Topic-based study materials for Mathematical and Numerical Methods in Engineering in the Biomedical Engineering degree programme at Politecnico di Milano. The document covers: 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 +
Topic-based study materials for Mathematical and Numerical Methods in Engineering in the Biomedical Engineering degree programme at Politecnico di Milano. The document covers: 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 +
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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
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