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Partial Differentiation Taylor series for functions of 2 variables Stationary points Exact First Order ODEs

MATHEMATICS FOR ELECTRICAL

ENGINEERING I

Dr. G.A. Pavliotis

Department of Mathematics Imperial College London

Spring term 2007/08

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We consider functions of two variables

u=u(x, y).

The theory that we will develop also applies to functions of more

variables. We want to define the rate of change of the function

u(x, y)with respect to the variablesxandy. The definition ofthe derivative for a function of a single variableu=u(x)is

du

dx = lim

h0

u(x+ h) u(x)h

.

In the case of a function of two variables we can use a similar

definition, but now we have to consider derivatives with respect

to eitherxory. We define thepartial derivativesthrough theformulas

u

x = lim

h0

u(x+ h, y) u(x, y)h

and

uy

= limh0

u(x, y+ h) u(x, y)h

.

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To calculate partial derivatives, we use the standard

differentiation rules while keeping the other variable fixed.

Example

Calculate the partial derivatives of

u(x, y) =x2 sin(y) + y3. (1)

We have

ux

=2xsin(y), uy

=x2 cos(x) + 3y2.

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P i l Diff i i T l i f f i f i bl S i i E Fi O d ODE

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Example

Calculate the first and second order partial derivatives of

u(x, y) =x2y4.

We calculate

ux =2xy4, uy=4x

2y3,

uxx=2y4, uyy=12x

2y2,

uxy =8xy3, uyx=8xy

3.

P ti l Diff ti ti T l i f f ti f 2 i bl St ti i t E t Fi t O d ODE

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Example

Calculate the first and second order derivatives of

u(x, y) =ex2+2xy+3y2 .

The first derivatives are

ux=ex2+2xy+3y2 (2x+ 2y), uy=ex2+2xy+3y2 (2x+ 6y).

The second derivatives are

uxx =ex2+2xy+3y2

(2x+2y)2+2, uyy=e

x2+2xy+3y2

(2x+6y)2+6The mixed derivatives are

uxy =ex2+2xy+3y2

(2x+ 2y)(2x+ 6y) + 2

,

uyx=ex2+2xy+3y2(2x+ 6y)(2x+ 2y) + 2.


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Theorem

Let u(x, y) C2(R2). Then2u

xy =

2u

yx.

We can also define higher order partial derivatives:

3u

x3,

3u

x2y,

3u

xy2,

3u

y3.

We have similar symmetry properties between higher orderpartial derivatives:

uxxy =uxyx=uyxx, uyyx=uyxy=uxyy,

and similarly for higher order derivatives.


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the matrix of second order derivatives of a functionu(x, y)iscalled theHessian matrixofu:

D2u=

uxx uxyuyx uyy

. (2)

Sinceuxy =uyx, the Hessian is a symmetric matrix. The

determinantof the Hessian is

det(D2u) =uxxuyy (uxy)2.

We can also consider partial derivatives of functions of several

variables,f =f(x1, x2, . . . , xn). The partial derivatives satisfy:

2f

xixj=

2f

xjxi, i,j=1, . . . n.

In this case the HessianD2f is ann nsymmetric matrix.

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We use now the fact thatht=

cto calculate:

u

t =

df

dh

h

t =f(h(x, t))(c),

and2u

t2 =

t f(h(x, t))(c)=c2f(h(x, t)).Consequently,

uxx 1c2

utt=f f =0.

Thus, the functionu(x, t) =f(x ct)is a solution of thewaveequation

2u

x2 =

1

c22u

t2.


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The wave equation is an example of apartial differentialequation (PDE). PDEs are used as models for most

phenomena in physics and engineering. The wave equation is

one of the most important PDEs. It governs the evolution of

waves, and electromagnetic waves in particular (Maxwells

equations). The constantcis thewave speed.

Exercise

Show that the function u(x, t) =h(x+ ct)where h is anarbitrary function is a solution of the wave equation

uxx= 1

c2utt.


