(المعادالت التفاضلية الجزئية) ريض 425

تمارين

The exercises from :

Introduction to partial differential equations and boundary value problems, by Rene

Dennemeyer, California State College at San Bernardino

:إعداد

فواز بن سعود العتيبي. أ

جامعة الملك سعود- قسم الرياضيات

2D Heat equation

1. 2D Poisson equation

A solution for a second-

order wave equation

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( lb)

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·----···-----.. ··-·- · -.-·-·---·.--·h--•H•-<

( ~S)

~WI ~l.i:i.ll ~'iJla.JI :.::Jllll ~I

~1~1 JIJJ.ll (.)41~ .w._,....l~ J CJI'~'i UlJIA.. : 3.2 ~I

AI Determine conditions under which the following functions are harmonic in three dimensions.

(c) u = a11x1 + G!l.f + auZ2 + 2a1.X}' + 2ataXZ + 2a13yz

'ltere the a,s 8.;fC constants.

1 (C)· au + 0.U Dat = 0

(e) u A: sin ax c<>sh by + B cos ax sinh by

(e) b =±a

3 A harmonic function in three-dimensional space which depends only on the distance fromafixed poinHscalledapurelyradiallydependentpotentiaJ. Letr = (x• + y + z1)~. Determine aU twice differentiable functions f such tbat u = f(r) is a potential. Do this also in two dimensions.

3 In three dimensionsf(r) =a+ hfr, a, h constants; two dimensions f(r) = a + h logr.

8 (a) Let~be the set of all points in the plane such thatx~ + y' :$; 1. Let u = r -3xy'. Find the maximum and minim.um values of u on. !II.

8 (a) max u(x,y) = 1, min u(x,y) = - 1; max occurs at the points (1 ,0}, ( - i, v3/2), ( - !,-V3j2); min occurs at tb.e points (!,v'3i2), ( - 1>0), (i,-'\1'3/2).

(2..9 )

~Ull ~li:ill ~'iJla..JI :I.!J\lll ~I

~'i ~Jl.a.o t) ~1~1 J..=i : 3.3 .l.i.:JI

' .. \ - · - -- --:"'~J!)IA thin rectangula~,.._homogeneotl! th~llJ., conducting plate lies in the xy plane

and <f'ccupies the rectanglc~ 10 < x "< a;:p ~ y < b. The faces of the pl.ate are insulated. and no internal sources or smki"ii:-e present.·· Theedge y = 0 is held at 100Q while the J.t_maining edges are held at 0~ Find the steady Temperature u(x,y) in th~e::.Jpt::.!l::.:at:.::::.e·:..,_..---

11 (a) 400 (X) .

U=--- " --------~--------~~--~--~~~---'TT k• l

(e} Find x if the edge y = 0 is held at temperature Tx(x - a), where Tis a constant. Hint: Since no heat sources are present in the plate, the steady-state temperature u must satisfy

(e)

(a) A thin h.omogen.eou pl~te ~upies th . region

y ~ O

in the xy plane, There are no heat sources in the plate, and the faces are in ulated. The temperature i held at .~ 'alann .!he edg__~- .x:_:: 0, x a, while u • _ !along the edge y = 0 (T constant). AJso Jim u(x,y) = 0 uniformly fortf s x < a. Derive the ert repre-

v-(1) sentation for the temperature in the plate. Obtain a closed~form expression for the temperature with the aid of the eq,uation . - .,. - -- -- -. ..._ - .. "'--- ..

(X) ln [(2k - l)x]e-tn-ar tan-1 (sin x{sinh y)

2k - l = 2

----21 (a)

2Ttan- 1 [sin 1rx/a)(sinh 'Tt}'/tf] u = ·-··· ~ . . .. . . . ......... · 1T

-{b) Dtrive the scrie expression for the temperature in the plate described in a if in tead the boundary condition along the edge x = a is

ou ox + hu. = 0

where h is a positive constant, the remaining boundary condition being the same,

(1( u =2Th

where {~~~:l denotes the equence of positive roots of the transcend.e.ntal equation tan aE """' -~jh.

(;0 )

~Llll ~Ulll ~~Jl.a..JI :~1 ~~

~~ t!Jts.. ~ ~~~~ J..=.! : 3.3 ~~

23 Sol e the boundary-value problem

Au = ex + dy in fit u = 0 on the boundary

where c and d are constants and fll denotes the interior of the rectangle 0 :$;' x .:::;; a, 0 .:::;; y < b.

