fall09_lecture_set6

# fall09_lecture_set6 - 1& Reviewing We are considering...

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Unformatted text preview: 1 & Reviewing: We are considering the use of Multiple Scales method on the Duffing’s oscillator We assumed a solution of the form: Substituting in the equation, and collecting terms Here A has to be determined using secularity conditions in u(0) = & 2 1 2 1 2 u u (T ,T ,T ) u u = + ε +ε 2 3 u u u 1 u(0) a , u(0) + ω +ε = ε << = = && & } 2 2 O( ) : D u u u (0,0,0) a , D u ε + ω = = = 0 0 i T 1 2 The solution is : u A(T ,T ) e C.C . ω = + 1 O( ) EOM! ε 0 0 0 0 i T 1 2 2 2 1 1 1 3i T 3 O( ) : D u u [2i D A 3A A] e 3A e C.C. ω ω ε + ω = - ω +- + &¡¢£ ¤¡¥£ ¦§ ¨£©ª«¬¤ ¦£¤­ ¥ ¡® ¥§«ª¦¡§® ¡ Then, a first-order approximation means where is A determined by equation. We then use initial conditions to determine a and 2 1 2 2 1 2 1 2 3 a(T ,T ) a(T ) ; (T ,T ) a T (T ) 8 = φ = +φ ω u u O( ) = + ε 2 a(T ) const! & = 2 1 Sec.producing terms 2i D A 3A A 0. = & ω + = 1 O( ) ε 2 (T ) const! φ = φ i 1 Letting A a e , we get 2 φ = 2 & For the first-order nonlinear approximation, we have but What do we need to satisfy initial conditions? Imposing these on the solution u u O( ) = + ε 2 1 i(3a T /8 ) i T . 1 a e e C.C 2 ω +φ ω = + 0 0 i T u A e C.C. ω = + 1 u (T 0,T 0) a , D u (0,0) = = = = i 1 u (0,0) a a e C.C. a 2 φ & = & + = a cos a & φ = ¡ The velocity condition gives:- & = D u (0,0) = & φ = & φ = a a asin ω ω ∴ = + 2 1 0 0 i(3a T /8 ) i T 1 u a e e C.C 2 i 1 a (i ) e C.C. 2 φ & ω + = 3 & Going back to the solution in terms of original time: & We capture frequency-amplitude relationship with the first-order approximation. ω ω = + 2 1 0 0 i(3a T /8 ) i T 1 u a e e C.C 2 ω + ε ω = + 2 i( 3a /8 )t 1 a e C.C 2 = ω + ε + ε ω 2 3 or u(t) a cos( a )t O( ) 8 ¡ Ex 2: Multiple scales method for the van der pol equation (Note : Read section 3.1 in text on damping mechanisms) Step 1: Let = + ε + ε + + & 2 2 2 2 1 1 2 2 d D 2 D D (D 2D D ) dt = + ε + ε + ε 2 1 2 1 1 2 2 1 2 3 u u (T ,T ,T ) u (T ,T ,T ) u (T ,T ,T ) O( ) +- ε- = 2 2 2 d u du The van der Pol equation is : u (1 u ) dt dt = + ε + ε & 2 1 2 d set D D D dt 4 & Step 2: Step 3: = + iT 1 2 The solution is : u A(T ,T )e C.C = D u (0,0,0) ε + = = 2 O( ): D u u u (0,0,0) a ε ε ε 1 2 [Please confirm the O( ), O( ) and O( ) equations!] ε + = - +- 1 2 2 1 1 1 O( ): D u u 2D D u (1 u )D u + =- +- + iT 2 2 1 1 1 Then D u u i( 2 D A A A A) e C.C- + 3iT 3 i A e C.C ¡ Now, Eliminate terms that will result in secular terms in the solution: Separate real and imaginary parts =- = 2 1 1 2 2D A A A A, where A A(T ,T )! φ φ & ¡ ∂ ∂φ ¢ + £ ¤ ∂ ∂ ¥ ¦ i i 1 1 1 a a 2 e i e 2 T 2 T φ ¢ = i 1 set A a e 2 φ φ- φ =- i 2 2i i 1 1 a e a e a e 2 8 5 & Try to work out qualitative behavior using velocity functions Interlude: if we are only interested in first-order approximation, then- ∂ & =- ¡ = ∂ ¡ ¢ + £ ∂φ ¡ = φ = φ ¡ ∂ ¤ 1 3 2 1 T 2 2 1 a 1 1 4 a a a T 2 8 1 c(T )e (T ) T φ = = const...
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## This note was uploaded on 12/28/2011 for the course ME 580 taught by Professor Na during the Fall '10 term at Purdue.

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fall09_lecture_set6 - 1& Reviewing We are considering...

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