Analysis III (BAUG) Assignment 1 Prof. Dr. Alessandro Sisto Due 29 of September 2017

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1 Aalysis III (BAUG) Assigmet 1 Prof. Dr. Alessadro Sisto Due 29 of September Verify that the each of the followig fuctios satisfy the PDE u rr + u tt =. u(r, t) = e 2r cos 2(t) u rr + u tt = 2 (e 2r cos(2t)) + 2 (e 2r cos(2t)) = r (2e2r cos(2t)) + t ( 2e2r si(2t)) = 4e 2r cos(2t) + ( 4e 2r cos(2t)) = u(r, t) = 3r 2 t t 3 u rr + u tt = 2 (3r 2 t t 3 ) + 2 (3r 2 t t 3 ) = (6rt) + r t (3r2 3t 2 ) = 6t + ( 6t) = u(r, t) = si(r) cosh(t). u rr + u tt = 2 (si(r) cosh(t)) + 2 (si(r) cosh(t)) = (cos(r) cosh(t)) + (si(r) sih(t)) r t = si(r) cosh(t) + si(r) cosh(t) = u(r, t) = log(r 2 + t 2 ) for r 2 + t 2 u rr + ( u tt = 2 (log(r 2 + t 2 )) + 2 (log(r 2 + t 2 )) = 2r ) ( r r 2 +t + 2t ) 2 t r 2 +t 2 = 2t2 2r 2 + 2r2 2t 2 =. (r 2 +t 2 ) 2 (r 2 +t 2 ) 2 The coditio r 2 + t 2 is eeded so that the expressio log(r 2 + t 2 ) makes sese. u(r, t) = e r cos(t) + 3r 2 t t 3 + si(r) cosh(t) Notice that the PDE is liear. Sice the fuctio i part (v) is the sum 1

2 of the fuctios i the first three parts, by the superpositio priciple, it satisfies the PDE. 2. Verify that the each of the followig fuctios satisfy the PDE u θθ c 2 u rr =, where c is a real costat. u(r, θ) = si(r cθ) u θθ c 2 u rr = 2 (si(r cθ)) c 2 2 (si(r cθ)) = ( c cos(r cθ)) θ c2 (cos(r cθ)) r = c 2 si(r cθ) c 2 ( si(r cθ)) = u(r, θ) = log(r + cθ) for r + cθ > u θθ ( c 2 u rr = 2 (log(r + cθ)) c 2 2 (log(r + cθ)) = c ) ( θ r+cθ c 2 1 ) ( r r+cθ) = c2 c 2 1 = (r+cθ) 2 (r+cθ) 2 u(r, θ) = cos(ar) si(caθ) for ay real costat a u θθ c 2 u rr = 2 (cos(ar) si(caθ)) c 2 2 (cos(ar) si(caθ)) = (ca cos(ar) cos(caθ)) θ c2 ( a si(ar) si(caθ)) r = (ca) 2 cos(ar) si(caθ) c 2 ( a 2 cos(ar) si(caθ)) = u(r, θ) = e r+cθ + e r cθ u θθ c 2 u rr = 2 (e r+cθ + e r cθ ) c 2 2 (e r+cθ + e r cθ ) = θ (cer+cθ ce r cθ ) c 2 r (er+cθ + e r cθ ) = (c 2 e r+cθ + c 2 e r cθ ) c ( 2 e r+cθ + e r cθ) = 3. Verify that the each of the followig fuctios satisfy the PDE u t ku xx =, where k is a real costat. u(x, t) = x 2 + 2kt 2

