A question from www.wisfaq.nl My favourite formula A moving string/belt wi4243AP: Complex Analysis week 8, Monday K. P. Hart Faculty EEMCS TU Delft Delft, 20 october, 2014 K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Outline A question from www.wisfaq.nl 1 My favourite formula 2 A moving string/belt 3 K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt The question http://www.wisfaq.nl/show3archive.asp?id=60954&j=2009 I have following integral � ∞ d x √ x ( x 2 + 1) 0 Sing points in the upper half plane are x = i (pole order 1) and x = 0 (pole order 1 / 2). I can rewrite the integral: � − ε � R � � � f ( z ) d z = 2 π i Res( f , i ) = . . . + . . . + . . . + + − R γ ε ε γ R K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt The question The second and fourth integral (on the right-hand side) tend 0 (can be proved via an estimate, I did that WAS OK. If I want to calculate the given integral then I have � 0 � ∞ 2 π i Res( f , i ) = π d x d x √ = √ x ( x 2 + 1) + √ x ( x 2 + 1) i −∞ 0 My problem now is: How do I get the desired integral d x / √ x ( x 2 + 1) � ∞ 0 Well . . . ? K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Answer √ √ √ √ √ √ i = 1 2 + i i = π 2 − i π 2 and so π/ 2 2 2 2 2 For x < 0 we have √ x = i � 0 � ∞ � | x | so −∞ becomes − i 0 We get √ � ∞ √ x ( x 2 + 1) = π 2 d x 2 0 K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt One of Euler’s best ∞ n 2 = π 2 1 � 6 n =1 K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt A useful function and contour We use f ( z ) = π cos π z z 2 sin π z and large squares for contours. Γ N is the square with vertices ( ± 1 ± i )( N + 1 2 ). K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Here’s Γ 3 The contour Γ 3 Pole of order 3 at 0 Poles of order 1 at all nonzero integers K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Residues The residues at nozero n are easy: Res( f , n ) = π cos π n n 2 π cos π n Pole of order 1, so we substitute n in π cos π z z 2 (sin π z ) ′ K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Residues We calculate a few terms of the Laurent series of cos z sin z : ( a − 1 + a 0 + a 1 z + · · · )( z − 1 6 z 3 + · · · ) = 1 − 1 2 z 2 + · · · z or a − 1 + a 0 z + ( a 1 − 1 6 a − 1 ) z 2 + · · · = 1 − 1 2 z 2 + · · · so: a − 1 = 1, a 0 = 0, a 1 = − 1 2 + 1 6 a − 1 = − 1 3 . K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Residues We find z 3 − π 2 f ( z ) = π 1 z + · · · 3 and so Res( f , 0) = − π 2 3 K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt The integrals We find that, for every N , N � f ( z ) d z = − 1 1 � 3 π 2 + 2 2 π i n 2 Γ N n =1 Now we show � lim f ( z ) d z = 0 N →∞ Γ N K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Estimates Length of Γ N : 4 × (2 N + 1) = 8 N + 4. 1 1 On Γ N we have N 2 . z 2 � Remember: � sinh 2 y + cos 2 x | cos z | = and � sinh 2 y + sin 2 x | sin z | = K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Estimates If x = ± ( N + 1 2 ) then � sinh 2 π y | cotan π z | = � 1 � sinh 2 π y + 1 K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Estimates If y = ± ( N + 1 2 ) then � sinh 2 π ( N + 1 2 ) + cos 2 π x | cotan π z | = � sinh 2 π ( N + 1 2 ) + sin 2 π x � sinh 2 π ( N + 1 2 ) + 1 � � sinh 2 π ( N + 1 2 ) � 2 (when N � 1). K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Finishing up Everything combined: � � � � � (8 N + 4) × π × 2 � � f ( z ) d z � � N 2 � Γ N which suffices to have the limit be equal to zero. K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt A differential equation