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y y p

Proposition

The solution of the wave equation

2u

x2 =

1

c22u

t2 (3)

isu(x, t) =f(x ct) + h(x+ ct), (4)

where f and h are arbitrary functions. These functions are

determined from the initial conditions

u(x, 0) =u0(x), ut(x, 0) =v(x).

The solution (4) represents two waves, one propagating to the

left and one to the right with speed c.


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y y p

The four most important PDEs are:

Thewave equation

2u

x2 =

1

c22u

t2.

Shrdingers equation

iu

t =

2

2m

2u

x2+ V(x)u.

Laplaces equation

2ux2

+ 2u

y2 =0.

Theheat equationu

t =

2u

x2 .


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Example

The function u(x, y) =tan1 yx

solves Laplaces equation.

We use the chain rule to calculate the partial derivatives of

u(x, y) =tan1(f(x, y)), f =y/x:

ux = d

df

tan1 f

fx=

1

1 + f2

y

x2

= y

x2 + y2,

uy =

x

x2 + y2 , uxx=

2xy

(x2 + y2)2 , uyy = 2xy

(x2 + y2)2 .

Consequently

uxx+ uyy= 2xy

(x2

+ y2

)2

2xy

(x2

+ y2

)2

=0.

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Example

The functionu(x, t) =

1t

ex2

4t

solves the heat equation.

We calculate

ut = 12

t3e

x2

4t +

x2

4t2

1

te

x2

4t,

ux = 1

tx

2t e

x2

4t,

uxx = 1

t

1

2t

e

x2

4t + 1

t

x2

4t2

e

x2

4t

=

1

2t3

ex2

4t + x2

4t2

1

te

x2

4t.


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Exercise

Find the solution of the heat equation

ut=Duxx,

where D>0is a positive constant.

Theinitial value problemfor the heat equation is

ut=Duxx, u(x, 0) =f(x).

Fourier developed the method of Fourier series in his attempt to

solve the heat equation in 1822.


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We can also define the partial derivatives of functions of several

variables. For example, letu=u(x, y, t). The partial derivatives

are calculated through ordinary differentiation while keeping theother variables fixed:

u

x =

du

dx

y=const,t=const

and similarly for the other derivatives.

Example

The function

u= 1

2x2 +

1

2y2 + 2t

is a solution of the heat equation in 2 dimensions.

ut=2, ux=x, uy=y, uxx=1, uyy =1.

Consequently, 2= ut=uxx+ u =2.


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Exercise

Show that the function

u=1

te

x2+y2

4t

solves the heat equation

ut=uxx+ uyy.

in2dimensions.

We introduceLaplaces operator

= 2

x2+

2

y2.

Laplaces equation and the heat equation in 2dcan be written

as

u=0, ut= u.


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Differentials

Consider a function of two variablesu=u(x, y). Assume thatxchanges by an amountxandyby an amounty,

x x+x, y y+y. What is the resulting change inu(assuming thatxandyare small)?We have:

u = u(x+x, y+y) u(x, y)= u(x+x, y+y) u(x+x, y) + u(x+x, y) u(x, y)=

u(x+x, y+y) u(x+x, y)y

y

+

u(x+x, y)

u(x+x, y)

x x

uy

(x+x, y)y+u

x(x, y)x

u

y

(x, y)y+u

x

(x, y)x.


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Differentials

We take the limitsx, y

0. We calculate thedifferentialof

u(x, y):

du=u

xdx+

u

ydy. (5)

Similarly, we can calculate the differential of a function

u=u(x, y, z)of 3 variables:

du=u

xdx+

u

ydy+

u

zdz.

Notice the difference with the 1dcase:

du=du

dxdx.


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Differentials

Example

Calculate the differential of

u(x, y) =xy2 + cos(y).

We calculate the partial derivatives:

ux=y2, uy=2xy sin(y).

The differential is

du=y2 dx+ (2xy sin(y)) dx.


