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# A bar with initialtemperature profle 0 withends heldat 0

1 o C willcool 1 and approach a steady-state temperature 0o x 201 which is a linear df dx 1 fdx 0 012 df 0 1 0253a bdx 1 2 0a b The frst two equations yield the same thing 30 Substituting this into

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### Document on Subject : "A bar with initialtemperature profle 0 withends heldat 0"— Transcript:

1. A bar with initialtemperature pro\fl
1. A bar with initialtemperature pro\fle 0, withends heldat 0 o C, willcool !1 , and approach a steady-state temperature 0 o x  1, which is a linear df dx (1), f dx (0) =0 12 = = df 0= (1) = � (3a + b) dx 1 2= � a + b The \frst two equations yield the same thing, 3 � . Substituting this into the last equation gives 13 a = ;b X   (b) Solve the problem ;t) =0 = (1;t); 0) = Note: you can just write down the solution we had in class, but make sure you know how to get it! This is the basic heat problem we considered in class, with 1 x;t)= sin (nx �n 2  2 t (2) n=1 where Z 1 =2 )sin (nx 0 ) is given in (1). The form of (1) is already a sine series, with =3 � 2 and =0 for all other . You can check this for yourself by computing integrals in (3) for )given by (1),

from the orthogonality of sin nx.
from the orthogonality of sin nx. Therefore, 31 x;t)= sin (x � 2 t sin (3x �9 2 t (4) 2 �2 (c) Show that for some , 0  0) is positive and for others x  it is negative. How is the sign of 0) related to the shape of the initial temperature pro\fle? How is the sign of x;t), t � 0, related to subsequent temperature pro\fles? Graph the temperature pro\fle for =01 on the same axis (you may use Matlab). Di erentiating x;t)in time gives 39 x;t)= �  2 sin (x � 2 t sin (3x �9 2 t 2 �2 =0 gives 3 0) =  2 (sin (x � 3sin (3x � 2 Note that     ut 1 6;0 = 15 4 2 � 0; ut 1 2;0 = �62 0 Thus at =1is positive and for =1is negative. From the PDE, ut = uxx 2  

  2 1.8 u(x,t 0 ) 1.6 1.4
  2 1.8 u(x,t 0 ) 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 t=0 t=0.2 t=0.5,1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x Figure 1: Plots of x;t)for =0 and hence the sign of gives the concavity of the temperature pro\fle x;tconstant. Note that for x;t0, the pro\fle x;t)is concave up, and for x;t0, the pro\fle x;t)is concave down. At =0, the sign of 0) give the concavityofthe initialtemperature 0)= ). In Figure 1, x;t)is plotted for =0 2. Initial temperature pulse. Solve the inhomogeneous heat problem with Type I boundary conditions: @u @ 2 u = ;t)=0 = (1;t); 0) = @t @x 2 t � 0, 0  x  1, and\r 8 w 0 if 0 x 1 2 � 2 w 1 + w (5) )= w 2 � u 0 if 1 x 22 � : 0 if 1 + w x 22 Note: we derived the form of the solution in class. You may simply use this and replace ) with

(a) Show that the temperature at the mi
(a) Show that the temperature at the midpoint of the rod when =1 2 (dimensionless) is approximated by 11 2sin (w= u ; 2 2 e w= 3 � X X  Can you distinguish between a pulse withwidth =11000 from one with =12000, say, by measuring 11  ? 2 ; 2 (b) Illustrate the solution qualitatively by sketching (i) some typical tempera­ture pro\fles in the � plane (i.e. = constant) and in the � x plane (i.e. = constant), and (ii) some typical level curves x; t) = constant in � plane. At what points of the set f x; t): 0  x  ; t  0 g x; t)discontinuous? This is the Heat Problem with Type I homogeneous BCs.The solution we derived in class is, with ) replaced by 11 x; t)= x; t)= sin (nx �n 2  2 t (6) n=1 n=1 where the 's are the Fourier coecients of )= ), giv

