MA6351 TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS L T P C

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1 MA635 TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS L T P C 3 4 OBJECTIVES: To itroduce Fourier series ysis which is cetr to my ppictios i egieerig prt from its use i sovig boudry vue probems? To cquit the studet with Fourier trsform techiques used i wide vriety of situtios. To itroduce the effective mthemtic toos for the soutios of prti differeti equtios tht mode sever physic processes d to deveop Z trsform techiques for discrete time Systems. UNIT I PARTIAL DIFFERENTIAL EQUATIONS 9+3 Formtio of prti differeti equtios Sigur itegrs -- Soutios of stdrd types of first order prti differeti equtios - Lgrge s ier equtio -- Lier prti differeti equtios of secod d higher order with costt coefficiets of both homogeeous d o-homogeeous types. UNIT II FOURIER SERIES 9+3 Dirichet s coditios Geer Fourier series Odd d eve fuctios Hf rge sie series Hf rge cosie series Compex form of Fourier series Prsev s idetity Hrmoic ysis. UNIT III APPLICATIONS OF PARTIAL DIFFERENTIAL 9+3 Cssifictio of PDE Method of seprtio of vribes - Soutios of oe dimesio wve equtio Oe dimesio equtio of het coductio Stedy stte soutio of two dimesio equtio of het coductio (excudig isuted edges). UNIT IV FOURIER TRANSFORMS 9+3 Sttemet of Fourier itegr theorem Fourier trsform pir Fourier sie d cosie trsforms Properties Trsforms of simpe fuctios Covoutio theorem Prsev s idetity. UNIT V Z - TRANSFORMS AND DIFFERENCE EQUATIONS 9+3 Z- trsforms - Eemetry properties Iverse Z - trsform (usig prti frctio d residues) Covoutio theorem - Formtio of differece equtios Soutio of differece equtios usig Z - trsform. TOTAL (L:45+T:5): 6 PERIODS. TEXT BOOKS:. Veerrj. T., "Trsforms d Prti Differeti Equtios", Tt McGrw Hi Eductio Pvt. Ltd., New Dehi, Secod reprit,.. Grew. B.S., "Higher Egieerig Mthemtics", 4d Editio, Kh Pubishers, Dehi,. 3. Nry.S., Micvchgom Piy.T.K d Rmih.G "Advced Mthemtics for Egieerig Studets" Vo. II & III, S.Viswth Pubishers Pvt. Ltd.998. REFERENCES:. Bi.N.P d Mish Goy, "A Textbook of Egieerig Mthemtics", 7th Editio, Lxmi Pubictios Pvt Ltd, 7.. Rm.B.V., "Higher Egieerig Mthemtics", Tt Mc Grw Hi Pubishig Compy Limited, NewDehi, Gy Jmes, "Advced Moder Egieerig Mthemtics", 3rd Editio, Perso Eductio, Erwi Kreysig, "Advced Egieerig Mthemtics", 8th Editio, Wiey Idi, Ry Wyie. C d Brrett.L.C, "Advced Egieerig Mthemtics" Tt Mc Grw Hi Eductio Pvt Ltd, Sixth Editio, New Dehi,. 6. Dtt.K.B., "Mthemtic Methods of Sciece d Egieerig", Cegge Lerig Idi Pvt Ltd, Dehi, 3.

2 CONTENTS S.NO TOPICS PAGE NO UNIT-I PARTIAL DIFFERENTIAL EQUATIONS. INTRODUCTION. FORMATION OF PARTIAL DIFFERNTIAL EQUATIONS.3 SOLUTIONS OF PARTIAL DIFFERENTIAL EQUATIONS 7.4 LAGRANGE S LINEAR EQUATIONS 3.5 PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER WITH CONSTANT 9 CO-EFFECIENTS.6 NON-HOMOGENOUS LINEAR EQUATIONS 36 UNIT-II FOURIER SERIES. INTRODUCTION 4. PERIODIC FUNCTIONS 4.3 EVEN AND ODD FUNCTIONS 54.4 HALF RANGE SERIES 6.5 PARSEVAL S THEOREM 68.6 CHANGE OF INTERVAL 69.7 HARMONIC ANALYSIS 76.8 COMPLEX FORM OF FOURIER SERIES 8.9 SUMMARY 83 UNIT-III APPLICATIONS OF PARTIAL DIFFERENTILA EQUATIONS 3. INTRODUCTION SOLUTION OF WAVE EQUATION SOLUTION OF THE HEAT EQUATION SOLUTION OF LAPLACE EQUATIONS UNIT-IV FOURIER TRANSFORMS 4. INTRODUCTION INTEGRAL TRANSFORMS FOURIER INTEGRAL THEOREM FOURIER TRANSFORMS AND ITS PROPERTIES CONVOLUTION THEOREM AND PARSEVAL S THEOREM FOURIER SINE AND COSINE TRANSFORMS 54 UNIT-V Z-TRANSFORMS AND DIFFERENCE EQUATIONS 5. INTRODUCTION LINEAR DIFFERENCE EQUATIONS Z-TRANSFORMS AND ITS PROPERTIES INVERSE Z-TRANSFORMS CONVOLUTION THEOREM APPLICATIONS OF Z-TRANSFORMS TO DIFFERENCE EQUATIONS FORMATION OF DIFFERENCE EQUATIONS 99 BIBLIOGRAPHY

3 UNIT I PARTIAL DIFFERENTIAL EQUATIONS This uit covers topics tht expi the formtio of prti differeti equtios d the soutios of speci types of prti differeti equtios.. INTRODUCTION A prti differeti equtio is oe which ivoves oe or more prti derivtives. The order of the highest derivtive is ced the order of the equtio. A prti differeti equtio cotis more th oe idepedet vribe. But, here we sh cosider prti differeti equtios ivovig oe depedet vribe d oy two idepedet vribes x d y so tht = f(x,y). We sh deote = p, = q, = r, = s, = t. x y x xy y A prti differeti equtio is ier if it is of the first degree i the depedet vribe d its prti derivtives. If ech term of such equtio cotis either the depedet vribe or oe of its derivtives, the equtio is sid to be homogeeous, otherwise it is o homogeeous.. Formtio of Prti Differeti Equtios Prti differeti equtios c be obtied by the eimitio of rbitrry costts or by the eimitio of rbitrry fuctios. By the eimitio of rbitrry costts Let us cosider the fuctio ( x, y,,, b ) = () where & b re rbitrry costts Differetitig equtio () prtiy w.r.t x & y, we get x y + p = () + q = (3) Eimitig d b from equtios (), () d (3), we get prti differeti equtio of the first order of the form f (x,y,, p, q) =

4 Exmpe Eimite the rbitrry costts & b from = x + by + b Cosider = x + by + b () Differetitig () prtiy w.r.t x & y, we get x y = i.e, p= () = b i.e, q = b (3) Usig () & (3) i (), we get = px +qy+ pq which is the required prti differeti equtio. Exmpe Form the prti differeti equtio by eimitig the rbitrry costts d b from = ( x + ) ( y + b ) Give = ( x + ) ( y + b ) () Differetitig () prtiy w.r.t x & y, we get p = x (y + b ) q = y (x + ) Substitutig the vues of p d q i (), we get 4xy = pq which is the required prti differeti equtio.

