Review SOUTIONS: Exm. True or Flse? (And give short nswer ( If f(x is piecewise smooth on [, ], we cn find series representtion using either sine or cosine series. SOUTION: TRUE. If we use sine series, the series will converge to the odd extension of f on [, ], then to the periodic extension of tht over the rels (with the usul cvet bout points t which the periodic extension hs jump discontinuity. If we use cosine series (with the constnt term, the series will converge to the even extension of f on [, ], then to the periodic extension of tht over the rels (gin with the cvet bout the jump discontinuities. (b If f(x is piecewise smooth on [, ], we cn find series representtion using either sine or cosine series. SOUTION: FASE. On the full intervl [, ] we need both sines nd cosines to get complete set of functions (tht is, both sines nd cosines re needed to mke bsis for this vector spce. In other words, we must ssume the form: f(x + n cos + b n sin (Note tht we re ssume on these problems tht the rguments for the sine nd cosine hve nπx/ in them... Without tht ssumption, the sttement could be true. (c The sine series for f(x on [, ] will converge to the odd extension of f. SOUTION: FASE. The sine series for f(x will converge to the odd prt of f, which is given by: f odd = (f(x f( x If f itself ws odd, the sttement would be true (but TRUE would men the sttement is true for ll f. A similr rgument cn be mde bout cosines nd the even prt of f. The key point of this question is to be sure you know the difference between the even/odd prt of f nd the even/odd extension of f. (d The Gibbs phenomenon (n overshoot of the Fourier series occurs only when we use finite number of terms in the Fourier series to represent function tht is discontinuous. SOUTION: TRUE. If we use n infinite number of terms, there is no overshoot, nd the series converges to f(x where f is continuous, nd (f(x+ + f(x where f is discontinuous. Therefore, the only time tht you cn get this overshooting phenomenon is when you use finite number of terms in the sum. (e The functions sin(nx for n =,, 3, re orthogonl to the functions cos(mx for m =,,, 3, on the intervl [, π]. SOUTION: FASE. For exmple, consider the following integrl of the product of sin(x with (or cos(x: π sin(x dx = cos(x π = Similrly, (you wouldn t need to compute this without tble: π sin(x cos(x dx = 3 We should note, however, tht if the intervl is chnged to [ π, π], then the sttement would hve been TRUE.
. Short Answer: ( Questions bout when the Fourier series will be continuous: i. et x. For wht functions f cn we gurntee tht the Fourier series of f will be continuous (t every rel number? SOUTION: For the Fourier series to be continuous in the interior of (,, the function f must be s well. For the series to be continuous t every rel number, the periodic extension of f must be continuous s well- Which mens tht f( = f(. ii. How does the previous nswer chnge if we hve x for f nd use Fourier cosine series? SOUTION: We require f to be continuous in (,, nd we need the even extension of f to be continuous t x = (which it lwys is if f is continuous on [, ], then we need the periodic extension to be continuous on the rels- For the even extension, we lwys hve f( = f(. Therefore, in this cse, the Fourier cosine series for f will be continuous t every rel number s long s f is continuous on [, ]. iii. How does the first nswer chnge if we hve x for f nd use Fourier sine series? SOUTION: As usul, f must first be continuous on (,. Then the odd extension needs to be continuous on [, ]. This occurs if f( =. The odd extension would then need to be continuous s periodic extension, which only hppens if f( = s well. (b et f(x = 3x + 5. Compute the even nd odd prts of f. SOUTION: The odd prt is f odd = (f(x f( x = 3x The even prt is f even = (f(x + f( x = 5 Side note: If we hd the full Fourier series for 3x + 5 on the intervl [, ], then the sine series would converge to 3x nd the cosine series to 5 (in fct, the cosine series is just the number 5. (c Differentition nd the Fourier