en by Z 1 =2 )sin (nx 0 Breakin
en by Z 1 =2 )sin (nx 0 Breaking the integral into three pieces and substituting for ) from (5) gives Z w= Z 2+w==2 )sin (nx+ 2 )sin (nxw= 1 )sin (nx2+w= Z 2+w= u0 = 0 + 2 sin (nx+ 0 w= w cos (nx  2+w= = w � n w= � n � n  cos 2 (1 � w) � cos 2 (7) (1 + = u0 wn= We applythe cosine rule cos ( � s) � cos ()=2sin n=nw=2to Eq.(7), n nw = sin sin wn 22 is even (and nonzero), i.e. =2for some integer 2u0 = sin mw =0 wm 4 X                   is odd, i.e. =2 � 1 for some integer 2u 0 ( � 1) m+1 sin ((2 � w= B2m�1 = : (8) (2m � w= (a) The temperature at the midpoint of the rod

, =12, at scaled time =1 2 is, from (6)
, =12, at scaled time =1 2 is, from (6) and (8), 1  2u 0 ( � 1) m+1 sin ((2 � w=  (21) 2 x;t) = (2m � w= sin (2 � 1) e 2 m=1 1  X sin ((2 � w= = : 1) 2 (2m � w= m=1 For  1= 2 , the \frst term gives a good approximation to x;t 11 112sin (w= u ; ; = : 2 2  u 1 2 2 e w= To distinguish between pulses with =11000 and =12000, note that lim w!0 w= =1, and so for smaller and smaller , the corresponding w= � 1  temperature 1 gets closer and closer to 2 2 ; 2 11 112 2 " 2 u ; ; = + ;w  1: 2 2  u 1 2 2 e 1 2 3!   In particular, 111 111 ; ;w = ; ;w = u1 2 2 1000 � u 1 2 2 2000 sin (=2000) sin (= = e == 2u

0  3:1  10 �7  �
0  3:1  10 �7  � Thus it is hardto distinguish these two temperature distributions, at least by measuring the temperature at the center of the rod at time =1 2 By this time, di usion has smoothed out some of the details of the initial (b) The solution x;t) is discontinuous at = 0 at the points  2. That said, x;t) is piecewise continuous on the entire in-terval[01]. Thus, the Fourier series for 0) converges everywhere on the interval and equals 0) at all points except =(1  2. The temperature pro\fles ( � t plane, u � plane), 3D solution and levelcurves are shown. 5 u-t plot 10 u(x 0 ,t)/u 0 8 6 4 2 0 00.02 0.04 0.06 0.08 0.1 t Figure 2: Time temperature pro\fles ;t) at =05, 04 and 01 (from top to bottom).The -axis is the time pro\fle corresponding to =0, 1. Recall that to draw the lev

el curves, it is easiest to already have
el curves, it is easiest to already have drawn the spatialtemperature pro\fles.Draw a few horizontal broken lines across your plot. Suppose you draw a horizontal line . Suppose this line crosses one of your pro\fles x;t) at position Then (;t)is a pointon the levelcurve x;t)= . Now plot this point in your level curve plot. By observing where the line crosses your various spatial pro\fles, you \fll in the level curve x;t)= .Repeat this process for a few values of to obtain a few representative level curves. Plot also the special case level curves: x;t)= x;t) = 0, etc. Come and see me if you still have problems. 3. Consider the homogeneous heat problem with type II BCs: @u @ 2 u @u @u = ; (0;t) =0 = (1;t); 0) = ) (9) @t @x 2 @x @x t � 0, 0  x  1 and is a piecewise smooth function on[0 (a) Find the eigenvalues and t