CX(X - ll) dy(y - fJ) 23 u = v + w1 + w1 where v

2 +

2

. = ~. A sinh [ 2n - 1)1T(a - x)/b] + inh [(2n - 1}trx/b1 . (2n - 1)1ry w1

;:1

8 sinh [(2n - l)na/b] 810

b

w1 = CQ B., sinh [ 2n - l)~b ~/}Jal . l) s~nh

1u2n - l)1r /h] in (2n - 1 )nx

1 sm · · · · 2n - 1r fa b

tt-

where

27 An infinitel long bar of homogeneous material occupie . the region 0 < x :$;' a, 0 < y S b, 0 < z < oo. There are no heat sources within the bar. The base z = 0 i held at the temperature Txy(x - a (y - b • where Tis a constant. while the sides are held at 0". At o the temperature satisfies the condition lim u(x;y;z = 0 uniformly in x y

;r ~

for 0 < x < a, 0 S y S b. Find the steady temperature in the solid.

Z7 u = 64T<l2b

where

/lit = ,2Jc - l k - 2 .. . , =

~Li.ll ~illll ~'iJt.a...l l :~1 ~I

~'i 4JJlL. ~ ~1~1 ~ : 3.3 ~I

29 The disk of Prob. 28a has the following prescribed temperature on the rim r = a: u = c, c a constant, 0 < 6 < ex, u = O,cx < 8 < 27T, where ex is a given angle, 0 < ex < l1r. Find the series expression. for temperature at interior points of tbe disk. In particular consider the case where c = 100 and ex = v/2.

For this case use Poisson's integral (3 .. 35) to derive a closed-form expression for the temperature inside the disk. Use the closed-form expression to show that

lim u{rjlJ) = 100 ~

Jim u(rO) = 0

1T 0 < 8 < -

2

11'

2 < 0 < 211'

What is the temperature at the center of the di k?

29 e [« oo u = - ... - + .

1T 2 tt -

f'

a

31 Find a function u harmonic in the region 0 ::;; r < a, 0 < 8 < 1r/2 such that u = 1 on the edge 8 = 0, 0 < r < a, u = 0 on the edge 8 = 'IT'/2; 0 < r < a, and u 0 on the rim r = a, 0 < 6 < Tr/2. Hint: Consider that the harmonic fun.ction tJ = l - 20/'tt satisfies the boundary conditions along 0 = 0, 8 = r./2. ow construct a harmonic function w in the region such that the superpo ition u = v + w satisfies the conditions of the probJem.

31 u = 1r - 20 _ ~ «l (· !:..).tm in 2n6 1r 1 a n

33 Determine the temperature distribution in the annulus of Prob. 32 if instead the temperature on the outer rim r """' h is held at u ·= 1· cos (6/2). 0 < 8 < 211", where Tis a constant.

32 A thin annulu occupies the region 0 < a S r ::;; h, 0 < 8 < 211", where h > a. The faces are insulated, and along the inner edge the temperature is, maintained at 0°, while along the outer edge the temperature is held at 100°. Show that the temperature in the annulus is given by

u ..... 100

log (r/a) Jog(h/a)

33 in n8

(J2.)

~Ull ~li:i.ll ~'iJlA..JI :~1 J...::!,iJI

~'i ~Jla.. <.) ~~~~ J..d : 3.3 .i.i:JI

35 A thin thermally conducting sheet occupies the region 0 < a s; r < b, 0 s fJ s ex, in the xy plane, where r, 8 are polar coordinates, a and b are given numbers such that a < b, and ex is a given angle, 0 < ex < 21r. The edges r = a, r = bare held at oo, as is also the edge (} = 0. The edge 8 = ex is held at 100°, Find the steady temperature in the sheet.

. 0 35 u = 100-

(X [ J sin /' 8

where

n= l 2 ...

37 (a} A homogeneousmerrilaftY conductjng cylinder occupies the region 0 < r < a, 0 s (} s 21r, 0 :::;; z s h, where r, fJ, z are cylindrical coordinates. There are no sources of heat within the cylinder. The top z = h and .the lateral surfacer = a are held at oo, while the base z = 0 is held at 100°. Find the steady-temperature distribution within the cylinder.

37 (a)

whete { · k) denotes the sequen of po iti e zeros of J o(e)

39 A wedge-shaped solid occupies the region described by the inequalities 0 :S r s a. 0 s 8 :::;; {J, 0 s z :::; h, where fJ is a given angle, 0 < {J < 21r. The top z = h, the lateral surface r - a, and the faces 0 ""' 0 and 8 ...,.. fJ are insulated. The base z 0 is held at temperature f(r,O ). Derive the expression for the steady temperature in the solid if there are no sources of heat within. Consider the special case f(r 6) - JOO.

