Materiale diviso per argomento di Mathematical and Numerical Methods in Engineering per il corso di Biomedical Engineering presso Politecnico di Milano. Il documento tratta: MATHEMATICAL YETHODS Introduction Newton's 2 Caw => F m a * = . GMEmm. i Gravitational LaW E Fo IIRI 115112 from this law you can find a relation describing the rotation of the moon around the Earth, providing you know the position of the moon Newton-Leibniz = ) · r(t) _ = x +…
Materiale diviso per argomento di Mathematical and Numerical Methods in Engineering per il corso di Biomedical Engineering presso Politecnico di Milano. Il documento tratta: MATHEMATICAL YETHODS Introduction Newton's 2 Caw => F m a * = . GMEmm. i Gravitational LaW E Fo IIRI 115112 from this law you can find a relation describing the rotation of the moon around the Earth, providing you know the position of the moon Newton-Leibniz = ) · r(t) _ = x +…
Qualità dell’importazione: il testo è stato estratto direttamente dal documento originale.
Passaggi rappresentativi riconosciuti nelle diverse parti del materiale. Il testo completo resta presente nella pagina per la ricerca, mentre l’anteprima compatta rende più semplice la lettura.
Introduction Newton's 2 Caw => F m a * = . GMEmm. i Gravitational LaW E Fo IIRI 115112 from this law you can find a relation describing the rotation of the moon around the Earth, providing you know the position of the moon Newton-Leibniz = ) · r(t) _ = x + y(tj + zn - vectorz depends on time 。 11 rl 1 = XItI t Y ( ' + ZIE? _ · alt) r "It)= Fo = MrEHH The acceleration can be written as " = -Gr { . ME 6 or equivalently X " ( XItYI + E ) X (H) with a system of Differential equations which has an unique solution , . ME 6 given ゾニー · YIt) initial conditions , such as Alta ,Ito). 1x + y2 + z232 " . ME 6 で = 1+ ) in other words if we know the position 1 × 2 ry 라 Za / (고 3 , and velocity of the moon at a given time to, then we can know its position w rt . . . the earth at any tre+> to { 北 シシッ州 ) This system is just a particular case of the General System = 0 Ju \ t シジ ' ッソ州 ) , = 0 Several phisical phenomena were modeled through Differential equations, but the unknown Functions were usually dependent on more than one variable . 1) Motion of incompressible Non-viscous fluids (Euler 2) Wave Motion ID'Alembert) 3) Heat Diffusion (Fourier) al Gravitational or Electrostatic Potential (a place & Poisson ) 5) Motion of incompressible Viscous fluids (Navier & Stokes) d Electromagnetic wave Propagation I Maxwell re All the equation contain Partial Derivatives of the unknown functions> Partial Differential Equations * When the unknow functions depend only on 1 variable IPDEs) = Ordinary Differential Equations lODEs) Partial Differential Equations => F(x, y , Du,, ..., DPu) = 0 - F= given function - X = vector XEM & RV M open set = - u= unknown function - Du = Partial Derivative of M * j 1 , jn) Multi-index = ..., = D'M = ·"'M * (j) j1 + + jn = length of Multi-index .. . = dx , ....* - R…
Transport Equation · Consider a substance le g pollutant) which is transported by a fluid with a given velocity field . . and subject to a source/sink E f(x +) = , b ( x) , =β = ( 3 x ' , / × 2 , ×3) EIR さ "The polutant can increase - souce decrease - > sink · If we indicate the mass density 5 x,t an consider a bounded open set r ch, with a "nice boudary" M .R the CONSERVATION OF MASS in & gives un this identity F _ r0(X H)ax (rf(x ,+)dX Sp(x,+ ) b(x) V(x)dt , = - . . * Sep Hay Setdy , = · The variation of mass in time is due to 2 factors 1) Presence of a source/sink V 2) Velocity of the fluid * Problem : we have the sum of Sr-volume integral and Sr-Surface integral , that cannot ↳ this is the mass flowing through be grouped under the same integral the boundary Recall : Gauss Divergence Theorem Let McMV IN 2 3) be an open bounded set with a smooth closed boundary M =M · = , . · If FeC"(M ; RV) n Co ; MV) - (-my Vector function is continuous and has 1storder PD continuous in M . . an each component is continuous up to the boundary => Then we have SadivEd =+ SrEar Sr misura la il flusso netto del campo uscente attraverso la superficie M = cioè la quantità di una grandezza che attraversa M nell'unità di tempo Su somma la divergenza puntuale è una misura locale di quanto il campo si sta espandendo concentrando in quel punto dello spazio - · To convert the boundary flux integral into a volume integral over we apply the Gauss Divergence theorem to the Flux Vector field E Ph = Setting F(pbi pbz pb3) and using the gauss theorem we get segni!! · * * cneck * = : , , , ·rP(X t) B(x) ·or , . < SrEX t) Vldr , · = Sr divEx HdX , · Substituting in the conservation of mass we get St) => - divE H-f(X, t)dX 0, = this is true for any bounded open set ret with a nice boundary. · We can…