he eigenfunctions )for this problem. Wri
he eigenfunctions )for this problem. Write the formal solution of the problem (a), and express the constant coecients as integrals involving 6 0 0.2 0.4 0.6 0.8 1 0 2 4 6 8 10 12 xu(x,t0)/u0u-x plot Figure3:Spatialtemperaturepro\flesx;t)at=0(dash),0001,001,01.The-axisfrom0to1isthelimitingtemperaturepro\flex;t)as!1 0 0.5 1 0 0.02 0.04 0.06 0.08 0.1 0 5 10 x 3D plot of u(x,t) t u(x,t)/u0 Figure4: 0 0.2 0.4 0.6 0.8 1 0 0.02 0.04 0.06 0.08 0.1 xtLevel curves u(x,t)=const 4210.50.50.250.25 Figure5:Levelcurvesx;tforvariousvaluesoftheconstant.Numbersadjacenttocurvesindicatethevalueof.Thelinesegment(1(1+2at=0isthelevelcurvewith=1=10.Thelines=0and=1arealsolevelcurveswith=0.(b)Findaseriessolutioninthecasethat)=aconstant.Findanapproximatesolutiongoodforlargetimes.Sketchtemperaturepro\fles()fordi erenttimes.(c)Evaluatelimx;t)forthesolution(a)

when)=)with)de\fnedin(5).Illustratetheso
when)=)with)de\fnedin(5).Illustratethesolutionqualitativelybysketchingtemperaturepro\flesandlevelcurvesasinProblem2(b).Itisnotnecessaryto\fndthecompleteformalsolution.Solution:(a)To\fndaseriessolution,weuseseparationofvariables,x;t)=ThePDEin(9)givestheusual X=T0 isconstantsincethelefthandsideisafunctionofonlyandthemiddleisafunctionofonly.Substituting(10)intotheBCsin(9)gives(0)=(1)=0 p p p   X � The Sturm-Liouville boundary value problem for )is thus\r 00 00 =0; (0) = (1) =0 Let us try 0. Then the solutions are )= � p jjx + c2e p jjx and imposing the BCs gives 0 0= (0)= cc jj p jj 0 0= (1)= � p j  j c1e � p jj + j  j c2e �jjThe \frst equation gives � and substituting this into the second, we have �pjj pjj 0= c jj 0, the bracketed ex

pression is positive. Hence =0, i.e. ) m
pression is positive. Hence =0, i.e. ) must be the trivial solution, and we discard the case  For =0, )= and both BCs are satis\fed by taking =0. Thus )= (we'll use by convention -it's just another way to name the constant). Hence, the case = 0 is allowed and yields a non-trivial solution. For � 0, we have X = c1 sin px + c2 cos px 0 The BC (0)=0 implies =0.The other BC implies 0 0= (1)= � c2psin p For a non-trivial solution, must be nonzero. Since � 0 then we must have sin =0, which implies the eigenvalues are n = n 2  2 ;n =1;::: andthe eigenfunctions are ) =cos (nx For each , the solution for )is )= � n t . Hence the series solution x; t)is 1 x; t)= cos (nx)exp � n 2  2 t  (12) n=1 9\r X ( 6( 6 At =0, the initial condition giv

es 1 )= 0) = cos (nx The orthogon
es 1 )= 0) = cos (nx The orthogonality conditions are found using the identity 2cos (nx)cos (mx)=cos (( � x)+cos ((x Note also that for m; n =1, we have Z 1 =0 cos (( � x 0 0 m = n Z 1 cos ((x=0 0 The last integral follows since cannot be zero for any positive integers . Combining the three previous equations gives the orthogonality Z 1 cos (nx)cos (mx 1=2 m = n 6 =0 (14) 0 0 m = n Multiplying each side of (13)by cos (mx), integrating from =0 to 1, and applying the orthogonality condition (14) gives Z 1 A0 = f (x)dx (15) 0 Z 1 = 2 cos (mx 0 (b) )= into (16) and (15) gives Z 1 =2cos (nx=0; n � 0 Z 1 A0 = u0dx = u0 0 It is no surprise that =0 for n � 0 since the IC )= is one of the eigenfunctions, )= . The series solution is simpl

