SECTION x2 x > 0, t > 0, (8.19a)

Transcription

1 SECTION Application of aplace Tranform to Partial Differential Equation In Section 8.2 and 8.3 we illutrated the effective ue of aplace tranform in olving ordinary differential equation. The tranform replace a differential equation in y(t) with an algebraic equation in it tranform ỹ(). It i then a matter of finding the invere tranform of ỹ() either by partial fraction and table (Section 8.1) or by reidue (Section 8.4). aplace tranform alo provide a potent technique for olving partial differential equation. When the tranform i applied to the variable t in a partial differential equation for a function y(x t) the reult i an ordinary differential equation for the tranform ỹ(x ). The ordinary differential equation i olved for ỹ(x ) and the function i inverted to yield y(x t). We illutrate thi procedure with five phyical example. The firt two example are on unbounded patial interval; invere tranform are found in table. The lat three example are on bounded patial interval; invere tranform are calculated with reidue. Problem on Unbounded Interval Example 8.18 A very long cylindrical rod i placed along the poitive x-axi with one end at x = (Figure 8.22). The rod i o long that any effect due to it right end may be neglected. It ide are covered with perfect inulation o that no heat can enter or ecape therethrough. At time t = the temperature of the rod i C throughout. Suddenly the left end of the rod ha it temperature raied to U and maintained at thi temperature thereafter. The initial boundary-value problem decribing temperature U(x t) at point in the rod i t = 2 U x2 x > t > (8.19a) U(t)=U t > (8.19b) U(x ) = x > (8.19c) where i a contant called the thermal diffuivity of the material in the rod. Ue aplace tranform on variable t to find U(x t). Inulation x= Inulation Figure 8.22 x Solution When we apply the aplace tranform to the partial differential equation and ue property 8.1a { 2 } Ũ(x ) U(x ) = U x 2. Since the integration with repect to t in the aplace tranform and the differentiation with repect to x are independent we interchange the order of operation on the right Ũ(x ) = 2 Ũ x 2 where we have alo ued initial condition 8.19c. Becaue only derivative with repect to x remain we replace the partial derivative with an ordinary derivative

2 434 SECTION 8.5 Ũ = d2 Ũ x >. (8.2a) dx2 When we tae aplace tranform of boundary condition 8.19b we obtain Ũ()= U (8.2b) a boundary condition to accompany differential equation 8.2a. A general olution of 8.2a i Ũ(x ) =Ae /x + Be /x. Becaue U(x t) mut remain bounded a x become infinite o alo mut Ũ(x ). We mut therefore et A = in which cae 8.2b require B = U /. Thu Ũ(x ) = U e /x. (8.21) The invere tranform of thi function can be found in table { U(x t) =U 1 e } ( ) /x x = U erfc 2 (8.22a) t where erfc (x) i the complementary error function Example 8.19 erfc (x) =1 erf (x) =1 2 x e u2 du = 2 e u2 du. π π Notice that for any x> and any t> temperature U(x t) i poitive. Thi indicate that the abrupt change in temperature at the end x = from C to U i felt intantaneouly at every point in the rod. We have hown a plot of U(x t) for variou fixed value of t in Figure h( x t) x t t t > t > t t t = = 3 Figure 8.23 t t = 2 (8.22b) A very long taut tring i upported from below o that it lie motionle on the poitive x-axi. At time t = the upport i removed and gravity i permitted to act on the tring. If the end x = i fixed at the origin the initial boundary-value problem decribing diplacement y(x t) of point in the tring i 2 y t 2 = c2 2 y g x2 x > t > (8.23a) y(t)= t > (8.23b) y(x ) = x > (8.23c) y t (x ) = x > (8.23d) x where g =9.81 and c> i a contant depending on the material and tenion of the tring. Initial condition 8.23d expree the fact that the initial velocity of the tring i zero. Ue aplace tranform to olve thi problem.

