Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
vuoi
o PayPal
tutte le volte che vuoi
Numerical Modelling of Differential Problems
Introduction to Numerical Analysis
Formulation of a generic problem: F(x, d) = 0
- F can be any type of eqt: linear, int, partial diff...
- F, d known → direct problem
- F, x known → inverse problem
- x, d known → identification problem
Problem is said to be WELL-POSED when the pb admits a solution, UNIQUE and continuously depending on the data.
If introduced a small perturbation on data is expected to have a small perturbation on the solution F(x+δx, d+δd)=0
∇y∃δ : 3K(y, d) : ||δd|| < ε ⇒ ||δx|| < K(d, y, d) ||δd||
Can be defined the CONDITION NUMBER as the number that quantifies the well-posed property of a problem
Relative Condition Number:
K(d) = supδdε ||δx||/||x|| / ||δd||/||d||
"Which is the entity of the relative perturbation when introduced a pert. on the data"
Absolute Condition Number:
Kabs(d) = supδdε ||δx|| / ||δd||
For big K (≥1000) the pb is ILL-CONDITIONED and the solution is totally different
small K (≤1-100) the pb is WELL-CONDITIONED
NUMERICAL METHODS
Discrete numerical counterpart Fn(xn,dn)=0 n>1 of the continuous pb:
A method can be said CONSISTENT if: Fn(x, d) = Fn(x, d) - F(x, d) ⇒0
∇n→∞
If this property is true for every n the method is said STRONGLY CONSISTENT
Fm(x, d) = F∀n
A method is said to be STABLE (or well-posed) if ∇ fixed n, ∃! solution
xn fort d and xn depends continuously on the data, similarly to the continuous pb.
Fn(xn+δxn, dn+δdn)=0
If the continuous pb is well-posed, passing to the numerical method we want to maintain this property, unless it's useless.
Remark. Sobolev spaces are Banach spaces for 1 ≤ p ≤ ∞; for p = 2, W1,2(Ω) is also a Hilbert space. The case for p = 2 is the most common in FE analysis and we will use the notation H1(Ω) = W1,2(Ω). For k = 1 we have
H1(Ω) = {u ∈ L2(Ω); ∇u ∈ L2(Ω)}
with the following inner product and norm
(u,v)H1(Ω) = (u,v)L2(Ω) + (∇u, ∇v)L2(Ω)
‖u‖H1(Ω) = ‖u‖L2(Ω) + ‖∇u‖L2(Ω)
Remark. For d = 1 it can be shown that its functions are continuous for d = 1, may lack values at certain isolated points for d = 2 and be discontinuous along a curve for d = 3.
How to define the boundary values of a function belonging to a Sobolev space?
IDEA: make a smooth approximation from within the domain, and then evaluate this approximation on the boundary. This is called the trace of the function. We can define the trace operator
γ : W1,p(Ω) → Lp(∂Ω)
and the space
H01(Ω) = {v ∈ H1(Ω) : (γv)|∂Ω = 0}
Classification of PDEs
The equations of elasticity (no inertial terms) or the Laplacian are elliptic PDEs.
∇⋅σ + F = 0, ε = 1/2 [∇u + (∇u)T], σ = C : ε, -△u = f
The heat conduction equation is an example of a parabolic PDE.
∂T/∂t - ∇⋅(k∇T) = 0
Hyperbolic PDEs describe wave propagation and transport phenomena.
∂2u/∂t2 - c2△u = 0, ∂u/∂t + a ⋅ ∇u = 0
FINITE DIFFERENCES
f′(xi) = limh→0 [f(xi + h) - f(xi)]/h
uiFD = [f(xi+1) - f(xi)]/h, uiBD = [f(xi) - f(xi-1)]/h, uiCD = [f(xi+1) - f(xi-1)]/2h
f(xi+1) = f(xi) + h f′(xi) + h2/2 f″(ξi), f′(xi) = f′(xi) - uiFD = - h/2 f″(ξi)
f′(xi) - uiBD = h/2 f″(ξi), f′(xi) - uiCD = - h2/6 f‴(ξi)
∫01 w''v'dx = ∫01 fvdx - ∫01 r''v'dx
the solution of the pb is a translation of the solution obtained for w
A non-homo Dirichlet problem can be recast into an homo one,
where the pb can be solved for w then recall u = w+r
MINIMIZATION FORMULATION of the pb
A third formulation of the DIFFERENTIAL PROBLEM can be introduced
J(v) = ½ ∫01 v'v'dx - ∫01 fvdx
the solution u to the diff. pb is the one which minimizes the internal
energy of the system F(u)
(H) min J(v) = min ½ ∫01 v'v'dx - ∫01 fvdx
(D) - u'' = f in (0,1)
u(0) = u(1) = 0
(V) find u ∈ H10(0,1) s.t.
∫01 u'v'dx = ∫01 fvdx ∀v ∈ H10(0,1)
(M) find u ∈ H10(0,1) s.t.
min F(v) = min ½ ∫01 v'v'dx - ∫01 fvdx
v ∈ H10(0,1) v ∈ H10(0,1)
⇒ (D) ⇔ (V) ⇔ (M)
⇒ if u'' ∃ and is continuous
(V) ⇒ (M) ∀ε H10(0,1)
if u is solution
of (V) then is
solution of (M)
Now we want to check if our pb is well-posed, so Lax-Milgram lemma's hypothesis must be satisfied.
-Δuⁿ = f in Ω
u = 0 on ∂Ω
a(u,v) = ∫Ω∇v·∇v dx
- V is an Hilbert space
- Continuity of a(u,v),
|a(u,v)| = |∫Ω∇v·∇v dx| ≤ ∫Ω||u||L2||v||L2 ≤ ||u||L2||v||L2
Continuity of F(v):
|F(v)| = |∫Ωf·v dx| ≤ ||f||L2||v||L2 ≤ ||F||L2||v||L2
- Coercivity of a(v,v)
a(v,v) ≥ α||v||H0¹² >0
∫Ω||v||L2
a(v,v) = ∫Ω∇v·∇v dx = ||∇v||L2
||v||L2 = ||v1||L2 + ||v2||L2
≤ c∑||v1||L2 + ||v2||L2
≤ (r+c2)∑||v||L2
||v||H0¹² =
a(v,v) = ∫Ω∇v·∇v dx = ||∇v||² ≥
||v||H0¹
Proved these hy it is possible to say that the continuous problem: find u∈Γ a(u,v)=F(v) ∀ v∈ V in well-posed → solution ∃, it's unique and depends continuously on data
Using the considerations made with ii the discrete pb (c) is well-posed too
Strong Form
-Δu = f in Ω u = 0 on ∂Ω
Weak Form: find u∈Γ a(u,v)=F(v) ∀ v∈ V
Algebraic pb
Au=f where A contains the coeff of uh in Vh
a(∫φ∫φ) = ∫0 ∫0 ∫φ∫φ'