2103-601 Advanced Engineering Mathematics Lecture Note 2.0: Integral Curve and Surface of A Vector Field ABJ

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Lecture Note 2.0: Integral Curve and Surface of A Vector Field

Contents: 1. Integral Curve and Surface of A Vector Field

1.1. Integral Curve of A Vector Field 1.2. Integral Curve of A Vector Field Passing Through A Given Point at A Given Value of The Curve Parameter 1.3. Integral Surface of A Vector Field Containing A Given Curve

2. Exercise . Note: The note emphasizes underlying ideas of the current topic with only a few simple examples (e.g., constant

coefficients, etc.). Basic Ideas of The Note: 1. In this note we shall investigate some geometrical aspects of a vector field, namely its integral curve and surface. 2. Integral curve of the vector field V is the curve that is everywhere tangent to V . 3. Integral surface of V is the surface which at each point x on the surface V lies in its tangent plane). 4. Problems Statement: Given a vector field 3: R→ΩV , 3R⊂Ω , Ω∈x ,

))(),(),(()( xcxbxax =V , or in Cartesian coordinates )),,(),,,(),,,((),,( zyxczyxbzyxazyx =V ,

where Rcba →Ω:,, , we want to find (or to construct): 1. An integral curve / A family of integral curves. 2. An integral curve that passes through a given point 3Rxo ∈ at a given value of the curve

parameter oss = .

3. An integral surface that contains a given curve 3: RR →χ , ( ))(),(),()( rzryrxr χχχ=χ , at the value of the integral curve parameter oss = .

5. General Ideas and Approach to The Solutions:

Integral curve: In order to construct an integral curve of a given vector field V , the main ideas and steps are as follows.

5.1. Define a curve c and its parameteric representation )(sx . 5.2. Find the tangent of the curve )()( ss xT &= . 5.3. Since we want to construct a curve that is everywhere tangent to V , we set ))(()( ss xVx =& . 5.4. The resulting equation ))(()( ss xVx =& is then the parametric equation for the integral curve of V . 5.5. Given )(xV , the equation ))(()( ss xVx =& can then be integrated to find the integral curve.

5.5.1. In the integration, if we integrate from ),( oo xs to ),( xs , in principle we have the solution

),;( oo xssx , where 3Rxo ∈ is a given point, which is an integral curve that passes through the point ox at the value of the parameter of the curve

oss = . This is a one-parameter (vector parameter) family of curves.

Integral surface: 5.6. An integral surface of a vector field V that contains a given curve 3: RR →χ , )(rχ , (at the value of the

integral curve parameter oss = ) is basically a union of all the integral curves ),;( oo xssx that pass through points )(rxo χ= of the curve χ . Therefore, it is given by ))(,;( rss o χx . Note that there are two parameters, s is the parameter of an integral curve and r is the parameter of the given curve χ .

5.7. In computation, we allow the integration constant ox to be function of r , )(rx oo x= , and find the integration constant (which is now a function of r ) from the condition of the integral surface containing the given curve )(rχ at oss = . That is, from the equation )())(,;( rrsss ooo χxx == .

2103-601 Advanced Engineering Mathematics Lecture Note 2.0: Integral Curve and Surface of A Vector Field ABJ

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Notation: 1. =: , define the left with the right; := define the right with the left. 2. Vectors 2.1. A vector point will not be written as boldface as commonly done. However, the following notations are used.

1) If it is a point in nR , it will be declared and denoted using italic letter as, e.g., nRx∈ . 2) If it is a vector-valued function, it will be declared and denoted using bold face letter as, e.g., nRR →:x .

2.2. Caution: When Cartesian coordinates are employed, there can be multiple uses of a letter such as Rx∈ for the x -component of the point, and 3Rx∈ for a vector point x in 3R . The context should be clear to which one is referred.

2.3. Subscript – Superscript for components of vectors 1) Subscript is used for component of a point, e.g., nRx∈ , as in ix ( ni ,...,1= ) for component of a point

( nxx ,...,1 ) in nR .