3 SECTION Solution When we apply aplace tranform to the partial differential equation and ue property 8.1b { 2 ỹ(x ) y(x ) y t (x ) = c 2 2 } y x 2 g. We now ue initial condition 8.23cd and interchange operation on the right 2 ỹ = c 2 d2 ỹ dx g 2 or d 2 ỹ dx 2 2 c 2 ỹ = g x >. (8.24a) c 2 Thi ordinary differential equation i ubject to the tranform of 8.23b ỹ()=. (8.24b) A general olution of differential equation 8.24a i ỹ(x ) =Ae x/c + Be x/c g 3. For thi function to remain bounded a x we mut et A = in which cae condition 8.24b implie that B = g/ 3. Thu ỹ(x ) = g 3 + g 3 e x/c. (8.25) Property 8.4b give y(x t) = gt2 2 + g ( t x ) 2 ( h t x ) (8.26) 2 c c where h(t x/c) i the Heaviide unit tep function. What thi ay i that a point x in the tring fall freely under gravity for <t<x/c after which it fall with contant velocity gx/c [ince for t > x/c y(x t) =(g/2)( 2xt/c + x 2 /c 2 )]. A picture of the tring at any given time t i hown in Figure It i parabolic for <x<ct and horizontal for x>ct. A t increae the parabolic portion lengthen and the horizontal ection drop. y gt - 2 ct 2 - y gx = x ct 2c ( -2 ) 2 y gt 2 = 2 x Problem on Bounded Interval Figure 8.24 The next three example are on bounded interval; we ue reidue to find invere tranform.

4 436 SECTION 8.5 Example 8.2 A cylindrical rod of length ha it end at x = and x = on the x-axi. It ide are inulated o that no heat can enter or ecape therethrough. At time t = the left end (x = ) ha temperature C and the right end (x = ) ha temperature C and the temperature rie linearly between the end. Suddenly at time t = the temperature of the right end i reduced to C and both end are held at temperature zero thereafter. The initial boundary-value problem decribing temperature U(x t) at point in the rod i t = 2 U x2 <x< t> (8.27a) U(t)= t > (8.27b) U( t) = t > (8.27c) U(x ) = x <x< (8.27d) again being the thermal diffuivity of the material in the rod. Ue aplace tranform to find U(x t). Solution or When we tae aplace tranform of 8.27a and ue property 8.1a Ũ(x ) x = d2 Ũ dx 2 d 2 Ũ dx 2 Ũ = x <x<. (8.28a) Thi ordinary differential equation i ubject to the tranform of 8.27bc Ũ()= Ũ( ) =. (8.28b) (8.28c) A general olution of differential equation 8.28a i Ũ(x ) =A coh x + B inh x + x. For Example 8.18 on an unbounded interval we ued exponential in the olution of 8.2a. On bounded interval hyperbolic function are preferable. Boundary condition 8.28bc require From thee =A =A coh + B inh +. ( Ũ(x ) = 1 x inh ) /x inh. (8.29) / Although / denote the principal quare root function Ũ(x ) in 8.29 i a olution of problem 8.28 for any branch of the root function. It remain now to find the invere tranform of Ũ(x ). We do thi by calculating reidue of e t Ũ(x ) at it ingularitie. To dicover the nature of the

5 SECTION ingularity at = we expand Ũ(x ) in a aurent erie around =. Provided we ue the ame branch for the root function in numerator and denominator of 8.29 we may write that { Ũ(x ) = 1 x [ /x +( } /x) 3 /3! + ] / +( /)3 /3! + = 1 [ x x + ] x3 /(6) /(6)+ = 1 [ x( 2 x 2 ] ) + 6 = x(2 x 2 ) 6 + term in 2. Since thi i the aurent erie of Ũ(x ) valid in ome annulu around = it follow that Ũ(x ) ha a removable ingularity at =. To invert Ũ(x ) with reidue the function cannot have a branch cut in the left half-plane Im <. Suppoe we chooe a branch of / in 8.29 with branch cut along the poitive real axi. We continue to denote thi branch by / notwithtanding the fact that it i no longer the principal branch. Singularitie of Ũ(x ) in the left half-plane then occur at the zero of inh /; that i when / = nπi or = n 2 π 2 / 2 n a poitive integer. Becaue the derivative of inh / doe not vanih at = n 2 π 2 / 2 thi function ha zero of order 1 at = n 2 π 2 / 2. It follow that Ũ(x ) ha imple pole at thee ingularitie. Reidue of e t Ũ(x ) at thee pole are [ Re e t Ũ(x ) n2 π 2 ] ( 2 = lim + n2 π 2 ) ( e t n 2 π 2 / 2 2 x inh ) /x inh / π 2 t/ 2 nπxi = e n2 inh n 2 π 2 /2 Hôpital rule yield [ Re e t Ũ(x ) n2 π 2 ] 2 = i3 2 n 2 π 2 e n π 2 t/ 2 in nπx = 2i2 π 2 t/ 2 n 2 π 2 in nπx e n2 = 2 nπ ( 1)n+1 e n We um thee reidue to find the invere tranform of Ũ(x ) U(x t) = 2 π + n 2 π 2 / 2 lim n 2 π 2 / 2 inh /. 1 lim n 2 π 2 / 2 2 coh / 1 coh nπi nπi 2 π 2 t/ 2 in nπx. ( 1) n+1 e n2 π 2 t/ 2 in nπx. (8.3) n Reader familiar with other method for olving partial differential equation will be well aware that thi olution can alo be obtained by eparation of variable.