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Differentials

Area of a rectangle

Example

Consider the area of a rectangle of length xandy. Find the

change in the area when the length of the sides changes by xandy.

The area is given byA= xy. We have that

dA=Axdx+ Aydy=y dx+ x dy.

Alternatively:

A = (x+x)(y+y) xy=yx+ xy+xy yx+ xy,

byneglecting higher order terms(asymptotic analysis).


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The chain rule

Consider the functionu=f(x)wherex=h(t). uis a functionof the variablet,u=f(h(t)). We can calculate the derivative of

uwith respect totby using thechain rule:du

dt =

du

dx

dx

dt =f(h(t))h.

Consider now the functionu(t) =u(x(t), y(y)). We use thechain rule to calculate du

dtdu

dt =

u

x

dx

dt +

u

y

dy

dt =uxx+ uyy.

To prove this, note that

u uxx+ uyy,where

x xt, y yt u ux(xt) + uy(yt) u

tu

x

x+ uy

y

u=ux

x+ uy

y.

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The chain rule

Proof

Newtons equation can be written as a system of two first order

differential equations

x=y, my= V.

We use the chain rule to calculate the rate of change of the

total energy:

dE

dt

= E

x

x+E

y

y

= dV

dxy+ myy=

dV

dxydV

dxy

= 0.


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The chain rule

Example

The damped harmonic oscillator

x+ 2x+2x=0

dissipates energy.

The equation of motion can be written as a system of first orderordinary differential equations:

x=y, y= 2y 2x.The energy is

E(t) = 12

y2 + 12

2x2.

The rate of change is

dE

dt =

E

xx+

E

yy=2xy+ y(2x 2y) = 2y2

0.


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The chain rule

The chain rule applies to functions of 3 or more variables: if wehave a functionu=u(x(t), y(t), z(t), t)thetotal derivativewithrespect tot is

du

dt =

u

t +

u

x

dx

dt +

u

y

dy

dt +

u

z

dz

dt.

Suppose the functionsx, y, zare functions of two variablesr, s: x=x(r, s), y=y(r, s), z=z(r, s). Then

ur = uxxr+ uyyr+ uzzr,ur = uxxs+ uyys+ uzzs.

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The chain rule

Example

Calculate the derivatives with respect to s, tof the function

u=x2y withx=st, y=s+ t.

First method:

us = uxxs+ uyys=2x(s, t)y(s, t)xs+ (x(s, t))2ys

= 2st(s+ t)t+ s2t2 =3s2t2 + 2s2t2 =3s2t2 + 2st3,

ut = uxxt+ uyyt=2s2t2 + 2s3t.

Second method:

u(t, s) = (st)2(s+ t) =s3t2 + s2t3 us=3s

2t2 + 2st3, ut=2s3t+ 3s2t2.

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I li i F i

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Implicit Functions

In order to calculate the derivatives of a function defined

implicitly we note that from (6) it follows that

F =0 F =Fxx+ Fyy+ Fzz=0.

Or, by taking differentials,

Fxdx+ Fydy+ Fzdz=0.

In order to calculate the partial derivative of zwith respect tox,

say, we notice that in this casey is held fixed and consequently

we havedy=0. Consequently:

Fxdx+ Fzdz

y=const=0 z

x =

dz

dx

y=const

= FxFz

.


I li it F ti

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Implicit Functions

Similarly:

z

y = Fy

Fz,

x

y = Fy

Fx,

x

z = Fz

Fx,

y

z = Fz

Fy,

y

x = Fx

Fy.

Notice thatx

y =

1yx

.

This is because whenzis kept fixed then the relation betweenxandyis that of a single variable, i.e., for fixed zthe equation

F(x, y, z) =0 defines a function of single variabley=y(x)orx=x(y)and dy/dx=1/(dx/dy).


Implicit Functions

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Implicit Functions

LetF(x, y, z) =0. We have

yx zy xz = FxFyFyFzFzFx= 1. (8)Equation (8)is very useful in thermodynamics.