2) Superscript is used for component of a vector-valued function, e.g., ix ( ni ,...,1= ) for component of a vector –valued function nRR →:x .

2103-601 Advanced Engineering Mathematics Lecture Note 2.0: Integral Curve and Surface of A Vector Field ABJ

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1. Integral Curve and Surface of A Vector Field

Problem Statement: Given a vector field 3: R→ΩV , 3R⊂Ω , Ω∈x ,

))(),(),(()( xcxbxax =V , or in Cartesian coordinates )),,(),,,(),,,((),,( zyxczyxbzyxazyx =V , (1.1) where Rcba →Ω:,, , we want to find (or to construct):

4. Integral Curve: a curve, or a family of curves, that is everywhere tangent to the field. 5. Integral Curve: a specific curve in the family that passes through a given point at a given value of

the curve parameter. 6. Integral Surface: a specific integral surface in the family that cotains a given curve.

Note that an integral curve of the vector field V is the curve that is everywhere tangent to V , while an integral surface of V is the surface which at each point x on the surface V lies in its tangent plane.

Note that we will deal only with vector fields in which the following two conditions are satisfied:

1. V is nonvanishing in Ω . 2. 1,, Ccba ∈ .

We shall find that this problem is closely related to first-order pde. Nonetheless, in this note we shall focus and view it as a problem of the geometry of a vector field. Where do we find this?: For a vector field V , an integral curve of V is often referred to as a field line. If V is a

force field (e.g., magnetic force field), the integral curves of V are lines of force. If V is a velocity field, the integral curves of V are called trajectories in a dynamical system, or streamlines in fluid.

For students who are further interested in dynamical system, the following book can be suggested: Hirsch, M. W., and Smale, S., 1974, Differential equations, dynamical systems, and linear algebra, Academic Press, New York.

x

y

z

)),,(),,,(),,,((),,( zyxczyxbzyxazyx =V

Integral curves

Integral surfaces

x

y

z

Fig. 1.1. Integral curves and integral surfaces of a vector field V .

2103-601 Advanced Engineering Mathematics Lecture Note 2.0: Integral Curve and Surface of A Vector Field ABJ

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1.1. Integral Curve of A Vector Field Problem Statement: Given a vector field 3: R→ΩV , 3R⊂Ω , Ω∈x ,

))(),(),(()( xcxbxax =V , or in Cartesian coordinates )),,(),,,(),,,((),,( zyxczyxbzyxazyx =V , (1.1) where Rcba →Ω:,, , we want to find (or to construct) a curve, or a family of curves, that is everywhere tangent to the field. General Ideas and Approach:

In order to construct an integral curve of a given vector field V (say, that passes through a given but not yet specieified point 3Rxo ∈ at a given value of the curve parameter oss = ), the main ideas and steps are as follows.

1) Define a curve c and its parameteric representation )(sx . 2) Find the tangent of the curve )()( ss xT &= . 3) Since we want to construct a curve that is everywhere tangent to V , we set ))(()( sks xVx =& , where k is

constant (which shall later be set to 1). 4) The resulting equation ))(()( ss xVx =& is then the parametric equation for the integral curve of V . 5) Given )(xV , the equation ))(()( ss xVx =& can then be integrated to find the integral curve.

Details: 1 1. Let c be an integral curve of V with parameteric representation given by 3: RI →x , RI ⊂ , Is∈ ,

( ))(),(),()( 321 sxsxsxs =x , or in Cartesian coordinates ( ))(),(),()( szsysxs =x , (1.2) where s is the parameter of the curve.

