ENGI 9420 Engineering Analysis Solutions to Additional Exercises

Transcription

1 ENGI 940 Engineering Analsis Solutions to Additional Exercises 0 Fall [Partial differential equations; Chapter 8] The function ux (, ) satisfies u u u + = 0, subject to the x x u x,0 = u x, =. Classif the partial differential boundar conditions = 0 equation (hperbolic, parabolic or elliptic) and find the complete solution ux. (, ) Compare this PDE with the standard form for a d Alembert solution u u u A + B + C = 0 x x A = B = C = D = B AC = = > This PDE is therefore,, hperbolic everwhere Solving the characteristic (or auxiliar) equation: B ± D + ± λ = = = or A This PDE is homogeneous, so the complementar function is also the general solution: ux, = f + x + g + x, where f and g could be an twice-differentiable functions of their arguments. Imposing the boundar conditions: u x,0 = f x + g x = x ( A) (, ) = ( + ) + ( + ) (,0) = + ( ) = ( B) u x f x g x u x f x g x ( A) : f ( x) + g ( x) = 0 ( C) ( C) ( B ) : ( ) ( A ) f ( x) = g( x) = x k g x = g x = g x = x+ k

2 ENGI 940 Solutions to Additional Exercises Page of. (continued) Therefore the complete solution is (, ) ( ) u x = f + x + g + x = + x k + (( + x ) + k ) [Note that the arbitrar constant k alwas cancels out in this tpe of PDE solution.] ux, =. Classif the partial differential equation and find its general solution. u u u = 0 x x A = B = C = D = B AC = = This PDE is therefore parabolic everwhere 4,, Solving the characteristic (or auxiliar) equation: B ± D ± 0 λ = = = or A 8 General solution: ux (, ) = f( + ( ) x) + hxg (, ) ( + ( ) x), or equivalentl ux (, ) = f( x) + hxg (, ) ( x), where f and g could be an twice-differentiable functions of their arguments and h is an convenient non-trivial linear function of x and except a multiple of x. Arbitraril choose h = x. Then the general solution can be written as (, ) = ( ) + ( ) u x f x x g x [The functional forms of f and g remain arbitrar in the absence of an further information.]

3 ENGI 940 Solutions to Additional Exercises Page of. A disturbance on a ver long string causes a vertical displacement ( xt, ) at a distance x from the origin at time t. The string is released from rest at time t = 0 with initial displacement f ( x) 8x. xt,. (a) Find the subsequent motion of this string The general solution to the wave equation with initial displacement f ( x ) and initial velocit g( x ) is f ( x + ct ) + f ( x ct ) x+ ct ( x, t) g( u) du = + c x ct g x = at all times. Release from rest 0 f ( x) = + 8x. The complete solution follows quickl: (, ) xt = ( x ct ) ( x ct ) (b) Sketch or plot the wave form at time t = 0 and at an two subsequent times. There is an animation available from this Maple file.

4 ENGI 940 Solutions to Additional Exercises Page 4 of u u 4. Classif the partial differential equation x and find the complete solution given the additional information u ux x ( 0, ) = 0, (, ) = x = 0 u = + = x A= C =, B= 0 D = 0 4 = 4 < 0 This PDE is therefore elliptic everwhere Solving the characteristic (or auxiliar) equation: 0± 4 λ = = ± j Complementar function: uc ( x, ) = f ( + ( j) x) + g ( + jx), Particular solution: The right side is a first order polnomial. The second derivatives of u P must match that first order polnomial. Therefore tr up = ax + bx + cx + d u = ax + bx + c + 0, u = 0 + bx + cx + d ( P) x ( P) u P xx = 6 ax + b + 0, u P = 0 + cx + 6 d Substitute into the PDE: u + u = ax + b + cx + d = x + ( P) xx ( P) Matching coefficients of x: 6a+ c= c= 6 a = ( a) Matching coefficients of : b+ 6d = 0 b= d There are two free parameters (a and d). Leave them unresolved for now. General solution: u ( x, ) = f ( + ( j) x) + g ( + jx) + ax dx + ( a) x + d ux ( x, ) = j f ( jx) + j g ( + jx) + ax 6dx + ( a) + 0

5 ENGI 940 Solutions to Additional Exercises Page 5 of 4. (continued) Imposing the boundar conditions, u 0, = = 0 (A) x ( 0, ) = j f j g 0 0 ( a) f g d u = (B) d ( A ) : f + g + d = 0 (C) d (B) + j (C) : ( d a ) 0 + j g ( a+ d) = g = (D) j (A) f = g d (E) Now select values for a and d that simplif this problem as much as possible. If we choose a = and d = 0 then g = 0 g = 0 and (D) (E) f ( ) = 0 so that the complete solution becomes u x, = x 0x + x + 0 (, ) = + u x x x Note that the solution is not a harmonic function on, because u 0 (except on the -axis). However it is subharmonic on an domain entirel within the right half-plane (x > 0) and it is superharmonic on an domain entirel within the left half-plane.