5 Exmpe 3 Fid the prti differeti equtio of the fmiy of spheres of rdius oe whose cetre ie i the xy - pe. The equtio of the sphere is give by ( x ) + ( y- b) + = () Differetitig () prtiy w.r.t x & y, we get (x- ) + p = ( y-b ) + q = From these equtios we obti x- = -p () y -b = -q (3) Usig () d (3) i (), we get p + q + = or ( p + q + ) = Exmpe 4 Eimite the rbitrry costts, b & c from x y + + = d form the prti differeti equtio. b c The give equtio is x y + + = () b c

6 Differetitig () prtiy w.r.t x & y, we get x p + = c y q + = b c Therefore we get x p + = () c y q + = (3) b c Agi differetitig () prtiy w.r.t x, we set (/ ) + (/ c ) ( r + p ) = (4) Mutipyig ( 4) by x, we get x x r p x + + = c c From (), we hve p xr p x + + = c c c or -p + xr + p x = By the eimitio of rbitrry fuctios Let u d v be y two fuctios of x, y, d Φ(u, v ) =, where Φ is rbitrry fuctio. This retio c be expressed s u = f(v) ()

7 Differetitig () prtiy w.r.t x & y d eimitig the rbitrry fuctios from these retios, we get prti differeti equtio of the first order of the form f(x, y,, p, q ) =. Exmpe 5 Obti the prti differeti equtio by eimitig f from = ( x+y ) f ( x - y ) Let us ow cosider the equtio = (x+y ) f(x - y ) () Differetitig () prtiy w.r.t x & y, we get p = ( x + y ) f ' ( x - y ). x + f ( x - y ) q = ( x + y ) f ' ( x - y ). (-y) + f ( x - y ) These equtios c be writte s Hece, we get p - f ( x - y ) = ( x + y ) f '( x - y ). x () q - f ( x - y ) = ( x + y ) f '( x - y ).(-y) (3) p - f ( x - y ) q - f ( x - y ) = - x y i.e, py - yf( x - y ) = -qx +xf ( x - y ) i.e, py +qx = ( x+y ) f ( x - y ) Therefore, we hve by(), py +qx = Exmpe 6 Form the prti differeti equtio by eimitig the rbitrry fuctio f from = e y f (x + y) Cosider = e y f ( x +y ) ( ) Differetitig () prtiy w.r. t x & y, we get p = e y f ' (x + y) q = e y f '(x + y) + f(x + y). e y Hece, we hve q = p +

8 Exmpe 7 Form the PDE by eimitig f & Φ from = f (x +y ) + Φ ( x y) Cosider = f (x +y ) + Φ ( x y) () Differetitig () prtiy w.r.t x &y, we get p = f '(x +y ) + Φ' (x y) () q = f ' (x +y ). + Φ' (x y) ( -) (3) Differetitig () & (3) gi prtiy w.r.t x & y, we get r = f "( x+y) + Φ "( x y) t = f "( x+y). + Φ"( x y) (-) i.e, or t = { f"( x + y) + Φ"( x y)} t = r Exercises:. Form the prti differeti equtio by eimitig the rbitrry costts & b from the foowig equtios. (i) = x + by (ii) x + y + = b (iii) = x + by + + b (iv) x + by + c = (v) = x + b y + b. Fid the PDE of the fmiy of spheres of rdius hvig their cetres ie o the xy pe{hit: (x ) + (y b) + = } 3. Fid the PDE of spheres whose cetre ie o the (i) xis (ii) x-xis 4. Form the prti differeti equtios by eimitig the rbitrry fuctios i the foowig cses. (i) = f (x + y) (ii) = f (x y ) (iii) = f (x + y + ) (iv) (xy, x + y + ) =

9 (v) = x + y + f(xy) (vi) = xy + f (x + y ) (vii) = f xy (viii) F (xy +, x + y + ) = (ix) = f (x + iy) +f (x iy) (x) = f(x 3 + y) +g(x 3 y).3 SOLUTIONS OF A PARTIAL DIFFERENTIAL EQUATION A soutio or itegr of prti differeti equtio is retio coectig the depedet d the idepedet vribes which stisfies the give differeti equtio. A prti differeti equtio c resut both from eimitio of rbitrry costts d from eimitio of rbitrry fuctios s expied i sectio.. But, there is bsic differece i the two forms of soutios. A soutio cotiig s my rbitrry costts s there re idepedet vribes is ced compete itegr. Here, the prti differeti equtios coti oy two idepedet vribes so tht the compete itegr wi icude two costts.a soutio obtied by givig prticur vues to the rbitrry costts i compete itegr is ced prticur itegr. Sigur Itegr Let f (x,y,,p,q) = () be the prti differeti equtio whose compete itegr is (x,y,,,b) = () where d b re rbitrry costts. Differetitig () prtiy w.r.t. d b, we obti = (3) d = (4) b The eimit of d b from the equtios (), (3) d (4), whe it exists, is ced the sigur itegr of ().

10 Geer Itegr I the compete itegr (), put b = F(), we get (x,y,,, F() ) = (5) Differetitig (), prtiy w.r.t., we get F'() = (6) b The eimit of betwee (5) d (6), if it exists, is ced the geer itegr of (). SOLUTION OF STANDARD TYPES OF FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS. The first order prti differeti equtio c be writte s f(x,y,, p,q) =, where p = /x d q = / y. I this sectio, we sh sove some stdrd forms of equtios by speci methods. Stdrd I : f (p,q) =. i.e, equtios cotiig p d q oy. =. Suppose tht = x + by +c is soutio of the equtio f(p,q) =, where f (,b) Sovig this for b, we get b = F (). Hece the compete itegr is = x + F() y +c () Now, the sigur itegr is obtied by eimitig & c betwee = x + y F() + c = x + y F'() =. The st equtio beig bsurd, the sigur itegr does ot exist i this cse. To obti the geer itegr, et us tke c = ().

11 The, = x + F() y + () () Differetitig () prtiy w.r.t., we get = x + F'(). y + '() (3) Eimitig betwee () d (3), we get the geer itegr Exmpe 8 Sove pq = The give equtio is of the form f (p,q) = The soutio is = x + by +c, where b =. Sovig, b = The compete itegr is Z = x y + c () Differetitig () prtiy w.r.t c, we get =, which is bsurd. Hece, there is o sigur itegr. To fid the geer itegr, put c = () i (), we get Z = x y + () Differetitig prtiy w.r.t, we get = x y + () Eimitig betwee these equtios gives the geer itegr.

12 Exmpe 9 Sove pq + p +q = The give equtio is of the form f (p,q) =. The soutio is = x + by +c, where b + + b =. Sovig, we get b = Hece the compete Itegr is = x y+c () + Differetitig () prtiy w.r.t. c, we get =. The bove equtio beig bsurd, there is o sigur itegr for the give prti differeti equtio. To fid the geer itegr, put c = () i (), we hve = x y + () () + Differetitig () prtiy w.r.t, we get = x y + () (3) ( + ) Eimitig betwee () d (3) gives the geer itegr. Exmpe Sove p + q = pq The soutio of this equtio is = x + by + c, where + b = b. Sovig, we get + ( 4) b =

13 Hece the compete itegr is + 4 =x y + c () Differetitig () prtiy w.r.t c, we get =, which is bsurd. Therefore, there is o sigur itegr for the give equtio. To fid the geer Itegr, put C = (), we get + 4 = x y + () Differetitig prtiy w.r.t, we hve + 4 = x y + () The eimit of betwee these equtios gives the geer itegr Stdrd II : Equtios of the form f (x,p,q) =, f (y,p,q) = d f (,p,q) =. i.e, oe of the vribes x,y, occurs expicity. (i) Let us cosider the equtio f (x,p,q) =. Sice is fuctio of x d y, we hve d = dx dy x y or d = pdx + qdy Assume tht q =. The the give equtio tkes the form f (x, p, ) = Sovig, we get Therefore, p = (x,). d = (x,) dx + dy. Itegrtig, = (x,) dx + y + b which is compete Itegr.

