ch3 review of partial differential operators

8/13/2019 Ch3 Review of Partial Differential Operators

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3. Review of Partial Differential Operations

1. Partial Derivatives

Given a certain multidimensional function, ),,,( t z y x A , a partial derivative at a specific point defines the local rate of change of that function in a particular direction. For the4-dimensional variable, ),,,( t z y x A , the partial derivatives are expressed as

( ) ( ) x

t z y x At z y x x A x A

t z y x

t z y +=

,,,,,,limconstant,,

0 ,,

= slope of A in the x direction

( ) ( ) y

t z y x At z y y x A y A

t z x y

t z x +=

,,,,,,limconstant,,

0 ,,

= slope of A in the direction

( ) ( ) z

t z y x At z z y x A z A

t y x z

t y x +=

,,,,,,limconstant,,

0 ,,

= slope of A in the ! direction

( ) ( )t

t z y x At t z y x At A

z y xt

z y x +=

,,,,,,limconstant,,0

,,

= the local time rate of change of A

"he subscripts on the brac#ets indicate that those dimensions are held constant.

$otice that the definition of a partial derivative of a multi-variable function is the same as

derivatives of functions of a single variable, but %ith the other variables of the function beingheld constant. &henever ou see the 'bac#%ard-six notation for the derivative, ou shouldthin# about %hat variable ou are operating on, as indicated in the denominator of theexpression, %hile holding the other variables constant.

t is common convention that the directions being held constant are implied and not explicitel%ritten %ith subscripts.

2. Higher order partial derivatives

*


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&e can appl the partial derivative multiple times on a scalar function or vector. Forexample, given a multivariable function, ( ) y x f , , there are four possible second order partialderivatives+

x y f

x f

y y x f

y f

x y f

y f

y x f

x f

x = =

=

=

- - -

"he last t%o partial derivatives, y x

f

,

and x y

f

,

are called 'mixed derivatives. n

important theorem of multi-variable calculus is the mixed derivative theorem . "he proof is be ond the scope of this course and onl the results are stated.

Mixed derivative Theorem: f a function ( ) y x f , is continous and smooth to second order,then the order of operation of the partial derivatives does not matter. n other %ords+

x y f

y x f

=

,,

for a continous and smooth (to second order) function ( ) y x f ,

xample: For the function ( ) ( ) y x xy y x f exp, += , sho% x y

f y x

f

=

,,

!nswer Provided:

( )( ) ( )( ) ( ) ( )( ) y x yx x y y x x xy x

y x xy y x y

f x y x

f exp*expexp ++=+

=

+

=

=

( )( ) ( )( )( ) ( ) ( )( ) y x yx x y y x xy y y

y x xy x y x

f

y x y

f exp*expexp ++=+

=

+

=

=

&e can see that the order of operation of the partial derivative on a continous and smooth scalarfunction does not matter.

3. Del operator:

"he del operator is a linear combination of spatial partial derivatives. n rectangularcoordinates, it is expressed as

zk

y j

xi ++=

///

(*)

$otice the second e ualit above is missing the vector arro%. is al%a s a vector operatorand thus it is common convention to 1ust leave off the vector arro%.


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"he anal sis of the del operator on various ob1ects such as scalar functions or vectors can be rather complex. n rectangular coordinates, ho%ever, the rules %e learned about in chapter on 'multipl ing vectors appl to the del operator as %ell. t is important to notice ho%ever thatthe order is extremel important in the use of e uation (*). "he del operator acts on all ob1ects tothe right of it. "t is #r$i#ial to note that the del operator is not #omm$tative when applied to

s#alars or ve#tors% &o$ onl' appl' del operators on what is to the right in the term andnever on the o()e#ts to the left.

*. +radient Operator

ppl ing the gradient operator, , on a scalar function ( ) z y x ,, = , simpl re uiresscalar multiplication. "he gradient of ields the follo%ing+

k z

j y

i x z

k y

j x

i 222222+

+=

+

+= ( )

$otice that e uation ( ) is a linear combination of vector components and basis vectors. nother %ords the gradient of a scalar ields a vector. &o$ will (e tested on the appli#ationleading to e,$ation -2 as well as the fa#t that the res$lt of is a ve#tor. 3ince thegradient of a scalar function is a vector, it obe s all the rules that %e learned about in chapter .