series: i. Generlly speking, if f is defined on [, ], under wht conditions cn we differentite the generl Fourier series to obtin the series for f (x? SOUTION: We need the Fourier series to be continuous everywhere (tht mens f is continuous on [, ] nd f( = f(, nd f is PWS (which will gurntee the convergence of its Fourier series. Further, note tht if f is not continuous t point x, the Fourier series for f will converge, s usul, to (f (x + + f(x ii. Does our nswer chnge if we use only cosine series on [, ]? SOUTION: The generl prt of the solution does not- Tht is, we need (i the Fourier cosine series to converge to f (so f is PWS, (ii the series should be be continuous everywhere (so in this cse, f just needs to be continuous on [, ], nd (iii f must be PWS so tht we know it hs series representtion. iii. Does our nswer chnge if we use only sine series on [, ]? SOUTION: Agin, the generl sttement does not chnge, just the conditions under which the sttements will be true chnge. Tht is, we need (i the Fourier sine series to converge to f (so f is PWS, (ii the series should be be continuous everywhere (so in this cse, f just needs to be continuous on [, ] AND f( =, f( =, nd (iii f must be PWS so tht we know it hs series representtion. 3. et f(x = { x for < x < for < x < ( Write the even extension of f s piecewise defined function. The even extension of f on the intervl [, ] would be defined s:
for < x < x for < x < f(x = x for < x < for < x < (b Write the odd extension of f s piecewise defined function. Similrly, the odd extension on [, ] is defined s: for < x < x for < x < f(x = x for < x < for < x < (c Drw sketch of the periodic extension of f. SOUTION: (d Find the Fourier sine series (FSS for f, nd drw the F SS on the intervl [ 4, 4]. NOTE: The verticl lines don t belong in the grph, nd in the plces where there is jump discontinuity (t -6, -,, 6, we ought to drw point to indicte tht the series converges to zero there. The lgebric form of the series is: f(x b n sin b n = f(x sin dx Therefore, with = : b n = x sin dx + sin dx = 4 ( nπ ( nπ n π ( sin + nπ cos 4 nπ ( + ( n It is possible to simplify tht bit, but tht is unnecessry for the exm. (e Find the Fourier cosine series (FCS for f, nd drw the F CS on the intervl [ 4, 4]. SOUTION: NOTE: The verticl lines don t belong in the grph, the series would continue out in continuous fshion. 3
The lgebric form of the series is: f(x + n cos n = The formul for is slightly different, so do tht one first: And, for n =,, 3,...: = f(x dx = x dx + n = x cos dx + 4 ( nπ ( nπ n π ( + cos + nπ sin cos f(x cos dx = 3 dx = 4 nπ sin nπ It is possible to simplify tht bit, but tht is unnecessry for the exm. 4. Suppose tht x, nd f(x is represented by the Fourier sine series, f(x B n sin Then we know tht f (x hs Fourier cosine series, f (x A + A n cos dx ( If we differentite the series for f term by term, wht is nother cosine series for f (x? SOUTION: Another wy of expressing f should be f (x (b Use integrtion by prts to show tht SOUTION FOR A : A = ( nπ B n cos A = (f( f( f (x dx A = (f( f( Continuing: A n = f (x cos = (( n f( f( + nπ dx f(x sin dx = (( n f( f( + nπ B n SOUTION: Esy to see if you build the tble to do integrtion by prts. 4
(c Put (, (b together to get formul for the series of the derivtive of f, SOUTION: This summrizes the formul- Given f(x B n sin Then the derivtive hs the series: f (x (f( f( + (( n f( f( + nπ B n cos 5. Consider u t = ku xx subject to the conditions: u x (, t =, u x (, t = nd u(x, = f(x. Solve in the following wy: ook for solutions s Fourier cosine series, nd ssume tht u, u x re continuous, nd u xx, u t re PWS. SOUTION: et u(x, t = A (t + A n (t cos We cn differentite in time s long s u is continuous nd u t is PWS: u t = A (t + A n(t cos We re told tht u x is continuous nd u xx is PWS, so we cn differentite twice in x: u xx = n π A n(t cos Now, u t = ku xx, so we cn equte coefficients of the Fourier series. First, for n = : Similrly, for n =,, 3,...: so tht A n(t = n π A (t = A (t = u(x, t = + Finlly, we require tht u(x, = f(x, or: A n(t A n (t = n e (nπ/ n e (nπ/ t cos f(x = + n cos If we multiply both sides by, then integrte from x = to x =, we get (by orthogonlity: f(x dx = dx + + + +... = And if we multiply both sides by cos ( kπx ( kπx f(x cos t f(x dx nd integrte, we get (gin by orthogonlity: dx = + +... + k ( kπx cos dx + + +... 5