39 u = A00 + I !An.t cosh (~ .. c/1/a)Jy,. (enkrfa) cos v,.(J where, for each fixed n, {~til} n-o k- I 1

is the sequence of po itive zeros of J "(¢) , v,. = n11;/{J, and

n = 0, 1.. .. k = 1, 2, ...

(33)

~_,.JI 4JJls..ll :&I.JI J-ill ..bo.IJ ~ <) ~_,.JI 4JJI£All 4JI~'il ~I U...... : 4.2 ~I

(d) Construct a nontrivial real-valued wave function which satisfies the stated condition. (:i) u is harmonic time-dependent and u(O,t) = 0 all 1.

(ii) t1. is hann.onic time-dependent, u(O,t) = 0 all r, and u(x,O) = 0 all x. (iii) u(O,t) = 0, Jim u(x,t) = 0, lim u(x,t) = 0.

, GO ~«;!

(iv) u:t(O,t) = 0 all t and u,(x,O) = 0 all x. (v) (u!j; + IXU)I,-o = 0 all t, a. a real cons~t.

1 (d) (i) u = sin k · cos wt w = k

('ui) u = e p [- kx - w.t ~;J - e p [- k + wt .]

(v) u ...;_ ·k c·os kx - (X sin kx) cos ())t kc

1 3 }Use J?'Alem.· ~rt's ?l~!ion (4-15) to con ruct the oluti.on oft~e initial- .alue p~ob~~ ~17) With the gtVen Jmtlal data. ore that f. g may not satJSfy the dlfferenttabiiJty

conditions assumed in the text at every point. evertheless erify that the functjon obtained by applying o• Al.embert' formula satisfies the initial conditions. Show also that the wave equation is satisfied except po i.bly at point along the characteristic curves of Eq. (4-S). Sketch the solution at times t = 0, t = 1/c, and t = 2fc.

~ f(x) - e-,g(x) == 0 - oo < x < ro (b) /(x) = 1/(1 + x').g(x) = 0, -«> < x < «>

@J<x . = A sin wx, ~x) = Bcospx, - oo < x < oo (d) f(x) = 0, g(x) = A inh ax -co < x < oo ~ /(x) = 1, lxl::; l,f(x) = 0, !xl > l;g(x) = 0, -oo < x < «>

..

3 (a) u = e co h ct A . B •

(c) u = · sm wx cos met -- cos p.x sm pet Cf'

(e) u U(l- x + ct)- U(- l - x + ct) + U(1 - x - ct) - U(-1 - x - ct)

2

where U(E) is the unit step futtcfion defined by

U(~ ..,.. 1 E > 0 U(E) = 0 l < 0

(x - ct)

[U('"/2- x- ct) - U( -11/2 - x- ct)] cos (x + ct) + . 2

W Construct the solution of the initial-value problem (4-ll) if the data are as described.

~ F x,t) = l,f x) sin wx.g(x) = 0 (®I F(x,t) = xt j(x) """' g(x) = 0

(c) Flx,t) = 4x ' t, f(x) = O,g(x) = coshbx (d) F< x,t) = A sin wx sin f't,f(x) = g(x) """ 0

(e) F(x,r) == A sin (kx - wt),f(x) = g(x) = 0, k me

~.JAll ~Jla.JI :&1)1 ~I ..u.I.J ~ c) ~.JAll ~Jt-ll 4ii~'JI ~I ~ : 4.2 ~I

,, 5 (a) u = sin wx cos wet + 2 . ~c) cosh bx sinh bet .

1 , ..

~ u = . be . + 2xl + 6

(.t) = kc

4_,...ll t!JUr....ll :~1)1 ~I ~IJ ~ <) 4_,...ll t!Jla...ll ~l~'il ~I U...... : 4.2 .l.i:JI