14 (ii) Let us cosider the equtio f(y,p,q) =. Assume tht p =. The the equtio becomes f (y,, q) = Sovig, we get q = (y,). Therefore, d = dx + (y,) dy. Itegrtig, = x + (y,) dy + b, which is compete Itegr. (iii) Let us cosider the equtio f(, p, q) =. Assume tht q = p. The the equtio becomes f (, p, p) = Sovig, we get p = (,). Hece d = (,) dx + (, ) dy. d ie, = dx + dy. (,) Itegrtig, d = x + y + b, which is compete Itegr. (,) Exmpe Sove q = xp + p Give q = xp + p () This is of the form f (x,p,q) =. Put q = i (), we get = xp + p i.e, p + xp =. -x +(x + 4) Therefore, p =

15 x ± x + 4 Itegrtig, = dx + y + b x x x Thus, = ± (4 + x )+ si h y + b 4 4 Exmpe Sove q = yp This is of the form f (y,p,q) = The, put p =. Therfore, the give equtio becomes q = y. Sice d = pdx + qdy, we hve d = dx + y dy y Itegrtig, we get = x b Exmpe 3 Sove 9 (p + q ) = 4 This is of the form f (,p,q) = The, puttig q = p, the give equtio becomes 9 (p + p ) = 4 Therefore, p = ± ( + ) d q = ± ( + ) Sice d = pdx + qdy,

16 d = ± dx ± dy Mutipyig both sides by +, we get + d = dx dy, which o itegrtio gives, 3 3 (+ ) 3/ = x y + b. 3/ 3 3 or ( + ) 3/ = x + y + b. Stdrd III : f (x,p) = f (y,q). ie, equtios i which is bset d the vribes re seprbe. Let us ssume s trivi soutio tht f(x,p) = g(y,q) = (sy). Sovig for p d q, we get p = F(x,) d q = G(y,). But d = dx dy x y Hece d = pdx + qdy = F(x,) dx + G(y,) dy Therefore, = F(x,) dx + G(y,) dy + b, which is the compete itegr of the give equtio cotiig two costts d b. The sigur d geer itegrs re foud i the usu wy. Exmpe 4 Sove pq = xy The give equtio c be writte s p y = = (sy) x q

17 p Therefore, = impies p = x x y y d = impies q = q Sice d = pdx + qdy, we hve y d = xdx dy, which o itegrtio gives. x y = b Exmpe 5 Sove p + q = x + y The give equtio c be writte s p x = y q = (sy) p x = impies p = ( + x ) d y q = impies q = (y ) But d = pdx + qdy ie, d = + x dx + y dy Itegrtig, we get x x y y = ----x sih y cosh b Stdrd IV (Cirut s form) Equtio of the type = px + qy + f (p,q) () is kow s Cirut s form.

18 Differetitig () prtiy w.r.t x d y, we get p = d q = b. Therefore, the compete itegr is give by = x + by + f (,b). Exmpe 6 Sove = px + qy +pq The give equtio is i Cirut s form. Puttig p = d q = b, we hve = x + by + b () which is the compete itegr. To fid the sigur itegr, differetitig () prtiy w.r.t d b, we get = x + b = y + Therefore we hve, = -y d b= -x. Substitutig the vues of & b i (), we get = -xy xy + xy or + xy =, which is the sigur itegr. To get the geer itegr, put b = () i (). The = x + ()y + () () Differetitig () prtiy w.r.t, we hve = x + '() y + '() + () (3) Eimitig betwee () d (3), we get the geer itegr.

19 Exmpe 7 Fid the compete d sigur soutios of = px + qy + + p + q The compete itegr is give by = x + by b () To obti the sigur itegr, differetitig () prtiy w.r.t & b. The, = x b b = y b Therefore, x = () ( + + b ) b d y = (3) ( + + b ) Squrig () & (3) d ddig, we get + b x + y = b Now, x y = b i.e, + + b = x y Therefore, ( + + b ) = (4) x y Usig (4) i () & (3), we get

20 x = x y d y = b x y -x -y Hece, = d b = x y x y Substitutig the vues of & b i (), we get - x y = x y x y x y which o simpifictio gives = x y or x + y + =, which is the sigur itegr. Exercises Sove the foowig Equtios. pq = k. p + q = pq 3. p +q = x 4. p = y q 5. = p + q 6. p + q = x + y 7. p + q = 8. = px + qy - pq 9. { (px + qy)} = c + p + q. = px + qy + p q EQUATIONS REDUCIBLE TO THE STANDARD FORMS Sometimes, it is possibe to hve o ier prti differeti equtios of the first order which do ot beog to y of the four stdrd forms discussed erier. By chgig the vribes suitby, we wi reduce them ito y oe of the four stdrd forms. Type (i) : Equtios of the form F(x m p, y q) = (or) F (, x m p, y q) =. Cse(i) : If m d, the put x -m = X d y - = Y.

21 X Now, p = = = (-m) x -m x X x X Therefore, x m p = (-m) = ( m) P, where P = X X Simiry, y q = (-)Q, where Q = Y Hece, the give equtio tkes the form F(P,Q) = (or) F(,P,Q) =. Cse(ii) : If m = d =, the put og x = X d og y = Y. X Now, p = = = x X x X x Therefore, xp = = P. X Simiry, yq =Q. Exmpe 8 Sove x 4 p + y q = The give equtio c be expressed s (x p) + (y q) = Here m =, = Put X = x -m = x - d Y = y - = y -. We hve x m p = (-m) P d y q = (-)Q i.e, x p = -P d y q = -Q. Hece the give equtio becomes P Q = () This equtio is of the form f (,P,Q) =. Let us tke Q = P.

22 The equtio () reduces to P P = Hece, d Sice ( + 8) P = ( + 8) Q = d = PdX + QdY, we hve ( + 8) ( + 8) d = dx dy i.e, d ( + 8) = (dx + dy) Itegrtig, we get + 8 og = (X + Y) +b ( + 8) Therefore, og = b which is the compete soutio. x y Exmpe 9 Sove x p + y q = The give equtio c be writte s (xp) + (yq) = Here m =, =. Put X = og x d Y = og y.