· We observe that : z'Is) = 3 DM(x+ bs, t+ S) + M+ (x 35 t+s . + , and since M+ + 3 VM f . = < z'(s) = f(x(s), t(s)) s = 0 (z(0) = M(x + 0b , t + 0) = M(X , t) S = - t > EL t )_ = Uld tb, tt) = - U ( X - tb , ) O - →= 0 ( × bt 1 - - g ( x tbl - · We can now apply the Fundamental Theorem of calculus wat S = z10) z-t . - = S z1s) 티야타t)- 0 레 (s) > - Substituti 210), 71 -t) z1s) , < MIxct) = g(x - + b) + Sif(x sb t s)ds + , + change of variables o-trs -s--t c0 0 > - - 5 = 0 > - 0 = t do ds = and s = o t - > X + Sb = x + 10 - t)b UNIQUE SOLUTION TO THE PROBLEM MIxct) = g(x - + b) + So f(x 10 t)b 0)do+ - , HOMOGENEOUS SOLUTION The initial : PARTICULAR INTEGRAL The accumulated contribution of all internal sources/sinks : state transport along the f encountered along that characteristic line from o to time - , characteristic line { * Consider a Transpor Equation ut + bMx = 0 in a Domain 10 , + a) x 10 , + M) or 10 , L)x10, + x)0 , 0) u(X M0(X) = What kind of assumptions we need to find aque solution? M(X1 , ti) . . . . ^ … - Observe that … ⑧ M(Xo, to) le . g b> 0 · λ t 1 × ,← ) = y βt ) ( X- M0 EC' (10, +A]) T we need to know a Boundary Condition ge ('(20, A]) + namely ulo,+ ) git) to where g is given , = ㉚ uo(d) g(d) = 1010) =g'lol ☆ { Mt + bMx = 0 in 10 , + y) - More precisly we have IP: M(x, 0) = Mo(X) XE ( O , TN) M10, t) g(t) te (0, +d) = It is not difficult, using the characteristics that a candidat solution is : - , ) µ 1 ×+ )= 온오사업 Mo(X bt) - 0 h ×< bt x >, b+ assuming Suo ge + , C 120 tall , Mold = g(0) , Mold = g'10) - compatibility conditions we have that Me C'120,+x]) is a classical solution to the problem (P) μ
· Now we have to solve the other equation : { TH T = O in 10, +∞ ) Xn(x) [(0) Mo(x)= X ( 10,it) If we want to impose the initial condition to v we get w(x, a) XIX) TIO) Mo(X) = = but we cannot impose an itial condition like that and preserve the separation of values. Therefore we take Twid 1 Inamely we impose un(X 0) Xn(x)) = = , { T +T e-met = = 0 => the solution is Tnlt) = . 「 10) =± and our Unix, +) = e-*'SinInx { Vt Exx in 10 , n)x10 , +x) following problem 0 This solutionIn solves the - = V10 ,t ) = V 11+ t) to I which is different from the original prob , v(X, 0) = SiN1nx) Xe10 , iT) Since the equation and the conditions ar linear, the candidat I formall solution in Sininxet where EBnB neNo solvesformally) the following problem { + ux - = 0 in 10,H) x 10, + x) [10 , H) = u (it,t) to u(x, 0) = Ensin(nx)x((0 T) , We now need to choose the coefficientsbn , in order to fulfill the initial condition x, 0 Mox = ↳ Let's set Mix { Mox x[0, it] where we assumed that no is defined at the endpoints = - Mo(x) XC[- π, 0] Y NB: Mo is the odd extension of Mo to -,] Conditions > Mold MoIIT O ↓M ' * , × compatibilily - = Then we consider the zt-periodic extension of Mo to R . Ifbn , welNo, are the Fouriers coefficients , then the formall sum of theseries MixH = entInsinwx) is a candidate solution to our problem . Fourier series- > MolX) = Sen siniwx) where en = So Mo sininxlax Remarks : · The fact that is becomes a solution depends on the assumptions on No First of all no must have Fouriers coefficients lie ) it must be . intergrable in10 T , the series must converge Moreover, it is easy to observe that if we want a solution which is continuous up to the boundary then the compatibility condition must be satisfied Mod MON O following = Consider now, for instace No e C120 )…
MAXIMUM Principles Let's consider the fol lowing Cauchy-Diricet Problem in Bounded domain REM with a "nice" a · boundary M dr = h = (P(Tx(0 ,T)) { Mt DM = in QT = M x (0 ,T) - 1 P) : u(x, t) _ = h(x, t)(x, t) = Mx(0,T) supposing : - Moe (o(r) U 1 * O) = MOl * E ) R - compatibility condition Mo(X) 2(X , 0) M = We als suppose that IP) has a solution MeCGt) and with dem , M E CGT) VaEN ; 12122 * x(0,T] + v + Set now the Parabolic Boundary PF = (ix 503) U(Mx 10 ,T) a h 3 X O ho L Def . VeCQ) St . Ut , DYreCiQT) IVeIV (12) is a Super (sub) solution to the heat Equation : in QT if V+ - AV), O in QT 의 The followin result holds Theorem Iweak Maximum Principle) if v is a Super Isub) solution to the heat Equation in at then 2 max v max u = QT PT minu ) v e (min = Corollary : Let v be a super lsub) solution to the heat Eq. in QT S t . . MIO ( 3) on PT then MSO in Q ) The followin result holds Theorem Iweak Maximum Principle) if v is a SUB solution to the Heat Equation in QT and 71x0 , to) EQT St . VIXo, to) max v M = = ↳ then u = M in QT Remark : As a consequence, a NON constant subsolution attains its maximum on PT . A similarResult is true for supersolutions and their minima Non constant Hence, we can conclude that a strong solution to the heat Equation attains its minimun and its maximum values on PT . Remark : If Mi Mz are strong solutions to the problem , then V M , us is a solution to the heat Eg = . So that maxU is on PT and minU is on Pr as well, U O is on Pr = Hence, U 0 in QT = - > Thus the problem has a unique solution
Prima pagina del documento.
