MAS 315 Waves 1 of 8 Answers to Examples Sheet 1. To solve the three problems, we use the methods of 1.3 (with necessary changes in notation).

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1 MAS 35 Waves of 8 Answers to Exampes Sheet. From.) and.5), the genera soution of φ xx = c φ yy is φ = fx cy) + gx + cy). Put c = : the genera soution of φ xx = φ yy is therefore φ = fx y) + gx + y) ) To sove the three probems, we use the methods of.3 with necessary changes in notation). by the chain rue. Thus ) gives ) φ y = f x y) + g x + y) φx, 0) = fx) + gx); dφ dy x, 0) = f x) + gx) ) i) ) fx) + gx) = 0; f x) + g x) = 4x Integrate the second of these, obtaining fx) gx) = A x, where A is a constant. Thus gx) = fx) + x A. Then fx) + gx) = 0 fx) = x + A, gx) = +x A. Thus, from ), φ = x y) + x + y), i.e. i) φ = 4xy ii) ) fx) + gx) = coskx), f x) + g x) = k sinkx) Integrate the second of these, obtaining fx) gx) = A + coskx), where A is a constant. Thus gx) = fx) coskx) A. Then fx) + gx) = coskx) fx) coskx) + A, gx) = A Thus, from ), ii) φ = cos kx y) A traveing wave, traveing in positive x direction) iii) ) fx) + gx) = 0, f x) + g x) = k sinkx) As in ii), fx) gx) = A + coskx), gx) = fx) coskx) A Then fx) + gx) = 0 fx) = coskx) + A, gx) = coskx) A Thus, from ), φ = [cos{kx y)} cos{kx + y)}], i.e. since cos A cos B = sin ) B A sin A+B ) ) iii) φ = sinkx) sinky) a standing wave)

2 MAS 35 Waves of 8 Answers to Exampes Sheet. This PDE is precisey.5) whose GS is.), i.e. From ) Thus we need fx) and gx) to satisfy fx) + gx) = y = fx ct) + gx + ct) y t = cf x ct) + cg x + ct). 0 < x < a) a x ) a x a) f x) = g x) 0 a < x < ) ) The second gives fx) = gx) + A, so fx) A = yx, 0), i.e. fx) = 0 < x < a) A + a x ) a x a) gx) = fx) A 0 a < x < ) Thus fx ct) = A+ gx + ct) = A+ 0 < x < ct a) [a x ct) ] ct a x ct + a) 0 ct + a < x < ) 0 < x < ct a) [a x + ct) ] ct a x ct + a) 0 ct + a < x < ) ct = 0 ct = a ct = a and fx ct) = y = fx ct) + gx + ct) 0 [ < x < a) a ] x a) a x 3a) 0 3 a < x < ) 0 [ < x < 3a) gx + ct) = a ] x + a) 3 a x a) 0 a < x < ) 0 [ < x < a) fx ct) = a x a) ] a x 3a) 0 3a < x < ) 0 [ < x < 3a) gx + ct) = a x + a) ] 3a x a) 0 a < x < ) We have ignored A which cances. Hence the sketch graphs which foow:

3 MAS 35 Waves 3 of 8 Answers to Exampes Sheet ct = 0 a y a a x ct = a OA = 3 4 a OB = a OC = 3 8 a A B C y 3a a a 3 a x ct = a a y 3a a a 3a x 3

4 MAS 35 Waves 4 of 8 Answers to Exampes Sheet 3. i) ii) This is zero for a x, t c = a u t = u = Akc sin{kx ct)} t u x = u = Ak sin{kx ct)} x u t + au x = Ak sin{kx ct)} c a) A and k are arbitrary) u tt = u t = Ak c cos{kx ct)} u xx = u x = Ak cos{kx ct)} u tt au xx + bu = A cos{kx ct)} [ k c + k a + b ] This is zero for a x, t k 0 and c = a + bk c a NB The vaue of A is irreevant because both PDEs are inear. 0 k 4