3 u t ku xx = t (x2 + 2kt) k 2 (x 2 + 2kt) = 2k k (2x) x = 2k 2k = u(x, t) = e kt si(x) u t ku xx = t (e kt si(x)) k 2 (e kt si(x)) = ke kt si(x) k x (e kt cos(x)) = ke kt si(x) k( e kt si(x)) = u(x, t) = e kt cosh(x) u t ku xx = t (ekt cosh(x)) k 2 (e kt cosh(x)) = ke kt cosh(x) k x (ekt sih(x)) = ke kt cosh(x) ke kt cosh(x) = u(x, t) = e a2 kt cos(ax) for ay real costat a u t ku xx = kt t (e a2 cos(ax)) k 2 (e a2kt cos(ax)) = a 2 ke a2kt cos(ax) k kt x ( ae a2 si(ax)) = a 2 ke a2kt cos(ax) k( a 2 e a2kt cos(ax)) = 4. Classify each of the followig 2d order PDEs accordig to their homogeeity (homogeeous or ohomogeeous), liearity (liear or o liear), coefficiets (costat or o costat) ad, whe the PDE is liear, type (parabolic, hyperbolic or elliptic). Note: A equatio may have differet types i differet regios of the domai of the fuctio. (a) (1 + y 2 )u xx + e x2 2 u yy xu x + (3 x 2 )u y = [Past exam questio] The PDE is homogeeous, liear (sice oe of the coefficiets is a fuctio of u) ad has o costat coefficiets (because the coefficiets are fuctios that deped o x ad y). Moreover, sice 2 4(1 + y 2 )e x2 2 < for every possible values of (x, y), we coclude that the PDE is elliptic. 3

4 (b) u xx + 4xy + e y u y = (x + y) 2 u x [ Past exam questio ] The PDE is ot homogeeous (because of the term 4xy), liear ad has o costat coefficiets. Moreover, sice 4( 1) =, we coclude that the PDE is parabolic. (c) u rθ 6u rr + uu θθ e r u r = 4x The PDE is ot homogeeous (because of the term 4x), ot liear (because of the term uu θθ ) ad has o costat coefficiets. (d) 6u tt + 12u rr 24u r + 2u t = 42 The PDE is ot homogeeous (because of the o-zero term 42), liear ad has costat coefficiets. Moreover, sice 4(6 12) < for every possible values of (r, t), we coclude that the PDE is elliptic. (e) u xx + si(x)u x + e πy u y 1xyu yy = The PDE is homogeeous, liear ad has ot costat coefficiets. Moreover, sice 4(1 ( 1xy)) = 4xy we coclude that the PDE is hyperbolic for xy >, parabolic for xy = ad elliptic for xy <. 4

5 5. (Calculus review) Compute the followig itegrals. (a) cos(x) dx, for a positive iteger cos(x) dx = 1 si(x) π = 1 si(π) 1 si() = = (b) π x si(x) dx, for a positive iteger. [Past exam questio] Note first of all that, sice x ad si(x) are odd fuctios, the fuctio x si(x) is eve. Thus, we have: π x si(x) dx = 2 x si(x) dx Itegratig by parts ad usig the result i a) we get: π x si(x) dx = x cos(x) π π cos(x) + dx = π cos(π) π( 1) +1 Thus we coclude that x si(x) dx = 2π( 1)+1. π + = (c) cos(2x) cos(x) dx Itegratig by parts twice, we get: 2 2 cos(2x) cos(x) dx = cos(2x) si(x) 2π 2π 2 si(2x) si(x) dx = si(2x) si(x) dx = 2 si(2x) cos(x) 2π 2 cos(2x) cos(x) dx = 4 Thus, we coclude that cos(2x) cos(x) dx cos(2x) cos(x) dx =. + (d) e x si(x) dx, for a iteger 5

6 If = the the itegral is sice si() =. Cosider ow. Itegratig by parts twice, we get: e x si(x) dx = e x cos(x) 2π 2π 2π + e x si(x) dx = 1 e 2π 2 e x si(x) 2 Ad so we ca coclude that e x cos(x) 1 2 e x si(x) dx = (1 e 2π ) dx = 1 e 2π e x si(x) dx (e) si(3x) 2 dx Itegratig by parts twice, we get: 2π cos(3x) 2 dx = si(3x) 2 dx. Thus, we ca coclude that cos(3x) 2 dx = si(3x) 2 dx = si(3x) 2 dx = π. si(3x) cos(3x) 3 2π + 1 si(3x) 2 dx = 2π 6

EXERCISE - BINOMIAL THEOREM

BINOMIAL THOEREM / EXERCISE - BINOMIAL THEOREM LEVEL I SUBJECTIVE QUESTIONS. Expad the followig expressios ad fid the umber of term i the expasio of the expressios. (a) (x + y) 99 (b) ( + a) 9 + ( a) 9