Implicit Functions

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Implicit Functions

Example

Equation of state for an ideal gas The equation of state for an

ideal gas is

PV =NkBT, (9)

whereP is the pressure,V is the volume, Nis the number of

particles,T is the absolute temperature andkBis Boltzmannsconstant. The equation is of the form (6):

F(P, V, T) =PV NkBT =0.

We use (8) to obtain the very important formulaP

V

T

T

P

V

V

T

= 1.


Change of Coordinates

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Quite often it is useful to change coordinates, i.e. to go from the

cartesiancoordinatesx, y zto new coordinatesr, s, t.

Example

Plane Polar Coordinates Thepolar coordinateson the plane

are defined through the transformation

x=rcos(), y=rsin(). (10)

We also have

r=

x2 + y2, =tan1

y

x. (11)

The equation of a circle or radius ain cartesian coordinates is

x2 + y2 =a2.

In polar coordinates it becomes

r=a.



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We can calculate the partial derivatives ofx, ywith respect tor, and of, wrtx, y. From (10) we have

xr =cos(), x = rsin(),yr =sin(), y =rcos().

On the other hand, from(11) we obtain

rx= xx2 + y2

=x

r =cos(), ry=sin(),

x= 1

1 + y2/x2 y

x2= y

r2 = sin()

r , y =

cos()

r .

Notice that

xr= 1rx

, x= 1x

, yr= 1ry

, y= 1y

.

WHY?



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C a ge o Coo d ates

Notice thatx, yare written as functions ofr, ,x=x(r, ), y=y(r, ). Similarly, we have that

r=r(x, y), =(x, y). When we calculatexr, we considerxas a function ofrand and we keepfixed, whereas when wecalculaterxwe considerras a function of xandyand we keep

yfixed:x

r =

dx

dr =const = r

x =

dr

dxy=const.Let nowu=u(r, ) =u(r(x, y), (x, y))). We want to calculateux, uyin terms ofur, u. We have:

u

x

=urrx+ ux =cos()ur

sin()

r

u.

u

y =urry+ uy =sin()ur+

cos()

r u.

Notice that we can write

x =cos()

rsin()

r

,

y =sin()

r +

cos()

r

.



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g

Example

Calculate the partial derivatives wrtxandyof the function

u(r, ) =r2 cos(2),

wherer, are the planar polar coordinates.

First method. We have

ux = cos()ur sin()r

u

= cos()2rcos(2)

sin()

r

(

2r2 sin(2))

= 2r

cos() cos(2) + sin() sin(2)

= 2r

cos()

cos2() sin2() + 2cos() sin2()= 2rcos()cos2() + sin2()=2rcos().

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Exercise

Let u=u(x, y) =u(x(r, ), y(r, )). Show that

u2x+ u2y =u

2r +

1

r2u2 .



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ExerciseThe Laplacian is polar coordinates Let

u=u(x, y) =u(x(r, ), y(r, )). Show that

u=uxx+ uyy=1

r

rrur+ 1r2 2u

2.

In particular,for functions that depend only on the distance from

the origin, i.e. u=u(r), the Laplacian becomes

u(r) = 1r

ddrrdu

dr . (12)

Functions that depend only onrare calledradially symmetric.



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Example

Show that the function

u(x, y) =1

2ln(y2 + x2) (13)

is a solution of Laplaces equation in 2d.

We use polar coordinates:

u(r) =1

2ln(r2) =ln(r).

From(12)we get

ln(r) = 1

r

d

dr

r

1

r

=

1

r

d

dr (1) =0.

The functionu(r) =ln(r)is called thefundamentalsolution of

Laplaces equation in 2d.



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There are many different coordinate frames that we can use in

2dand 3d. We choose the one that is more convenient for our

problem. When studying radially symmetric problems, then

polar coordinates are a good choice.For other problems it might be more convenient to use

x=rcosh(), y=rsinh().


