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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