y x; t)= Temperature pro\fles are simp
y x; t)= Temperature pro\fles are simply a plot of )= , i.e. the temperature alongthe rod does notchange.This is reasonable, since the rod is initially 10 u-t plot 10 u(x 0 ,t)/u 0 8 6 4 2 0 00.02 0.04 0.06 0.08 0.1 t Figure 6: Time temperature pro\fles ;t) at =05, 04 and 01 (from top to bottom). at a constant temperature and is completely insulated -so nothing will happen. (c) Takingthe limit !1 of (12) and using (15) gives Z 1 Z 1 Z 2+w= u0 x;t)= dx = u0 t!1 w= w Thus the temperature alongrodeventually becomes the constant . Lastly, we illustrate the solution qualitatively by sketching temperature pro\fles and level curves in Figures 6 to 9. 4. Consider the homogeneous heat problem with type III (mixed) BCs: @u @ 2 u @u = ; ;t) =0 = ;t);0)= @t @x 2 @x t � 0, 0  x  1 and is a piece

wise smooth function on[0 (a) Find the
wise smooth function on[0 (a) Find the eigenvalues and the eigenfunctions )for this problem. Write the formal solution of the problem (a), and express the constant coecients as integrals involving (b) Find a series solution in the case that )= a constant. Find an approximate solution good for large times. Sketch temperature pro\fles ()for di erent times. 11 0 0.2 0.4 0.6 0.8 1 0 2 4 6 8 10 12 xu(x,t0)/u0u-x plot Figure7:Spatialtemperaturepro\flesx;t)at=0(dash),0001,003,01.Thex;t=1isthelimitingtemperaturepro\fleas!1.Notethattheendsoftherodheatup! 0 0.5 1 0 0.02 0.04 0.06 0.08 0.1 0 5 10 x 3D plot of u(x,t) t u(x,t)/u0 Figure8: 0 0.2 0.4 0.6 0.8 1 0 0.02 0.04 0.06 0.08 0.1 xtLevel curves u(x,t)=const 42110.50.50.250.25 Figure9:Levelcurvesx;tforvariousvaluesoftheconstant.Numbersadjacenttocurvesindicatethevalueof.Thelinesegment(1(1

+2at=0isthelevelcurvewith=1=10.(c)Evalua
+2at=0isthelevelcurvewith=1=10.(c)Evaluatelimx;t)forthesolution(a)when)=)with)de\fnedin(5).Illustratethesolutionqualitativelybysketchingtemperaturepro\flesandlevelcurvesasinProblem2(b).Itisnotnecessaryto\fndthecompleteformalsolution.Solution:(a)To\fndaseriessolution,weuseseparationofvariablesasbefore(Eq.(10)),butnowobtaintheSturmLiouvilleproblem=0;(0)=(1)=0Letustry0.Thenthesolutionsare)= jjx+c2ep andimposingtheBCsgives0=(0)= jjc1+p 0=(1)= jj+c2ep The\frstequationgivesandsubstitutingthisintothesecond,wehave0= jj1�e2p jj13   !     ( 6  0, then 2 p jj 1 and the bracketed expression is negative.\r=0, i.e. ) must be the trivial solution, and we discard\rthe case  For =0, )= and imposingthe BCs gives =0.Thus\rwe discard the =0 case.\rFor  � 0,

we have\r X = c1 sin px + c2 cos p&
we have\r X = c1 sin px + c2 cos px 0 The BC (0)=0 implies =0.The other BC implies 0= (1) = For a non-trivial solution, must be nonzero. Since � 0 then we must have cos =0, which implies the eigenvalues are (2n � 1) 2  2 ;n =1;::: andthe eigenfunctions are ) =cos 2n � 1 x 2 For each , the solution for )is )= � n t . Hence the series solution x;t)is  X (2n � 1) 2  2 x;t)= 2n � 1 x 2 � 4 n=1 At =0, 1  )= 0)= X An cos (2n � 1) x 2 n=1 The orthogonality conditions are found using the identity 2cos 2n � 1 x 2m � 1 x =cos (( � x)+cos (( � x 22 Note also that for m;n =1, we have Z 1 =0 cos (( � x 0 0 m = n Z 1 cos (( � x=0 0 14    