7 Consider the boundary- and initial-value problem

Uu = c1u~ x > 0; t > 0

u(x,O) f(x) u,(x.O) = g(x) x ;;::: 0

u(O,t) == 0 t ~ 0

Note that in order for the data to agree at (0,0) it is necessary that /(0) = 0, g(O) = 0 (see Fig. P4-1). The given functions f, g are defined only for x ~ 0; however assume for the moment that f. g are defined in some manner on (- co oo), and apply D'Alembert's formula (4-lS) to the problem. Impose the condition a1ong x = 0. and obtain

t[/( -ct ) + f(ct)] + 2

1 Jct. g(~) de == 0 r > 0 c - ct

This equation will hold provided f. g are defined such that

f(ct) = - f( - ct) ft g(~) de = 0 I> 0 J_t Hence, the appropriate procedure is to exte.nd [and gas odd functions on (-co.co):

f(x) = -[( - x) g(x) === - g( -x) x < 0

ow D'Alembert's formula defines the olution of the problem

(

i{/(x - ct) + f(x + ct)] + 2~ l. : . c:1

g(~)d~ 0 .s; ct $ x u(x,t) =

1 i=+et U - {I ct - x) + f(x + ct)] + 2 . g f) d~ 0 < x :s; ct C ct- z

Verify that u satisfies the boundary co.odition along x - 0 as well as the initial conditions. Note that u atisfies the wave equation except possibly along the characteristics x - ±ct

Figure P~l

u=f(r) u1 =g(x)

~_,...ll ~Jls.JI :&1)1 J,..:illl

~I.J ~ ~ ~_,...ll ~Jl.a.Al.l ~l~'il ~I ti....... : 4.2 ~I

in the xt plane. Use this formula to construct the solution of the boundary· and initia -value problem correl ponding to the following prescribed data. Sketch the solution at t = 0 t = 1/c, 1 = 3/2c, I ""' 2/c. (a) f(x) = l;g(x) = - cosx, x > 0

7 {a) u 1 l

- -co c

- 1

sin cr

sin cos ct

0 < ct .S x

8 Consider the following boundary· and initial-value problem

Uti = c"-u= X > 0; t > 0

u(x,O) = f(x) u,(x,O) = g(x) x 2: 0

u(O,t) = h(t) t :;::: o Assume f, h are twice continuously differentiable and g is continuously differentiable, on [O, +co). Note that if the boundary and initial values agree at the corner (0,0) in the xt plane, the necessary conditioDS h(O) = j(O), h' (O) = g(O) are implied in the data. Let v denote the function constructed as the solution in Prob. 7. Suppose w is a solution of the wave equation (except possibly along the characteristic x = ct) such that

w(x,O) = 0 wt(x,O) = 0 x ~ 0 w(O,t) = h(t) t ~ 0

Then u = v + 111 satisfies the conditions of the boundary- and initial-value problem first proposed, except that the wave equation may not be satisfied along the characteristic. To construct 111 assume w = q;(x- ct), x ::1: ct. Impose the boundary condition at x = 0; then

q;(- ct)=ht) t ?! O

Let$ = -ct, so that q;($) = h( -$/c) and

cp{x - ct)"" h(t - ~) Define was

w(x,t) = 0 0 < Cl ~ X w(x,l) = h(t - ~) 0 ~ x < cl

Verify that w has the desired properties at t = 0 &Jld satisfies tbe wave equation, except possibly along x = ct, t ?:: 0. Use the results of these considerations to construct the solution of the boundary- and initial- alue problem for each of the following cases. (a) f(x = e- "',g(x) = l , h(t) = 1

(a) {

tr"' co h ct .. + t u - x

- e-ttt 'inb x + -c

1 0 < x < ct < ~ - -

4_,..ll AJJ\.a.JI :&1)1 J..,.aiJI ~IJ ~ ~ 4_,..ll AJJ~ ~~~~~ ~~ U........ : 4.2 ~~

~_,.JI ~Jls...JI :&1)1 ~I ( ~1~1 ~ ~..)= ) ]+JI &....ll : 4.3 .¥1

19 The string with fastened ends at x = 0 x = b executes free vibrations after release from th.e initial displacement f(x) and with initial. speed rf(x). Determine the series expression for the subsequent displacement u(x.t).

(•) f (x) = 4/rx(b- x)/b1 g(x) """ 0 0 ~ x .:S b (h = const) (b) f(x) = J 0 sin 7TX{b), rf(x) == 0, 0 ~ x ~ b (c) f(x) = A sin wx g(x) - 1. 0 .:S x S b (w i a constant and is not an int.egral multiple of 7T/b) (d) f(x) = 0, 0 S x S b; g(x) = v0, b/2 - t ~ x S b/2 + £, g(x) = 0 elsewhere (e is a small positive constant)

19 (a) 32h ~ sin [(2n - 1)7Tx/·b.l co [(2n - 1)7Tct/b]

u = - k. 1T' 11- 1 (2n - 1)#

21 (a) The string with fastened ends at x == 0, x = b e~tecutes forced vibrations under the external force per uni.t mass

F(t ) == F0 sin wt

where F0, w are given positive constants. The string is released f1:om rest with zero dis­placement. Use Eq. (4-41) to obtain an expression for the displacement u(x,t). DJs.. tinguish the cases (i) w ':/:: wk. an k ; (ii) w = w.,, some k. Case ii illustrates resonance for the vibrating string with fastened ends.