6 ENGI 940 Solutions to Additional Exercises Page 6 of u u u 5. Classif the partial differential equation = x x and find the complete solution, given the additional information u x,0 = x + 4, u x,0 = x A=, B= 7, C = 0 D= = 9 > 0 This PDE is therefore hperbolic everwhere Solving the characteristic (or auxiliar) equation: + 7± 9 λ = = or 5 The complementar function is therefore uc ( x, ) = f( + x) + g( + 5x) The right side of the PDE is a constant and the left side includes onl second derivatives. Therefore tr a pure second order polnomial function as the particular solution: up = ax + bx+ c u = ax + b + 0, u = 0 + bx + c ( P) x ( P) ( P) ( P) ( P) ( u ) ( u ) ( u ) a b c u = a , u = 0 + b+ 0, u = c xx x = = xx x P P P There is onl one constraint on three parameters. Leave the choice of the two free parameters unresolved for now. The general solution to the PDE is: u( x, ) = f ( + x) + g( + 5x) + ax + bx+ c (where a 7b+ 0c= ) u = f ( + x) + g ( + 5x) + 0+ bx + c Including the two additional items of information: u( x,0) = x + 4 f ( x) + g( 5x) + ax = x + 4 (A) u ( x,0) = x f ( x) + g ( 5x) + bx + 0 = x (B) d ( A ) : f ( x) + 5g ( 5x) + ax = 4x (C) dx ( + b a) x (C) (B): 0+ g ( 5x) + ( a b) x = ( 4 ) x g ( 5x) = (D) (B) f ( x) = ( b) x g ( 5x) (E) This problem is simplified if we choose b = and a =. Then

7 ENGI 940 Solutions to Additional Exercises Page 7 of 5. (continued) (D) g ( 5x) = 0 g( 5x) = 0 and (A) f ( x) x x f ( x) f ( x) + 0+ = + 4 = 4 = 4 With b = and a = we have c= c = 0 Therefore the complete solution is u x, = x + x+ 0 u x, = x + x + 4 It is fairl straightforward to verif that this solution does satisf the partial differential equation and both conditions: u x, = x + x + 4 u = 4 x +, u = x u = 4, u =, u = 0 xx x u 7u + 0u = = xx x (,0) = x (, ) (,0) u x u x = x u x = x x u u 6. Classif the partial differential equation = 4 and find its complete solution on t x the interval 0 x 00 for all positive time t, given the additional information u 0, t = 0 and u 00, t = 00 t 0 x and u x, 0 x x 0, 00 0 Also write down the stead state solution. = [ ] t is plaing the role of. A = 4, B = 0, C = 0 D = 0 4(4)(0) = 0 This is also the PDE for heat diffusion. Therefore the PDE is parabolic.

8 ENGI 940 Solutions to Additional Exercises Page 8 of 6. (continued) From page 8.6 of the lecture notes (Example 8.06.), the complete solution of the heat equation (with the boundar conditions of constant temperatures T and T at the ends x = 0 and x = L respectivel and an initial temperature profile on [0, L] of u(x, 0) = f (x) ) is T T u( xt, ) = x + T + L L T T nπz nπx n π kt f ( z) z T sin dz sin exp L n 0 L L L L = In this case k = 4, L = 00, T = 0, T = 00 and f (x) = x (x/0). T T 00 0 x + T = x + 0 = x L 00 T T z z f ( z) z T = z z = z L 0 0 The coefficients in the Fourier sine series become 00 z nπ z z sin dz = z nπ z z + cos 0 nπ 50 nπ 00 z 00 nπ z + sin 50 nπ = 0 + cos( nπ ) nπ = 50 nπ 50 nπ 00 ( n odd) bn = 5 nπ 0 ( n even ) 0 n ( ) Integration b parts:

9 ENGI 940 Solutions to Additional Exercises Page 9 of 6. (continued) Also L 50 the index of summation n, the complete solution becomes =. After some simplification of constants and substitution of ( ) 800 k πx k π t u( xt, ) = x + sin exp π ( ) k = k The Fourier series converges rapidl with increasing k. The stead state solution is lim u( xt, ) t = x An animated version of the graph of this solution is available at Two snapshots of u (x, t): k for [The Maple file used to generate these diagrams is available here.]

10 ENGI 940 Solutions to Additional Exercises Page 0 of 7. An ideal perfectl elastic string of length m is fixed at both ends (at x = 0 and at x = ). The string is displaced into the form x,0 = f x = x x and is released from rest. Waves travel without friction along the string at a speed of m/s. Find the displacement (x, t) at all locations on the string (0 < x < ) and at all subsequent times (t > 0). Write down the complete Fourier series solution and the first two non-zero terms. From page 8.05 of the lecture notes, with L = and c =, the complete solution can be quoted as nπu nπ x nπt ( x, t) = f ( u) sin du sin cos n = 0 ( π ) 0 0 ( π ) c = f u sin n u du = u u sin n u du n 0 ( 4 ) sin ( π ) = u u + u n u du 4 u u + u u+ u 4 = + nπ ( nπ ) ( nπ ) u 6u + 4u + 4u + 4 sin ( nπ) ( nπ) 5 cos ( nπ u) ( nπ u) 4 n 4 = 0+ 5 ( ) 0+ 5 ( nπ) ( nπ) ( nπ) ( nπ) 0 4 n odd ( ) ( nπ) ( nπ) 0 even = = ( nπ) ( nπ) Let n= k, ( k ) ( n ) ( n )

11 ENGI 940 Solutions to Additional Exercises Page of 7. (continued) The complete solution is 8 ( xt, ) = sin k x cos k t π k = ( k ) ( ( k ) π ) The series is also ( π ) ( π ) 8 ( xt, ) = sin ( π x) cos( πt) + sin ( π x) cos( 6πt) + π π 7 ( π ) Plotted here are x,0 = f x = x x (in blue) on top of 8 S x = sin π x sin ( π x) π π + (in red) 7 ( π ) The convergence of the Fourier series is so rapid that there is excellent agreement between the two plots, even when using just the first two non-trivial terms of the series! A Maple animation of the wave is available at a7/q7graph.html. Return to the index of assignments