MA6351-TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS SUBJECT NOTES. Department of Mathematics FATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY

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1 MA635-TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS SUBJECT NOTES Deprtmet of Mthemtics FATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY MADURAI 65, Tmildu, Idi

2 Bsic Formule DIFFERENTIATION &INTEGRATION FORMULAE Fuctio f ( ) Differetitio d d log 3 si cos 4 cos si 5 e e 6 C (costt) 7 t sec 8 sec sec t 9 cot cos ec cos ec cos ec cot si 3 cos 4 t 5 sec 6 cosec 7 cot 8 log Pge Deprtmet of Mthemtics FMCET MADURAI

3 9. If uv, the d du dv v u d d d. If du dv v u u v, the d d d d v Pge3 Deprtmet of Mthemtics FMCET MADURAI

4 . d e. e d e, e d & e d e 3. si d cos & si d 4. cos d si & cos d cos si 5. t d log sec log cos d 6. sec t d 7. d t d 8. d log d 9. d si. d d sih. d d cosh. d si 3. d sih 4. d cosh d 5. log d 6. log Pge4 Deprtmet of Mthemtics FMCET MADURAI

5 7. log d log 8. d d 3 3. d e. e cosbd cosb bsi b b e. e si bd si b bcosb b 3. udv uv u v u v u v f ( ) d f ( ) d whe f() is eve 5. f ( ) d whe f() is odd 6. e cos bd b b 7. e sibd b TRIGNOMETRY FORMULA. si A si Acos A.cos cos si A A A si A A cos Pge5 Deprtmet of Mthemtics FMCET MADURAI

6 3. cos cos cos & si 4. si( A B) si Acos B cos Asi B si( A B) si Acos B cos Asi B cos( A B) cos Acos B si Asi B cos( A B) cos Acos B si Asi B 5.si Acos B si( A B) si( A B) cos Asi B si( A B) si( A B) cos Acos B cos( A B) cos( A B) si Asi B cos( A B) cos( A B) A 4 A A A 4 A A 3 6. si 3si si 3 3 cos 3cos cos 3 7.si A A A si cos cos A A A cos si A A si cos A si LOGRATHEMIC FORMULA Pge6 Deprtmet of Mthemtics FMCET MADURAI

7 log m log m log log m log m log log m log m log log log e log UNIT - PARTIAL DIFFRENTIAL EQUATIONS PARTIAL DIFFERENTIAL EQUTIONS Nottios p q r s t Formtio PDE b Elimitig rbitrr fuctios Suppose we re give f(u,v) = The it c be writte s u = g(v) or v = g(u) LAGRANGE S LINEAR EQUATION (Method of Multipliers) Geerl form Pp + Qq = R Subsidir Equtio d d d P Q R Pge7 Deprtmet of Mthemtics FMCET MADURAI

8 d d d m P Q R P mq R Where (, m,) re the Lgrgi Multipliers Choose, m, such tht P + mq + R = The d + m d + d = O Itegrtio we get solutio u = Similrl, We c fid other solutio v = for other multiplier The solutio is (u, v) = TYPE (Clirut s form) Geerl form Z = p + q + f(p,q) () Complete itegrl Put p = & q = b i (), We get () Which is the Complete itegrl Sigulr Itegrl Diff () Prtill w.r.t We get (3) Diff () Prtill w.r.t b We get (4) Usig (3) & (4) Fid & b d sub i () we get Sigulr Itegrl REDUCIBLE FORM F( m p, q) = () (or) F( m p, q, )= () If m & the X = -m & Y = - m p = P(-m) & q = Q(-) Usig the bove i ()we get F(P,Q) = (or) F(P,Q,) = F( k p, k q)= If k the Z = k+ k Q q k Usig the bove i () We get F(P,Q) = Pge8 Deprtmet of Mthemtics FMCET MADURAI

9 If m= & = the X= log & Y= log p = P & q = Q Usig the bove i () we get F(P,Q) (or) F(P, Q, ) = If k =- the Z = log p P & q Q Usig the bove i () we get F(P,Q) = Pge9 Deprtmet of Mthemtics FMCET MADURAI

10 STANDARD TYPES TYPE TYPE 3() TYPE 3(b) TYPE 3(c) TYPE 4 Geerl form F(p,q) = () Complete Itegrl Put p = d q = b i () Fid b i terms of The sub b i = + b + c we get () which is the Complete Itegrl Sigulr Itegrl Diff () prtill w.r.t cwe get, = (bsurdthere is o Sigulr Itegrl Geerl form F(,p,q) = () Complete Itegrl Put q = i () The, fid p d sub i d = p d + q d Itegrtig, We get () which is the Complete itegrl Sigulr Itegrl Diff () prtill w.r.t cwe get, = (bsurdthere is o Sigulr Itegrl Geerl form F(,p,q) = () Complete Itegrl Put p = i () The, fid q d sub i d = p d + q d Itegrtig, We get () which is the Complete itegrl Sigulr Itegrl Diff () prtill w.r.t cwe get, = (bsurdthere is o Sigulr Itegrl Geerl form F(,p,q) = () Complete Itegrl Put q = p i () The, fid p d sub i d = p d + q d Itegrtig, We get () which is the Complete itegrl Sigulr Itegrl Diff () prtill w.r.t cwe get, = (bsurdthere is o Sigulr Itegrl Geerl form F(,,p,q) = () Complete Itegrl () C be writte s, f(,p) =f(,q) = The, fid p d q sub i d = p d + qd Itegrtig, We get () which is the Complete itegrl Sigulr Itegrl Diff () prtill w.r.t cwe get, = (bsurd There is o Sigulr Itegrl Geerl Itegrl Put c = () i ()We get (3)Diff (3) prtill w.r.t We get (4)Elimitig from (3) d (4) we get Geerl Itegrl Geerl Itegrl Put c = () i ()We get (3)Diff (3) prtill w.r.t We get (4)Elimitig rom (3) d (4) we get Geerl Itegrl Geerl Itegrl Put c = () i ()We get (3) Diff (3) prtill w.r.t We get (4) Elimitig from (3) d (4) we get Geerl Itegrl Geerl Itegrl Put c = () i ()We get (3)Diff (3) prtill w.r.t We get (4)Elimitig from (3) d (4) we get Geerl Itegrl Geerl Itegrl Put c = () i ()We get (3) Diff (3) prtill w.r.t We get (4)Elimitig from (3) d (4) we get Geerl Itegrl Pge Deprtmet of Mthemtics FMCET MADURAI