(It s quicker just to recll tht the integrl on the left is / thn to go through the double ngle formul- Tht s fine. Therefore, k = ( kπx f(x cos dx 6. Solve the following nonhomogeneous problem: u t = ku xx + e t + e t cos ( 3πx where we hve insulted ends t x = nd x =, nd u(x, = f(x, nd we cn ssume tht k(3π/. Use the following method: ook for the solution s Fourier cosine series. SOUTION: We hve the eigenvlues nd eigenfunctions: λ = n π φ n (x = cos n =,, 3, Therefore, we ssume solutions to the non-homogeneous eqution re in the form: u(x, t = (t + n (t cos We lso note tht we cn write the function q(x, t = e t + e t cos ( 3πx s cosine series: e t + e t cos ( 3πx = q (t cos(x + q (t cos ( πx + q (t cos ( πx Therefore, we get q (t = e t, q 3 (t = e t, nd q n (t = for ll other n. + q 3 (t cos ( 3πx + Now, we substitute our series into the PDE. The prime nottion is the derivtive in time: ( n (t + π ( 3πx n(t cos = k n (t cos + e t + e t cos This leds to the (infinite system of ODEs, one for ech n, which we will lso solve: n = : (t = e t (t = C e t We cn write the solution in terms of ( s they do in the book: Therefore, ( = C C = + ( (t = + ( e t For n = 3, we get something similr. Use n integrting fctor to solve: 3(t = 9π k 3(t + e t ( 3 (te (9π k/ t = e t e (9π k/ t = e ( +9π k/ t Antidifferentiting, 3 (te (9π k/ t = ( + 9π k/ e( +9π k/ t + C 6
This is where we need to be sure tht 9π k/ (if this ws true, the right side would reduce to. Simplifying, we get: 3 (t = We cn express this in terms of 3 ( s before: 3 ( = ( + 9π k/ e t + Ce (9π k/ t ( + 9π k/ + C C = 3( ( + 9π k/ Therefore, 3 (t = ( ( + 9π k/ e t + 3 ( ( + 9π k/ e (9π k/ t For ll other n: The solution for ech of these is: n(t = n π k n (t n (t = n (e (n π k/t 7. et f(x be given s below. f(x = { x if < x < + x if < x < ( Find the Fourier series for f (on [, ], nd drw sketch of it on [ 3, 3]. SOUTION: I ll leve the sketch to you. The min purpose here is to hve you recll the formuls for the series coefficients. In this cse, with the formuls: nd, n = nd similrly b n = f(x + n cos(nπx + b n sin(nπx = f(x dx = ( x dx + f(x cos(nπx/ dx = f(x sin(nπx/ dx = x cos(nπx dx + x sin(nπx dx + ( + x dx = ( + x cos(nπx dx = ( + x sin(nπx dx = nπ ( ( n ( n (NOTE: If you subtrcted / from your function f(x, it becomes n odd function- Tht s why the cosine terms ended up being zero. 7
(b Find the Fourier sine series for f on [, ] nd drw sketch of it on [ 3, 3]. SOUTION: Agin, the min point here is to hve you recll the formuls nd set up the integrls: f(x b n sin with b n = f(x sin(nπx/ dx = ( + x sin(nπx dx = ( n nπ (c Find the Fourier cosine series for f on [, ] nd drw sketch of it on [ 3, 3]. SOUTION: The formuls: f(x + n cos with n = = f(x dx = f(x cos(nπx/ dx = 8. Put the folowing BVP in Sturm-iouville form: on the intervl < x <. ( + x dx = 3 ( + x cos(nπx dx = ( x φ xφ + ( + λxφ = φ( = φ( = TYPO: The y in the eqution should hve been φ (correct bove. We ll recll tht we sid tht, given: φ + α(xφ + β(xφ = We cn put this in Sturm-iouville form by computing the integrting fctor: Then, multiplying both sides by it, we hve: µ(x = e α(x dx e α(x dx (φ + α(xφ = (e α(x dx φ So in this prticulr cse, first we ll put in our stndrd form: φ x x φ + + λx x φ = ( + ( n n π The integrting fctor is µ = e x/( x dx = x And therefore, the eqution, in stndrd form, looks like: (( x φ + ( + λxφ = (( x φ + φ = λxφ I like to write the nswer in eigenvlue form, but you could hve left your expression without putting λ on the right. 8
