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EUT 203 1
Chapter 1
Partial Differential Equations• Introduction
• Small Increments and Rates of Change
• Implicit Function
• Change of Variables • Inverse Function: Determine Partial Derivatives
• Stationary Point
• Applied Partial Differential Equations
UniMAPUniMAP
EUT 203 2
Introduction
• Consider the following function
f (x1, x2,…, xn)
where x1, x2,…, xn are independent variables.
• If we differentiate f with respect to variable xi , then we assume
a) xi as a single variable,
b) as constants.nxi xxxxx ,,,,,, 1121
UniMAPUniMAP
EUT 203 3
Write down all partial derivatives of the following functions:
22
22
32
32
322
343
)3()2(2)1(3
26
)1()1(2)2(3
23)
xyxyx
yxyxxy
f
yyxy
yyxyx
f
xyxyyxfi
Example 1.1
32
3)
sin)zxyefiii
xyzfii
UniMAPUniMAP
EUT 203 4
1.1 Small Increments and Rates of Change
Notation for small increment is δ. Let z f (x1,x2,…, xn). a) A small increment δz is given by
where δx1, δx2,, δxn are small increments at variables stated.
b) Rate of change z with respect to time t is given by
where are rates of change for
the variables with respect t.
nn
xx
zx
x
zx
x
zz
22
11
dt
dx
x
z
dt
dx
x
z
dt
dx
x
z
dt
dz n
n
... 2
2
1
1
dt
dx
dt
dx
dt
dx n,,, 21
UniMAPUniMAP
EUT 203 5
.hh
Vr
r
VV
.1.0 and 2.0
,10 ,5
cmhcmr
cmhcmr
Example 1.2 Given a cylinder with r cm and h cm.
Determine the small increment for its volume when r increases to cm and h decreases to cm.
Solution
Volume of a cylinder is given by V r2h.
Small increment is
Given
KUKUMKUKUM
EUT 203 6
.5.17 increases Volume
.5.17
)1.0)(25()2.0)(100(
Hence,
.25)5(
and
100)10)(5(2
2
have, We
3
3
2
2
cm
cm
V
rh
V
rhr
V
KUKUMKUKUM
EUT 203 7
Exercise 1.1 The radius r of a cylinder is increasing at the rate of
0.2 cm s while the height, h is decreasing at 0.5 cm s. Determine the rate of change for its volume when r 8 cm and h 12 cm.
.1.20or 4.6
of rate at the increasescylinder of Volume1313 scmscm
Answer
KUKUMKUKUM
EUT 203 8
1.2 Implicit Functions
• Let f be a function of two independent variables x and y, given by
• To determine the derivative of this implicit function, let z f (x, y) c.
• Hence,
.constant,),( ccyxf
dx
dy
.0..
dx
dy
y
z
dx
dx
x
z
dx
dz
.yzxz
dx
dy
KUKUMKUKUM
EUT 203 9
Solution
Let
.32
)(2
.32
)(222
.02
2
2
32
yx
yx
dx
dy
yxy
z
yxyxx
z
yxyxz
yzxz
Example 1.3
. ,02 32
dx
dyfindyxyxGiven
KUKUMKUKUM
EUT 203 10
.1)0,0(
.(0,0) ,2
dx
dy
atdx
dyevaluateyxeGiven xy
Exercise 1.2
Answer
KUKUMKUKUM
EUT 203 11
1.3 Change of Variables
• Let z be a function of two independent variables x and y. Here x and y are functions of two independent variables u and v.
• We write the derivatives of z with respect to u and v as follows
...and
..
v
y
y
z
v
x
x
z
v
z
u
y
y
z
u
x
x
z
u
z
KUKUMKUKUM
EUT 203 12
Let z exy, where x = 3u2 + v and y = 2u + v3. Find
.v
zand
u
z
.3 and
26
2 xyxy
xyxy
xevyev
z
xeuyeu
z
Answer
Example 1.4
KUKUMKUKUM
EUT 203 13
Let where
Find
.sinandcos ryrx
.
z
andr
z
,22 yxz
.0 and 2
z
rr
z
Answer
Exercise 1.3KUKUMKUKUM
EUT 203 14
Let be n number of functions of n variables
i.e. ,,,,, 321 nxxxx
.,,,
,,,
,,,
21
2122
2111
nnn
n
n
xxxff
xxxff
xxxff
nffff ,,,, 321
Jacobian for this system of equations is given by the following determinant:
Definition 1.1 (Jacobian) KUKUMKUKUM
EUT 203 15
.TJJ
n
nnnn
n
n
n
n
n
nnn
n
n
n
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
xf
JorJ
321
3
3
3
2
3
1
3
2
3
2
2
2
1
2
1
3
1
2
1
1
1
321
33
3
3
2
3
1
22
3
2
2
2
1
11
3
1
2
1
1
Note that both Jacobians will give the same answer, since
KUKUMKUKUM
EUT 203 16
,sincos yevandyeu xx
.2cos
sincos
sinsincos)cos(
cossin
sincos
22
y
yy
yeyeyeye
yeye
yeye
J
xxxx
xx
xx
yv
xv
yu
xu
Example 1.5
Given determine the
Jacobian for the system of equation?