11 HOMOGENEOUS LINEAR EQUATION Geerl form 3 3 ( ) (, ) D bd D cdd dd f () To Fid Complemetr Fuctio Auilir Equtio Put D = m & D = i () Solvig we get the roots m, m, m 3 Cse () If the roots re distict the C.F. = ( m ) ( m) 3( m3 ) Cse () If the roots re sme the C.F. = ( m) ( m) ( m ) 3 Cse (3) If the two roots re sme d oe is distict, the C.F = ( m) ( m) 3( m 3 ) Fuctio F(,) = e +b F(,)= si(+b)(or) Cos (+b) F(,) = r s PI = (, ) F( D, D ) F Put D = & D = b Put D ( ), DD ( b)& D ( b ) PI= F( D, D ) r s Epd d opertig D & D o r s F(,) = e +b f(,) Put D = D+ & D = D +b Pge Deprtmet of Mthemtics FMCET MADURAI

12 Prticulr Itegrl F(,)=e + cosh(+) F(,)= e e F(,)=e + sih(+) F(, ) = e e F(,)=si cos F(, ) si( ) si( ) F(,)= cos si F(, ) si( ) si( ) F(,)= cos cos F(, ) cos( ) cos( ) F(,)= si si F(, ) cos( ) cos( ) Note: D represets differetitio with respect to D represets differetitio with respect to D represets itegrtio with respect to D represets itegrtio with respect to Pge Deprtmet of Mthemtics FMCET MADURAI

13 PARTIAL DIFFRENTIAL EQUATIONS. Elimite the rbitrr costts & b from = ( + )( + b) Aswer: = ( + )( + b) Diff prtill w.r.to & here p & q p = ( + b) ; q = ( + ) ( + b) = p/ ; ( + ) = q/ = (p/)(q/) 4 = pq. Form the PDE b elimitig the rbitrr fuctio from = f() Aswer: = f() Diff prtill w.r.to & here p & q p = f ( ). q = f ( ). p/q = / p q = 3. Form the PDE b elimitig the costts d b from = + b Aswer: = + b Diff. w.r. t. d here p & q p = - ; q = b - Pge3 Deprtmet of Mthemtics FMCET MADURAI

14 p ; b q p q p q 4. Elimite the rbitrr fuctio f from f d form the PDE Aswer: f Diff. w.r. t. d here p & q p f p. q f q. p p. q q p pq q pq p q 5. Fid the complete itegrl of p + q =pq Aswer: Put p =, q = b p + q =pq +b=b b b = - b The complete itegrl is = + +c Pge4 Deprtmet of Mthemtics FMCET MADURAI

15 6. Fid the solutio of p q Aswer: = +b+c ----() is the required solutio give p q -----() put p=, q = b i () b b b ( ) ( ) c 7. Fid the Geerl solutio of p t + q t = t. Aswer: d d d t t t cot d cot d cot d tke cot d cot d cot d cot d log si log si log c log si log si log c c si si c si si si si, si si 8. Elimite the rbitrr fuctio f from f d form the PDE. Aswer: f p f q f p q p ; ( ) q 9. Fid the equtio of the ple whose cetre lie o the -is Aswer: Geerl form of the sphere equtio is Pge5 Deprtmet of Mthemtics FMCET MADURAI

16 c r () Where r is costt. From () +(-c) p= () +(-c) q = (3) From () d (3) p q Tht is p -q = which is required PDE.. Elimite the rbitrr costts Aswer: b b p ; q b p q p q b b d form the PDE.. Fid the sigulr itegrl of p q pq Aswer: The complete solutio is b b b ; b b ; ( ) ( ) (. ). Fid the geerl solutio of p+q= Aswer: The uilir equtio is d d d Pge6 Deprtmet of Mthemtics FMCET MADURAI

17 From d d o simplifig c. Itegrtig we get d d c log = log + log c Therefore, is geerl solutio. 3. Fid the geerl solutio of p +q = Aswer: d d d The uilir equtio is d d From d d Also Itegrtig we get c Itegrtig we get c Therefore, is geerl solutio. 4. Solve Aswer: Auilir equtio is D DD 3D m m 3 m 3 m m, m 3 The solutio is f f 3 Pge7 Deprtmet of Mthemtics FMCET MADURAI

18 5. Solve Aswer: Auilir equtio is D 4DD 3D e PI m m 4 3 m 3 m m, m 3 The CF is CF f f 3 D 4DD 3D e Put D, D Deomitor =. PI e D 4D e Z=CF + PI f f 3 e 6. Solve. D 3DD 4D e Aswer: Auilir equtio is m m C.F is = f ( + 4) + f ( - ) 3 4 m 4 m PI e e e D 3DD 4D Fid P.I D 4DD 4D e Aswer: PI e D 4DD 4D Put D, D PI D D e e e 6 Pge8 Deprtmet of Mthemtics FMCET MADURAI

19 8. Fid P.I Aswer: PI D D DD 6D D D D D D D 6D D Fid P.I Aswer: si PI Si Put D DD Si Si D DD, ()( ). Solve Aswer: Auilir equtio is 3 3 D 3DD D Z 3 m m 3 m m m m m m m, m The Solutio is CF f f f3 FOR PRACTICE:. Elimitig rbitrr costts b c. Solve si 3. Fid the complete the solutio of p. d.e 4. Form p.d.e elimitig rbitrr fuctio from p q pq 4, Pge9 Deprtmet of Mthemtics FMCET MADURAI

20 5. Fid the sigulr sol of p q p q. (i) Solve (ii) Solve p q p q. (i) Solve m l l m (ii) Solve 3 4 p 4 q 3 3. (i) Solve p q (ii) Solve p q 4. (i) Solve p q (ii) Solve p q 5. Solve 6. Solve 7. Solve 3 D DD D e 3 si(3 ) cos cos D DD 6D cos 8. Solve D DD 3D e 6 9. Solve D 6DD 5D e sih. Solve D 4DD 4D e. Solve 3 3 D D D DD D e cos( ). Solve (i) p q p q (ii) p q p q 3. Solve p q Pge Deprtmet of Mthemtics FMCET MADURAI