xample: For scalar function xyz = sho% that

( ) ( ) x x

xample Given a velocit vector

///

k w jviuu ++= and the gradient of a scalar function, as

defined in e uation ( ), expand out u in rectangular coordinates+

!nswer provided: 5sing e uation (*0) from chapter ,

zw

yv

xuk

z j y

i x

k w jviuu+

+=

++

++= 222

///

6$otice the result is a scalar as re uired for the dot product of t%o vectors.66$otice that there %ere no parantheses given for the application of the operation above. &etoo# the dot product of the vector u %ith the vector . &e could have 1ust as %ell ta#en thedot product of the vector u %ith the operator and then applied that on the scalar function+

7


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( ) zw

yv

xu

zw

yv

xuk

z j y

i x

k w jviuu+

+=

++

=+

+++= 222

///

.

n other %ords, ( ) = uu . "his e ualit is onl relavent %hen %e are operating on ascalar .

n this course, %e %ill onl ta#e gradients of s#alar f$n#tions . t is possible to ta#e gradientsof vectors but ou obtain a 8 element matrix called the Dyadic product 1 of the vector field, xu .

For example, given the vector

///

k w jviuu ++= , the gradient of u is

=+

+

=

zw

zv

zu

yw

yv

yu

xw

xv

xu

k zu

j yu

i xu

u///

9ou can see %h %e %ant to avoid operations li#e this.

+radient properties: magnit$de

: uation ( ) is a vector since it has a magnitude and direction. For a function( ) z y x f f ,,= , the magnitude of f is simpl found using the rules of chapter .

+

+

==

z f

y f

x f

f f f (7)

* "he operation is also related to the transpose of the ;acobian


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+radient properties: dire#tion

"he direction of f is a bit more complicated. From the previous chapter %e can see that

the direction of f can be expressed b the unit vector, f f

, but %e also can interpret the

direction of f in a more geometric or ph sical %a . First %e need to use the differential of f , %hich is labeled df . differential is an infintesimal (meaning reall small) change in thevalue of the multivariable function f and has components+

dz z

f dy

y f

dx x f

df +

+=

f %e define the vector line element,

///

k dz jdyidxd ++= , then %e can see b inspection that thedifferential ta#es the simple form

d f df =

$o% let us appl the geometric definition of the dot product d f +

( ) cos d f d f df == %here is the coplanar angle bet%een the vector f andd .

f d is perpendicular f then o80= and 0=df . n other %ords, d is along lines ofconstant f %hen it is perpendicular to f . lternativel , %e find that df is a maximum%hen d is parallel to f . "his means that df is maximum %hen d is in the samedirection as f (and also perpendiculal to contours of constant f ). "his also means that f must al%a s be in the direction that leads to the greatest df . "he direction of f is alsocalled the asecendant of f. Figure *, on page >, sho%s ou a picture relating the direction of

f to lines of constant f .

/: The #hange of a ,$antit' in the dire#tion of the velo#it' field -!dve#tion

&e can find the change of a scalar, ( ) z y x f ,, , in an arbitrar direction,/

7

/

,

/

*

/

k u juiuu ++= %here*7* =++ uuu , b ta#ing the dot product of

/

u %ith f . "he results is+

?


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f udt df

=/

(4)

"o derive e uation (4), parametri!e the spatial curve, ( ) ( ) ( )///

k t z jt yit x ++= %ith respect to the variable t +

( ) ** ut x

tu xt x o =

+=

( ) ut y

tu yt y o =

+=

( ) 77 ut z

tu z t z o =

+=

"hen, using the #hain r$le , %e obtain e uation (4)+

/

u f t

z z f

t y

y f

t x

x f

dt df

=

+

+

=

Finding variations in a specific direction often occurs %hen %e tr to find that variationof a ph sical uantit in the direction of the flow field , u . &e usuall discuss the rate of changeof the scalar uantit , ( ) z y x f ,, due to variations in f along the flo% field, u . "his isrepresented mathematicall as+

f udt df

=

"he term on the right side of the e ualit is called the advective term and is one of t%ocontributions to the total or material derivative that %e %ill learn more about later in thesemester. @ften %e are interested in determining if there is an variation in the direction of flo%.

f one obtainAs the result+

0= f u

&e sa that the function, f , is spatiall constant along the flo% field, u . For example, if ourscalar uantit is a time-independent pressure field, ( ) z y x p ,, , then the e uation 0= pu ,tells us that isobars are constant along the flo% field %hich also means that isobar contours areever %here parallel to the velocit vector field.

&hat %e have done here is to create a curve that is parameteri!ed along the direction of differentiation. "hisho%ever, is exactl %hat %e %ant to find the change in the function f along a specific direction. f ou find these

points confusing, ou ma %ant to revie% the definition of a curve from a calc textboo#.

>


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0$mmar':* Find the gradient of the multivariable function ( ) z y x f ,, , f , is a vector %ith magnitude

+

+

==

z f

y f

x f f f f and unit vector f

f

.

