( cA ) LN(~) ~-~:
\)J
l
0) ::;.
Sl)< --S) J'I-
0
l ::;
""')
t?
-\
CT
C
~: - ~olx
-=-
C
~
c - -t
t
): -
fr o(t =
C-
~
-
t (i - i):.-
C-
Y><
;
~ )(
t:
I
l
C -- ..L
Y
J..
4
I---~
.L-
-------------------~
-
i2
>
J..
2-
3
3
x
2. Keep the same equation -(eu')' = o(x -1j3) as Problem 1 with jump in e(x).
(a) What is the weak form of this problem? (Tell me the equation that holds for every
test function v(x).) Also tell me what boundary conditions are imposed on u(x) and
v(x).
(b) Draw the graphs of three piecewise linear finite elements <Po(x) , <p,(x), ifJ2(x) with step
'/3, satisfying the appropriate boundary conditions on trial functions for tills
problem. Describe the graph of Uo <Po(x) + U, <p, (x) + U2 <P2(X).
size h =
(c) These <Po, <p" <P2 will be trial functions in U(x) = Uo <Po(x ) + U, <Pl(X) + U2 <P2(X). The
same <Po, <P" <P2 will be the test functions Vo, V" II.,. From the weal, form in part (a),
with jump from e = 2 to c = 4 at x = 2/3, find the matrix J( and right side F in the
finite element equation J(U
"') - I'Ie III') I
-=
= F.
(Not necessary to solve for U = (Uo, U" U2).)
S- ()( - 3) ~ J\'C vi I v I VI.)(
J
-:.
11 (' ( ')
<f x -~
o
for"
v&lx
0
~ -ks;+ fv--.c..DOf"'\ viI<)
Such. 4/"0,+
VI
(I) "" \Ill) ~ b
ot
U, ti, (K) +
Uz/ P,&('f.) I' S ~ FE-(VI C,Pf"'Ox,'IVlr;"n o ,-"
-I-v til ( )() S\/(,[" +t-.. ",-l- OJ + e~ .. /."
""'.L-~l,po, · r.+ X" 0 I t / t -/-/"e 8~t"~
eel vOl I.> u" J V, / OJ Vlvl U.z.. ) t"e>f€ c.. h·I.A I'if
C.I IAO( -~ 5 ~c, f f-."
,'s
( / /\ eCj('
~/\ I -::f.v, s /"-'o,f- I U ( x.)
-rl...t. 'y"'p'"
C,
Vo(,zUi<.)
-I-
...
~~ c
::
.5
¢: ¢:cAlC s: L¢: rf:~«
-t
3.
'I--------~ --~
.b
-l:,
-10
0
/ 2 -f:,
o -("
('6
~-------------o
Fz.-::;
r ~,~()(-J)c;()C
o
=
(
_
3. (a) Find the gradient v
= grad u
of u(x,y)
= x + x2 -
y2. Then set w(x,y)
= v(x,y).
Then compute the divergence div w(x, y). Then find a stream function s(x, y) that
solves EJs/EJy = wJ(x,y) and - EJs/EJx
=
W2(X,y).
Verify that u and s solve the
Cauchy-Riemann equations. If u + is is a function F(x
+ iy) , find
F.
(b) (Connected to part (a)) Suppose Laplace's equation -Uxx - U yy = a holds inside the
unit circle x 2 + y2 = 1, and on the boundary u = uo(O) = cos 0 + cos 20. What is the
solution u(r, 0) at all points inside? What is that same solution written as u(x, y)?
Find the flux
J W· nds out of the circle, either directly (the vector n =
(x, y) points
radially out) or by the Divergence Theorem.
-2..'()
I +- LX
;3s
=-
", Yo
2:L
dX
~
d'd
:;:::.
.. 1
VV 2
;;:
1+ '2..-y
:::
d5
7ft
- L-'rJ - _os
)x
tA -+ t' "> -=
X~ T X
+ L(l-
J
+2. x ~l- - (J 2.
F :::: ( 'f--fL(J)2. + [ Xtl
O
)
: Il )( -rl~)( )< -t (' ;:0/
ViL,/e)
-=-
1J.1>l)~) ~
r(.PSe + (2- l2-r...os'e
'i -t
2-/- - /- -
~\di\lW dxJ'1r _
",rtle
'(J 2.
-I ) - ,0056 +
= \~ + )«~~
@] -: : PIJY
00
2.-CjWse)'--- j'-
c:rt 0"rc.l~
4. (a) Suppose </>(x, y) is a piecewise linear function with </>(0, 0) = 1 at t he center of this
Square of side 2, and </> = 0 around the boundary. Find t he formul a a
+ bx + cy
for
</>( x, y) in each of the four parts of the square.
</> =0
(-1,1 )
(1, 1)
1"1
</>=0
~
</>=0
(-1, -1)
(1, -1 )
</> =0
(b) Find the weak form of Poisson's equation with u
8'u
- 8
0
x-
-
= 0 along t hat square:
8'u
1
8 0 = 51/,,0 = point load at x = -2' Y = O.
y-
Write this weak form neatly (symmetrically in u and v on the left side, explicitly on
the right side) .
(c) Use only that one trial function </>(x, y) (it is also the test function) from part (a).
Substitute U, </> for u in the weal, form , and substitute </> for v. Integrate over the four
parts of the square to find the equation for the number U, (here J(U = F is only one
equation). Then U(x, y) = U, </>(x, y) is the finite element approximation to the exact
solut ion u(x, y).
..)
~(. ~vr
' T, ~
(X1(+>
1+ '1-
Tz.. : I - X
T3 : 1
T~I
-Ij-
: 1+ 'f
q
r{.
I C,!::e-lec( i,. ., +L--uL
oI 'c'fjrc,,,,,,,
r
...;) I )~ 'v1"V x T
ulJvlf ol.x cA~ =
== v( 1.
z--,
~
V;
Pr"..-..J- ~ of
+-k
OvV T , :
$tt U'" re, ""re..
.
'/"'Ibe Ie&'!
,-~
,
$&1
,Dc.
I~
c,s
T,
1-
r+- (cd. ,
t1'(d(J-"
+
-T
H'2. tJ ,¢/-C/J'I.+ U ,0';) rlJ-a otxpl~
0)
'
V\<..e..
~r t),~//J)l+ UI(~'IJ {J~rJ.x,).'J- ~ f)s~,o ¢
T,
T2..:
vO{XCll ~
Sf XLo
~tl ~f"{..
SC"IVo.. \""Q...
-=-f~ ~~,o
¢ ol)C v(cy
2.
T:J,
Ove.s
~ ~rT3 u, ¢x ¢ x + U 1¢li¢-a cA'(d'a ~ IT
L ¢ ,,{ /< vl'f
73
,- , 0
~
1"
Ovv
T '1
~ ~-r4 U
I
IA. ¢)< -t V,¢ 'd ¢'a of 1<01 (J
~
;:, [ .'( t, 0
~
\J;
U, ( I) ( \) -+ U I
(\H
I) + U I ( I )l~
y- U 1
=
¢di<d;;t.-
~
l; I
(
(H t) -
tH-
~
-t
0 + 0