21 4. Solve 5. Solve (i) (ii) ( p q ) ( p q ) ( p q ) UNIT - FOURIER SERIES f b ( ) cos si (, ) ( -, ) Eve (or) Hlf rge Fourier co sie series Odd (or) Hlf rge Fourier sie series Neither eve or odd Pge Deprtmet of Mthemtics FMCET MADURAI

22 f ( ) d f ( ) d f ( ) d f ( )cos d f ( )cos d f ( )cos d b f ( )sid b = b f ( )sid b f ( )sid f b ( ) cos si (, ) ( -, ) Eve (or) Hlf rge Fourier cosie series Odd (or) Hlf rge Fourier sie series Neither eve or odd f ( ) d f ( ) d f ( ) d f ( )cos d f ( )cos d f ( )cos d b f ( )si d b = b f ( )si d b f ( )si d Eve d odd fuctio: Eve fuctio: f(-)=f() eg : cos,,,, si, cos re eve fuctios Odd fuctio: Pge Deprtmet of Mthemtics FMCET MADURAI

23 f(-)=-f() eg : si, 3,sih, t re odd fuctios For deductio I the itervl (, ) if = or = the f() = f( ) = f() f( ) I the itervl (, ) if = or = the f() = f( ) = f() f( ) I the itervl (-, ) if = - or = the f(- ) = f( ) = f( ) f( ) I the itervl (-, ) if = - or = the f(- ) = f( ) = f( ) f( ) HARMONIC ANALYSIS f()= + cos +b si + cos + b si for form cos, cos b si b, si f()= + cos +b si + cos + b si ( form) Pge3 Deprtmet of Mthemtics FMCET MADURAI

24 cos cos b si, b si. Defie R.M.S vlue. If let f() be fuctio defied i the itervl (, b), the the R.M.S vlue of f() is defied b b f ( ) d b. Stte Prsevl s Theorem. Let f() be periodic fuctio with period l defied i the itervl (c, c+l). l 4 c l o f ( ) d b c Where o, & b re Fourier costts 3. Defie periodic fuctio with emple. If fuctio f() stisfies the coditio tht f( + T) = f(), the we s f() is periodic fuctio with the period T. Emple:- i) Si, cos re periodic fuctio with period ii) t is re periodic fuctio with period 4. Stte Dirichlets coditio. (i) f() is sigle vlued periodic d well defied ecept possibl t Fiite umber of poits. (ii) f () hs t most fiite umber of fiite discotiuous d o ifiite Discotiuous. (iii) f () hs t most fiite umber of mim d miim. 5. Stte Euler s formul. Pge4 Deprtmet of Mthemtics FMCET MADURAI

25 Aswer: I ( c, c l) o f cos bsi where o c l c c l l f ( ) d f ( )cos d c c b f ( )si d c 6. Write Fourier costt formul for f() i the itervl (, ) Aswer: o f ( ) d f ( )cos d b f ( )si d 7. I the Fourier epsio of f() =,, i (-π, π ), fid the vlue of b Sice f(-)=f() the f() is eve fuctio. Hece b = 8. If f() = 3 i π < < π, fid the costt term of its Fourier series. Aswer: Give f() = 3 Hece f() is odd fuctio f(-) = (- ) 3 = - 3 = - f() The required costt term of the Fourier series = o = Pge5 Deprtmet of Mthemtics FMCET MADURAI

26 9. Wht re the costt term d the coefficiet of cos i the Fourier Epsio Aswer: Give f() = 3 f() = 3 i π < < π Hece f() is odd fuctio f(-) = - - (- ) 3 = - [ - 3 ] = - f() The required costt term of the Fourier series = =. Fid the vlue of for f() = ++ i (, ) Aswer: o f ( ) d 3 ( ) d (i)fid b i the epsio of s Fourier series i (, ) (ii)fid b i the epsio of si Fourier series i (, ) Aswer: (i) Give f() = f(-) = = f() Hece f() is eve fuctio I the Fourier series b = (ii) Give f() = si f(-) = (-)si(-) = si = f() Hece f() is eve fuctio I the Fourier series b = Pge6 Deprtmet of Mthemtics FMCET MADURAI

27 . Obti the sie series for f l / l l / l Give f l / l l / l Aswer: Give f l / l l / l Fourier sie series is l b f ( )si d l l l l f b si l si d ( l )si d l l l l l cos si cos si l l () l l l( l ) l l ( ) l l l l l cos l si l cos l l si l l si 4lsi l Fourier series is f 4l si si l 3. If f() is odd fuctio i ll,. Wht re the vlue of & Aswer: If f() is odd fuctio, o =, = Pge7 Deprtmet of Mthemtics FMCET MADURAI

28 4. I the Epsio f() = s Fourier series i (-. ) fid the vlue of Aswer: Give f() = Hece f() is eve fuctio f(-) = - = = f() o d 5. Fid hlf rge cosie series of f() =, i Aswer: o d cos si si d () Fourier series is cos o f cos cos 6. Fid the RMS vlue of f() =, Aswer: Give f() = Pge8 Deprtmet of Mthemtics FMCET MADURAI

29 R.M.S vlue l f ( ) d d l Fid the hlf rge sie series of f ( ) i (, ) Aswer: b f ( )si d cos si si d () ( ) ( ) Hlf rge Fourier sie series is ( ) f si 8. Fid the vlue of i the cosie series of f ( ) i (, 5) Aswer: o d Defie odd d eve fuctio with emple. Aswer: (i) If f ( ) f ( ) the the fuctio is eve fuctio. eg : cos,,, si, cos re eve fuctios (ii) If f ( ) f ( ) the the fuctio is odd fuctio. eg : si, 3,sih, t re odd fuctios Pge9 Deprtmet of Mthemtics FMCET MADURAI