23 The xp = P d yq = Q. Hece the give equtio becomes P + Q = () This equtio is of the form F(,P,Q) =. Therefore, et us ssume tht Q = P. Now, equtio () becomes, P + P = Hece P = (+ ) d Q = (+ ) Sice d = PdX + QdY, we hve d = dx dy. (+ ) (+ ) d i.e, (+ ) = dx + dy. Itegrtig, we get (+ ) og = X + Y + b. Therefore, (+ ) og = ogx + ogy + b, which is the compete soutio. Type (ii) : Equtios of the form F( k p, k q) = (or) F(x, k p) = G(y, k q). Cse (i) : If k -, put Z = k+, Z Z Now = = (k+) k = (k+) k p. x x x Z Therefore, k p = k+ x

24 Z Simiry, k q = k+ y Cse (ii) : If k = -, put Z = og. Z Z Now, = = p x x Z Simiry, = q. y Exmpe Sove 4 q p = The give equtio c so be writte s ( q) ( p) = Here k =. Puttig Z = k+ = 3, we get Z Z Z k p = d Z k q = k+ x k+ y Z Z i.e, Z p = d Z q = x 3 y Hece the give equtio reduces to i.e, Q 3P 9 =, Q P = 3 3 which is of the form F(P,Q) =. Hece its soutio is Z = x + by + c, where b 3 9 =. Sovig for b, b = ± (3 +9)

25 Hece the compete soutio is Z = x + (3 +9). y + c or 3 = x + (3 +9) y + c Exercises Sove the foowig equtios.. x p + y p =. (p +q ) = x + y 3. (p x + q ) = 4. x 4 p yq 3 = 5. p + x y q = x 6. x p + y q = 7. x /p + y /q = 8. (p q ) = 9. (p /x + q /y ) =. p x + q y =..4 Lgrge s Lier Equtio Equtios of the form Pp + Qq = R (), where P, Q d R re fuctios of x, y,, re kow s Lgrge s equtios d re ier i p d q.to sove this equtio, et us cosider the equtios u = d v = b, where, b re rbitrry costts d u, v re fuctios of x, y,. Sice u is costt, we hve du = (). But u s fuctio of x, y,, u u u du = dx + dy + d x y Comprig () d (3), we hve u u u dx + dy + d = (3) x y Simiry, v v v dx + dy + d = (4) x y

26 By cross-mutipictio, we hve dx dy d = = u v u v u v u v u v u v y y x x y x x y (or) dx dy d = = (5) P Q R Equtios (5) represet pir of simuteous equtios which re of the first order d of first degree.therefore, the two soutios of (5) re u = d v = b. Thus, ( u, v ) = is the required soutio of (). Note : To sove the Lgrge s equtio,we hve to form the subsidiry or uxiiry equtios dx dy d = = P Q R which c be soved either by the method of groupig or by the method of mutipiers. Exmpe Fid the geer soutio of px + qy =. Here, the subsidiry equtios re dx dy d = = x y Tkig the first two rtios, dx dy x = y Itegrtig, og x = og y + og c

27 or i.e, x = c y c = x / y From the st two rtios, dy d = y Itegrtig, og y = og + og c or i.e, y = c c = y / Hece the required geer soutio is Exmpe Φ( x/y, y/) =, where Φ is rbitrry Sove p t x + q t y = t The subsidiry equtios re dx dy d = = tx ty t Tkig the first two rtios, dx dy = tx ty ie, cotx dx = coty dy Itegrtig, og six = og siy + og c ie, six = c siy Therefore, c = six / siy Simiry, from the st two rtios, we get siy = c si i.e, c = siy / si Hece the geer soutio is

28 six siy Φ, =, where Φ is rbitrry. siy si Exmpe 3 Sove (y-) p + (-x) q = x-y Here the subsidiry equtios re dx dy d = = y- - x x y Usig mutipiers,,, dx + dy + d ech rtio = Therefore, dx + dy + d =. Itegrtig, x + y + = c () Agi usig mutipiers x, y d, ech rtio = xdx + ydy + d Therefore, xdx + ydy + d =. Itegrtig, x / + y / + / = costt or x + y + = c () Hece from () d (), the geer soutio is Φ ( x + y +, x + y + ) = Exmpe 4 Fid the geer soutio of (m - y) p + (x- )q = y - mx.

29 Here the subsidiry equtios re dx dy d = = m- y x - y - mx Usig the mutipiers x, y d, we get xdx + ydy + d ech frctio = ` xdx + ydy + d =, which o itegrtio gives x / + y / + / = costt or x + y + = c () Agi usig the mutipiers, m d, we hve dx + mdy + d ech frctio = ` dx + mdy + d =, which o itegrtio gives x + my + = c () Hece, the required geer soutio is Exmpe 5 Φ(x + y +, x + my + ) = Sove (x - y - ) p + xy q = x. The subsidiry equtios re dx dy d = = x -y - xy x Tkig the st two rtios, dx xy = d x

30 xy x ie, dy d = y Itegrtig, we get og y = og + og c or y = c i.e, c = y/ () Usig mutipiers x, y d, we get xdx + y dy + d xdx + y dy + d ech frctio = = x (x -y - )+xy +x x ( x + y + ) Comprig with the st rtio, we get xdx + y dy + d d = x ( x + y + ) x xdx + ydy + d d i.e, = x + y + Itegrtig, og ( x + y + ) = og + og c or x + y + = c x + y + i.e, c = () From () d (), the geer soutio is Φ(c, c ) =. x + y + i.e, Φ (y/), =

31 Exercises Sove the foowig equtios. px + qy =. py + qx = xy 3. xp yq = y x 4. y p + x q = y x 5. (x y) = px qy 6. ( x) p + (b y) q = c 7. (y p) /x + xq = y 8. (y + ) p xyq + x = 9. x p + y q = (x + y). p q = og (x+y). (x + y)p + (x y)q = x + y. (y )p (x + y)q = x +.5 PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER WITH CONSTANT COEFFICIENTS. Homogeeous Lier Equtios with costt Coefficiets. A homogeeous ier prti differeti equtio of the th order is of the form c c c = F (x,y) () x x - y y where c, c, , c re costts d F is fuctio of x d y. It is homogeeous becuse its terms coti derivtives of the sme order. Equtio () c be expressed s (c D + c D - D ' c D ' ) = F (x,y) or f (D,D ' ) = F (x,y) (),

32 where, D d D '. x y As i the cse of ordiry ier equtios with costt coefficiets the compete soutio of () cosists of two prts, mey, the compemetry fuctio d the prticur itegr. The compemetry fuctio is the compete soutio of f (D,D ' ) = (3), which must coti rbitrry fuctios s the degree of the poyomi f(d,d ' ). The prticur itegr is the prticur soutio of equtio (). Fidig the compemetry fuctio Let us ow cosider the equtio f(d,d ' ) = F (x,y) The uxiiry equtio of (3) is obtied by repcig D by m d D ' by. i.e, c m + c m c = (4) Sovig equtio (4) for m, we get roots. Depedig upo the ture of the roots, the Compemetry fuctio is writte s give beow: Roots of the uxiiry Nture of the Compemetry fuctio(c.f) equtio roots m,m,m 3.,m distict roots f (y+m x)+f (y+m x) +.+f (y+m x). m = m = m, m 3,m 4,.,m two equ roots f (y+m x)+xf (y+m x) + f 3 (y+m 3 x) +.+ f (y+m x). m = m =.= m = m equ roots f (y+mx)+xf (y+mx) + x f 3 (y+mx) x - f (y+mx) Fidig the prticur Itegr Cosider the equtio f(d,d ' ) = F (x,y). Now, the P.I is give by F (x,y) f(d,d ' ) x +by Cse (i) : Whe F(x,y) = e P.I = e x+by f (D,D ' ) Repcig D by d D ' by b, we hve P.I = e x+by, where f (,b).