6 438 SECTION 8.5 Exponential enhance convergence for large value of t. For intance uppoe the rod i 1/5 m in length and i made from tainle teel with thermal diffuivity = m 2 /. Conider finding the temperature at the midpoint x =1/1 m of the rod at the four time t =2 5 3 and 1 minute. Serie 8.3 give U(1/1 12) = 2 ( 1) n+1 e n2 in nπ 5π n 2.1 C; U(1/1 3) = 2 5π U(1/1 18) = 2 5π U(1/1 6) = 2 5π ( 1) n+1 e n2 in nπ n 2.92 C; ( 1) n+1 e n2 in nπ n 2.23 C; ( 1) n+1 e n2 in nπ n 2.4 C. To obtain thee approximation we ued four nonzero term in the firt erie three in the econd and one in each of the third and fourth. aplace tranform alo yield a repreentation for temperature in the rod that i particularly valuable when t i mall. Thi repreentation i not available through eparation of variable. We write tranform 8.29 in the form Ũ(x ) = x inh /x inh / = x = x e (e / /x e /x ) 1 e 2. / e /x e /x e / e / If we regard 1/(1 e 2 / ) a the um of an infinite geometric erie with common ratio e 2 / then Ũ(x ) = x = x [ e /(x ) e ] /(x+) n= [ e /[(2n+1) x] Table of aplace tranform indicate that { } 1 e a ( ) a = erfc 2 t n= e 2n / e /[(2n+1)+x] where erfc (x) i the complementary error function in equation 8.22b. Hence U(x t) may be expreed a a erie of complementary error function { [ ] [ ]} (2n +1) x (2n +1) + x U(x t) =x erfc n= 2 erfc t 2 t { [ ] [ ]} (2n +1) + x (2n +1) x = x erf 2 erf t 2 t n= ].

7 SECTION where we have ued the fact that erfc (x) =1 erf (x). Thi erie converge rapidly for mall value of t (a oppoed to olution 8.3 which converge rapidly for large t). To ee thi conider temperature at the midpoint of the tainle teel rod at t = 3 : { [ ] [ ]} U(1/1 3) = (2n +1)/5+1/1 erf 5 2 (2n +1)/5 1/1 erf n= 6 (3) (3) For n> all term eentially vanih and U(1/1 3) [erf (4.42) erf (1.467)] =.92 C. For t = 18 { [ ] [ ]} U(1/1 18) = (2n +1)/5+1/1 erf 5 2 (2n +1)/5 1/1 erf n= 6 (18) (18) Once again only the n = term i required; it yield U(1/1 18) [erf (.1797) erf (.5991)] =.23 C. Even for t a large a 6 we need only the n = and n = 1 term to give { [ ] [ ]} U(1/1 6) = (2n +1)/5+1/1 erf 5 2 (2n +1)/5 1/1 erf (6) (6) n= [erf (.9843) erf (.3281) + erf (2.297) erf (1.641)] 5 =.4 C. Example 8.21 The end of a taut tring are fixed at x = and x = on the x-axi. The tring i initially at ret along the axi and then at time t = it i allowed to drop under it own weight. The initial boundary-value problem decribing diplacement y(x t) of point in the tring i 2 y t 2 = c2 2 y g x2 <x< t> (8.31a) y(t)= t > (8.31b) y( t) = t > (8.31c) y(x ) = <x< (8.31d) y t (x ) = <x< (8.31e) where g =9.81 and c> i a contant depending on the material and tenion of the tring. Initial condition 8.31e expree the fact that the initial velocity of the tring i zero. Ue aplace tranform to olve thi problem. Solution 8.31de or When we tae aplace tranform of 8.31a and ue initial condition 2 ỹ(x ) =c 2 2 ỹ x 2 g