30 . Write the first two hrmoic. Aswer: The first two hrmoics re o f cos b si cos b si FOURIER SERIES. Epd f( ) (, ) (, ) s Fourier series d hece deduce tht Fid the Fourier series for f() = i (-. ) d lso prove tht (i) (ii) (i) Epd f() = cos s Fourier series i (-. ). (ii) Fid cosie series for f() = i (, Show tht (i) Epd f() = si s Fourier series i (, ) ) use Prsevls idetit to (ii) Epd f() = s Fourier series i (-. ) d deduce to 5. If f( ) , (,) si, (, ) Fid the Fourier series d hece deduce tht (i) Fid the Fourier series up to secod hrmoic X Y (ii)fid the Fourier series up to third hrmoic Deprtmet of Mthemtics FMCET MADURAI Pge3

31 X π/3 π/3 π 4π/3 5π/3 π F() (i) Fid the Fourier epsio of f ( ) ( ) i (, ) d Hece deduce tht (ii). Fid Fourier series to represet i the rge (,3) f ( ) with period 3 (ii) Fid the Fourier series of f e i (, ). (ii) Fid the Fourier series for f i (, ) i (, ) d hece show tht (i) Fid the the hlf rge sie series for f i the itervl (, ) d deduce tht (ii) Obti the hlf rge cosie series for f i (,) d lso deduce tht (i) Fid the Fourier series for f() = i (-. ) d lso prove tht (use P.I) (ii) Fid the Fourier series for f() = i (-. ) d lso prove tht (use P.I) Pge3 Deprtmet of Mthemtics FMCET MADURAI

32 (i)obti the sie series for f l c, l c l, l (ii). Fid the Fourier series for the fuctio f k, l k l l, l.(i).fid the Fourier series for the fuctio f i (, ) d lso deduce tht (ii) Fid the Fourier epsio of f() =,, i (-π, π ), d lso deduce tht Pge3 Deprtmet of Mthemtics FMCET MADURAI

33 UNIT - 3 APPLICATIONS OF P.D.E S. N O VELOCITY MODEL STEP- Oe Dimesiol wve equtio is t ONE DIMENSIONAL WAVE EQUATION INITIAL POSITION MODEL STEP- Oe Dimesiol wve equtio is t STEP- Boudr coditios. (,t) = for t. (, t) = for t 3. (,) = for < < 4. t = f() for < < t STEP- Boudr coditios. (,t) = for t. (, t) = for t 3. t = for < < t 4. (,) = f() for < < Pge33 Deprtmet of Mthemtics FMCET MADURAI

34 3 STEP-3 The possible solutios re (,t) = (A e + B e - ) (C e t + D e - t ) (,t) = (A cos + B si )( C cos t + D si t) (,t) = (A + B) ( Ct + D) STEP-3 The possible solutios re (,t) = (A e + B e - ) (C e t + D e - t ) (,t) = (A cos + B si )( C cos t + D si t) (,t) = (A + B) ( Ct + D) 4 STEP-4 The suitble solutio for the give boudr coditio is (,t) = (Acos +B si )(Ccos t+d si t) () 5 STEP-5 Usig Boudr coditio (,t) = The () becomes, (,t) = (A cos +B si ) ( C cos t + Dsi t) = (A) ( C cos t + D si t)= A = Usig A = i () (,t) = ( B si ) ( C cos t + D si t) (3) STEP-4 The suitble solutio for the give boudr coditio is (,t) = (Acos +B si )(Ccos t+d si t) () STEP-5 Usig Boudr coditio (,t) = The () becomes, (,t) = (A cos +B si ) ( C cos t + D si t) = (A) ( C cos t + D si t)= A = Usig A = i () (,t) = ( B si ) ( C cos t + D si t) (3) 6 STEP-6 Usig Boudr coditio (,t) = The (3) becomes, (,t) = (B si ) ( C cos t + D si t)= (B si ) ( C cos t + D si t)= STEP-6 Usig Boudr coditio (,t) = The (3) becomes, (,t) = (B si ) ( C cos t + D si t)= (B si ) ( C cos t + D si t)= = The (3) becomes, t t (, t) Bsi( ) C cos( ) Dsi( ) (4) = The (3) becomes, t t (, t) Bsi( ) C cos( ) Dsi( ) (4) 7 STEP-7 Usig Boudr coditio 3 (,) = The (4) becomes, STEP-7 Usig Boudr coditio 3 Pge34 Deprtmet of Mthemtics FMCET MADURAI

35 (, t) Bsi( ) C cos Dsi Bsi( ) C C = The (4) becomes, t (, t) Bsi( ) Dsi( ) The most geerl solutio is t (, t) B si( )si( ) (5) = t t = The (4) becomes, Differetitig (5) prtill w.r.to t d put t = t t B si( ) C si( ) D cos( t t Bsi( ) D D = The (4) becomes, t (, t) Bsi( ) C cos( ) The most geerl solutio is t (, t) B si( )cos( ) (5) 8 STEP-8 Differetitig (5) prtill w.r.to t t B si( )cos( ) t Usig Boudr coditio (4), = f() t t f ( ) B si( ) This is the Hlf Rge Fourier Sie Series. B f ( )si( ) B f ( )si( ) d 9 STEP-9 The required solutio is t (, t) B si( )si( ) Where B f ( )si( ) d STEP-8 Usig Boudr coditio (4), (,) = f() (,) B si( )cos() f ( ) B si( ) This is the Hlf Rge Fourier Sie Series. B f ( )si( ) STEP-9 The required solutio is t (, t) B si( )si( ) Where B f ( )si( ) d Pge35 Deprtmet of Mthemtics FMCET MADURAI