33 f (,b) Cse (ii) : Whe F(x,y) = si(x + by) (or) cos (x +by) P.I = si (x+by) or cos (x+by) f(d,dd ',D ' ) Repcig D = -, DD ' = -b d D ' = -b, we get P.I = si (x+by) or cos (x+by), where f(-, - b, -b ). f(-, - b, -b ) Cse (iii) : Whe F(x,y) = x m y, P.I = x m y f(d,d ' ) = [f (D, D ' )] - x m y Expd [f (D,D ' )] - i scedig powers of D or D ' d operte o x m y term by term. Cse (iv) : Whe F(x,y) is y fuctio of x d y. P.I = F (x,y). f (D,D ' ) Resove ito prti frctios cosiderig f (D,D ' ) s fuctio of D oe. f (D,D ' ) The operte ech prti frctio o F(x,y) i such wy tht F (x,y) = F(x,c-mx) dx, D md ' where c is repced by y+mx fter itegrtio Exmpe 6 Sove(D 3 3D D ' + 4D '3 ) = e x+y The uxiry equtio is m=m 3 3m + 4 =

34 The roots re m = -,, Therefore the C.F is f (y-x) + f (y+ x) + xf 3 (y+x). e x+y P.I.= (Repce D by d D ' by ) D 3 3D D ' +4D '3 e x+y = ()() + 4() 3 e x+y = Hece, the soutio is = C.F. + P.I e x+y ie, = f (y-x) + f (y+x) + x f 3 (y+x) Exmpe 7 Sove (D 4DD ' +4 D ' ) = cos (x y) The uxiiry equtio is m 4m + 4 = Sovig, we get m =,. Therefore the C.F is f (y+x) + xf (y+x). P.I = cos (x-y) D 4DD ' + 4D ' Repcig D by, DD ' by d D ' by 4, we hve P.I = cos (x-y) (-) 4 () + 4(-4) cos (x-y) =

35 Soutio is = f (y+x) + xf (y+x) Exmpe 8 Sove (D DD ' ) = x 3 y + e 5x The uxiiry equtio is m m =. Sovig, we get m =,. Hece the C.F is f (y) + f (y+x). x 3 y P.I = D DD ' = (x 3 y) D ' D D D ' = (x 3 y) D D D ' 4D ' = (x 3 y) D D D 4 = (x 3 y) D ' (x 3 y) D ' (x 3 y) D D D 4 = (x 3 y) (x 3 ) () D D D P.I = (x 3 y) (x 3 ) D D 3 x 5 y x 6 P.I =

36 e 5x P.I = (Repce D by 5 d D ' by ) D DD ' e 5x = x 5 y x 6 e 5x Soutio is Z = f (y) + f (y+x) Exmpe 9 Sove (D + DD ' 6 D ) = y cosx. The uxiiry equtio is m + m 6 =. Therefore, m = 3,. Hece the C.F is f (y-3x) + f (y + x). y cosx P.I = D + DD ' 6D ' y cosx = (D + 3D ' ) (D D ' ) = y cosx (D+3D ' ) (D D ' ) = (c x) cosx dx, where y = c x (D+3D ' ) = (c x) d (six) (D+3D ' ) = [(c x) (six) (-) ( - cosx)] (D+3D ' ) = [ y si x cos x)] (D+3D ' ) = [(c + 3x) six cosx] dx, where y = c + 3x

37 = (c + 3x) d( cosx) cosx dx = (c + 3x) ( cosx) (3) ( - six) six = y cosx + six Hece the compete soutio is = f (y 3x) + f (y + x) y cosx + six Exmpe 3 x +y Sove r 4s + 4t = e Give equtio is = e x + y x xy y i.e, (D 4DD ' + 4D ' ) = e x + y The uxiiry equtio is m 4m + 4 =. Therefore, m =, Hece the C.F is f (y + x) + x f (y + x). e x+y P.I. = D 4DD ' +4D ' Sice D 4DD ' +4D ' = for D = d D ' =, we hve to ppy the geer rue. e x+y P.I. = (D D ' ) (D D ' ) = e x+y (D D ' ) (D D ' ) = e x+c x dx, where y = c x. (D D ' )

38 = ec dx (D D ' ) = e c.x (D D ' ) = xe y+x D D ' = xe c-x + x dx, where y = c x. = xe c dx = e c. x / x e y+x = Hece the compete soutio is = f (y+x) + f (y+x) x e x+y.6 No Homogeeous Lier Equtios Let us cosider the prti differeti equtio f (D,D ' ) = F (x,y) () If f (D,D ' ) is ot homogeeous, the () is o homogeeous ier prti differeti equtio. Here so, the compete soutio = C.F + P.I. The methods for fidig the Prticur Itegrs re the sme s those for homogeeous ier equtios. But for fidig the C.F, we hve to fctorie f (D,D ' ) ito fctors of the form D md ' c. Cosider ow the equtio (D md ' c) = ().

39 This equtio c be expressed s p mq = c (3), which is i Lgrgi form. The subsidiry equtios re dx dy d = = (4) m c The soutios of (4) re y + mx = d = be cx. Tkig b = f (), we get = e cx f (y+mx) s the soutio of (). Note:. If (D-m D ' C ) (D m D ' -C ) (D m D ' -C ) = is the prti differeti equtio, the its compete soutio is = e c x f (y +m x) + e c x f (y+m x) e c x f (y+m x). I the cse of repeted fctors, the equtio (D-mD ' C) = hs compete soutio = e cx f (y +mx) + x e cx f (y+mx) x - e cx f (y+mx). Exmpe 3 Sove (D-D ' -) (D-D ' ) = e x y Here m =, m =, c =, c =. Therefore, the C.F is e x f (y+x) + e x f (y+x). e x-y P.I. = Put D =, D ' =. (D D ' ) (D-D ' ) e x-y = ( ( ) ) ( ( ) )

40 e x-y = e x-y Hece the soutio is = e x f (y+x) + e x f (y+x) Exmpe 3 Sove (D DD ' + D ' ) = cos (x + y) The give equtio c be rewritte s (D-D ' +) (D-) = cos (x + y) Here m =, m =, c = -, c =. Therefore, the C.F = e x f (y+x) + e x f (y) P.I = cos (x+y) [Put D =,DD ' = -,D ' = 4] (D DD ' + D ' ) = cos (x+y) ( ) + D ' = cos (x+y) D ' si (x+y) = Hece the soutio is = e -x f (y+x) e x f (y) Exmpe 33 Sove [(D + D ' ) (D + D ' 3)] = e x+y x +6y Here m =, m =, c =, c = 3. si(x+y)

41 Hece the C.F is = e x f (y x) + e 3x f (y x). e x+y P.I = [Put D =, D ' = ] (D+D ' ) (D + D ' 3) e x+y = (+ ) (+4 3) e x+y = P.I = (4 + 3x + 6y) (D+D ' ) (D + D ' 3) = (4 + 3x + 6y) D + D ' 3 [ (D+D ' )] D + D ' - = [ (D + D ' )] (4 +3x+6y) D + D ' = ----[ + (D + D ' )+ (D+D ' ) +...] (D+D ' ) +...] (4 + 3x +6y) 4 5 = D D ' (4 + 3x + 6y) 3 3 3

42 4 5 = x + 6y (3) (6) = x + y + 6 Hece the compete soutio is e x+y = e x f (y-x) + e 3x f (y x) x + y Exercises () Sove the foowig homogeeous Equtios = cos (x + y) x xy y = si x.cos y x xy 3. (D + 3DD ' + D ' ) = x + y 4. (D DD ' + D ' ) = xy + e x. coshy e y + e y e x+y + e x-y Hit: e x. coshy = e x = (D 3 7DD ' 6D ' 3 ) = si (x+y) + e x+y 6. (D + 4DD ' 5D ' ) = 3e x-y + si (x y) 7. (D DD ' 3D ' ) = xy + e 6x+y 8. (D 4D ' ) = cosx. cos3y 9. (D DD ' D ' ) = (y )e x

43 3x y. 4r + s + 9t = e (b) Sove the foowig o homogeeous equtios.. (DD ' + D ' 3D ' ) = 3 cos(3x y). (D + DD ' + D ' ) = e -x 3. r s + p = x + y 4. (D DD ' + D ' 3D + 3D ' + ) = (e 3x + e -y ) 5. (D D ' 3D + 3D ' ) = xy + 7.