2. The tangent of c , 3:)( RRs →T , is then given by

( ))(),(),()()( szsysxss &&&& == xT (1.3)

3. Since c is everywhere tangent to V , we have ))(()( sks xVx =& , where k is a nonvanishing scalar.2 4. Without loss of generality, we set 1+=k . Thus, we set the direction of the rate of change of )(sx with increasing

s , or the direction of the tangent of c , to be the same as the direction of V . Finally, we have the equation for the integral curve of V as

Integral Curve: ))(()( ss xVx =& , or ( )( )( ))(),(),())(()(

)(),(),())(()()(),(),())(()(

szsysxcscszszsysxbsbsyszsysxasasx

======

xxx

&

&

&

. 1.4(A,B,C)

5. Note that

5.1. The system of equations (1.4) is autonomous. There is no explicit dependency on s . 5.2. From ode, in general the integration of each of the equations in (1.4) results in one arbitrary parameter.

Hence, the system has three arbitrary parameters, and the solution may be written as );( oxsx , where 3Rxo ∈ is an arbitrary vector point, (1.5)

or );( oxsx , );( oysy , and );( ozsz , where Rzyx ooo ∈,, are arbitrary constants. 1 Here, due to familiarity, we switch from the constructive tone to derive/proof tone. 2 )(sx is the parametric representation of the integral curve c .

)(xV is the value of the vector field function V evaluated at a (spatial) point x .

)(sx is the value of the vector function x evaluated at a point s , and )(sx& is the value of the tangent vector function

x& evaluated at a point s . We want to construct a curve c such that the vector V at a point on the curve is parallel to the tangent of the curve at that

point. o That is, we want to construct a curve c such that the vector field function V evaluated at )(sx x= is parallel

to the vector x& at )(sx x= , or to the tangent vector function x& evaluated at s .

We evaluate the vector field function V at )(sx x= to be ( ) ( ))()( ssx xVxV == .

We evaluate the tangent vector function x& at the corresponding point )(sx x= to be )(sx& .

We then set the two vectors at the same point )(sx x= proportional: ))(()( sks xVx =& .

2103-601 Advanced Engineering Mathematics Lecture Note 2.0: Integral Curve and Surface of A Vector Field ABJ

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5.3. The solution );( oxsx , Rs∈ , 3Rxo ∈ , then constitutes a family of curves, which is indexed by the

parameter 3Rxo ∈ (or the family is parametrized by the parameter 3Rxo ∈ ) and whose curves are parametrized by the parameter s (or curves in the family are parametrized by the parameter s ).

6. Recognizing that Eq. (1.4) can be rewritten as dsscsdzds

sbsdyds

sasdx

===))(()(,

))(()(,

))(()(

xxx, thus we can also rewrite

Eq. (1.4) as

),,(),,(),,( zyxc

dzzyxb

dyzyxa

dx== , 1.6(A,B)

Note that by getting rid of s above we have reduced the number of independent equations to two (any two

pairs of the above equalities), and the system of equations (1.6) is a system of two equations in three variables. Hence, typically we can solve for two variables in terms of the third. The system (1.6) is the spatial representation of the integral curve of V .

Example 1.1: Find an integral curve of a vector field ),,( cba=V , where cba ,, are constants. Solution: 1. Let ( ))(),(),()( szsysxs =x be the parametric representation of an integral curve of the vector field ),,( cba=V .

Note that the vector field ),,( cba=V is a constant field whose vector at any point x points in the direction ),,( cba .

2. Then, from the parametric representation of an integral curve, Eq. (1.4), we have

oo

oo

oo

zcszszcdsdz

ybsysybdsdy

xasxsxadsdx

+=→=

+=→=

+=→=

);(

);(

);(

[or, vectorially, ),,()( cbasxx o =− ], (Ex. 1.1)

where ooo zyx ,, are arbitrary constants. Note from ode that, without further conditions, it is expected that each of the equations above contains one arbitrary parameter. Hence, the system contains three arbitrary parameters.