- "he direction of the gradient of f is al%a s in the direction of the greatest increase in f and perpendicular to the contours of constant f. "he direction of f is also called the ascendant of

f . 7 - &e can find the spatial rate of change of a function in a specific direction/

u b ta#ing the

dot product of/

u %ith f , f u /

. f 0/

= f u then f is constant along/

u .

xample: Find the gradient of the function ( ), y x y x f =

graph of the surface ( ), y x y x f = is sho%n in figure *.

"he vector field of the gradient is sho%n in figure .

-1-0.8

-0.6-0.4

-0.20

0.20.4

0.60.8

1

-1

-0.5

0

0.5

1-2

-1.8

-1.6-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

distance (x)

graph of the function f(x,y)=-x2-y2

distance (y)

f ( x

, y )

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

distance (x)

Overhead view of the function f(x,y)=-x 2-y2

d i s

t a n c e

( y

)

Figure * Graph of the function ( ), y x y x f = in 7-B and from an overhead vie%.

C


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-1 -0.8 -0 .6 -0.4 -0 .2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

distance (x)

d i s

t a n c e

( y )

Overhead view of the vector field, grad(f(x,y)=grad(-x2-y2)

Figure @verhead vie% of the vector field, ( ) y x f = . $otice that the arro%s are alldirected to%ard the maximum increase in the function, f.

. Divergen#e of a e#tor $antit'

"here are t%o possible %a s to appl the del operator to a vector. "he first, called thedivergence, results in a scalar function. "he second operation, the curl, results in a vector field.

For a vector field,

///

k w jviuu ++= , the divergence of u is defined as

z

w

y

v

x

ukw jviu

zk

y j xiu

+

+

=++

+

+

= //////

(?)

n index notation, e uation (?) is expressed as

D


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ii uu =

6 4oti#e that the divergen#e of a ve#tor is a s#alar ,$antit'5 )$st li6e the dot prod$#t.

Eh sicall , the divergence is a measure of the addition or removal of a vector uantit .magine a sin# full of %ater. f %e examine the flo% of the %ater near the drain of the sin# %e

%ill notice it is directed radiall in%ard indicating a net loss of the fluid. "his %ould result in anegative divergence. f %e attached a hose to the drain, so %e are adding %ater to the s steminstead of removing it, then the flo% %ould be radiall out%ard, indicating a net outflo% and a

positive divergence. f the divergence is !ero, 0= u , then there is no net inflo% or outflo%. fluid field %here 0= u is called solenoidal or divergenceless .

Figure 7 (left) ector field near a sin# drain indicating a negative divergence 0 u

xample: For the flo% field/

,/

,

j yi xu += as seen in figure 4

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

distance (x)

d i s

t a n c e

( y

)

vector flow field u=(-x2,y2)

Figure 4 ector flo% field/

,/

, j yi xu +=

3in# drain - ;et or hose outflo% -

8


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a) Find u at x=0, =0.?+ b) Find u at x=0.?, =0+

. The 7$rl of a e#tor $antit'

"he other %a to appl the del operator on a vector field is the curl . For a vector field,///

k w jviuu ++= , the curl of u is defined as

///

///

k yu

xv j

xw

zui

zv

yw

wvu z y x

k ji

u + + ==(>)

"he curl is a measure of the rotational properties of a vector field about a point. For a velocitfield, u , the curl is a measure of the rotation of a fluid parcel about its center of mass and iscalled the vorticity . "he vorticit is usuall denoted b the vector omega, . @ne %a to

imagine the vorticit is to place a small compass arro% in a fluid and to see ho% the arro%rotates about its center as it travels throughout the medium. f the vorticit of a fluid is !ero, it iscalled irrotational .

n oceanographic and atmospheric applications, one has particular interest in the vertical

vorticit component, ( )

== yu

xv

k u /

. "his is simpl a measure of the hori!ontal shear of

the fluid medium.

xample: Find the vorticit of the velocit field+

//

j xi yu += sho%n in figure ?+

*0


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-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

distance (x)

d i s t a n c e

( y )

velocity field (-y,x)

Figure ? Bepiction of the velocit field -

//

j xi yu +=

8. 9apla#ian of a s#alar f$n#tion:

n certain circumstances, it is possible to relate a velocit field to a scalar function called

the velocit potential, k z j

yi

xu 222

+

+

== . &hen %e examine the divergence of the

velocit field, %e obtain a ne% operation on the scalar called the aplacian.