Answer
KUKUMKUKUM
EUT 203 17
Given u xy and v = x y, determine the Jacobian for the system of equations?
Answer
J y – x
Exercise 1.4
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EUT 203 18
1.4 Inverse Functions: Determine Partial Derivatives
• Let u and v be two functions of two independent variables x and y, i.e.
uf (x, y) and v g(x, y).
• Partial derivatives are given by
where J is the Jacobian of the system of equations.
vy
uy
vx
ux
and ,,
Jv
y
Jv
x
Ju
y
Ju
x
xu
yu
xv
yv
,
,
KUKUMKUKUM
EUT 203 19
.2cos
cos
2cos
sin
2cos
sin
2cos
sin
2cos
sin
2cos
cos
.2cos
y
ye
Jv
y
y
ye
y
ye
Jv
x
y
ye
y
ye
Ju
y
y
ye
Ju
x
yJ
x
xx
xx
x
xu
yu
xv
yv
Example 1.6
Given
evaluate
Answer
,sincos yevandyeu xx
. ,, vy
vx
uy
ux and
KUKUMKUKUM
EUT 203 20
Let z 2x2 + 3xy + 4y2, u x2 + y2 and
v x + 2y. Find
Solution
.v
zand
u
z
)2( ..
)1( ..
v
y
y
z
v
x
x
z
v
z
u
y
y
z
u
x
x
z
u
z
Example 1.7
KUKUMKUKUM
EUT 203 21
Hence,
yxy
zyx
x
z83,34
The Jacobian is
).2(2
24
22
12
yx
yx
y
xJ
yv
yu
xv
xu
KUKUMKUKUM
EUT 203 22
Therefore,
.2)2(2
2
2)2(2
2
)2(2
1
2
1
)2(2
2
yx
x
yx
x
Jv
y
yx
y
yx
y
Jv
x
yxJu
y
yxyxJu
x
xu
yu
xv
yv
KUKUMKUKUM
EUT 203 23
From (1) and (2), we have
.2
343
2)83(
2)34(
,)2(2
25
)2(2
1)83(
2
1)34(
22
and
yx
yxyx
yx
xyx
yx
yyx
v
z
yx
yx
yxyx
yxyx
u
z
KUKUMKUKUM
EUT 203 24
1. Let z x3 + 2xy + 3y2, u 3x2 + 4y2 and v 2x + 5y.
Find
2. Let z x3 + 2xy + 3y2, x 3u2 + 4v2
and y 2u + 5v. Find
.v
zdan
u
z
.v
zdan
u
z
Exercise 1.5
KUKUMKUKUM
EUT 203 25
Answer
).62(5)23(8
and
)3(4)23(6)2
.815
812186
and
1630
2415 )1
2
2
222
2
yxyxvv
z
yxyxuu
z
yx
yyxxyx
v
z
yx
yxx
u
z
KUKUMKUKUM
EUT 203 26
.
2
2
3
2
2
2
1
2
3
2
23
2
23
2
13
2
2
2
32
2
22
2
12
2
1
2
31
2
21
2
21
2
nnnn
n
n
n
x
fxxf
xxf
xxf
xxf
x
fxxf
xxf
xxf
xxf
x
fxxf
xxf
xxf
xxf
x
f
H
Let f be a function of n number of variables
Hessian of f is given by the following determinant :
.,,,, 321 nxxxx
Definition 1.2 (Hessian)KUKUMKUKUM
EUT 203 27
Hessian of a Function of Two Variables
Let f be a function of two independent variables x and y. Then the Hessian of f is
...22
2
2
2
2
2
22
2
2
2
xy
f
yx
f
y
f
x
f
y
f
xy
f
yx
f
x
f
Η
KUKUMKUKUM
EUT 203 28
Since x and y are independent, we have
The Hessian becomes
.22
xy
f
yx
f
..22
2
2
2
2
yx
f
y
f
x
fΗ
KUKUMKUKUM
EUT 203 29
.128
Answer
?
.2),(
222
222
yxyH
fof Hessian the is What
yyxxyxffunction a Given
Example 1.8
KUKUMKUKUM
EUT 203 30
1.5 Stationary Point
• Given a function f f (x, y).
• The stationary point of f f (x, y) occurs
when
and0
x
f.0
y
f
KUKUMKUKUM
EUT 203 31
Properties of Stationary Points :
1) If
This stationary point is a SADDLE POINT.
022
2
2
2
2
yx
f
y
f
x
f
Figure 1.1 Saddle point
KUKUMKUKUM
EUT 203 32
2) If
and whether
a) and
This Stationary point is a MAXIMUM POINT.
0.22
2
2
2
2
yx
f
y
f
x
f
02
2
x
f0
2
2
y
f
Figure 1.2 Maximum point
KUKUMKUKUM
EUT 203 33
b) This stationary point is a MINIMUM POINT.