8 44 SECTION 8.5 d 2 ỹ dx 2 2 c 2 ỹ = g c 2 <x<. (8.32a) Thi ordinary differential equation i ubject to the tranform of 8.31bc ỹ()= ỹ( ) =. (8.32b) (8.32c) A general olution of 8.32a i ỹ(x ) =A coh x c + B inh x c g 3. Boundary condition 8.32bc require =A g 3 =A coh c + B inh c g 3. Thu ỹ(x ) = g 3 coh x c ( g coh inh /c c ) inh x c g 3. (8.33) We invert thi tranform function by finding reidue of e t ỹ(x ) at it ingularitie. To determine the type of ingularity at = we expand ỹ(x ) in a aurent erie around =: ỹ(x ) = g 3 1 (1+ 2 x 2 2c x 4 ) ( 2 24c c ) x 24c 4 + c + 3 x 3 6c 3 + c c + 3 )] = g ( [ 2 x 2 3 2c x 4 ) ( 2 24c c )( x 24c x(x ) 6c 2 + gx( x) = 2c 2 + Thi how that ỹ(x ) ha a imple pole at = and from the product e t ỹ(x ) = (1+t + 2 t 2 )[ ] gx( x) + 2! 2c 2 + gx( x) = + 2c 2 the reidue of e t ỹ(x ) at =i gx( x)/(2c 2 ). The remaining ingularitie of ỹ(x ) occur at the zero of inh (/c); that i when /c = nπi or = nπci/ n an integer. Becaue the derivative of inh (/c) doe not vanih at = nπci/ thi function ha zero of order one at = nπci/. When n i even = nπci/ i a imple zero of 1 + coh (/c) and therefore thee are removable ingularitie of ỹ(x ). When n i odd = nπci/ i not a zero of 1 + coh (/c) and thee are therefore imple pole of ỹ(x ). Reidue of e t ỹ(x ) at thee pole are

9 SECTION Re [ e t ỹ(x ) nπci ] ( = lim nπci ) ge t [1 nπci/ 3 coh xc ( coh c ) ] inh (x/c) inh (/c) g (nπci/) 3 enπcti/ ( 1 + coh nπi) inh nπxi nπci/ lim nπci/ inh (/c) = g3 i n 3 π 3 c 3 enπcti/ ( 1 + co nπ)i in nπx lim nπci/ 1 (/c) coh (/c) = g3 n 3 π 3 c 3 enπcti/ [1 + ( 1) n+1 ] in nπx 1 (/c) coh nπi = g2 n 3 π 3 c 2 enπcti/ [1 + ( 1) n+1 ] in nπx. Since n i odd we may write that reidue of e t ỹ(x ) at the pole =(2n 1)πci/ are Re [ e t ỹ(x ) ] (2n 1)πci 2g 2 (2n 1) 3 π 3 c 2 e(2n 1)πcti/ in Diplacement of point in the tring are given by gx( x) y(x t) = 2c 2 + n= 2g 2 (2n 1) 3 π 3 c 2 e(2n 1)πcti/ in (2n 1)πx. (2n 1)πx. We eparate the ummation into two part one over poitive n and the other over nonpoitive n and in the latter we et m =1 n gx( x) y(x t) = 2c 2 + 2g2 1 (2n 1)πx π 3 c 2 (2n 1) 3 e(2n 1)πcti/ in + 2g2 π 3 c 2 n= gx( x) = + 2g2 2c 2 π 3 c 2 + 2g2 π 3 c 2 m=1 1 (2n 1)πx (2n 1) 3 e(2n 1)πcti/ in 1 (2n 1)πx (2n 1) 3 e(2n 1)πcti/ in 1 [2(1 m) 1] 3 e[2(1 m) 1]πcti/ in [2(1 m) 1]πx. If we now replace m by n in the econd ummation and combine it with the firt gx( x) y(x t) = + 2g2 1 (2n 1)πx 2c 2 π 3 c 2 (2n 1) 3 e(2n 1)πcti/ in + 2g2 π 3 c 2 gx( x) = 2c 2 + 2g2 π 3 c 2 gx( x) = 2c 2 + 4g2 π 3 c 2 1 (2n 1)πx (2n 1) 3 e (2n 1)πcti/ in e (2n 1)πcti/ + e (2n 1)πcti/ (2n 1)πx (2n 1) 3 in 1 (2n 1)πct co (2n 1) 3 in (2n 1)πx.(8.34)