( ) z y xu

+

+

==

"he aplacian, a scalar operation, is defined generall as

z y x +

+ (C)

n index notation, it ta#es the form

( )iii == - $otice that i is a dumm or repeating index .

t consists of the divergence of the gradient and thus is a measure of the spatial rate of change of

the gradient on a scalar function.n alculus, %e learned for *-B functions that, b setting the first derivative of a function

e ual to !ero %e can find the extrema of the curve. &e can resolve if the extrema points arelocal maximums or minimums b observing the sign of the second derivative. f the secondderivative is less than 0, the local extrema is a maximum. f the second derivative is greater than0, the local extrema is a minimum. f the second derivative is e ual to !ero, then %e have aturning point .

**


12/16

For multivariable functions, there are similar results. f the gradient of a scalar function,, is 0, then %e have a local extrema in the surface. &e can then use the aplacian to measure

the concavit of the surface and %hether the local extrema is a maximum 0 or a saddle point 0= .

xample: Hecall from our second example, that %e too# the gradient of the surface,( ), y x y x f = . &e can see from figure *, that %e have an extrema at the point x=0, y=0. t

is easil verified b finding %here the gradient of the surface is !ero, 0,,//

== j yi x + "his isclearl at x=0, y=0. &e can also see from figure *, that the extrema is a maximum. "his can beconfirmed b ta#ing the aplacian of + 04


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*22

,

,,//

=+=++== r y xr j yi xr j yi xrrr

7. For the vector, ( )//

,

/

, exp k xyz j xyi yxu += find u

From e uation (?)

zw

yv

xukw jviu zk y j xiu ++=++

++=

//////

For ( )//

,/, exp k xyz j yi xu +=

, %e obtain

( )( ) ( ) ( ) ( ) xyz xy xyz xy xy xy

xyz z

xy y

yx x z

w yv

xu

u

==

+

=

+

+

=

expexp

exp

&hich part of the functionAs domain is the divergence positive, negative or 0.

*7


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0> u for xJ0, K0 and xK0, J0 ( the second and fourth uadrant of the x, plane for everfinite value of !)

0


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From e uation (C), z y x +

+ and since %e have no ! or vertical dependence %e

onl need to consider,

( ) ( ) ( ) ,*

,,

,

,

,

*,,

,

,

,

*,,

,

,

,

,, * +

++=+

+

=

y x y

y x x

y x y xr h

First examine ( ) ( ) ( )

+=+

+

,

7,,

,

*,,

,

*,,

,

,

+ y x x y x x

y x x

3o ( ) ( ) ( ) ( )( ) ,

?,,

,,

,

?,,,

,

7,,

,

7,,

,

*,,

,

, ,7

y x

y x y x x y x y x x

x y x

x +=+++=

+

=+

&here %e simplified in the last step b finding a least common denominator. &e can seeimmediatel b s mmetr of the problem that %e can obtain the second partial %ith respect to y

b s%apping the xAs and As above+

( )( ) ,

?,,

,,

,

*,,

,

,

,

y x

x y y x y +

=+

dding the t%o terms together %e obtain our solution+

( ) ( ) ( ) 7

,

?,,

,,

,

?,,

,,

,

?,,

,,, *,,*

r y x

y x

y x

x y

y x

y x

r h =

+

+=+

++

=

4ote: &e can see that evaluating

r *

in artesian coordinates is tedious. f one use polarcoordinates, ho%ever, it is rather simple. oo# up the aplacian in polar coordinates and see

ho% simple it is to evaluate

r *

.

!ppendix !: 3ome useful vector calculus identities are (memori!e the first five)+

". The gradient prod$#t r$le of two s#alar f$n#tions:( ) g f f g fg +=

"". The divergen#e prod$#t r$le with a ve#tor and a s#alar:( ) uuu +=

""". The divergen#e of the gradient of a s#alar The 9apla#ian: =

" . The #$rl of the gradient of a s#alar:

*?


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( ) ( )0,0,00 == $otice the solution is the vector !ero, 0 , of %hich each component is !ero. t is commonnotation to impl the vector s mbol of the vector !ero since the curl is al%a s a vector result.

. The divergen#e of the #$rl of a ve#tor:0= u

$otice that this is 1ust the scalar number 0 since the divergence al%a s results in a scalarfunction or number.

". The #ross;prod$#t prod$#t r$le with a ve#tor and a s#alar:( ) = uuu

"". The divergen#e of the #ross prod$#t:( ) ( ) ( )!aa!!a =

""". The #$rl of the #ross prod$#t of a ve#tor:( ) ( ) ( ) ( ) ( )a!!a!aa!!a +=

"

ch3 review of partial differential operators

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