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Consider a function of a single variable u=u(x). We can wecan expand it into a Taylor series at the point x0:

u(x) = u(x0) + (x x0) dudx

(x0) + 12!

(x x0)2 d2udx2

(x0)

+1

3!(x x0)2

d3u

dx3(x0) +. . .

=

n=0

u(n)(x0)n! (x x0)n.

By settingx x0 =hwe can rewrite this in the form

u(x0+ h) =

n=0u(n)(x0)

n! h

n

.

Consider the case of a function of 2 variables. We have a

similar expansion. To obtain this expansion, we first do a Taylor

series expansion inxwill keepingyfixed and we then do an

expansion inywhile keepingxfixed.


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u(x0+ h, y0+ k) = u(x0, y0+ k) + hux(x0, y0+ k)

+h2

2uxx(x0, y0+ k) +. . .

= u(x0, y0) +

hux(x0, y0) + ku0(x0, y0)

+1

2h2uxx(x0, y0) + 2hkuxy(x0, y0)+k2uyy(x0, y0)

+. . .

We introduce the differential operatorD=h x + k y.We have

that

D2f = D(Df) =

h

x + k

y

h

f

x + k

f

y

= h2fxx+ 2hkfxy+ k

2fyy.


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We can calculateD3, D4 etc in a similar fashion. Using thisnotation we can write the Taylor series expansion for a function

of 2 variables in the form

u(x0+ h, y0+ h) =

n=0

Dnu(x0, y0)

n! . (14)

Let nowu=u(x, y, z)be a function of 3 variables. Using thenotation

D=h

x + k

y +

z,

we can write the Taylor series expansion foru(x, y, z)at

(x0, y0, z0)in the form

u(x0+ h, y0+ h, z0+) =

n=0

Dnu(x0, y0, z0)

n! .


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Exercise

Let

D=h

x + k

y +

z,

Show that

D2 = h2 2

x2+ k2

2

y2+2

2

z2

+2hk

2

xy + 2h

2

xz + 2k

2

xz


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Example

Calculate the first three terms in the Taylor series expansion of

u(x, y) =e2xy at (x0, y0) = (0, 0).

We have

u(0, 0) =1, ux =2e2xy

ux(0, 0) =2.

uy= e2xy uy(0, 0) = 1uxx=4e

2xy uxx(0, 0) =4,uxy = 2e2xy uxy(0, 0) = 2,uyy =e

2xy

uyy(0, 0) =1.

Consequently

u(h, k) 1 + 2h k+ 12

4h2 4kh+ k2

1 + (2h k) +

12h k2 = (2h k)n.


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Letu=u(x)be a function of a single variable. Astationarypointis defined as a point x

0where the first derivative vanishes:

du

dx(x0) =0.

It is amaximumwhenu(x0)< 0 and aminimumwhen

u

(x0)>0.Our goal is to develop a similar classification for functions of

several variables.

Definition

Letu=u(x, y)be a function of 2 variables. We will call(x0, y0)a critical point ofuprovided that

ux(x0, y0) =uy(x0, y0) =0.


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We want to determine whether a critical point is a maximum, a

minimum or a saddle. For this, we will use the Taylor series

expansion. We will use the notation

A= uxx(x0, y0), B=uxy(x0, y0), C=uyy(x0, y0). Let(x0, y0)be

a critical point (and consequently the first derivatives vanish).The Taylor series expansion about(x0, y0)is

u(x0+ h, y0+ k) =u(x0, y0) +1

2Ah2 + 2Bkh+ Ck2

+. . . .


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The change inu, when we changexbyhandybyk, is

u=u(x0+h, y0+k)u(x0, y0) =1

2

Ah2 + 2Bhk+ Ck2

(15)

Ifu>0 for allh, ksufficiently small, then

u(x0+ h, y0+ k)>u(x0, y0) (x0, y0) is a minimum.