3. In order to obtain a spatial curve, we must get rid of s from the above system. Solve for s from the x -component

equation, we have axx

s o−= . Then, substitute this into the remaining two equations, we have the spatial

repreenstaion of the integral curve

oooo

oooo

zxxaczxxz

yxxabyxxy

+−=

+−=

)(),;(

)(),;(. (Ex. 1.2)

Figure (1.1) below shows some of the integral curves for ),,( cba=V . ANS

),,(),,( cbazyx =V

x

y

z

ox

Fig. 1.2. Integral curves of a vector field V .

2103-601 Advanced Engineering Mathematics Lecture Note 2.0: Integral Curve and Surface of A Vector Field ABJ

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1.2. Integral Curve of A Vector Field Passing Through A Given Point at A Given Value of The Curve Parameter Problem Statement: Given a vector field 3: R→ΩV , 3R⊂Ω , Ω∈x ,

))(),(),(()( xcxbxax =V , or in Cartesian coordinates )),,(),,,(),,,((),,( zyxczyxbzyxazyx =V , (1.1) where Rcba →Ω:,, , we want to find (or to construct) the integral curve of V that passes through a specified point

3Rxo ∈ at a given value of the parameter oss = . General Ideas and Approach: In this case, the general ideas and approach are the same as in Sec. 1.1. Only this time, in the integration of Eq. (1.4) we integrate not from ),0( ox but from ),( oo xs to ),( xs , and we rewrite Eq. (1.5) as

),;( oo xssx , where 3Rxo ∈ is a given point, (1.7) or ),;( oo xssx , ),;( oo yssy , and ),;( oo zssz .

Example 1.2: Find the integral curve of a vector field ),,( cba=V , where cba ,, are constants that passes through

a point 3Rxo ∈ at a given value of the curve parameter oss = . Solution: 1. The process is the same as in Example 1.1. Only this time, in the integration of Eq. (1.4), we supply the given

condition of passing through ox at a given value of the curve parameter oss = in the integration. Hence, Eq. (Ex 1.1) is modified to

oooo

oooo

oooo

zssczsszcdsdz

yssbyssybdsdy

xssaxssxadsdx

+−=→=

+−=→=

+−=→=

)(),;(

)(),;(

)(),;(

[or, vectorially, ),,)(()( cbassxx oo −=− ], (Ex. 1.3)

2. Note that the spatial representation remains the same, i.e.,

oooo

oooo

zxxaczxxz

yxxabyxxy

+−=

+−=

)(),;(

)(),;(. (Ex. 1.4)

ANS

),,(),,( cbazyx =V

x

y

z

oo xs ,

Fig. 1.3. The integral curve of the vector field V that passes through a given point ox at a given curve parameter oss = .

2103-601 Advanced Engineering Mathematics Lecture Note 2.0: Integral Curve and Surface of A Vector Field ABJ

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1.3. Integral Surface of A Vector Field Containing A Given Curve Problem Statement: Given a vector field 3: R→ΩV , 3R⊂Ω , Ω∈x ,

))(),(),(()( xcxbxax =V , or in Cartesian coordinates )),,(),,,(),,,((),,( zyxczyxbzyxazyx =V , (1.1) where Rcba →Ω:,, , we want to find (or to construct) the integral surface of V that contains a given curve χ ,

defined parametrically by 3: RR →χ , ( ))(),(),()( rzryrxr χχχ=χ , at the value of the integral curve parameter

oss = . General Ideas and Approach:

In order to construct an integral surface of a vector field V that contains a given curve χ at the value of the integral curve parameter oss = [we denote this surface by ),;,( χx osrs or for short ),( rsx ], we simply construct it from the union of all the integral curves that each passes through a point on χ at oss = , ))(,;( rss o χx . The main ideas and steps are as follows.

1. Start from a point, any point, )(rχ on χ , then 2. construct an integral curve of V that passes through the point )(rχ at oss = , i.e., ))(,;( rss o χx ,

Eqs. (1.5) and (1.7), then 3. union these curves to get the integral surface ),;,( χx osrs , or for short ),( rsx .

