Download Laboratorio di Metodi Numerici per Equazioni alle Derivate Parziali I

Laboratorio di Metodi Numerici per
Equazioni alle Derivate Parziali I
a.a.2016/2017
Francesca Fierro
Email: [email protected]
Ricevimento:
Venerdı̀ 10.30 - 12.30
(o su appuntamento via email)
Example 1: a model of pressure sensor
Practical application: aneroid barometer
This is an instrument for measuring pressure without involving any liquid, invented in
France by Lucien Vidi in 1844.
It uses a small cilindrical metal box with a
flexible basis (aneroid capsule). Inside the
capsule there is vacuum so that small changes
in the external pressure causes the capsule to
expand or contract. With the help of gears
and levers these little movements are amplified and dispalyed over a graduate scale on
the front of the barometer.
A model problem
Elastic membrane
Elastic
Membrane
fixed Pressure

−div(A ∇u) = f
u = g
in Ω
on ∂Ω
where we chose Ω = (−1, 1)2 ⊂ R2
"
#
1 0
, c ∈ R material dependent
0 1
f = Pi − Pe difference of pressure
u = displacement of the membrane from the
reference situation
g=0
A=c
The solution u of the model problem depends
linearly on the data f .
Let u be the solution of problem (1) with
data f
−div(c ∇(2u)) = −c∆(2u) = −2c∆u = 2f
i.e. 2u solves the same problem with data 2f
Example 2: electric potential and
electric field in a condenser
A condenser is an electric component used
to store energy elettrostatically in an electric
field.
Condensers may have different shapes, but
always contain at least two conductors separated by a dielectric.
Once set a potential difference across the
conductors, positive charge collects on one of
them and negative on the other one. Moreover an electric field develops through the
dieletric.
The model problem
We consider a cave square section condenser
and determine the electric potential inside
it, supposing that on the external boundary
Γ0 and on the internal boundary Γ1 different
electric potential are fixed.
Γ0 is in blue, Γ1 is in red and the dieletric is in light blue.

−∆u = 0
u = g
in Ω
on Γ0 ∪ Γ1
where Ω = (−2, 2)2 \ (−1, 1)2 ⊂ R2
u is the electric potential
g(x) =

0
1
on Γ0
on Γ1
For simmetry reasons we restrict our study
to the upper left quadrant.
The light blue domain Ω is the dieletric. The blue line is Γ0 ,
the red one is Γ1 . In green the artificial boundary Γ2.
As a side effect we have to set conditions on
the new generated artificial boundary Γ2.
A natural choice would be to set homogeneous Neumann boundary conditions.
∂u
=0
on Γ2
∂n
where n is the outer normal to Γ2
Example 3: Heat diffusion in a
refrigerator at steady state
We want to study the distribution of the temperature inside a refrigerator at steady state
(i.e. no change in time). To construct a
model for the refrigerator we consider the
following pattern
The light blue region, Ω0 , is the interior of the refrigerator. The green one, Ω1 , corresponds to the sides
and the door of the fridge. Γ0, in blue, is the refrigerated wall, Γ1 is the boundary at room temperature
and Γ2 , in green, is part of the boundary where we
suppose the temperature varies linearly.
The model problem


−div(A(x) ∇u) = 0




u = 5

u = 20




u(x , x ) = a x + b
0 1
1
A(x) =

1
0.1
in Ω = Ω0 ∪ Ω1
on Γ0
on Γ1
on Γ2
if x ∈ Ω0
if x ∈ Ω1
A(x) describes the thermal conductivity, i.e.
the property of the material to conduct heat.
u is the temperature.
For symmetric reason we can restrict our
model to the upper half of the previous scheme.
here the yellow line correspond to an artificially created boundary,Γ3, where we assume
homogeneous Neumann condition in order to
ensure thermal insulation (i.e. no heat flux).
The problem reads now



−div(A(x) ∇u) = 0






u = 5
u = 20




u(x0, x1) = 15 x1 + 5




 ∂u = 0
∂n
in Ω = Ω0 ∪ Ω1
on Γ0
on Γ1
on Γ2
on Γ3
where Ω = (−1, 1)2 = Ω0 ∪ Ω1
and Ω0 = (−1, 0)2