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NUMERICAL METHODS
DISCRETIZZAZIONE DELLE EQUAZIONI DIFFERENZIALI ORDINARIE
REAL PHENOMENON
like BLOOD IN ARTERIES
We choose a model (this is often driven by what we want to see)
There are lots of continuous models that could be chosen.
PARTIAL DIFFERENTIAL EQUATION
PDE
We choose a discretization model like NAVIER-STOKES
MODEL ERROR
We obtain FINITE ELEMENTS
FEM code
also in this case we can choose several implementations
ROUND-OFF ERROR
(errore di arrotondamento)
It's important that the discretization doesn't change our model.
We have to quantify the error made when we choose a PDE to model a real phenomenon.
We don't know xh and so this generates the need to estimate the error.
The numerical analysis transforms the PDE into a discretized problem.
During the computations we are introducing other errors. The discretized
problem is correctly solved only by hand, while using the PC we made a
ROUND-OFF ERROR.
Approximation of Derivatives
Let xj, for some j's, be points in ℝ and suppose h = xj - xj-1.
We know the values of a function v(x) in these points, so we know v(xj-1), v(xj), ...
How could I approximate v'1(xj) = ?
Example
- Forward Euler Formula: v'1(xj) ≈ D+v(xj) = (v(xj+1) - v(xj)) / h
- This is an easy way to approximate derivatives.
The formula is convergent and v's of the first order.
We introduce the error:
Ej = |v'1(xj) - D+v(xj)| = |v'1(xj) - (v(xj+1) - v(xj))/h| = Θ(h)
Taylor Expansion
v(x) = v(y) + v'1(y) (x-y) + v''1(y) / 2 (x-y)2 + v'''1(y) / 6 (x-y)3 + Θ(|x-y|4)
A quantity q is said to be Θ(ε) if limε→0 q/ε < +∞
Take x = xj+1, y = xj; v(xj+1) = v(xj) + v'1(xj)h + Θ(h2)
So we have h = x-y where x = xj+1 & y = xj.
⇒ Θ(h2) = v(xj+1) - v(xj) - v'1(xj)h ⇒ Θ(h) = v'1(xj);
⇒ RICORDIAMO che IL SEGNO DI Θ(h) E' INDIFFERENTE
This problem admits a unique solution provided that there exists L > 0 such that | f(t, y1) - f(t, y2) | ≤ L | y1 - y2 | ∀ t, ∀ y1, ∀ y2.
This is the result of a continuous analysis. Of course, from the numerical point of view I assume that this is the case because it would be useless to search for a numerical solution of a problem which does not admit a continuous solution.
The main difference between the previous case and now is that in the previous case I know the function and I want an approximation of derivative, while now I know the derivative and I want to know an app. of the function.
Apply The FEF (Forward E. F.) to my ODE
Δt = tn - tn-1
We assume that Δt does not depend by n
Write ODE at t = tn
d y(tn) / dt = f(tn, y(tn))
FEF → (y(tn+1) - y(tn)) / Δt ≈ f(tn, y(tn))
because it's an approximation
04/10/2016
dy/dt = f(t, y(t))
In case of FEM we have always a linear equation independent of t while BEM depends on the linearity. We cannot conclude anything about the convergence of FEM and BEM even if the formulas obtained are convergent.
DEF. of convergence
Given a numerical method to approx our continuous problem, the numerical solution un, n=1,..., Δt→0, this method is said to be CONVERGENT if |yn - y(tn)| → 0, ∀n
If |um - y(tm)| = θ(Δtp), ∀n, the method is of order p.
DEF
CONSISTENCY: Write a general numerical method:
um+1 = Φ(un+1)
where ui = {um, um-1, ..., u1}
EX: FEM Φ(un) = um + Δt f(tm, un)
BEM Φ(un+1) = um + Δt f(tn+1, um+1)
Φ is a general method.
The consistency lets us introduce the TRUNCATION ERROR when the general method and used definition.
τm+1 = 1/Δt (y(tn+1) - Φ(y(tm+1))), y(tm) = {y(tm), ..., y(t1)}.
The ERROR is the error done by the numerical solution when the latter fails to solve the continuous problem, while the TRUNCATION ERROR is the error committed when the real continuous solution pretends to solve the numerical one.
the same is true for BEM
BEM: μm+1 = μn + Δt μm+1 → μn+1 = 1/1 - λΔt vn
(λt < 0 (se negativo quindi tale rapporto non può andare all ∞))
We have so proved that these numerical methods are convergent. When Δt goes to 0 our numerical solution is in fact the continuous one so we are happy because we have something which approximates the continuous solution in a good way. (Dal punto di vista puram. matematico da dim. finisce qui.)
From the engineering point of view we want to know what happens for a finite Δt. All works very well for Δt → 0, thus doesn't imply that something could be wrong for a finite Δt. In fact from an engineering point of view this is false.
Example 1
- y'(t) = -10 y(t)
- y(0) = 1
limt→0 y(t)/e2t = 0 so we proved that the method for an increasing time goes to 0 but the convergence doesn't guarantee this because the convergence tells us something about the behaviour when t is small.
While here I'm looking for what happens to the solution for an increasing time but with a finite Δt.
The convergence alone is not enough to ensure that the numerical solution doesn't grow up for an increasing time.
Also I want that the numerical solution for an increasing time because decreases.
Crank-Nicolson
is another method used for PDE. Θ = 1/2
The problem is y'(t) = f(t, y(t))
=> un+1 = un + Δt/2 [f(tn, un) + f(tn+1, un+1)]
Θ is a number chosen by the user between 0 < Θ < 1
I take a linear combination of f evaluated in an implicit way and in an explicit way.
Θ method
(um+1 - un)/Δt = Θ f(tn+1, un+1) + (1 - Θ) f(tn, un)
Θ ∈ [0,1]
For Θ = 0 → FEM
For Θ = 1 → BEM
but I have a family of methods which can be obtained varying Θ. In
particular Crank-Nicolson is obtained with Θ = 1/2
The Θ method is again an implicit method because if f is non linear
I have to use Newton.
What about the convergence and the A.S of Θ method?
For Θ ≠ 1/2 Θ method is of order 1 (|u(tn) - un| = Θ(Δt))
For Θ = 1/2 Θ method is of order 2 (|u(tn) - un| = Θ(Δt2))
this means that the error decrease
as the square of Δt and so the Crank-
Nicolson is the fastest among the Θ methods
(is the most accurate among the Θ methods)
What about the A.S?
For Θ < 1/2 : Θ method is conditionally, absolute stable
=> Δt <
2/λ(2Θ - 1)
un+1 = un + Δt(1 - Θ)λun + ΔtΘλun+1
un+1 =
(1 + Δt(1 - Θ)λ)/
1 - ΔtΘλ · un
= cT
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