10 442 SECTION 8.5 Example 8.22 The firt term i the tatic poition that the tring would occupy were it lowly lowered under the force of gravity. The erie repreent ocillation about thi poition due to the fact that the tring wa dropped from a horizontal poition. The technique of eparation of variable and eigenfunction expanion lead to the following olution of problem 8.31 y(x t) = 2g2 [1 + ( 1) n+1 ( ] π 3 c 2 n 3 1 co nπct ) in nπx. We can ee the advantage of aplace tranform. They have rendered part of the erie olution in cloed form namely the term gx( x)/(2c 2 ) in olution Example 8.18 and 8.21 contained pecific nonhomogeneitie and/or initial condition. Although aplace tranform can handle nonhomogeneitie and initial condition with arbitrary function they do not do o particularly efficiently. Our final example i an illutration of thi. Solve the following heat conduction problem in a rod of length when the initial temperature ditribution i an unpecified function f(x) Solution or t = 2 U x2 <x< t> (8.35a) U(t)= t > (8.35b) U( t) = t > (8.35c) U(x ) = f(x) <x<. (8.35d) When we tae aplace tranform of both ide of PDE 8.35a Ũ f(x) = d2 Ũ dx 2 d 2 Ũ dx 2 Ũ = f(x) <x< ubject to tranform of 8.35bc Ũ()= Ũ( ) =. Variation of parameter lead to the following general olution of the differential equation Ũ(x ) =A coh x + B inh x 1 x f(u) inh (x u) du. The boundary condition require =A =Acoh + B inh 1 f(u) inh ( u) du. Thu Ũ(x ) = inh x inh f(u) inh 1 x ( u) du f(u) inh (x u) du

11 where = 1 f(u) p(x u ) du 1 x SECTION f(u) inh (x u) du p(x u ) = inh x inh ( u) inh. The econd integral in Ũ(x ) ha a removable ingularity at = and therefore thi term may be ignored in taing invere tranform by reidue. The function p(x u ) ha ingularitie when / = nπi or = n 2 π 2 / 2. Since [ x ( x ) 3 ( u) ( ( u) ) ][ ] p(x u ) = [ ( ) 3 + ] = x( u) + it follow that p(x u ) ha a removable ingularity at =. The ingularitie = n 2 π 2 / 2 are pole of order one and reidue of e t p(x u ) at thee pole are ( lim + n2 π 2 ) n 2 π 2 / 2 2 e t inh x inh ( u) inh = e n2 π 2 t/ 2 nπ i/ inh nπxi inh nπi( u) 1 lim n 2 π 2 / 2 2 coh = 2 2 nπi e n π 2 t/ 2 in nπx nπ( u) 1 in nπ coh nπi i = 2 2 e n π 2 t/ 2 in nπx ( 1)n+1 in nπu 1 ( 1) n = 2 2 e n π 2 t/ 2 in nπx nπu in. Reidue of e t time the firt integral in Ũ(x ) can now be calculated by interchanging limit and integration in to obtain [ 1 lim n2 π 2 2 lim n2 π 2 2 ( + n2 π 2 2 ) e t f(u) p(x u ) du ( + n2 π 2 ) ] 2 e t f(u) p(x u ) du [ = 1 f(u) = 1 f(u) = 2 e n2 π 2 t/ 2 [ lim n2 π 2 2 ( + n2 π 2 ) ] e t p(x u ) du 2 [ 2 2 e n π 2 t/ 2 in nπx ] f(u) in nπu du ] nπu in du in nπx.