21 (a) For the case of nonresonance

4cF0 co w sin W111- 1t - w:,.- 1 sin wt . (2n - 1)7TX u - - · ·· · · sm----b n-l w:11- 'l(cot - w!.- 1) b

If w = a>,. (resonance. case), m even, solution is given: by the above series.. If w == w J<t-h

then in the series the term involving We.t - 1 is replaced by the term

sin w~_1t - w211- 1 t co w --.11 . (2n - l )1rx w:.-t sm . b

23 A pair of concentrated impulsive forces, of equal magnitude F0 but oppositely directed, is applied transversely to the interior points of trisection of the string with fastened ends. Obtain the formal series expression for the subsequent displacement~ Show that the mid,. point of the string remains at rest.

23 u :=; _ ~~0 I cos (tt1r/2)sin n7r/(i). sin (n1r:r/b

'lTC tt==l n

~_,..J1 ~Jla..J1 :&1)1 ~1 ( ~1~1 ~ ~~ ) jif...J11ill..J1 : 4.3 ~1

(d) :Wive the bOundary- ane1 mmal·vaiue problem

0 < X < b; t > 0

u(O.t)=h0 ubt)=h1 t > O

u(x 0) = x{b - x) ut(x,O) = 0 0 < x < b .

here F. 11, w h0, h1 are gi en real p itive constants. Distinguish the cases

(i) w =I= n1rcfb n = 1, 2, ...

(il) w = m"cfb m a positive integer.

25 (d) If w + mrc/b n = 1, - , . .•• then u = v + w where

v = X(x) cos wr

X = h0 sin [w(b -:- x)/c] + h1 sin w /c) + X,.(x) sin whfc in (wbfc)

• x) = -Fo [C'V sin (J)X/c) - w sin vx] j c v2 - w•)

. . w (X)

~ • mrx n'1t'Ct w= ,L;A,.stn-cos-1 b b

2ib A,. =- [x(b -b 0

( . • n'JT: dx - ,'Xi x)] s1n b

17 Derive the formal ties solution f he problem

u t- clu:r: O < x b•t > O ....... ...... -0 u bt)+hubt - 0 t>O

ut(x 0 0 S ..;_ b

wbe h1 and h~ ar gi n p · iti e constant

(Lft' )

~_,..31 4JJla.&ll :&l.;ll J-ill ( ~~~~ ~ ~~ ) _J:lf...JIIill..JI : 4.3 ~~

27 Let (It,.} be tbe sequence of positive roots (in increasing order of magnitude) of the transcendental equation

tan hfl> = p(hl + hJ p' - hlht

Let ru,. = cp.,., n .... 1, 2, . . . . Then (\()

u = L (A,. cos w,.t + B,. sin w,.t) sin (f',.X + fJ,.) n-l

A,. = ~ ('' f(x) in {llx. + 8,.) dx a..Jo B. = .-1-· fb g(x) sin (px,. e.) dx

«,.w,. Jo

19 a) Solve the boundar . ... and initi 1· alue problem

Uu - c%ua -c u(O,t) = h0 in wt

0 < < b• t > 0 ~ - J -

u(b,t) = 0

w reg, Ito Q) are given rea1 ositi · con· tants and

nne (,tJ :p -

b n = l, 2 ...

29 (a} u = (1 - i)ho sin rut+ w1 + w,

Wt = 2who I ( - 1)" 1 sin (n,.,xfb) sin (n,.,crfb)

11 n-1 nw,.

.,. n=l nw, w

1 = _ ~ }: {g[l - ( - 1)"}(1

1

- cos w,.t)

h 2 w, sin wl - w sin w,.t} . n11x + ow sm-n( w• - w,.')w.. h

31 Deri · the formal seri

tlu + 2')/Ut - cau'" - g u{O t) = 0 u(b t) = 0

solution. 0 < X < b• t > 0 -..- - ._

b -x u1 xO · = 0 O< x _ h

n1TC cu,.-T

4_,..11 4JJ~I :&1)1 ~I ( ~l..):!i:WI ~ ~.._>6 ) ~I oill...JI : 4.3 .l.i:JI

35 Derive tbe formal series soluti.on

Utt. + 2yu-e - c'f4:~ - F0 si.n vt

u 0 t) = 0 U:r(b,t) h1 in mt

u(z,O)· = 0 u1(z,O) = 0

0 < X < b; .t ~ 0

t ~ O

O<x< h ~ ~

n real positive constants.