9. Given the differentil eqution: φ + λφ =, determine the eigenvlues λ nd eigenfunctions φ if φ stisfies the following boundry conditions (nlyze ll three cses; you my ssume the eigenvlues re rel. ( φ( =, φ(b = NOTE: To solve this prt, we need to use trig identity: sin(a B = sin(a cos(b cos(a sin(b It slipped pst me s I ws putting this together- You will NOT need to memorize this for the exm (lthough for other things, it wouldn t hurt. SOUTION: The solutions to the chrcteristic eqution re r = ± λ. Cse : λ =. The solution is φ(x = C +C x. Using the boundry conditions, C = C =, nd we hve the trivil solution. Cse : λ <, or two rel distinct solutions to the chrcteristic eqution. In this cse, writing the solutions in exponentil form, we hve Using the boundry conditions, we hve: φ(x = C e λx + C e λx C e λ + C e λ = C e λb + C e λb = These lines re not the sme (unless = b, so the only solution for this set is C = C =, nd gin we hve trivil solution. Finlly, if λ >, we hve our usul solution: ( ( φ(x = A n cos λx + B n sin λx With the boundry conditions, we hve: ( A n cos λ A n cos ( + B n sin λ ( λb ( λb + B n sin = = Therefore, we look for non-trivil solutions to this system. Using liner lgebr nd/or Crmer s rule, we know tht there is non-trivil solution if the determinnt of the coefficient mtrix is zero: ( ( ( ( cos λ sin λb cos λb sin λ = ( λ(b or, if sin =. Therefore, we hve: λ n = ( ( nπ φ n (x = sin nπ x ( sin λ b b (b φ ( = nd φ ( = NOTE: This one nd the next re firly stndrd types of problems... SOUTION: The solutions to the chrcteristic eqution re r = ± λ. 9
Cse : λ =. The solution is φ(x = C + C x. Using the boundry conditions, φ(x = C is possible solution. Cse : λ <, or two rel distinct solutions to the chrcteristic eqution. In this cse, writing the solutions in exponentil form, we hve Using the boundry conditions, we hve: φ(x = C e λx + C e λx λc λc = λc e λ λc e λ =. Solve The only solution to this system is C = C =, so this hs only the trivil solution. Finlly, if λ >, we hve our usul solution: ( ( φ(x = A n cos λx + B n sin λx With the boundry conditions, we hve: φ ( = λb n = B n = φ ( = ( ( nπ λa n sin λ = λ n = nd φ n (x = cos (And we don t wnt to forget λ = with φ (x = (c φ( = nd φ ( = SOUTION: With the sme solution to the chrcteristic eqution, we hve: λ = : φ(x = C + C x. With the two boundry conditions, φ(x = is the only solution. λ <, nd we get two distinct rel solutions. Putting in the boundry conditions will yield system for which the only solution is C = C =. The lst cse: λ > : φ n (x = A n cos( λ x + B n sin( λ x The first condition, φ( = mkes A n =, leving only the sine expnsion. The second condition is stisified if: λbn cos( λ = λ = n π (Tht is, we need odd multiples of π/ for the cosine. Therefore, we now hve ( ( (n π (n π λ n = φ n (x = sin x u tt = 4 r ( r u r r < r <, t > with u(r, = f(r, u(, t bounded nd u(, t =. You should ssume tht the (rdil eigenfunctions re known nd complete.
NOTE: We cn leve 4 with φ or with T - In the solution below, we group it with T, but tht isn t the only wy to solve this. SOUTION: Using seprtion of vribles with u = φ(rt (t, we hve: φ(rt (t = 4 r (rφ(rt (t r T 4T = φr (rφ (r r = λ Therefore, dividing everything by 4φT, we hve: (rφ (r = λrφ T = 4λT The rdil eqution is in S- form with p(r = r, q = nd σ(r = r. To solve the time eqution, it is good to first see if we hve ny negtive eigenvlues by checking the Ryleigh quotient: λ = R(φ = rφφ + r(φ dr φ r dr From the boundry conditions, we know φ( = nd φ(r is bounded t r =. quotient then simplifies to: λ = r(φ dr φ r dr The Ryleigh For λ =, we would require φ =, or φ(x = C. But φ( = would mke this the trivil solution. Therefore, λ >. Now proceeding to the time eqution with λ > : nd the generl solution is: with u(r, t = T (t = A n cos( λt + B n sin( λt φ n (r(a n cos( λ n t + B n sin( λ n t u(r, = f(r = A n φ n (r A n = f(rφ n(rr dr φ n(rr dr (Remember to multiply by r, which is required for orthogonlity to hold.. Given the BVP in regulr S- form, with pproprite boundry conditions, show tht the eigenfunctions corresponding to two distinct eigenvlues re orthogonl with respect to σ(x. Hint: Consider φ n (φ m φ m (φ n dx SOUTION: Recll tht for to be the S- opertor, we hve: (φ = λσ(xφ Therefore, φ n (φ m φ m (φ n dx = λ m φ n φ m σ(x + λ n φ m φ n σ(x dx