Ifu

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Procedure for finding and characterizing critical points

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Procedure for finding and characterizing critical points

1 Find all critical points by solving the equations

ux=0, uy=0.

2 At each critical point calculate

A= uxx(x0, y0), B=uxy(x0, y0), C=uyy(x0, y0).3 CalculateAC B2 at each stationary point.

AC B2 0, A> 0 MINIMUMAC B2 >0, A< 0 MAXIMUM


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Exercise

Find and characterize the critical points of

u(x, y) = (x y)(x2 + y2 1) =x3 + xy2 x yx2 y3 + y.


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Quite often we need to find the maxima and minima of a

function subject to constraints.

ExerciseMaximize the volume of a rectangular box subject to the

constraint that the surface area is given.


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We want to find the maximum of the function (volume)

V(x, y, z) =xyz (16)

subject to the constraint (surface area is given)

A(x, y, z) =2(xy+ xz+ yz) =A. (17)

Of course, we also have the "constraint"

x, y, z>0. (18)

We solve equation (17) forzand substitute into(16). This will

give us a function of two variables x, y. We then maximize thisfunction.


From (17) we obtain

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From(17)we obtain

z= A 2xy2(x+ y)

.

Equation (16) becomes

V(x, y, z) =(A 2xy)xy

2(x+ y) .

To calculate the critical points we need to solve the equations

Vx(x0, y0) =Vy(x0, y0) =0.

We have:

Vx = 1

2

(Ay 4xy2)(y+ x) (Axy 2x2y2)(y+ x)2

,

Vy = 1

2

(Ax 4x2y)(y+ x) (Axy 2x2y2)(y+ x)2

,


We obtain the equations

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We obtain the equations

x2(A

2y2

4xy) = 0,

y2(A 2x2 4xy) = 0.

The solution that satisfies (18) is

x=y= A61/2 .We can check that this leads to a maximum. From this we

calculate

z= A61/2and

Vmax=

A

63/2


Lagrange Multipliers

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Lagrange Multipliers

Suppose we want to find the extremum of V(x, y, z)subject tothe constraint ofF(x, y, z) =0. We form the function

V(x, y, z) =V(x, y, z) F(x, y, z),whereis called aLagrange multiplier. We calculate theextrema of the new functionV(x, y, z).


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We can use partial derivatives in order to obtain the solution of

a general class of first order ODEs, ofexact ODEs. Consider

the ODE

P(x, y) + Q(x, y)dy

dx =0. (19)

Assume that there exists a function F(x, y)so that

dF

dx =P(x, y) + Q(x, y)

dy

dx.

Then the ODE (19) becomes

dFdx

=0,

and the solution can be written implicitly asF(x, y(x)) =C.ODEs for which such a function exists are called exact.

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Criterion for an ODE to be exact

The nonlinear first order differential equations

P(x, y) + Q(x, y)dy

dx =0

is exact provided thatPy

=Q

x. (20)

When condition (20) is satisfied, the solution of the ODE is

defined implicitly through the formula

F(x, y(x)) =C,

whereF(x, y)is such that Fx=P, Fy=Q.


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Exercise

Find the solution of the equation

(2xy+ cos(x) cos(y)) +dy

dx(x2 sin(x) sin(y)) =0.


A non-exact first order ODE can become an exact through

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g

multiplication by an appropriate function. Consider the equation

P(x, y) + Q(x, y) dydx

=0.

Assume thatPy=Qx. We multiply the equation by a function(x, y):

(x, y)P(x, y) +(x, y)Q(x, y) dydx

=0. (21)

We choose the function(x, y)so that

y(P) =

x(Q) . (22)

Hence, the modified equation (21) is exact. In order to find

(x, y)we need to solve equation (22), which is a partialdifferential equation. Usually it is not possible to solve this

equation Partial Differentiation Taylor series for functions of 2 variables Stationary points Exact First Order ODEs

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Exercise

Find the solution of the equation

(xy 1) dx+ (x2 xy) dy=0.

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