!S Let p.. = (2n - l}rr/2h, w,. = (c!Pu" - 1"l"', n ;;;: l, 2., • • • • Then

u = lt1x. sin rut + 1.1

v(x,t) = e-'1•[111

•m i ...... ( -1): {sin m'flt sin Jl.X) + wJi. b n=l mJtw :

w(x t) = .I . t~ i·. ~.t e1s sin [m.,(t - !)JP.(~ deJ.• sin PnX. n - 1 m o

F.(t) = t. 2

.. • · t IP~:.Fo sin 1Jt + wh1(m sin wt - 21' cos wt)( -1) 1] t1Pn

~_,..JI 4JJLLJI :&1)1 ~~ ( ulj,!LWI ~ u.a~ ) jif.&ll &....11 : 4.3 ~~

37 Derive a formal series solntian of the problem

Eu - c'E.- = 0 0 :s; X 5, D~ t > 0

.l ;u(b,t) - EAf#:(btt) = 0 t(·O· ·)· ~ ,t == 0 t 2;: 0

f(x,O) = (b1 : b}x ~,(x,O) = 0 o s x<b

where M. E, A, b1 are given real positive constants.

37 ~ = I A,.X~(xj cos {J).t ,-1

wbere X.(x) = sin~ and f#n} denotes the sequence of positive roots of

EA tan bJ.L = - --

Mcf'J.L

A hl - b . b

n = b s Sln J.tn a.,.fln

1 ( _u_c_• s_in_. '_l'_nb) ct, = 2 b - EA

3Y Denve trte to.muu senes soruuon.

co" = cp .. ; n = 1,2 •••.

(a) uu.- c•u= - w'u = F0 sin vt 0 < x 5 b; t 2: 0 u(O.t) = 0 u(b,t) = 0 t ~ 0

x.O) = f(x) ul(x,O) = g(x)

where c, w, F0,. v are given real positive constants, w * v.

39 (a) Let Wn = (n'Tr2c3fb2-- wt)~, n = 1, 2,.... Then u = v + w wbtm~ co

c(x,t) = I (An cos (J),,./ + Bn sin Wnt) sin n?tbX n-1

. ( · ) 4F..·0 ~.. [v sin (J).".tn- lt - (J).:n- 1 sin. vt] • [(2n l) TrX ..... ] w x.t = - ~ . • sm - -' 11 11.;.1 (2n - l)Wsn-1(v2 - ()):,_1) b .

2lbfl) . nTrXd 2 i b ( :'\ • mtXd. "tt = - . \x ·sm-. x B .. =-- . gxJ sm-.. - .:x

b 0 b h(J)tt 0 b

('fJl

4__,..ll ~Jla..JI :&1)1 ~I (~l~l~~)~lclt..J1:4.3 ~I

1 Derive ~the forma} series solution of the problem

Uu - c!(xutll)tt = 0 0 < x < b; t > 0 lim u(x,t) exists u(b,t) = 0 t 2:. 0 :c-o

> O

.-~x~O) = f(x) u,(x,O) = g(x)

4t u - .~/·( •·t)ct~. cos"'·' + a. sm "'·'> 1 ib . (. ~·)· A .. =. y .....

0 Jo ~\}b f(x)dx

B. ~ . ! .... J.' J,( e.t) g(x) dx

O ~ x ~ h

where {'"} denotes the positive zeros of J0(~) arranged in increasing order, and w. =

ci./2Vb, n = 1, 2f ..•.

(Iff

o)~l ~Jla.o :U"""WI J...=,ill

41~'i l ~~ Jjl....... : 5.2 ~~

3 So1ve Prob. 2 if the initial temperature is gi n ,y

- 61 <X< <51

I > - c5t < X < ~61 Or l < X S t

el ewhere

h .r the nstants u1 u. b <51 are u h ha

0 < Ut < Ut 0 < 01 < 6t

1 The initial temperature in the infinitely long bar is

f (x) = u. f(x) = 0 lxl > ~

where Uo ~are given positive constant . There are no beat sources within the bar. Show that the subsequent temperature distribution for t > 0 is

u(x,t) = ~(erf~ - erf ~) 2 47TKI 4-rrKI

where the error function is defined by

etf x = -= e- TJ drJ 2 ill: t

..;7T o - ro < x <ro

3 u

5 The problem of heat flow in a. semi-infinite slab is mathematically identical to the prob­lem of heat flow in the semi-infinite bar. Assume that a homogeneous conductor occupies the half space x ;-::: 0 in x z space. The temperature on the face x = 0 is h(t) at time t > 0. The initial temperature in the slab is f(x) , x > 0. Thus the ub equent heat flow is one­dimensional. Assume there are no heat sources within the lab. Determine the subsequent temperature for each of the following cases.