( 6     
( 6      !   0 The last integral follows since � 1cannot be zero for any positive inte­.Combiningthe three previous equations gives the orthogonality Z 1 2n � 12 � =0 x x dx 22 0 m = n Multiplyingeach side of (20) bycos ((2 � x=2), integrating from 0 to 1, and applying the orthogonality condition (21) gives Z 1 =2 cos 2m � 1 x f 2 0 (b) )= into (16) and (15) gives "� #1 Z 1 sin 2n�1 x 2 =2 2n � 1 x dx =2 2n�1 2  0 2 0 4u 0 ( � 1) n+1 4u0 2n � 1 = sin  = (2n � 2 (2 � 1) Thus the series solution is  4u0 X ( � 1) n+1 2n � 1 (2n � 1) 2  2 x;t)= x  2n � 12 n=1 After t  4= 2 , we may approximate the series by the \frst ter

m,   2 4u0  x x
m,   2 4u0  x x;t  x;t)= cos exp t  2 �4 The temperature pro\fles ()for di erent times are given below in Figures 10to 13. (c) Takingthe limit !1 of (19) gives x;t) =0 t!1 Thus the temperature along rod eventually goes to zero. Lastly, we illus­trate the solution qualitativelyby sketchingtemperature pro\fles and level curves (Figures 14 to 17). 15\r u-t plot 0 0.2 0.4 0.6 0.8 1 u(x0 ,t)/u 0 0 0.2 0.4 0.6 0.8 1 t Figure 10: Time temperature pro\fles ; t) at =001, 04 and 09 (from topto bottom). u-x plot 0 0.2 0.4 0.6 0.8 1 u(x,t 0 )/u 0 0 0.2 0.4 0.6 0.8 1 x Figure 11: Spatialtemperature pro\fles x; t)at =0001, 001, 01, 05and 1 from topto bottom.The line x; t=1 is the initialtemperature pro\fle at =0. 16 0 0.2 0.4 0.6 0.8 1 0 0.5 1 0 0.5 1 x 3D plot of u(x,t) t u(x,t

)/u0 Figure12: 0 0.2 0.4 0.6 0.8 1 0 0.2
)/u0 Figure12: 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 xtLevel curves u(x,t)=const 0.990.90.50.250.1 Figure13:Levelcurvesx;tforvariousvaluesoftheconstant.Numbersadjacenttocurvesindicatethevalueof u-t plot 0 0 2 4 6 8 10 u(x 0 ,t)/u 0 x0=0.5 0.1 0.4 0.1 0.2 0.3 0.4 0.5 t Figure 14: Time temperature pro\fles ;t) at =05, 04 and 01 (from topto bottom). u-x plot 0 2 4 6 8 10 12 u(x,t 0 )/u 0 t=0 0.001 0.030.1 0.5 00.2 0.4 0.6 0.8 1 x Figure 15: Spatialtemperature pro\fles x;t) at =0 (dash), 0001, 003, 01 and 0:5. 18 0 0.5 1 0 0.1 0.2 0.3 0.4 0.5 0 5 10 x 3D plot of u(x,t) t u(x,t)/u0 Figure16: 0 0.2 0.4 0.6 0.8 1 0 0.1 0.2 0.3 0.4 0.5 xtLevel curves u(x,t)=const 4210.50.50.250.25 Figure17:Levelcurvesx;tforvariousvaluesoftheconstant.Numbersadjacenttocurvesindicatethevalueof.Thelinesegment(1(1+2at=0isthelevelcurvewith=1=1