A string or belt runs on two wheels one unit apart at a speed of v units. The function w ( x , t ) describes the lateral displacement of the string at position x (0 < x < 1) and time t ( t > 0). It satisfies the following (partial) differential equation ∂ 2 w ∂ t 2 + 2 v ∂ 2 w ∂ x ∂ t + ( v 2 − 1) ∂ 2 w ∂ x 2 = 0 If v = 0 then we have the wave equation. K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Boundary conditions We impose boundary conditions: w (0 , t ) = 0 and w (1 , t ) = f ( t ) One wheel is stable, the other one wiggles a bit. We start at rest: w ( x , 0) = 0 and w t ( x , 0) = 0 and w x ( x , 0) = 0 K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Apply Laplace transform Transform the equation: � � s 2 W ( x , s ) − s · w ( x , 0) − w t ( x , 0) + 2 v sW x ( x , s ) − w x ( x , 0) + ( v 2 − 1) W xx ( x , s ) = 0 this can be rewritten as s 2 2 vs W xx + v 2 − 1 W x + v 2 − 1 W = 0 . Note: linear with constant coefficients K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Solve the differential equation Characteristic equation: s 2 2 vs λ 2 + v 2 − 1 λ + v 2 − 1 = 0 with solutions − s − s λ 1 = v + 1 and λ 2 = v − 1 Hence W ( x , s ) = C 1 ( s ) e λ 1 x + C 2 ( s ) e λ 2 x The ‘constants’ C 1 and C 2 depend on s . K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt C 1 and C 2 W (0 , s ) = 0, hence C 1 ( s ) + C 2 ( s ) = 0. W (1 , s ) = F ( s ) (transform of f ), hence C 1 ( s )( e λ 1 − e λ 2 ) = F ( s ) F ( s ) and so C 1 ( s ) = e λ 1 − e λ 2 . We get exp( − sx v +1 ) − exp( − sx v − 1 ) W ( x , s ) = F ( s ) · exp( − s v +1 ) − exp( − s v − 1 ) � �� � H ( x , s ) If we can find the inverse transform, h ( x , t ), of H ( x , s ) then we are done: � t w ( x , t ) = f ( t ) ∗ h ( x , t ) = f ( τ ) h ( x , t − τ ) d τ 0 K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Inversion Unfortunately H ( x , s ) has many singularities: the solutions to � − s � − s � � exp = exp v + 1 v − 1 or � � 2 s exp = 1 v 2 − 1 these are s n = ( v 2 − 1) n π i , with n ∈ Z . s 0 = 0 is a removable singularity, all others are poles of order 1. K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Inversion formula We know that � 1+ iR 1 H ( x , s ) e st d s h ( x , t ) = 2 π i lim R →∞ 1 − iR As we have infinitely many singularities we have no annulus; we can try a trick like with my favorite formula. K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Inversion formula The contour Γ 3 Removable singularity at 0 Poles of order 1 at ( v 2 − 1) n π i K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Slight problem The integrals along the three other sides do not converge to 0. What is true is that H ( x , s ) is bounded on those rectangles, 2 )( v 2 − 1) π i : provided the horizontals are at ± ( n + 1 So, we divide by s 2 : � H ( x , s ) � H ( x , s ) � � s 2 F ( s ) − sf (0) − f ′ (0) sf (0)+ f ′ (0) W ( x , s ) = + s 2 s 2 This makes inversion possible, at the cost of a more complicated convolution. Note that 0 is now a pole of order 2. K. P. Hart wi4243AP: Complex Analysis
A question from www.wisfaq.nl My favourite formula A moving string/belt Residues We have exp( − sx v +1 ) − exp( − sx v − 1 ) H ( x , s ) = exp( − s v +1 ) − exp( − s v − 1 ) At s n = ( v 2 − 1) n π i we have s n v +1 = ( v − 1) n π i and s n v − 1 = ( v + 1) n π i and so exp( − x ( v − 1) n π i ) − exp( − x ( v + 1) n π i ) Res( H ( x , s ) , s n ) = 1 1 − v +1 exp( − ( v − 1) n π i ) + v − 1 exp( − ( v + 1) n π i ) K. P. Hart wi4243AP: Complex Analysis
Recommend
More recommend
