DIFFERENTIABILITY OF REAL VALUED FUNCTIONS OF TWO...
Transcript of DIFFERENTIABILITY OF REAL VALUED FUNCTIONS OF TWO...
DIFFERENTIABILITY OF REAL VALUED FUNCTIONS OF TWO VARIABLES AND
EULER’S THEOREM
ARUN LEKHA
Associate Professor
G.C.G., SECTOR-11, CHANDIGARH
FUNCTION OF TWO VARIABLES
Definition: A variable Z is said to be a function of two independent variables x and y denoted by z=f (x,y) if to each pair of values of x and y over some domain Df ={(x,y): a<x<b,c<y<d} there corresponds a single definite value of Z.
e.g. The area A of a rectangle having sides of lengths x and y is xy.
i.e. A = xy is a function of two variables x and y. Domain of the function is Df={(x,y): x > 0 y > 0}.
PARTIAL DERIVATIVE (FIRST ORDER)
Partial derivative of Z = f(x,y) w.r.t.x, regarding y as constant is denoted by
z/x or f/x or fx and
provided it exists and is finite.
Similarly
provided it exists and is finite.
x
yxfyxxf
x
ltf x
),(),(
0
y
yxfyyxf
y
ltf y
),(),(
0
e.g. If z =e-x/y + tan-1 (x/y).
then
PARTIAL DERIVATIVES (SECOND ORDER)
The first order partial derivatives z/x or z/ y are generally functions of x and y and hence we can again find their partial derivatives w.r.t. X or y. The partial derivatives thus obtained are called second order partial derivatives and are denoted by fxy or fyx.
22
/
yx
y
y
e
x
z yx
HOMOGENEOUS FUNCTIONS
A function of two variables x and y of the form
f(x,y) = aoxn+a1x
n-1 y + ….an-1 xyn-1+anyn
in which each term is of degree n is called homogeneous function or if it can be expressed in the form yng(x/y) or xng(y/x).
e.g. f(x,y) = x2+y2 / x+y
is homogeneous function of degree 1
EULER’S THEOREM: If Z = f(x,y) is a homogeneous function of x and y of degree n , then
x z/x + y z/ y = nz
Ex. Show that Z =ax2 + 2hxy + by2 is homogeneous function of degree 2 and verify Euler’s theorem.
Sol. Z=ax2+2hxy + by2 = x2 [a+2h y/x + by2/x2] = X2 g (y/x).
Z is homo. Function of degree 2.
Verification of Euler’s theorem.
z/ x = 2ax + 2hy.
z/ y = 2hx + 2by
then x z/x + yz/y
= x(2ax+2hy) + y (2hx+2by)
=2ax2 + 4 hxy + 2by2
= 2z
Euler’s theorem for a homogeneous function of three independent variables.
If H is a homogeneous function or x,y,z of order n then x H/x + Y h/Y + z H/Z = nH
Differentiable Function: A function of (x,y) is said to be differentiable at (x,y) if z = f (x,y) can be expressed in the form
z = /x f(x,y). x + /yf (x,y). y
+ 1 y + 2 x
where 1 0, 2 0 as x, y 0.
Remark: Continuity of f, fx, fy at (x,y) are
sufficient conditions for differentiability. Total differential of a function If z = f (x,y), then total differential of Z is denoted
and defined by dz = z/x. dx + z/y.dy
Differentiability of f (x,y)
The function Z=f (x,y) is said to be differentiable at a point (xo,yo) if in a neighbourhood of (xo,yo), it can be represented in the form
f (xo+h, yo+k) – f (xo,yo) = Ah + BK + h +k
were A,B are independent of the variables h,k.
and , 0 as h,k 0 independently.
Thm: If a function f(x,y) is differentiable at a point (xo,yo), then it is continuous at that point.
Remark: Converse is not always true.
Example: f (x,y) = |xy| is not differentiable at (0,0) but continuous at (0,0).