36 ONE DIMENSIONAL HEAT EQUATION The oe dimesiol het equtio is u u t Boudr coditios.u(,t) = for t.u(,t) = fort 3.u(,t) = f() for << TWO DIMENSIONAL HEAT FLOW EQUATION The two Dimesiol equtio is u u Boudr coditios.u(,) = for <<.u(,) = for << 3.u(, ) = for << 4.u(,) = f() for << 3 The possible solutios re t u(, t) ( Ae Be ) Ce u(, t) ( Acos Bsi ) Ce u(, t) ( A B) C t The possible solutios re u(, ) ( Ae Be )( C cos Dsi ) u(, ) ( Acos Bsi )( Ce De ) u(, ) ( A B)( C D) 4, The most suitble solutio is t u(, t) ( Acos Bsi ) Ce () 5 Usig boudr coditio u(,t) = t u(, t) ( Acos Bsi) Ce = t ( A) Ce = A = The () becomes t u(, t) ( Bsi ) Ce (3) The most suitble solutio is (, ) ( cos si )( u A B Ce De ) () Usig boudr coditio u(,) = u(, ) ( Acos Bsi )( Ce De ) u(, ) ( A)( Ce Be ) A = The () becomes (, ) ( si )( u B Ce De ) (3) 6 Usig boudr coditio u(l,t) = t u(, t) ( Bsi ) Ce = t ( Bsi ) Ce The (3) becomes t u(, t) Bsi( ) Ce The most geerl solutio is u(, t) B si( ) e t (4) Usig boudr coditio u(l,t) = u(, ) ( Bsi )( Ce De ) ( si )( B Ce De ) The (3) becomes u(, ) ( Bsi )( Ce De ) (4) Pge36 Deprtmet of Mthemtics FMCET MADURAI

37 7 Usig Boudr coditio 3 u(,) = f() u(,) B si( ) f ( ) B si( ) This the Hlf rge Fourier sie series B f ( )si( ) d 8 The required solutio is u(, t) B si( ) e Where t B f ( )si( ) d Usig Boudr coditio 3 u(, ) = u(, ) ( Bsi )( Ce De ) ( Bsi )( C D ) C= the (3) becomes u(, ) ( Bsi )( De ) The most geerl solutio is u(, ) B si( ) e Usig Boudr coditio 4 (,) = f() u(,) B si( ) e f ( ) B si( ) (5 This the Hlf rge Fourier sie series B f ( )si( ) d The required solutio is u(, ) B si( ) e Where B f ( )si( ) d QUESTION WITH ANSWER. Clssif the Prtil Differetil Equtio i) Aswer: u u Pge37 Deprtmet of Mthemtics FMCET MADURAI

38 u u here A=,B=,&C=- B - 4AC=-4()(-)=4> The Prtil Differetil Equtio is hperbolic. Clssif the Prtil Differetil Equtio Aswer: u u u here A=,B=,&C= u u u B -4AC=-4()()=> The Prtil Differetil Equtio is hperbolic 3. Clssif the followig secod order Prtil Differetil equtio u u u u Aswer: u u u u B -4AC=-4()()=-4< here A=,B=,&C= The Prtil Differetil Equtio is Elliptic 4. Clssif the followig secod order Prtil Differetil equtio 4 u u u u u Aswer: 4 u u u u u here A= 4,B =4, & C = B -4AC =6-4(4)() = The Prtil Differetil Equtio is Prbolic 5. Clssif the followig secod order Prtil Differetil equtio i) u u u u 3u Pge38 Deprtmet of Mthemtics FMCET MADURAI

39 ii) u u u u 7 Aswer: i) Prbolic ii) Hperbolic (If = ) iii)elliptic (If m be +ve or ve) 6. I the wve equtio Aswer: c t t c wht does c stds for? here T m T-Tesio d m- Mss 7. I oe dimesiol het equtio u t = α u wht does α stds for? Aswer:- u t u = k c is clled diffusivit of the substce Where k Therml coductivit - Desit c Specific het 8. Stte two lws which re ssumed to derive oe dimesiol het equtio Aswer: i) Het flows from higher to lower temp ii) The rte t which het flows cross re is proportiol to the re d to the temperture grdiet orml to the curve. This costt of proportiolit is kow s the coductivit of the mteril. It is kow s Fourier lw of het coductio Pge39 Deprtmet of Mthemtics FMCET MADURAI

40 9. A tightl stretched strig of legth is fsteed t both eds. The midpoit of the strig is displced to distce b d relesed from rest i this positio. Write the iitil coditios. Aswer: (i) (, t) = (ii) (,t) = (iii) t t (iv) (, ) = b b ( ). Wht re the possible solutios of oe dimesiol Wve equtio? The possible solutios re Aswer: (,t) = (A e + B e - ) (C e t + D e - t ) (,t) = (A cos + B si )( C cos t + D si t) (,t) = (A + B) ( Ct + D). Wht re the possible solutios of oe dimesiol hed flow equtio? Aswer: The possible solutios re u(, t) ( Ae Be ) Ce t u(, t) ( Acos Bsi ) Ce u(, t) ( A B) C t. Stte Fourier lw of het coductio Aswer: Q u ka (the rte t which het flows cross re A t distce from oe ed of br is proportiol to temperture grdiet) Pge4 Deprtmet of Mthemtics FMCET MADURAI

41 Q=Qutit of het flowig k Therml coductivit A=re of cross sectio ; u =Temperture grdiet 3. Wht re the possible solutios of two dimesiol hed flow equtio? Aswer: The possible solutios re u(, ) ( Ae Be )( C cos Dsi ) u(, ) ( Acos Bsi )( Ce De ) u(, ) ( A B)( C D) 4. The sted stte temperture distributio is cosidered i squre plte with sides =, =, = d =. The edge = is kept t costt temperture T d the three edges re isulted. The sme stte is cotiued subsequetl. Epress the problem mthemticll. Aswer: U(,) =, U(,) =,U(,) =, U(,) = T 5. A isulted rod of legth 6cm hs its eds A d B mitied C d 8 C respectivel. Fid the sted stte solutio of the rod Aswer: Here = C & b=8 C I Sted stte coditio The Temperture u(, t) 8 6 b l u(, t) 6. Write the D Alembert s solutio of the oe dimesiol wve equtio? Aswer: Pge4 Deprtmet of Mthemtics FMCET MADURAI

42 t t t v( ) d here f g v f g 7. Wht re the boudr coditios of oe dimesiol Wve equtio? Aswer: Boudr coditios t. (,t) = for t. (, t) = for t 3. (,) = for < < 4. t = f() for < < t 8. Wht re the boudr coditios of oe dimesiol het equtio? Aswer: Boudr coditios.u(,t) = for t.u(,t) = fort 3.u(,t) = f() for << 9. Wht re the boudr coditios of oe dimesiol het equtio? Aswer: Boudr coditios.u(,) =.u(,) = 3.u(, ) = 4.u(,) = f() for << for << for << for << Pge4 Deprtmet of Mthemtics FMCET MADURAI