12 444 SECTION 8.5 The invere tranform of Ũ(x ) i therefore [ U(x t) = 2 ] f(u) in nπu du e n2 π 2 t/ 2 in nπx. Reader who are familiar with the method of eparation of variable for olving PDE will attet to the fact that eparation of variable i more efficient than aplace tranform in obtaining the above olution. Similar complication arie when nonhomogeneitie involve arbitrary function. In general aplace tranform are le appealing a an alternative to eparation of variable when nonhomogeneitie and initial condition contain unpecified function. A a reult exercie on bounded interval will be confined to problem containing pecific nonhomogeneitie and initial condition. EXERCISES 8.5 In thee exercie ue aplace tranform to olve the initial boundary-value problem. 1. A very long cylindrical rod i placed along the poitive x-axi with one end at x =. It curved ide are perfectly inulated o that no heat can enter or ecape therethrough. At time t = the temperature of the rod i C throughout. For t> heat i added at a contant rate to the left end. The initial boundary-value problem for temperature U(x t) in the rod i where > and C< are contant. t = 2 U x2 x > t > U x (t)=c t > U(x ) = x > 2. A very long cylindrical rod i placed along the poitive x-axi with one end at x =. It curved ide are perfectly inulated o that no heat can enter or ecape therethrough. At time t = the temperature of the rod i a contant U throughout. For t> the left end ha temperature U. The initial boundary-value problem for temperature U(x t) in the rod i t = 2 U x2 x > t > U(t)=U t > U(x ) = U x > where > i a contant. 3. Ue convolution to expre the olution to Exercie 1 in integral form when the boundary condition at x =iu x (t)=f(t) t>. 4. Ue convolution to expre the olution to Exercie 2 in integral form when the boundary condition at x =iu(t)=f(t) t>. 5. A very long tring lie motionle along the poitive x-axi. If the left end (x = ) i ubjected to vertical motion decribed by f(t) for t> ubequent diplacement y(x t) of the tring are decribed by the initial boundary-value problem

13 SECTION y t 2 = c2 2 y x2 x > t > y(t)=f(t) t > y(x ) = x > y t (x ) = x > where c> i a contant. 6. A very long tring lie motionle along the poitive x-axi. At time t = the upport i removed and gravity i permitted to act on the tring. If the left end (x = ) i ubjected to periodic vertical motion decribed by in ωt for t> ubequent diplacement y(x t) of the tring are decribed by the initial boundary-value problem 2 y t = 2 c2 2 y g x2 x > t > y(t) = in ωt t > y(x ) = x > y t (x ) = x > where g =9.81 and c> i a contant. 7. A cylindrical rod of length ha it end at x = and x = on the x-axi. It curved ide are perfectly inulated o that no heat can enter or ecape therethrough. At time t = the temperature of the rod i given by f(x) = in (mπx/) x where m> i an integer. For t> both end of the rod are held at temperature C. The initial boundary-value problem for temperature U(x t) in the rod i t = 2 U x2 <x< t> U(t)= t > U( t) = t > U(x ) = in mπx <x< where > i a contant. 8. A cylindrical rod of length ha it end at x = and x = on the x-axi. It curved ide are perfectly inulated o that no heat can enter or ecape therethrough. At time t = the temperature of the rod i given by f(x) =x x. For t> both end of the rod are inulated. The initial boundary-value problem for temperature U(x t) in the rod i where > i a contant. t = 2 U x2 <x< t> U x (t)= t > U x ( t) = t > U(x ) = x <x<