44 UNIT II FOURIER SERIES. INTRODUCTION The cocept of Fourier series ws first itroduced by Jcques Fourier (768 83), Frech Physicist d Mthemtici. These series becme most importt too i Mthemtic physics d hd deep ifuece o the further deveopmet of mthemtics it sef.fourier series re series of cosies d sies d rise i represetig geer periodic fuctios tht occurs i my Sciece d Egieerig probems. Sice the periodic fuctios re ofte compicted, it is ecessry to express these i terms of the simpe periodic fuctios of sie d cosie. They py importt roe i sovig ordiry d prti differeti equtios.. PERIODIC FUNCTIONS A fuctio f (x) is ced periodic if it is defied for re x d if there is some positive umber p such tht f (x + p ) = f (x) for x. This umber p is ced period of f(x). If periodic fuctio f (x) hs smest period p (>), this is ofte ced the fudmet period of f(x). For exmpe, the fuctios cosx d six hve fudmet period. DIRICHLET CONDITIONS Ay fuctio f(x), defied i the iterv c x c +, c be deveoped s Fourier series of the form ( cosx + b six) provided the foowig = coditios re stisfied. f (x) is periodic, sige vued d fiite i [ c, c + ]. f (x) hs fiite umber of discotiuities i [ c, c + ]. f (x) hs t the most fiite umber of mxim d miim i [ c,c+ ]. These coditios re kow s Dirichet coditios. Whe these coditios re stisfied, the Fourier series coverges to f(x) t every poit of cotiuity. At poit of discotiuity x = c, the sum of the series is give by f(x) = (/) [ f (c-) + f (c+)],

45 where f (c-) is the imit o the eft d f (c+) is the imit o the right. EULER S FORMULAE The Fourier series for the fuctio f(x) i the iterv c < x < c + is give by f (x) = ( cosx + b six), where = C + = f (x) dx. C C + = f(x) cosx dx. C C + b = f (x) six dx. C These vues of,, b re kow s Euer s formue. The coefficiets,, b re so termed s Fourier coefficiets. Exmpe Expd f(x) = x s Fourier Series (Fs) i the iterv [ -π, π] Let f(x) = o [ cos x + b si x ] () π = Here o = f (x) dx π -π π = x dx π -π π = x π - π π π = - = π o =

46 π = --- f(x) cosx dx π -π π si x = x d π -π = (x) si x - () -cos x π π - π = cos π cos π π = π b = f(x) si x dx π -π π -cos x = x d π -π π -cosx -si x π = (x) - () - π = - πcos π πcosπ π = -π cosπ π b = (-) + [ cos π = (-) ] Substitutig the vues of o, & b i equtio (), we get f(x) = (-) + si x = x = six - six + si 3x - 3

47 Exmpe Expd f(x) = x s Fourier Series i the iterv ( - x ) d hece deduce tht = = = Let f(x) = + [ cosx + b six ] = Here = f(x) dx - = x dx - = x = o = 3 = f(x) cosx dx - = x cosx dx -

48 = x d six - = (x ) six (x)-cosx + () six 3 - = cos + cos = 4 (-) b = f(x) six dx - = x d -cosx - = (x ) cosx (x) -six + () cosx 3 - = - cos cos cos cos b = Substitutig the vues of, & b i equtio () we get f(x) = + 4 (-) cosx 6 = i.e, x = + 4 (-) cosx 3 = i.e, x = + 4 (-) cosx

49 3 = = + 4 -cosx + cosx cos3x x = - 4 cosx + cosx + cos3x () 3 3 Put x = i equtio () we get = i.e, + - = (3) 3 Put x = i equtio () we get = i.e, - = i.e, = (4) 3 6 Addig equtios (3) & (4) we get = i.e, = i.e, = Exmpe 3 Obti the Fourier Series of periodicity π for f(x) = e x i [-π, π] Let f(x) = ( cosx + b six) () =

50 π = f(x) dx π - π π = e x dx π - π π = [e x ] π - π = {e π e π } π = si hπ π π = f(x) cos x dx π - π π = e x cos x dx π - π π = e x [cosx + si x] π (+ ) -π = e π (-) e π (-) π + + = (-) ( e π - e π ) (+ ) π = ( -) si hπ π(+ ) π b = f(x) si x dx π - π π = e x si x dx π - π

51 π = e x (six cosx) π (+ ) -π = e π {-(-) } e - π {-(-) } π + + = (-) + ( e π - e π ) π(+ ) b = (-) + si hπ π(+ ) f(x) = si hπ + (-) sihπ cosx + (-)(-) sihπ six π = π(+ ) π(+ ) e x = si hπ + si hπ (-) π π = + (cos x si x) ie, e x = si hπ + (-) (cos x si x) π = + Exmpe 4 Let f (x) = x i (O, ) ( - x) i (, ) Fid the FS for f (x) d hece deduce tht = = (-) 8 Let f (x) = cosx + b si x () = Here = f(x) dx + f (x) dx o = x dx+ ( - x) dx o

52 x ( - x) = = = i.e, o = = x cos x dx + ( - x) cos x dx o si x si x = ---- x d ( - x) d si x -cos x si x cosx = ---- (x) () (-x) ( ) o cos cos cos = cos = = [( ) -] b = f(x) si x dx + f(x) si x dx o cos x cos x = ---- x d ( - x) d o cos x six cos x six = ---- (x) () (-x) ( )

53 o cos cos = = i.e, b =. f (x) = [ ( ) ] cos x = 4 cosx cos3x cos5x = () 3 5 Puttig x = i equtio(), we get 4 = i.e, = i.e, = = ( ) 8 Exmpe 5 Fid the Fourier series for f (x) = (x + x ) i (- < x < ) of percodicity d hece deduce tht (/ ) = /6. = Let f(x) = ( cosx + b six) = Here, = (x + x ) dx x x 3 =

54 3 3 = o = = f (x) cos x dx si x = (x + x ) d si x cosx six = (x + x ) (+x) () ( ) ( ) = (+ ) ( ) ( ) = b = f (x) si x dx cos x = (x + x ) d cosx six cosx = (x + x ) (+x) ()

55 ( ) ( ) ( ) ( ) = ( ) + b = ( ) ( ) + f(x) = cos x six 3 =. cosx cosx cos3x six = si x Here x = - d x = re the ed poits of the rge. The vue of FS t x = is the verge of the vues of f(x) t x = d x = -. f ( - ) + f () f(x) = = = Puttig x =, we get = i.e, =

