Partial Differential Equation Exercise

Tpuo* $1, k,,

P^h* .*L V7 er&ia"

"tU^* 01wc;v"

hn"x 31o,*to"

tNru^*rao bvetrtr"x h<-,{&or s

b,W

Ex*reEses {SrSeati*FSE necture}

81jE$1f#iu 3 {WAVr SQl},&',rr#[\r}

A string of length L is stretched and fastened to two fixed point and displacement of the string is
given as

v{x,o) : I x, O. *.i L
IL-x,
L

i.r.L

The string is released with zero velocity. Applying the equation

)ztL 1 07tL

0x2 c2 0t2

with cz : 1 and using the separation of variable method, determine the subsequent motion u(x,t)

QUESTTBN 2 {WAVE EQUeTTSAii

An elastic string is stretched and fixed at two points J meter apart. The center of the string is
displaced vertically d meter upward from its rest position (as shown below), and then released with
initial zero velocity.

a{-r,0) lnitial Displacement, at llxed time r: 0

ilt' z/ -- x
6a
! I

2

Find the solution for this problem by obtaining the displacement u(x, t) at any time f with respect

to location x based on its rest position.

Use wave equation below with c = 1to obtain u(x,t)

azu ) d2u

dt, dxz

Given that

ft(x) :**, ir{r) :){t * r)

QUHSTTSTS 3 {WAVr EqUAYtSfU}

Suppose a L0 m long flexible string is stretched between two fixed points at the time f . lt is initially

:at rest with a point corresponding to n 5 displaced at 3 meters upwards and then released with

zero velocity. we describe the motion of the string by the 1D wave equation

A2u 1 Oztt
0x2 c? 1tz

where c2: L anduarethedtsplacementof thestringfromitsreststateatthepointsxandtimef.
since the string is fixed at r in [0,10], the boundary conditions are as foilow:

u (0, r) : 0 and u (10, t) : 0 for all times r

Consider the initial conditions for the string are u(x,0) : f (x)) where the function f (x) is given by

(2h 0<x<L/2

f(x): )tr
t?,L x), L/2<x<L

and ut{x,0) = 0.

Using the method of separation variable, find the solution to the wave equation

qufrsTroru €. {HEAT EQUATrorq}

Solvethe heatconduction problem, {}:i#, , < x <2, t> A,

Subject to the boundary condition u(0, r) = u{2, f) = 0

and initial condition u(x, 0) : 19

ftL' rsYF*fq $ i!.ifi&T fi ffi{",eTi#r.{}

Solve the heat conduction problem, :#:ff, O < x < 3, I > 0,

:Subject to the boundary condition U(0, t) U(3, t) : 0

and initial condition U(x,0) : Ssin4rcx

SUrSTr{3N 5 {F{EAT rguATtCIr{}

Solvetheheatequatior,:*:#,0 < x ( 5, r > 0,giventhatu(0,f):u(5,f):0,

uf5,t]:s

I

Ji

x=8 x=5

lf the initial temperature: u(x,0) : 4sinSrx - *B sin 3rcx 1-A sinZrx

Hmercis*s 2 {Dr$e**i_F*fi lectr*r*}: &JLJfWERIeAL METHOf

QUHSTISTU T

Solve the heat conduction probiem , * = k{} usingNUMER|CAL METHOD.

Subject to the boundary condition u$,t) : O,u(2,t') = 3t* 2 and initial conditionu(x,0) : +

Giventhat, L,x: A.25, Ar:0.1, k :0.1 and rrangesfrom 0to 1

ffi-#rs?r#eJ 3.

# k#Solve the heat conduction problem , = .

Subjecttothe boundarycondition u(0,t):u{l,1) = 0 and initialcondition u(x,0): f (x)

where, f (x):l|sin(T)- stn(T).ro Lx:0.25,Ar:0.1, k:0.1 andL: L

Use 5 iterations,

(i) using separation method
(ii) using numerical method
(iii) then, determine the absolute error

8{JESTISr1I3

: :The initial temperature of an iron bar is zero (?" 0"C ) for all points on the bar, except at x 0.0m
: I :as shown in Figure 1. The bar is heated at x 0.0m where the temperature is kept at
B oC at all

time. The bar is fully insulated. Therefore, the temperature at the other end of the bar is always

same with the temperature at its neighboring point (#: O at x : 1.0m)

T=B oC T=0 oC

0m x=1.0m

i'j

r*!l

Figure 1: lnitial condition of the bar

a

Please state the general formula to approximate the temperature T(x,t) using numerical method.

Then, by using calculator, calculate the value of temperature at any point r and any time f from t =
0 until t = 0.15 forAr : 0.025, Ax : 0.25 andthethermal conductivity k : 1.

Leave your answer in 4 decimal places.

'E*row : u\t sve €7vclu,

arA F<r,t neLv&'on .

(a."''*tr- f tr^e*c.oC)

Qoosko, ki

l/

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i CL )r L

}1

)n^-'

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O -tr Lor*) > o
e jY(vrt) > 0

tt ,f lrof (br,),,I;oo, : tI 14_ L.r,2
O {.$- ("* , o) > + rL) L-t
c
o ry 6 tu)
)* e -*o

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Jt"t >yt ctrr d TLv x aT

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ry y.'T jr'tn *rt ? f-'T fl

J+ !oz

* "1- ^\ f q

>\L ){c L

xf yl
I

'/ .l //

x =ts

T

$,ru .{e ?,.,J ov d-o-..' a De o*o-e

Y

r) ilt Y=D

d" =D -\ x>o n4x c E. i ti r&A

i,'/ if L is'
fos,ff"e

x >-a h,64 o e-ziit c"r+wy

firl ,+ tc ts' y"t,* w = -1

x A CEr f\ + B hn {*

I KT- =1) + D l-n(t

$'-;w* T LGyt

)L Lu 7't) * (fr) T c+)

(a G* (- t i5 l!i,, l*) c c crJ (t + D 9- (r)

b,r.o ,"v Ca"4irr"'

U '-* A vtWn )t =O UI ) =-0

a .2 (x cos b) L !ii'' ,o, ) c L tC-b fL "e D S' -gt)

4 *=o

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a: (f g-.{L) ( c (zr 4L + D 8n f')

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ac sti c \'*' o
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rt** ) 169 n fl li ftpt l- tt' Sln no(r4; ),
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n,tll L VL

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-U (ti.c) l,t-(-u1 '$ <'it <- i

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x

SU,u r-&u 2 hJ at dLzt ODE a'9;tt'4

5/

x l<-x =C n+r cg e trc lu-,y1

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s[(rrrt) - x(')'t) T(t)

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=

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$" t-

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cX Ll,
I b\7 -7JLT t

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C u( rC,1),L

ln'itr""rr- L<r,., Ab'a.: ' ,)e 14- A <- '14 -LJ /2 V:)i L:lD

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l .:n / a<.u.<5
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t\/

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o C

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c

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7

\*;\,r,r (*un*b,,-n

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u [r,t, o) &"i f () vi 11 t.a

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c,

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3 ?

q ('trt) .-L ( 4i'r g" ( r1
I )*

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@ u[5,t) -e

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x

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x

{t + -.2 =Q Y: A Cos -f r. "+ U 9i^ (u-

{Y

.L 'f/ ---t L T c-< ->f2 +

T ' , *a-l,2 T
1' * tlLr "o

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