5 MAS 35 Waves 5 of 8 Answers to Exampes Sheet 4. From.7), T = ρ ) y dx = t ρc {f x ct)} dx From.8), V = F ) y dx = x F {f x ct)} dx These are equa because F = ρc and both integras exist.) Likewise, y = gx + ct) T = ρc {g x + ct)} dx and V = F {g x + ct)} dx and these are equa because F = ρc y = fx ct) + gx + ct) and Hence y t = c{f x ct) g x + ct)} y x = {f x ct) + g x + ct)}. T = ρc [ [{f x ct)} f x ct)g x + ct) + {g x + ct)} ] dx V = F [ [{f x ct)} + f x ct)g x + ct) + {g x + ct)} ] dx F = ρc T V = ρc {f x ct)g x + ct)}dx 0 in genera 5

6 MAS 35 Waves 6 of 8 Answers to Exampes Sheet 5. i) r = x + y + z x r ) = x r ) = x by chain rue x r x = x x = xr ii) Thus φ x = φ φ chain rue) = xr x iii) φ x = ) φ xr x φ = r + x r = φ = r r φ φ r + x φ r ) φ r + x r ) φ r φ ) x Hence φ x + φ y + φ z = 3 φ r + x + y + z ) φ r ) φ = φ r + φ r. Thus PDE satisfied by φ = φr, t) is φ + φ r = φ c t ) ψ = rφ φ = r ψ φr = r ψ r r ψ and φ rr = r ψ rr r ψ r +r 3 ψ. Substitute into ) to get r ψ rr r ψ r + r 3 ψ + r ψ r r 3 ψ = c r ψ tt ψ rr = c ψ tt This is structuray identica to.5) so its genera soution is given by.) with the obvious changes in notion. Thus ψ = rφ = fr ct) + gr + ct) GS os spherica wave equation is φ = {fr ct) + gr + ct)} r 6

7 MAS 35 Waves 7 of 8 Answers to Exampes Sheet 6. Substitute y = Xx)T t) into the PDE to get X T = c X T + c τ X T Divide by XT to get X X = c ) T + τ T T ) The LHS of ) depends ony on x, and the RHS of ) depends ony on t. Thus ) can be true for a x and t ony if each side of ) is a constant. There are three cases to consider I: constant > 0, = p say). Thus X = p X X = A coshph) + B sinhpx). y0) = 0 A = 0 X = B sinhpx). y) = 0 B sinhp) = 0 B = 0 or p = 0 X = 0 for a x. REJECT II: constant = 0. Thus X = 0 X = Ax + B. y0) = 0 B = 0 X = Ax. y) = 0 A = 0 A = 0 X = 0 for a x. REJECT III: constant< 0, = p say). Thus X = p X X = A cospx) + B sinpx). y0) = 0 A = 0 X = B sinpx). y) = 0 B sinp) = 0 B = 0 or p = nπ. Reject B = 0 because that woud give X = 0 for a x nπx ) X = B sin n =,, 3,...) With each side of ) equa to p = n π, the equation for T is T + τ T + nπc ) T = 0. This is a inear ODE for T so try T e mt. This is a soution if m + nπc ) τ m + = 0 { m = ) } τ ± nπc τ 7

8 MAS 35 Waves 8 of 8 Answers to Exampes Sheet Again there are three cases to consider i) πcτ > τ ) nπc < 0 for a n. Let ) nπc τ = 4λ n τ ) ) t t λ n + β n sin λ n } and τ τ m = τ [ ± iλ n] T = e t τ {αn cos yx, t) = e t nπx ) ) ) t t τ sin {α n cos λ n +β n sin λ n } τ τ ii) πcτ < and there is no integer n with nπcτ =. Thus there wi be a finite number of integers {,,..., N} for which τ > ) nπc for each such n, m = m n, m n where m n and M n are rea and positive. For other n, we return to case i). The new types of soution are nπx ) {an yx, t) = sin e mnt + A n e Mnt} n =,,...N) iii) πcτ and there is an integer N with πncτ =. For n < N possiby an empty set) we get soutions as in ii), and for n > N, as in i). For n = N, we have m = τ as repeated root. Thus for this n = N ) Nπx yx, t) = sin e t τ At+B) [In cases ii) and iii) the friction is reativey so arge that some modes decay to zero as t increases without osciation.] 8