02
2
y
f0
2
2
x
fand
Figure 1.3 Minimum point
KUKUMKUKUM
EUT 203 34
3) If
0.22
2
2
2
2
yx
f
y
f
x
f
This test FAILS.
KUKUMKUKUM
EUT 203 35
)4( 02
)3( 033 22
xyy
z
yxx
z
Example 1.9
Determine the stationary points of z x3 – 3x + xy2 and types of the stationary points.
Solution
Step 1 : Determine the stationary points.
KUKUMKUKUM
EUT 203 36
• From (4), we have x 0 or y 0.
• When x 0, from (3);
• When y = 0, from (3);
• Hence, z has four stationary points, i.e.
.1
033 2
x
x
.0,1 and 0,1,3,0,3,0
.3
03 2
y
y
KUKUMKUKUM
EUT 203 37
• Step 2 : Compute the Hessian of z.
).3(4
412
226
.
22
22
2
22
2
2
2
2
yx
yx
yxx
yx
z
y
z
x
zΗ
KUKUMKUKUM
EUT 203 38
Step 3: Determine properties of the stationary points based on Hessian of z.
Point Hessian Conclusion
Saddle Point
Saddle Point
Maximum Point
Minimum Point
KUKUMKUKUM
3,0
3,0
0,1
0,1
012
012
0,0
012
2
2
2
2
y
z
x
z
0,0
012
2
2
2
2
y
z
x
z
EUT 203 39
Determine the stationary points of
f (x, y) x2 x y y2 5x – 5y 3 and types
of the stationary points.
Exercise 1.6KUKUMKUKUM
EUT 203 40
1.6 Partial Differential Equations
What is a PDE?
Given a function u u(x1,x2,…,xn), a PDE in u is an equation which relates any of the partial derivatives of u to each other and/or to any of the variables x1,x2,…,xn
and u.
Notation
xy
uu
x
uu xyx
2
,
EUT 203 41
Example of PDE
equation) (Wave )( 5.
equation)(Heat .4
.3
32 .2
0 .1
2
2
2
yyxxtt
xxt
uyyx
yx
x
uucu
uu
eyuxuu
uu
uu
EUT 203 42
Focus first order with two variables
PDEs We can already solve
Example. of ),( solutions all Find (a). 1. 22 yxuyxu x
Solution
)(3
)(
23
22
yfxyx
dxyxu
By integration
EUT 203 43
(b). Solve PDE in (a) with initial condition u(0,y) y
Solution
yxyx
yxu
yyf
yyf
yyu
yfxyx
yxu
23
23
3),( So,
)(
)(03
0
),0(
)(3
),(
EUT 203 44
Separation of Variables
Given a PDE in u u (x,t). We say that u is a product solution if
)()(),( tTxXtxu
for functions X and T.
How does the method work?
Let’s look at the following example.
EUT 203 45
t
t
etugiven x
u
t
uii
etu givendt
u
x
u i.
SolvetTxXtxuLet
2
2
6),0( ,23 .
10),0( ,
).()(),(
Example 1.10 KUKUMKUKUM
EUT 203 46
Solution
kT
T'k
X
X'
kT
T'
X
X'
T'XTX't
u
x
u
T'Xt
u TX'
x
u
X(x)T(t) u(x,t)
and or
, equation From
and
(5) Given
KUKUMKUKUM
EUT 203 47
bktT c kxX
bktbT ckxcX
k dtT
dT k dx
X
dX
k dtT
dT k dx
X
dX
kTdt
dT kX
dx
dX
kT T'kX X'
lnln
lnln
:sidesboth Integrate
and
Therefore,
2121
KUKUMKUKUM
EUT 203 48
. where
solution General )(
))(()(
:(5) into (7) and (6) Substitute
(7) )(
(6) )(
where
have wesides,both of lexponentia theTaking
ABD
Dex,tu
BeAex,t u
eB
eA
Be T Ae X
.eeT .ee X
e T e X
ktkx
ktkx
b
c
ktkx
bktckx
bktckx
KUKUMKUKUM
EUT 203 49
2 and 10
10
Thus solution. general theinto 0put
,10),0( satisfyingsolution obtain the toThen,
20.
2
kD
eDe
x
etu
tktk
t
txetxu
kD2210),( :solution general the
into 2 and 10 substitute Finally,
KUKUMKUKUM
EUT 203 50
xk
tk
BeXAeT
kX
X'-k
T
T'
kX
X'
T
T'
TX'T'Xx
u
TX'x
u T'X
t
u
tTxXx,tuii
23
and
2 and 3
Therefore,
23
23
:2t
u3 equation From
and
)()()(Given .
KUKUMKUKUM
EUT 203 51
xt
t
xk
tk
tk
tk
etxu
etu
CeBeAetxu
32
2
2323
6),(
,6),0(Given
),(
issolution general theSo,
KUKUMKUKUM
EUT 203 52
37 , (0, ) 10 .tu uSolve given u t e
x t
Answer 21 3( , ) 10 x tu x t e
Exercise 1.7
KUKUMKUKUM