56 Hece, = = 6 Exercises: Determie the Fourier expressios of the foowig fuctios i the give iterv.f(x) = ( - x), < x <.f(x) = i - < x < = i < x < 3.f(x) = x x i [-,] 4.f(x) = x(-x) i (,) 5.f(x) = sih x i [-, ] 6.f(x) = cosh x i [-, ] 7.f(x) = i < x < = i < x < 8.f(x) = -/4 whe - < x < = /4 whe < x < 9.f(x) = cosx, i - < x <, where is ot iteger.obti fourier series to represet e -x from x = - to x =. Hece derive the series for / sih.3 Eve d Odd fuctios A fuctio f(x) is sid to be eve if f (-x) = f (x). For exmpe x, cosx, x six, secx re eve fuctios. A fuctio f (x) is sid to be odd if f (-x) = - f (x). For exmpe, x 3, si x, x cos x,. re odd fuctios. () The Euer s formu for eve fuctio is f (x) = cosx = where o = ---- f(x) dx ; = f (x) cosx dx

57 () The Euer s formu for odd fuctio is f (x) = b si x = where b = f(x) si x dx Exmpe 6 Fid the Fourier Series for f (x) = x i ( -, ) Here, f(x) = x is odd fuctio. f(x) = b si x () = b = f (x) si x dx - cos x = x d cos x si x = (x) () cos = ( ) + b = ( ) + f (x)= si x = i.e, ( ) + x = si x = Exmpe 7

58 Expd f (x) = x i (-, ) s FS d hece deduce tht = Soutio Here f(x) = x is eve fuctio. o f (x) = cos x () = o = f(x) dx = x dx x = = = f (x) cos x dx si x = x d si x cos x = (x) () cos = = [( ) ] f (x)= [( ) ] cos x = 4 cos x cos3x cos5x

59 i.e, x = () 3 5 Puttig x = i equtio (), we get 4 = Hece, = Exmpe 8 x If f (x) = i ( -, ) x = i (, ) The fid the FS for f(x) d hece show tht (-) - = /8 = Here f (-x) i (-,) = f (x) i (,) f (-x) i (,) = f (x) i (-,) f(x) is eve fuctio o Let f (x) = cos x (). = x o = dx x = x =

60 x = cos x dx x si x = d x si x cosx = = [( ( ) ] 4 f (x)= [ ( ) ]cos x = 4 cos x cos3x cos5x = () 3 5 Put x = i equtio () we get or = = ( ) 8 Exmpe = ==> = Obti the FS expsio of f(x) = x six i (- < x<) d hece deduce tht =

61 Here f (x) = xsix is eve fuctio. o Let f (x) = cos x () = Now, o = xsi x dx = x d ( - cosx) = (x) (- cosx) () (- si x) = = f (x) cos x dx = x si x cosx dx = x [ si (+)x + si ( )x] dx cos (+)x cos ( ) x = x d cos (+)x cos ( ) x si (+)x si ( ) x = (x) () ( + ) ( ) cos (+) cos ( ) = [cos cos - si si] [cos cos - si si ]

62 = (+) ( ) + ( ) ( ) = ( ) = , Provided Whe = = x six cos x dx = x six dx - cosx = x d cos x -si x = (x) () Therefore, = -/ f (x)= cosx + cos x = ( -) = cosx cosx = - cosx cos3x cos4x ie, x six = cos x Puttig x = / i the bove equtio, we get

63 ----- = Hece, = Exercises: Determie Fourier expressios of the foowig fuctios i the give iterv: i. f(x) = / + x, - < x < / - x, < x < ii. f(x) = -x+ for + - < x < x+ for < x < iii. f(x) = six, - < x < iv. f(x) =x 3 i - < x < v. f(x) = xcosx, - < x < vi. f(x) = cosx, - < x < si si x si x 3si3x vii. Show tht for - < x <, si x = HALF RANGE SERIES It is ofte ecessry to obti Fourier expsio of fuctio for the rge (, ) which is hf the period of the Fourier series, the Fourier expsio of such fuctio cosists cosie or sie terms oy. (i) Hf Rge Cosie Series The Fourier cosie series for f(x) i the iterv (,) is give by f(x) = = cos x

64 where = f(x) dx d = f(x) cosx dx (ii) Hf Rge Sie Series The Fourier sie series for f(x) i the iterv (,) is give by f(x) = b six = where b = Exmpe f(x) six dx If c is the costt i ( < x < ) the show tht c = (4c / ) { six + (si3x /3) + si5x / 5) } Give f(x) = c i (,). Let f(x) = b six () = b = f(x) si x dx = c si x dx c - cosx = c -(-) = b = (c/) [ (-) ]

65 f(x) = (c / ) (-(-) ) six = Exmpe 4c si3x si5x i.e, c = --- six Fid the Fourier Hf Rge Sie Series d Cosie Series for f(x) = x i the iterv (,). Sie Series Let f(x) = b six () = Here b = f(x) sixdx = x d ( -cosx / ) - cosx - six = ---- (x) () (-) = (-) + b = f(x) = Cosie Series = ---- (-) + si x Let f(x) = cosx () = Here = f(x)dx

66 = xdx x = = = f(x) cosx dx = x d (six / ) six - cosx = ---- (x) () = (-) - f(x) = [ (-) - ] cosx = Exmpe 4 cosx cos3x cos5x => x = Fid the sie d cosie hf-rge series for the fuctio fuctio. f(x) = x, < x π/ = π-x, π/x< Sie series Let f (x) = b si x. =

67 π b = (/ ) f (x) si x dx =(/ ) / x si x dx + (-x) si x dx / / -cos x -cos x = (/ ) x.d + (-x) d / -cos x = (/ ) x -() -si x / cos x si x + (-x) - - -(-) - / -(/)cos (/) si (/) -(/)cos(/) si (/ ) = (/) + = (/) si(/) 4 = si (/) si(/) Therefore, f(x)= ( 4/ ) si x = si3x si5x ie, f (x)= ( 4/ ) six

68 Cosie series.let f(x) = ( o /) + cosx., where = o = (/) f(x) dx / =(/) x dx+ (-x) dx / 3 5 / =(/) (x /) + (x x /) = / / = (/) f(x) cosx dx =(/ ) / x cosx dx + (-x) cosx dx / / six six =(/ ) x d + (-x) d / six = (/ ) x -() -cosx / six cosx + (-x) -(-) - / ( /) si(/) cos (/) = (/) +

69 cosx ( /) si(/) cos (/) + + =(/) cos ( /) - {+(-) } cos ( /)- {+(-) } Therefore, f(x)= ( /4)+(/) cosx. = Exercises cos6x = ( /4)-(/) cosx Obti cosie d sie series for f(x) = x i the iterv < x <. Hece show tht / + /3 + /5 + = /8..Fid the hf rge cosie d sie series for f(x) = x i the rge < x < 3.Obti the hf-rge cosie series for the fuctio f(x) = xsix i (,).. 4.Obti cosie d sie series for f(x) = x (-x) i < x < 5.Fid the hf-rge cosie series for the fuctio 6.f(x) = (x) / 4, <x< (/) = (/4)(-x), / < x <. 7.Fid hf rge sie series d cosie series for f(x) = x i <x< (/) = i / < x <. 8.Fid hf rge sie series d cosie series for the fuctio f(x) == - x i the iterv < x <. 9.Fid the hf rge sie series of f(x) = x cosx i (,)