14 446 SECTION A cylindrical rod of length ha it end at x = and x = on the x-axi. It curved ide are perfectly inulated o that no heat can enter or ecape therethrough. At time t = the temperature of the rod i C throughout. For t> it left end (x = ) i ept at C and it right end (x = ) i ept at a contant U C. The initial boundary-value problem for temperature U(x t) in the rod i t = 2 U x2 <x< t> U(t)= t > U( t) =U t > U(x ) = <x< where > i a contant. 1. A cylindrical rod of length ha it end at x = and x = on the x-axi. It curved ide are perfectly inulated o that no heat can enter or ecape therethrough. At time t = the temperature of the rod i a contant U C. For t> it end x = i inulated heat i added to the end x = at a contant rate. The initial boundary-value problem for temperature U(x t) in the rod i t = 2 U x2 <x< t> U x (t)= t > U x ( t) =C t > U(x ) = U <x< where > and C>are contant. 11. A cylindrical rod of length ha it end at x = and x = on the x-axi. It curved ide are perfectly inulated o that no heat can enter or ecape therethrough. At time t = the temperature of the rod i a contant 1 C. For t> it end x = i ept at temperature and end x = ha temperature 1e t. The initial boundary-value problem for temperature U(x t) in the rod i t = 2 U <x< t> x2 U(t)= t > U( t) = 1e t t > U(x ) = 1 <x< where > i a contant. Aume that 2 /(n 2 π 2 ) for any integer n. 12. A cylindrical rod of length ha it end at x = and x = on the x-axi. It curved ide are perfectly inulated o that no heat can enter or ecape therethrough. At time t = the temperature of the rod i C and for t> the end of the rod continue to be held at C. When heat generation at each point of the rod i decribed by the function e αt where α i a poitive contant the initial boundary-value problem for temperature U(x t) in the rod i t = 2 U x + 2 e αt <x< t> U(t)= t > U( t) = t > U(x )= <x<

15 SECTION where > i a contant. Aume that α n 2 π 2 / 2 for any integer n. 13. A taut tring ha it end fixed at x =andx = on the x-axi. If it i given an initial diplacement at time t =off(x) =x( x) where >i a contant and no initial velocity the initial boundary-value problem for diplacement y(x t) of point in the tring i 2 y t = 2 c2 2 y x2 <x< t> y(t)= t > y( t) = t > y(x ) = f(x) <x< y t (x )= x > where c> i a contant. 14. Repeat Exercie 13 if the initial diplacement i zero and f(x) i the initial velocity of the tring. 15. A taut tring initially at ret along the x-axi ha it end fixed at x = and x = on the x-axi. If gravity i taen into account the initial boundary-value problem for diplacement y(x t) of point in the tring i 2 y t 2 = c2 2 y g x2 <x< t> y(t)= t > y( t) = t > y(x ) = <x< y t (x ) = <x< where g = A taut tring initially at ret along the x-axi ha it end at x = and x = fixed on the axi. For t it i ubjected to a force per unit x-length F = F in ωt where F i a contant a i ω nπ/ for any poitive integer n. The initial boundary-value problem for diplacement y(x t) of point in the tring i where c> and ρ> are contant. 2 y t 2 = c2 2 y x 2 + F in ωt ρ <x< t> y(t)= t > y( t) = t > y(x )= <x< y t (x )= <x< 17. A cylindrical rod of length ha it end at x = and x = on the x-axi. It curved ide are perfectly inulated o that no heat can enter or ecape therethrough. At time t = the temperature of the rod i C and for t> the end x = and x = of the rod are held at

16 448 SECTION 8.5 temperature 1 C and C repectively. The initial boundary-value problem for temperature U(x t) in the rod i t = 2 U x2 <x< t> U(t) = 1 t > U( t) = t > U(x ) = <x< where > i a contant. Find two olution one in term of error function and the other in term of time exponential. 18. A cylindrical rod of length ha it end at x = and x = on the x-axi. It curved ide are perfectly inulated o that no heat can enter or ecape therethrough. At time t = the temperature of the rod i C and for t> it left end x = continue to be ept at temperature C. If heat i added to the end x = at a contant rate the initial boundary-value problem for temperature U(x t) in the rod i t = 2 U x2 <x< t> U(t)= t > U x ( t) =C t > U(x )= <x< where > and C> are contant. Find two olution one in term of error function and the other in term of time exponential.