43 .T he eds A d B hs 3cm log hve their tempertures 3c d 8c util sted stte previls. If the temperture A is rised to4c d Reduced to 6C, fid the trsiet stte temperture Aswer: Here =3 C & b=8 C I Sted stte coditio The Temperture u(, t) Here =4 C & b=6 C b l u t PART-B QUESTION BANK APPLICATIONS OF PDE. A tightl stretched strig with fied ed poits = d = l is iitill t rest i its equilibrium positio. If it is set vibrtig givig ech poit velocit 3 (l-). Fid the displcemet.. A strig is stretched d fsteed to two poits d prt. Motio is strted b displcig the strig ito the form = K(l- ) from which it is relesed t time t =. Fid the displcemet t poit of the strig. 3. A tut strig of legth l is fsteed t both eds. The midpoit of strig is tke to height b d the relesed from rest i tht positio. Fid the displcemet of the strig. 4. A tightl stretched strig with fied ed poits = d = l is iitill t rest i its 3 positio give b (, ) = si. If it is relesed from rest fid the displcemet. l 5. A strig is stretched betwee two fied poits t distce l prt d poits of the c < < l strig re give iitil velocities where V Fid the c ( l ) < < l displcemet. 6. Derive ll possible solutio of oe dimesiol wve equtio. Derive ll possible solutio of oe dimesiol het equtio. Derive ll possible solutio of two dimesiol het equtios. 7. A rod 3 cm log hs its ed A d B kept t o C d 8 o C, respectivel util sted stte coditio previls. The temperture t ech ed is the reduced to o C d kept so. Fid the resultig temperture u(, t) tkig =. Pge43 Deprtmet of Mthemtics FMCET MADURAI

44 8. A br cm log, with isulted sides hs its ed A & B kept t o C d 4 o C respectivel util the sted stte coditio previls. The temperture t A is suddel rised to 5 o C d B is lowered to o C. Fid the subsequet temperture fuctio u(, t). 9. A rectgulr plte with isulted surfce is 8 cm wide so log compred to its width tht it m be cosidered s ifiite plte. If the temperture log short edge = is u (,) = si < <. While two edges = d = 8 s well s the other short 8 edges re kept t o C. Fid the sted stte temperture.. A rectgulr plte with isulted surfce is cm wide so log compred to its width tht it m be cosidered s ifiite plte. If the temperture log short edge = is give 5 b u d ll other three edges re kept t o C. Fid the sted ( ) 5 stte temperture t poit of the plte. Pge44 Deprtmet of Mthemtics FMCET MADURAI

45 Uit - 4 FOURIER TRANSFORMS FORMULAE. Fourier Trsform of f() is. The iversio formul F[ f ( )] f()e -is f ( ) F( s)e ds 3. Fourier cosie Trsform F c [f()] = F c (s) = 4. Iversio formul f() = - F ( )cos c s sds 5. Fourier sie Trsform (FST) F s [f()] = F s (s) = 6. Iversio formul f() = 7. Prsevl s Idetit Fs ( s)si sds f ( ) d F( s) ds - is d f ( )cos sd f ( )si sd 8. Gmm fuctio e d, & 9. e cos bd b e sibd b b. si d Pge45 Deprtmet of Mthemtics FMCET MADURAI

46 . e d & e d e e e e 3. cos & si i i i i ORKING RULE TO FIND THE FOURIER TRANSFORM Step: Write the FT formul. Step: Substitute give f() with their limits. Step3: Epd is e s cos s + isi s d use Eve & odd propert Step4: Itegrte b usig Beroulli s formul the we get F(s) WORKING RULE TO FIND THE INVERSE FOURIER TRANSFORM Step: Write the Iverse FT formul Step: Sub f() & F(s) with limit, i the formul Step3: Epd Step4: Simplif we get result is e s cos s -isi s d equte rel prt WORKING RULE FOR PARSEVAL S IDENTITY If F(s) is the Fourier trsform of f() the f ( ) d F( s) ds is kow s Prsevl s idetit. Step: Sub f() & F(s) With their limits i the bove formul Step: Simplif we get result WORKING RULE TO FIND FCT Step: Write the FCT formul & Sub f() with its limit i the formul Step: Simplif, we get F ( S ) C WORKING RULE TO FIND INVERSE FCT Pge46 Deprtmet of Mthemtics FMCET MADURAI

47 Step: Write the iverse FCT formul & Sub F ( ) Step: Simplif, we get f() WORKING RULE TO FIND FST C S with its limit i the formul Step: Write the FST formul & Sub f() with its limit i the formul Step: Simplif, we get F ( S ) WORKING RULE TO FIND INVERSE FCT S Step: Write the iverse FST formul & Sub Fs ( S) with limit i the formul Step: Simplif, we get f() WORKING RULE FOR f() = e Step: First we follow the bove FCT & FST workig rule d the we get this result F c (e - ) = s B Iversio formul, s F s (e - ) = s B Iversio formul, cos s ds s e s s si sds e TYPE-I : If problems of the form i) ii), the use Iversio formul TYPE-II: If problems of the form i) d d ii), the use Prsevl s Idetit TYPE-III d b, the use f ( ) g( ) d F f ( ) F g( ) d C C Pge47 Deprtmet of Mthemtics FMCET MADURAI

48 UNIT - 4 FOURIER TRANSFORM. Stte Fourier Itegrl Theorem. Aswer: If f( ) is piece wise cotiuousl differetible d bsolutel o, the, i( t) s f ( ) f t e dt ds. Pge48 Deprtmet of Mthemtics FMCET MADURAI

49 . SttedproveModultio F f cos F s F s Proof: is F f cos f cos e d theorem. i i e e is f e d i( s ) i( s ) f e d f e d F s F s F f cos F s F s 3. Stte Prsevl s Idetit. Aswer: If F s is Fourier trsform of f, the F s ds f d 4. Stte Covolutio theorem. Aswer: The Fourier trsform of Covolutio of f d g is the product of their Fourier trsforms. F f g F s G s 5. Stte d prove Chge of scle of propert. Pge49 Deprtmet of Mthemtics FMCET MADURAI