Theorem: If a function f (x,y) is differentiable at a point (xo,yo) then fx (xo,yo) and fy (xo,yo) both exist and
Ah
yxfyhxf
h
Ltyxf oooo
oox
),(),(
0),(
Bk
yxfkyx
k
Ltyxf oooo
ooy
),(),(
0),(
Que. Discuss the differentiability of
f (x1y) = |x|+|y| at (0,0)
Sol. f(x,y) is differentiable at (0,0) If f (0+h,0+k) – f(0,0) = Ah+Bk +h +k where ,0 as h, k0
01
0,1||
0
|)0||0(||0|||
0
)0,0()0,(
0)0,0(
hif
hif
h
h
h
Lt
h
h
h
Lt
h
fhf
h
LtfANow x
A does not exist.
Similarly B =fy (0,0) =
B does not exist.
Hence f(x,y) is not differentiable at (0,0).
YOUNG’S THEOREM
Let f(x) be defined in a domain D R2. Let (a,b) be an interior of D and let
(i) fx and fy exist in the neighbourhood of (a,b)
(ii) fx and fy are differentiable at the point (a,b)
then fxy = fyx at (a,b)
01
0,1||
0
)0,0(),0(
0
k
k
k
k
k
Lt
k
fkf
k
Lt
SCHAWARZ’S THEOREM If (a,b) be a point of the domain DR2 of
a function f (x,y) such that
(i) fx and fy exist in the neighbourhood of the point (a,b)
(ii) fxy is continuous at (a,b)
then fyx exist at (a,b)
and fyx =fxy at (a,b)
Change of Variables
Let Z = f(x,y), x = (u,v), y = (u,v)
Taking as constant,
Taking u as constant
and by solving the above equations in
z/x, z/y we get their values in terms of
z/u, z/v, u, v.
u
y
y
z
u
x
x
z
u
z
..
v
y
y
z
v
x
x
z
v
z
..
Composite functions:
Definition: Let Z = f (x,y) and let x = (t)
and y = (t), then z is called composite function of t.
Differentiation of composite functions:
Let Z = f (x,y) possess continuous partial derivatives and
X = (t), y = (t) possess continuous derivatives,
then dz/dt = z/x. dx/dt + z/y. dy/dt
Implicit functions:
Definition: Let f(x,y) be a function of two variables and
y = (x) be a function of x such that f (x,(x)) vanishes identically, then y = (x) is an implicit function defined by the functional equation
f(x,y) = 0
Differentiation of implicit functions
If f(x,y) = 0 or c be an implicit function then
(i) dy/dx = -f/x / f/y = -fx/fy, fy 0
(ii) d2y/dx2 =
Implicit function theorem (Two Variables)
Let f(x,y) be a function of two variables
x and y and (a,b) be a point of its domain of definition such that
(i) f(a,b) = 0
(ii) fx and fy exist and are continuous in certain nbhd. of (a,b).
0,)(
)(2)(3
22
fy
fy
ffffffyfxx xyyyxyx
(ii) fy (a,b) 0, then there exist a rectangle (a-h, a+h, b-k, b+k) about (a,b) such that for every x in the interval [a-h, a+h], f (x,y) = 0 determines one and only one value
y = (x) lying in the internal [b-k, b+k] with the following properties
(i) b = (a)
(ii) f(x, (x)) = 0 for every x in [a-h,a+h]
(iii) (x) is derivable and both (x) and ’(x) are continuous in [a-h, a+h].
e.g. f(x,y) = x2+y2-1 and a point (0,1)
So that f(0,1) = 0 and fy (0,1) = 2 0
Now of the two possible solutions
y = + 1 – x2
(i) y = + 1-x2 is implicit function in nbhd. of (0,1), where |x|<1, y>0.
(ii) y = - 1-x2 is implicit function in nbhd. of (0,-1) where |x|<1 , y<0.