(a) h(t ) = 0; f x) = u0 (u0 a positive constanc)

(b) h(t ) = uo; f(x) = 0

(c) h(t) = u1 ; f(x ) = u2 (u~> u: nonzero constants)

(d) h(t ) = u1, 0 < I < th h(l) = ul> 11 < t < 1:. h(t) = 0 t > t: ;f(x = 0

X 5 (a) u = uuerf-=

4Kt

X u1 - u1)erf-

4Kt

o).;a.JI (.]Jts.. :<..Jo"AWI J-ill 41~'i l ~I ~'1........ : 5.2 ~I

7 The model which describes the heat fiow in the infinitely long bar in which the insulated urface i replaced by a condition of radiation into the exterior region (held at temperature

zero) i

u,+bu - l(u= Fixt)

u .r.O) = f(x)

-eo < x < eo; t > 0

- eo < x <eo

here b i a po itive con tant. Show that if w is a solu1ion of problem {5-11), where F .r,1)i replaced by e •1F(x t), then

i a olution of the original problem. Thus the fundamental solution of the original problem i

where vi defined by Eq. 5-13). Find the temperature in the bar fort > 0 if there are no heal OUI" and the initial temperature i u0, a po itive con tant.

7 u u .X

'erf--

lJfbl

oJ~I (.JJla.o :(.)o"AI.i.ll ~I (4iljJI JIJ.l.ll ~_):,) o_;l~l (.JJI.a.J ~~~ J 4JI~')'I ~I Jjl.......: 5.3 ~I

11 A slender homogeneous conducting bar of uniform cross section lies along the x axis with ends at x = 0, x = L. The lateral surface is insulated. There are no heat sources within the bar. The bar has the indicated end conditions and initial temperature. Derive the series solution for the temperature distribution in the bar for t > 0. {a} Ends are at zero temperature; f x) = x , 0 ::5 :x < L/2, j(x) = L - x, L/2 .$ x :::;; L

(b) Ends are at zero temperature;f(x) = 3 sin 21rx/L, 0 ::5 x < L

(e) u(OJ) = Utt u(L,t) ""' u,, uh u, nonzero constants ; u(x,O) =f(x), 0 < x :::;; L

(d) Ends are insula.ted;f(x) = x(L - x) 0 :S x < L

(e) End x = 0 held at temperature zero at the end x = L there is a constant flux q0 ;

u :x 0} = J (x , 0 S x .$ L

11

· (x - ~ [-·B· n• l

l)"·u··. 1 - a1] • n .1r,X.· -n~ . IC•I " "' · · · · · sm - e 11 • "' .nrr L

2 i.•L nwx B = - · f· x\ in - d.x . L · I .. L 0 . .

~. r .. L . .. · .. , (2n - .l)rrx . L J£ f x sm 2L dx

13 The lateral surface of a slender homogeneous conducting bar of length L is insulated. The temperature at the end x = 0 is maintained at A sin w1t for 1 ~ 0, where A, <vt are positive constants. The end. x = L radiates heat into the exterior region . which i.s at temperature zero. The initial temperature in the bar is 110 , a constant. The heat·source dens1ty within the bar is

F(x,t) = F06(x - x0) cos wt

where x. is a given poin and F0 , w are positive constants. Determine the temperature in the bar for t > 0.

o.JI_~I ~.ll.a.4 :U""'W\ ~\ (~Ill\ JIJ.l.ll ~_;h ) o)~l ~Jla..J ~..u.l\ J ~l~'il ~\ J.'t........ : 5.3 ~\

sin w1t J3 u = A[l + h L - x)]

1 + hL + v(x,r)

00

v(x,t) = L [C,.(t) + B.] sin~ n=l

where {,u.) denotes the positive roots of tan fi.L = -~J/h, and

[F0 sin. ftnXo • .11 :t

C,.(t) = 2

, (~efi.,.• cos wt + w sm wt - tC!J.'e-/1( • "p.• + (I)

- .. (,I A~t •) (~<p., 1 coswlt + (l)l sin wlt - 1Cfi.·1e.-"P.•''>] •.. ;ux .. u• p.,. " p.,. + (/)1

lJX .. U'· = ~· + eos~.,.L B. = 11; .. 11* f.Lf(x) sinp,.x dx

15 The initial temperature of a so1id homogeneou c:onducting cylinder of radius a is

( r'

u0 l -­a'

where r is the distance from the axis. The temperature on the lateral surface is zero. Assume the cylinder is infinitely long. There are rto heat sources within the cylinder. Dete.rmine the temperature di tribution in the cylinder for t > 0.

