Advanced-numerical-method-2

University of Technical Education HCM City, 2013 Chapter – Numerical methods for ODE and PDE problems ODE 5.1 Euler Method 5.2 Runge-Kutta method 5.3 Laplace’s Transform 5.4 Analytic solution PDE 5.5 Examples  Initial Valuation Problem Mô hình thực tế Mô hình vật lí dy = f (t , y ); dt y (0) = y0 Phương pháp Cho h = t – t điều kiện ban đầu, y Euler = y(t0), tính : y1 = y + hf (t , y ) y = y1 + hf (t1 , y1 ) y j +1 = y j + hf (t j , y j ) Hoaëc y j = y j −1 + hf (t j −1 , y j −1 ) + O(h ) Dùng pháp Euler Tính dy = t − 2y dt Kết xác y(0) = 1 y = [2t − + 5e − 2t ] j tj f (t j −1 , y j −1 ) Euler yj=yj+hf(tj-1,yj-1) C.xaùc y(tj) 0.0 0.2 0.4 0.6 NaN 0-(2)(1) = - 2.000 0.2 – (2)(0.6) = -1.000 0.4 – (2)(0.4) = -0.4 (Đk ban đầu) 1.000 1.0 + (0.2)(-2.0) = 0.60 0.6 + (0.2)(-1.0) = 0.40 0.4 + (0.2)(- 0.4) = 0.32 1.000 0.688 0.512 0.427 Sai soá yj-y(tj) -0.0879 -0.1117 -0.1065 So sánh với đồ thò : Đối với h biết, sai số lớn kết số sai số rời rạc toàn cuïc max (∑ ( y j − y (t j )) ) j Đánh giá sai số : Sai số đòa phương bước là: ej = yj – y(tj) với y(tj) kết xác tj GDE = max( ej ) 1, … h max(ej ) 0.200 0.100 0.050 0.025 0.1117 0.0502 0.0240 0.0117 j= Giaûi baèng Matlab: function [t,y] = odeEuler(diffeq,tn,h,y0)] t = (0:h:tn)’; n = length(t); y = y0 + zeros(n , 1); for j = : n y(j) = y(j – 1) + h* feval(diffeq,t(j -1),y(j- 1)); >> rhs = inline(‘cos(t)’,’t’,’y’) ; end >> [t,Y] = odeEuler(rhs,2*pi,0.01, 0) ; >> plot(t,Y,’o’) ; Matlab code 2.4 % (i) Exact solution 2.2 sol = dsolve('Dz = z^2 * cos(2*t)', 'z(0)=1', 't'); pretty(sol) 1.8 t = linspace(0, 6, 601); z = eval(vectorize(sol)); 1.6 % (ii) Euler's method 1.4 tn = 6; z0 = 1; h = 0.125; [teul,zeul] =odeEuler(@f,tn,h,z0); 1.2 % Exact solution - Euler approximate solution plot(t,z, teul,zeul,'r-.') legend('Exact', 'Euler') 0.8 grid on 0.6 % 0.4 echo off; diary off % -function [t,z]=odeEuler(f,tn,h,z0) % Function M-file f.m t=(0:h:tn)'; function zp = f(t,z) n=length(t); zp = z.^2 * cos(2.*t); z=z0+zeros(n,1); % - for j=2:n z(j)=z(j-1)+h*feval(f,t(j-1),z(j-1)); end Exact Euler dz = z cos 2t z (0) = dt z (6) = 0.788465 Phương pháp RungeTính độ dốc vò trí ứng với bước laëp: Kutta k1 = f (t j , y j ) h h k = f (t j + , y j + k1 ) 2 h h k = f (t j + , y j + k ) 2 y j +1 k = f (t j + h, y j + hk )  k1 k2 k3 k4  = y j + h  + + + ÷+ O (h5 ) 6 3 6 Sử dụng hàm thư viện Matlab: Sử dụng ode45 Cú pháp : [t,Y] = ode45(diffep,tn,y0) [t,Y] = ode45(diffep,[t0 tn],y0) [t,Y] = ode45(diffep,[t0 tn],y0,options) [t,Y] = ode45(diffep,[t0 tn],y0,options,arg1,arg2,…) Giải Matlab >> rhs = inline(‘cos(t)’,’t’,’y’) ; >> [t,Y] = odeRK4(rhs,2*pi,0.01, function [t,y] = odeRK4(diffeq,tn,h,y0) >> plot(t,Y,’o’) ; t = (0:h:tn)’; n = length(t) ; y = y0+ zeros(n , 1) ; h2= h /2 ; h3= h /3 ; h6= h /6 ; for j = : n k1 = feval(diffeq, t(j -1), y(j-1)) ; k2 = feval(diffeq , t(j -1)+h2, y(j1)+h2*k1 ) ; k3 = feval(diffeq , t(j -1)+h2, y(j1)+h2*k2 ) ; k4 = feval(diffeq , t(j -1)+h , y(j1)+h*k3) ; y(j) = y(j – 1) + h6* (k1+k4) + Hàm thư vieän Matlab [t,Y] = ode45(diffeq,tn,y0) >> rhs = inline(‘cos(t)’,’t’,’y’) ; >> [t,Y] = ode45(rhs,[0 2*pi], >> plot(t,Y,’r’,’linewidth’,2) ; Apply the initial conditions (ii) a sin(0) + b cos(0) = a cos(σ l ) − b sin(σ l ) =   ⇒ b = and det  =0  cos(σ l ) − sin(σ l )  2n −  cos σ l = ⇒ σ n = 2l π , ⇒  X ( x) = a sin( (2n − 1)π x ), n  n 2l n = 1, 2,3,L n = 1, 2,3,L The time response equation 2  2n −  & & Tn (t ) + c  π ÷ T (t ) =  2l  (2n − 1)cπ (2n − 1)cπ ⇒ Tn (t ) = An