= (λ n λ m φ n φ m σ(x dx We lso know (proof using Green s formul tht if u, v stisfy the (regulr boundry conditions, then u(v v(u dx = And our φ n, φ m do indeed stisfy the BCs. Therefore, we conclude tht (λ n λ m φ n φ m σ(x dx = Since λ n λ m, the integrl must be zero (for ny n m.. Solve using seprtion of vribles: (Keep the constnts with the sptil eqution PDE u tt = u xx < x <, t > BCs u(, t = u(, t = ICs u(x, = sin(πx + sin(3πx + 3 sin(7πx u t (x, = sin(πx NOTE: In this cse, there weren t ny constnts to keep with the sptil eqution- Tht ws copy-pste error, so ignore it. SOUTION: We recognize tht we ll get the stndrd equtions for spce nd time, with the usul boundry conditions for the sine expnsion. Tht is: T = λt φ = λφ with λ n = (nπ φ n (x = sin(nπx We won t hve ny zero eigenvlues, so we cn lso solve the temporl eqution right wy: so the overll solution is: Now, u(x, = u(x, t = T n (t = A n cos(nπt + B n sin(nπt sin(nπx(a n cos(nπt + B n sin(nπt A n sin(nπx = sin(πx + sin(3πx + 3 sin(7πx Therefore, A =, A 3 = /, A 7 = 3 nd ll other A n re zero. Using the other initil condition, u t (x, = nπb n sin(nπx = sin(πx Therefore, ll B n = except for B. In this cse, B = frcπ. Putting ll the pieces together, the solution is: u(x, t = cos(πt sin(πx + cos(3πt sin(3πx + 3 cos(7πt sin(7πx + sin(πt sin(πx π
3. Consider (φ = φ. Find n expression like Green s Formul for this opertor on < x <. HINT: Use integrtion by prts for u(v dx until you get v(u dx on the right side of the eqution. Tht is, you should hve something in the form: u(v dx = (... + v(u dx SOUTION: Using integrtion by prts with tble, we hve the following: u(v dx + u v u v + u v u v + u v Putting this together, Therefore, u(v dx = (uv u v + u v u v + vu dx 4. Consider u(v v(u dx = (uv u v + u v u v ρu tt = T u xx + αu u(, t = u(, t = u(x, = f(x u t (x, = g(x with ρ(x >, α(x <, nd T constnt. Assume tht the pproprite eigenfunctions (in spce re known. Solve the PDE using seprtion of vribles. SOUTION: See 5.4.5, included below. Seprtion of vribles will led to our usul two ODEs, one in time, one in spce: (time h + λh = (spce φ + α T φ + λ ρ T φ = The sptil eqution is regulr Sturm-iouville problem with p(x = T (constnt, q(x = α < nd σ(x = ρ(x >. The Ryleigh quotient then becomes: λ = + T (φ αφ dx φ ρ dx Since α <, this quotient is greter thn zero (zero only when φ =. Therefore, we know λ n >. Therefore, the solutions to the time eqution re: h n (t = c n cos( λ n t + d n sin( λ n t 3
Using the superposition, we hve u(x, t = u(, t = φ n (x[ n cos( λ n t + b n sin( λ n t] n φ n (x = f(x n = (f, σφ n (φ n, σφ n where σ = ρ/t, nd (f, g is the inner product Also, for initil velocity (f, g = f(xg(x dx u t (x, t = u t (, t = b n = φ n (x[ n λn sin( λ n t + b n λn cos( λ n t] b n λn φ n (x = g(x λn (g, σφ n (φ n, σφ n 5. Use the Ryleigh quotient to obtin resonbly ccurte upper bound for the lowest eigenvlue of φ + (λ xφ = φ ( = φ ( + φ( = SOUTION: Find the simplest (nonzero! function tht stisfies the boundry conditions. A constnt or line will be trivil, so the next best thing is polynomil of degree : u(x = x + bx + c u ( = b = Now, with φ ( + φ( =, we hve: ( + ( + c = or 4 + b =, or b =. For exmple, we cn try u = x Now, with p(x =, σ(x =, nd q(x = x, the Ryleigh quotient is: λ + (x + x(x dx.87 (x dx 6. Use the lternte form of the Ryleigh quotient below to compute R(u, if λ n re the eigenvlues, φ n the eigenfunctions, nd u = φ + 3φ. To simplify our computtions, you my ssume tht the eigenfunctions hve been normlized so tht φ nσ(x dx =. In tht cse, the Ryleigh quotient simplifies to: R(φ = φ(φ dx NOTE: This exercise is to fmilirize you with the technique we used in proving the minimiztion principle... 4
SOUTION: u = φ + 3φ (u = λ σ(xφ 3λ σ(xφ so tht when we multiply u(u, we will hve mixed terms φ i φ j, so we integrte to mke them zero, nd we re left with: Similrly, Therefore, u(u du = λ φ σ(x dx + 3 λ φ σ(x dx = 4λ + 9λ u σ(x dx = φ σ(x dx + 3 φ σ(x dx = + 3 = 3 R(u = 4 3 λ + 9 3 λ 5