEAR ( (u : ∇u) u )
DN (u+) h =
limε→0 ((u+ + εB) . ∇ (u+ + εB) . (u+ + εB) - (u+ . ∇u+ ) u+) / ε
limε→0 (u+h) . ∇(u+ε) ub b + ε (Eb )b (u) + (u+h) . ∇ (D) u+
DN (u+) h2 = (∇ (u+) hb) h . (∇ (D) u+) h U+
-∇ E (u(n+1))
( u(n) . ∇) u(n) + (u(n+1) . ∇) u(n+1)) g U+ + ∇p ( u(n+1)) = f ( u(n) ), V(f)
∇ . U(n+1) = 0