17 A solid homogeneous circular cylinder of radius a and altitude h has its axis coincident with the z axis. The initial temperature is

u(r;O,t,O) f(r 8,z)

where r, 8 .z a.re cylindrical coordinates. The temperature of the base is held constant. at temperature u0• The top and lateral surface are insulated. Determine the temperature in the cylinder for t > 0.

co co !l()

+ ! ! (A""''V'~~41(r,O,z) + B.mJP~~(r,8,z)]e-~e.t.~~«<t n- o m- 1 <t- 1

VfSl

O)_po.ll ~Jis.. :(.)"'1\.i..ll J-i,ll

(~ljjl JIJ.l.ll ~.,_;.6 ) o)_po.ll ~JUt...! ~~I J 41~~1 ~I ~L....... : 5.3 ~I

where

n = 0, 1, • •• ;. m = 1, 2, . • • ;

q=l , 2:., ...

and, for each n, {~nml denotes the sequence of positive zeros of J~(E),

- 4u0 . . 2 J.a fbJ.b . . . . (2q = l)1rz . . .. Aq = (2q ....,. l)n-Saa + 1ra*b

0 J

0 0 f(r,6,z) m 2b r dr d(J dz

A,._ = f l ll' fa (21tr.• i(r,6,z)rp~;!(r,8,z)r dr d(J dz n = 0, 1, • •• ; m = 1, 2, . • • ; '~'··~ JoJo ··o

q = 1,2, •. •

1 iai2.ll' fb B.,,... = II V'nmt~ !l•

0 0 Jo f(r,fl,z)!p~~(r,O,z)r dr dO dz n = 1~ 2, • • • ; m = 1, 2, .•• ;

q = 1,2, • • ~

( ~-')· (?n - l)gz tp~ = I" (1 • · cos n8 sm .""'1 2b

20 A homogeneous solid sphere of radius a has the initial temperature distribution

0 :5 r :5 a

where r denotes distance measured from the center. The surface temperature is main­tained at zero. Show that the temperature in the phere for t > 0 is given by

u(r l) = lia• i ( - 1)"-t sin (n;r/a) e-Kn'r'tta'--.!Jr n ... l n

21. Solve Prob. 20 if instead of the surface temperature's being zero the surface radiates heat into the exterior region. The temperature of the exterior region is zero.

· tO

11 u = ! I An sin (pn,r)e-KP.n'~t whe.re {p"} denotes the sequence of positive roots of r n• l

tan ap ·~ ap/(1 - alt) and

a•pfl + (ah - t)• lh - a~t.' . A,. = 4 t .. • +·· t.( .L 1) , t smap, a 'f.l11 an an - 11.11 ..._

0) ..r-J I ~J\a.o : c.>'"" WI J-ill (~ljJI JIJJ.ll U..~ ) o.JI..r-JI ~JLI..J ~.b.ll J 41~')'1 ~I Jjt....... : 5.3 ¥1

23 A spherical conducting shell has inner radius a and outer radius b. Through the in.ner wall from the interior there is a constant flux of heat f/o• The outer waU radiates into the exterior region which is at temperature zero. The initial temperature of the shell is a constant u0• There are no heat sources within the shell materiaL Derive the formal series solution of the problem.

qoa• (bh - 1 1) 23 u = ~ • bsh - r .• + W(r~t

Ill()

w(r,t) = ~ A .... 'tP .. (r)e,...Ks.•t

where {;m} denotes the sequence of positive root& of the equation.

elj~(aE)y~(bl)- y;C~)j;(hE)J + hfj~(af)yo(b~) -y~.(a!>jo(b~)J = o

1 [·· . f.· b . . . .• .· q0a'J.• ··11 (bh - 1 . 1).' . . J A,. = - . . ' .. ":o ... · tp .. (r)rt dr - - . . . .. . - -.. •Y.m(r)r1 dr .. ... . a · · K b2h r · "' 4 a

«.. = f fAo'{r)r* dr }o(x) = .ji; Jw(x) y0(x) = .ji; Y3i(x)

(?Ol