50 Aswer: If F s F f, the is F f f e d F f F s i s t dt f t e ; where t F f F s d 6. Prove tht if F[f()] = F(s) the F f ( ) ( i) F( s) ds Aswer: is F s f e d Diff w.r.t s times d is F s f i e d ds is f ( i) ( ) e d d is ( ) F s f e d () i ds d is ( i) F s ( ) f e d ds d ds F f i F s 7. Solve for f() from the itegrl equtio Aswer: f ( )cos sd e s Pge5 Deprtmet of Mthemtics FMCET MADURAI

51 f ( )cos sd e s F f f cos s d c F f e c s f ( ) F f coss ds c e s cos s ds e cosb d b s e cos s ds, b 8. Fid the comple Fourier Trsform of f( ) Aswer: is F f f e d ; Pge5 Deprtmet of Mthemtics FMCET MADURAI

52 is F f e d (coss isi s) d si s (cos s) d s si s s [Use eve d odd propert secod term become ero] 9. Fid the comple Fourier Trsform of f( ) Aswer: is F f f e d is e d ; (coss i si s) d i cos s si s ( ( isi s) d () s s i s cos s si s s [Use eve d odd propert first term become ero] Pge5 Deprtmet of Mthemtics FMCET MADURAI

53 . Write Fourier Trsform pir. Aswer: If f( ) is defied i,, the its Fourier trsform is defied s is F s f e d If F s is Fourier trsform of f, the t ever poit of Cotiuit of f, we hve is f F s e ds.. Fid the Fourier cosie Trsform of f() = e - Aswer: F f f cos s d c F e e cos s d F c c e s e cos b d b. Fid the Fourier Trsform of Aswer: f( ) im e, b, otherwise Pge53 Deprtmet of Mthemtics FMCET MADURAI

54 is F f f e d b im is i m s e e d e d b i m s e i m s i m s b e i m s b e i m s 3. Fid the Fourier sie Trsform of. Aswer: F f f si s d F s s si s d 4. Fid the Fourier sie trsform of Aswer: e F f f si s d s F e e si s d F s s e s s e si b d b b 5. Fid the Fourier cosie trsform of Aswer: e e Pge54 Deprtmet of Mthemtics FMCET MADURAI

55 F f f cos s d c Fc e e e e coss d e cos s d e cos s d s 4 s s 4 s 6. Fid the Fourier sie trsform of f( ) Aswer: F f f si s d s, F f f si s d f si s d s cos s si s d s cos s s Pge55 Deprtmet of Mthemtics FMCET MADURAI

56 7. Obti the Fourier sie trsform of, o f ( ),., Aswer: F f f si s d s si s d si s d F f s cos s si s cos s si s s s s s cos s si s si s cos s si s s s s s s si s si s s 8. Defie self reciprocl d give emple. If the trsform of f is equl to f s, the the fuctio f is clled self reciprocl. e is self reciprocl uder Fourier trsform. Pge56 Deprtmet of Mthemtics FMCET MADURAI

57 9. Fid the Fourier cosie Trsform of f( ) Aswer: F f f coss d coss d c si s cos s cos s si s s s s s s s si s cos s s. Fid the Fourier sie trsform of. Aswer: L et f e F s e s s Usig Iverse formul for Fourier sie trsforms e s s si s ds s ( ie) si s ds e, s Pge57 Deprtmet of Mthemtics FMCET MADURAI

58 Chge d s, we get si s d e s F s si s d s e e s. (i)fid the Fourier Trsform of FOURIER TRANSFORM PART-B f( ) if if d hece cos si 3 deduce tht (i) cos 3 d 6 (ii) si cos 3 d 5 (ii). Fid the Fourier Trsform of f( ). hece si cos deduce tht 3 d 4. Fid the Fourier Trsform of f( ) if if d hece evlute i) si d ii) si d 4. Fid Fourier Trsform of f( ) if if d hece evlute i) si d ii) si 4 d Pge58 Deprtmet of Mthemtics FMCET MADURAI

59 5. Evlute i) d d ii) 6 i). Evlute () d b d ii). Evlute () 4 (b) (b) d b t dt t 4 t 9 7. (i)fid the Fourier sie trsform of f( ) (ii) Fid the Fourier cosie trsform of f( ) si ; whe o ; whe cos ; whe o ; whe 8. (i) Show tht Fourier trsform e is e s (ii)obti Fourier cosie Trsform of e 9. (i) Solve for f() from the itegrl equtio d hece fid Fourier sie Trsform e f ( )cos d e, t (ii) Solve for f() from the itegrl equtio f ( )si t d, t. (i) Fid Fourier sie Trsform of e (ii) Fid Fourier cosie d sie Trsform of e cos si tht ( i) d e ( ii) d e , t, > d hece deduce tht 4 si d, > d hece deduce.(i)fid F e & F e S c (ii) Fid F S e & F c e (iii) Fid the Fourier cosie trsform of f ( ) e cos Pge59 Deprtmet of Mthemtics FMCET MADURAI

60 Z - TRANSFORMS Defiitio of Z Trsform Let {f()} be sequece defied for Z Trsform is defied s =,, d f() = for < the its Z f ( ) F f ( ) (Two sided trsform) Z f ( ) F f ( ) (Oe sided trsform) Uit smple d Uit step sequece The uit smple sequece is defied s follows ( ) for for The uit step sequece is defied s follows Properties u ( ) for for Pge6 Deprtmet of Mthemtics FMCET MADURAI

61 . Z Trsform is lier (i) Z { f() + b g()} = Z{f()} + b Z{g()}. First Shiftig Theorem (i) If Z {f(t)} = F(), the t Z e f t F e T (ii) If Z {f()} = F(), the Z f F 3. Secod Shiftig Theorem If Z[f()]= F() the (i)z[f( +)] = [ F() f()] (ii)z[f( +)] = [ F() f()-f() ] (iii)z[f( +k)] = k [ F() f()-f() - f() ( k ) - f(k-) ] (iv) Z[f( -k)] = k F() 4. Iitil Vlue Theorem If Z[f()] = F() the f() = lim F ( ) 5. Fil Vlue Theorem If Z[f()] = F() the lim f ( ) lim( ) F( ) PARTIAL FRACTION METHODS Pge6 Deprtmet of Mthemtics FMCET MADURAI

