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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 .

    **

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    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 +=

    "