sin t + Bn cos t 2l 2l Thus the solution implies oscillation with Frequencies (2n − 1)cπ (2n − 1)π ωn = = 2l 2l E , ρ n = 1, 2,3L (6.63) w (x,t) BENDING VIBRATIONS OF A BEAM f (x,t) f(x,t) h1 h2 x M(x,t)+Mx(x,t)dx M(x,t) ·Q w(x,t) V(x,t)+Vx(x,t)dx V(x,t) dx A(x)= h1h2 bending stiffness = EI (x) E = Youngs modulus I (x) = cross-sect area moment of inertia about z ∂ w(x,t) M (x,t) = EI(x) ∂ x2 x x +dx Next sum forces in the y - direction (up, down) Sum moments about the point Q Use the moment given from stenght of materials Assume sides not bend (no shear deformation) Summing forces and moments  ∂V (x,t)  ∂ w(x,t) dx  − V (x,t) + f (x,t)dx = ρA(x)dx V (x,t) +   ∂x ∂t   ∂M(x,t)  ∂V (x,t)  dx  − M(x,t) + V (x,t) + dxdx  M(x,t) +     ∂x ∂x dx + f (x,t)dx =0 ∂M(x,t)  ∂V (x,t) f (x,t)  ⇒ + V (x,t)dx +  + (dx) =0   ∂x   ∂x  ∂M(x,t) ⇒ V (x,t) = − ∂x Substitute into force balance equation yields: ∂ M(x,t) ∂ w(x,t) − dx + f (x,t)dx = ρA(x)dx ∂x ∂t Dividing by dx and substituting for M yields ∂ w(x,t) ∂  ∂ w(x,t)  ρA(x) + EI(x)  = f (x,t) 2 ∂t ∂x  ∂x  Assume constant stiffness to get: ∂ w(x,t) ∂ w(x,t) EI +c = 0, c = ∂t ∂x ρA The possible boundary conditions Free end ∂ 2w bending moment = EI = ∂x ∂  ∂ 2w  shear force = EI  = ∂x  ∂x  Clamped (or fixed) end deflection = w = ∂w slope = =0 ∂x Pinned (or simply supported) end Sliding end deflection = w = ∂w slope = =0 ∂x ∂  ∂ 2w  shear force = EI  = ∂x  ∂x  ∂ 2w bending moment=EI = ∂x Solution of the time equation yields the oscillatory nature & & ′′′′ X ( x ) T (t ) c =− = ω2 ⇒ X ( x) T (t ) &(t ) + ω 2T (t ) = ⇒ T& T (t ) = A sin ωt + B cos ωt Two initial conditions: w( x, 0) = w0 ( x), wt ( x, 0) = w&0 ( x ) Spatial equation results in a boundary value problem (BVP) ω  X ′′′′(x) −   X(x) = c 2   ω ρAω Define β =   = c EI σx Let X(x) = Ae to get : X(x) = a1 sin βx + a2 cos βx + a3 sinh βx + a4 cosh βx Apply boundary conditions to get constants and the characteristic equation Example compute the mode shapes and natural frequencies for a clampedpinned beam At fixed end x = and X(0) = ⇒ a2 + a4 = X ′(0) = ⇒ β (a1 + a3 ) = At the pinned end, x = l and X(l ) = ⇒ a1 sin βl + a2 cos βl + a3 sinh βl + a4 cosh βl = EIX ′′(l ) = ⇒ β (−a1 sin βl − a2 cos βl + a3 sinh βl + a4 cosh βl ) = The boundary conditions in the constants can be written as the matrix equation:   a1  0 1      β β  a2  = 0  sin βl cos βl sinh βl cosh βl a3  0      2 βl −β cos βl β sinh βl β cosh βl a4  0 1−β4 sin 4 4 4 44 4 4 4 4 43 { a B Ba = 0,a ≠ ⇒ det(B) = ⇒ tan βl = βl The characteristic equation Solve numerically (fsolve) to obtain solution to transcendental (characteristic) equation β1l = 3.926602 β 2l = 7.068583 β 3l = 10.210176 β l = 13.351768 β5l = 16.493361 K n >5⇒ (4n + 1)π βnl = Next solve Ba=0 for of the constants: With the eigenvalues known, now solve for the eigen-functions Ba = yields constants in terms of the 4th : Choose a a1 = −a3 from the first equation a2 = −a4 from the second equation (sinh β n l − sin β n l )a3 + (cosh β n l − cos β n l )a4 = from the third (or fourth) equation Solving yields : cosh β n l − cos β n l a3 = − a4 sinh β n l − sin β n l cosh β n l − cos β n l  ⇒ X n (x) = (a4 ) n  (sinh β n l x − sin β n l x) − cosh β n l x + cos β n l x  sinh β n l − sin β n l  Plot the mode shapes to the system response X n, x cosh n cos n sinh n x sin n sinh n sin n x cosh n x cos n x 1.5 Mode Mode Note zero slope 0.5 X 3.926602 , x X 7.068583 , x X 10.210176 , x 0.2 0.4 0.6 0.8 0.5 1.5 Mode x Non zero slope