62 Model:I A B b b Model:II b A B C b ( b) Model:III A B C b b Covolutio of Two Sequeces Covolutio of Two Sequeces {f()} d {g()} is defied s { f ( )* g( )} f ( K) g( K ) K Covolutio Theorem If Z[f()] = F() d Z[g()] = G() the Z{f()*g()} = F().G() WORKING RULE TO FIND INVERSE Z-TRANSFORM USING CONVOLUTION THEOREM Step: Split give fuctio s two terms Step: Tke Step: 3 Appl both terms formul Step: 4 Simplifig we get swer Note:... Pge6 Deprtmet of Mthemtics FMCET MADURAI

63 ... ( ) Solutio of differece equtios Formul i) Z[()] = F() ii) Z[( +)] = [ F() ()] iii) Z[( +)] = [ F() ()- () ] iv) Z[( +3)] = 3 [ F() ()- () + () ] WORKING RULE TO SOLVE DIFFERENCE EQUATION: Step: Tke trsform o both sides Step: Appl formul d vlues of () d (). Step: 3 Simplif d we get F(Z) Step:4 Fid () b usig iverse method Z - Trsform Tble No. f() Z[f()].. 3. ( ) 4. ( ) 3 Pge63 Deprtmet of Mthemtics FMCET MADURAI

64 e.!. Cos. si log ( ) log ( ) log ( ) ( e ) e ( cos ) cos si cos 3. cos si 4. f(t) ( ) Z(f(t) t T ( ). t T ( ) 3 ( ) 3 e t 4. Si t T ( e ) si T cos T Pge64 Deprtmet of Mthemtics FMCET MADURAI

65 5. cos t ( cos T) cos T TWO MARKS QUESTIONS WITH ANSWER. Defie Z trsform Aswer: Let {f()} be sequece defied for =,, d f() = for < the its Z Trsform is defied s Z f ( ) F f ( ) (Two sided trsform) Z f ( ) F f ( ) (Oe sided trsform) Fid the Z Trsform of Aswer: Z f f Z ()... Z. Fid the Z Trsform of Aswer: Pge65 Deprtmet of Mthemtics FMCET MADURAI

66 Z f f Z Fid the Z Trsform of. Aswer: d Z Z Z d, b the propert, d d ( ) ( ) Stte Iitil & Fil vlue theorem o Z Trsform Iitil Vlue Theorem Fil Vlue Theorem If Z [f ()] = F () the f () = lim F ( ) If Z [f ()] = F () the lim f ( ) lim( ) F( ) 6. Stte covolutio theorem of Z- Trsform. Deprtmet of Mthemtics FMCET MADURAI Pge66

67 Aswer: Z[f()] = F() d Z[g()] = G() the Z{f()*g()} = F() G() 7. Fid Z Trsform of Aswer: Z f f Z Fid Z Trsform of cos d si Aswer: We kow tht Z Z f f cos cos cos Z cos cos cos Pge67 Deprtmet of Mthemtics FMCET MADURAI

68 Z si Similrl si cos Z si si cos 9. Fid Z Trsform of Aswer: Z f f Z log log log. Fid Z Trsform of Aswer:! Pge68 Deprtmet of Mthemtics FMCET MADURAI

69 Z f f Z!! 3...!!! 3! e e. Fid Z Trsform of Aswer: Z f f Z ( ) log log. Fid Z Trsform of Pge69 Deprtmet of Mthemtics FMCET MADURAI

70 Aswer: Z f f Z Stte d prove First shiftig theorem t Sttemet: If Z f t F, the Z e f () t F e Proof: Z e f ( t) e f ( T ) t T As f(t) is fuctio defied for discrete vlues of t, where t = T, the the Z-trsform is Z f ( t) f ( T ) F( ) ). t T T Z e f ( t) f ( T ) e F( e ) T 4. Defie uit impulse fuctio d uit step fuctio. The uit smple sequece is defied s follows: for ( ) for The uit step sequece is defied s follows: Pge7 Deprtmet of Mthemtics FMCET MADURAI

71 u ( ) for for 5. Fid Z Trsform of Aswer: t Z e t T T T Z e e e e T e [Usig First shiftig theorem] 6. Fid Z Trsform of Aswer: Z te t Z te t Z t e T T e T e Te T T [Usig First shiftig theorem] t 7. Fid Z Trsform of Z e cost Aswer: t Z e cos t Z cos t e T e T cos T T e e cost cos T cost e T e [Usig First shiftig theorem] Pge7 Deprtmet of Mthemtics FMCET MADURAI

72 8. Fid Z Trsform of Aswer: t T Z e Let f (t) = e t, b secod siftig theorem ( t T) Z e Z f t T F f ( ) ( ) () e e T e T T 9. Fid Z Trsform of Z si t T Aswer: Let f (t) = sit, b secod siftig theorem Z si( t T) Z f ( t T) F( ) f () si t si t t t cos cos. Fid Z trsform of Aswer: Z f f Z Z Z Pge7 Deprtmet of Mthemtics FMCET MADURAI

73 QUESTION BANK Z-TRANSFORMS. (i)fid Z (ii) Fid Z ( )(4 ) 8 ( )( b) & Z & Z ( )(4 ) 8 b covolutio theorem. b covolutio theorem ( )( 3). (i) Fid Z ( ) & Z ( ) b covolutio theorem 3. (i ) Stte d prove Iitil & Fil vlue theorem. (ii) Stte d prove Secod shiftig theorem (i) Fid the Z trsform of ( )( ) & 3 ( )( ) 4. (i) Fid Z ( 4) b residues. (ii) Fid the iverse Z trsform of 5. (i) Fid Z 6. (i)fid the Z trsform of (ii) Fid & Z f( )! 7 ( ) Hece fid b prtil frctios. Z d ( )! Z d lso fid the vlue of si( ) d cos( ).! 7. (i)solve 6 9 with & (ii) Solve 4 4 () =,() = 8. (i )Solve 3 4, give () 3& () (ii) Solve 3 3, 4, & 8, 9. (i)fid Z cos & Z si d lso fid Z cos & Z si Z. ( )! Pge73 Deprtmet of Mthemtics FMCET MADURAI