70 .Obti cosie series for f(x) = cos x, <x< (/) =, / < x <..5 Prsev s Theorem Root Me squre vue of the fuctio f(x) over iterv (, b) is defied s b [f(x)] dx [f (x)] r m s = b The use of r.m.s vue of periodic fuctio is frequety mde i the theory of mechic vibrtios d i eectric circuit theory. The r.m.s vue is so kow s the effective vue of the fuctio. Prsev s Theorem If f(x) defied i the iterv (c, c+π ), the the Prsev s Idetity is give by c+π [f (x)] c dx = (Rge) = ( π) o + ( + b ) 4 or o + ( + b ) 4 Exmpe 3 Obti the Fourier series for f(x) = x i π < x < π Hece show tht π = 9 π 4 (-) we hve o = 3, =, b =, for (Refer Exmpe ). By Prsev s Theorem, we hve π o

71 [ f(x)] dx = π + ½ ( + b ) = - π 4 π 4π 4 6(-) i.e, x 4 dx = π + / = - π 36 4 π π 4 i.e, x 5 = π + 8 = 5 -π 9 π 4 π 4 = + 8 = π 4 => = = Hece π = 9.6 CHANGE OF INTERVAL I most of the Egieerig ppictios, we require expsio of give fuctio over iterv other th. Suppose f(x) is fuctio defied i the iterv c< x < c+. The Fourier expsio for f(x) i the iterv c<x<c+ is give by x x f(x) = cos b si ---- = c+ where = f(x)dx c c+ = f(x) cos (x / ) dx & c c+ b = f(x) si (x / ) dx c

72 Eve d Odd Fuctio If f(x) is eve fuctio d is defied i the iterv ( c, c+ ), the x f(x) = cos ---- = where = f(x)dx = f(x) cos (x / ) dx If f(x) is odd fuctio d is defied i the iterv ( c, c+ ), the x f(x) = b si ---- = where b = f(x) si (x / ) dx Hf Rge Series Sie Series f(x) = b si ---- = x where b = f(x) si (x / ) dx Cosie series x f(x) = cos ---- =

73 where = f(x)dx = f(x) cos (x / ) dx Exmpe 4 Fid the Fourier series expsio for the fuctio f(x) = (c/)x i <x< = (c/)(- x) i <x< x x Let f (x) = cos b si = Now, = f(x)dx = (c/) x dx + (c/) ( - x) dx o = (c/) (x / ) + (c/) (x - x /) c = ---- = c = f(x) cos (x / ) dx x x = (c/)x cos dx + (c/)(- x) cos dx c si(x /) si(x /) = x d + (- x) d / /

74 x x si -cos c = ( x ) () x x si -cos + (- x) (-) c cos cos cos = + + c = { cos } c = { (-) } x b = f(x). si dx x x = (c/)x si dx + (c/)(- x) si dx c cos(x /) cos(x /) = x d - + (- x) d - / /

75 x x cos si c = (x) () x x cos si + (- x) (-) c cos cos = + =. c c { (-) } Therefore, f(x) = cos (x /) = Exmpe 5 Fid the Fourier series of periodicity 3 for f(x) = x x, i x 3. Here = 3. = 3 /. x x Let f (x) = cos b si = 3 3 where o = ( / 3) (x - x ) dx 3 = ( / 3) (x /) (x 3 /3) dx 3

76 =. 3 x = ( / 3) (x - x ) cos dx 3 3 si(x /3) = ( / 3) (x - x ) d (/3) si(x /3) cos(x /3) si(x/3) = ( / 3) (x - x ) ( -x) - + (-) - (/3) (4 /9) (8 3 3 /7) 3 = ( / 3) - ( 9 / ) ( 9 / ) = - 9 / 3 x b = ( / 3) (x - x ) si dx 3 3 cos(x /3) = ( / 3) (x - x ) d (/3) cos(x /3) si(x /3) cos(x/3) = ( / 3) (x - x ) ( -x) - (/3) (4 + (-) /9) (8 3 3 /7) 3 = ( / 3) ( 9 /) ( 7/ ) + ( 7/ ) x x Therefore, f (x) = - ( 9 / ) cos (3 / ) si = 3 3 Exercises = 3 /.Obti the Fourier series for f(x) = x i < x <..Fid the Fourier series to represet x i the iterv (-, ). 3.Fid Fourier series i (-, ), if f(x) =, - < x <

77 =, < x <. 4.Obti the Fourier series for f(x) = -x i < x < = i < x <. Hece deduce tht - (/3 ) +(/5) (/7) + = /4 & (/ ) + (/3 ) + (/5 ) + = ( /8) 5.If f(x) = x, < x < = (-x), < x <, Show tht i the iterv (,), cos x cos3x cos 5x f(x) = (/) (4/) Obti the Fourier series for f(x) = x i < x < = i < x < 7.Obti the Fourier series for f(x) = (cx / ) i < x < = (c/ ) ( - x ) i < x <. 8.Obti the Fourier series for f(x) = ( + x ), - < x <. = ( - x ), < x <. Deduce tht = ( ) 8 9.Obti hf-rge sie series for the fuctio f(x) = cx i < x < ( /) = c ( x) i (/) < x <.Express f(x) = x s hf rge sie series i < x <.Obti the hf-rge sie series for e x i < x <.

78 .Fid the hf rge cosie series for the fuctio f(x) = (x-) i the iterv < x <. Deduce tht = ( ) 8.7 Hrmoic Aysis The process of fidig the Fourier series for fuctio give by umeric vues is kow s hrmoic ysis. f(x) six & b = [me vues of f(x) six] = I (), the term ( cosx + b six) is ced the fudmet or first hrmoic, the term ( cosx + b six) is ced the secod hrmoic d so o. Exmpe 6 f (x) = ( cosx + b six), where = ie, f(x) = ( /) + ( cosx + b six) + ( cosx + b six) + ( 3 cos3x + b 3 si3x) () f(x) Here = [me vues of f(x)] = f(x) cosx = [me vues of f(x) cosx] = Compute the first three hrmoics of the Fourier series of f(x) give by the foowig tbe. x: π/3 π/3 π 4π/3 5π/3 π f(x): We excude the st poit x = π. Let f(x) = ( /) + ( cosx + b six) + ( cosx + b six) + To evute the coefficiets, we form the foowig tbe.

79 x f(x) cosx six cosx six cos3x si3x. π/ π/ π π/ π/ f(x) ( ) Now, = = = f(x) cosx = = f(x) cosx = = -. 6 f(x) cos3x 3 = =.33 6 f(x) six b = =.7 6 f(x) six b = = f(x) si3x b 3 = = 6 f(x) =.45.37cosx +.7 six.cosx.6 six +.33 cos3x+ Exmpe 7 Obti the first three coefficiets i the Fourier cosie series for y, where y is give i the foowig tbe: x: y: Tkig the iterv s 6 o, we hve : o 6 o o 8 o 4 o 3 o x: y: Fourier cosie series i the iterv (, π) is y = ( /) + cos + cos + 3 cos3 +.. To evute the coefficiets, we form the foowig tbe.