Details:

1. Let S be an integral surface of the vector field V with parametric representation 3: R→Ωx , 2R⊂Ω , Ω∈),( rs ,

( )),(),,(),,(),( 321 rsxrsxrsxrs =x , or in Cartesian coordinates ( )),(),,(),,(),( rszrsyrsxrs =x , (1.8)

where s is the parameter along the integral curve and r is the parameter along the given curve χ .

2. In order for S to be an integral surface of V , ),( rsx must be everywhere tangent to V . 3. Use the concept of the integral curve, we pick a point, any point, )(rχ and construct an integral curve through this

point at os . That is, we construct a curve ))(,;( rss o χx . This curve is given by, Eq. (1.4),

Integral Curve: ));(();( rsrs xVx =& , or ( )( )( ));(),;(),;());(();(

);(),;(),;());(();();(),;(),;());(();(

rszrsyrsxcrscrszrszrsyrsxbrsbrsyrszrsyrsxarsarsx

======

xxx

&

&

&

. 1.9(A,B,C)

Note that dsd /=⋅

and r is a parameter, resulted from the intital point )(rxo χ= in ))(,;( rxss oo χx = from integration (see Eq. (Ex. 1.3)). Also note that the integration of Eq. (1.9) is from ))(,( rso χ to ),( xs .

)),,(),,,(),,,((),,( zyxczyxbzyxazyx =V

Integral curves

oo xr =)(χ

Fig. 1.4. Construction of an integral surface of a vector field, ),( rsx , containing a given curve )(rχ at oss = .

)(rχ

running r

),( rsx running s

ooo xrs =),(x Given (initial) curve

2103-601 Advanced Engineering Mathematics Lecture Note 2.0: Integral Curve and Surface of A Vector Field ABJ

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4. Finally, we union these integral curves and obtain the equations for the integral surface

Integral Surface: )),((),( rsrs xVx =& , or ( )( )( )),(),,(),,()),((),(

),(),,(),,()),((),(),(),,(),,()),((),(

rszrsyrsxcrscrszrszrsyrsxbrsbrsyrszrsyrsxarsarsx

======

xxx

&

&

&

. 1.10(A,B,C)

5. To satisfy the condition of the surface containting a given curve:

5.1. The integration of Eq. (1.10) results in the arbitrary functions of integration, ( ))(),(),()( rzryrxr oooo =x , which are functions of the parameter r . That is, the integration results in

))(,;( rss oo xx (1.11)

5.2. These functions, ( ))(),(),()( rzryrxr oooo =x , can be determined from the condition of the integral surface containing the given curve )(rχ at oss = . That is, from the equation

)())(,;( rrsss ooo χxx == . (1.12) 6. To convert to a spatial representation of the surface: If, e.g., Eq. 1.10 (A) and (B) do not depend on ),( rsz , we

have after integration ),(),,( rsyrsx . In addition, if the transformation ),(),,( rsyrsx is invertible, i.e., ( ) ( )),(),,((),(),,( yxryxsrsyrsx ↔ , then ),( rs can be solved in terms of ),( yx , i.e., ( )),(),,(( yxryxs and the results substituted in ( )),(),,((:),( yxryxszrsz .

Example 1.3: Find the integral surface ),( rsx of a vector field

),,( cba=V , where cba ,, are constants,

that contains the curve 3: RR →χ , ( ))(),(,)( rzryrr χχ=χ , [or ( ))(),(,)( xzxyxx χχ=χ , if x is used as a parameter for χ ]

at the value of the integral curve parameter oss = . Solution: 1. Let ( )),(),,(),,(),( rszrsyrsxrs =x be an integral surface of the vector field ),,( cba=V , with s being parameter

of the integral curve and r of the given curve χ , containing the curve χ . 2. Then, from the parametric representation of an integral surface, Eq. (1.10), we have

)()(),(),(

)()(),(),(

)()(),(),(

rzsscrszcds

rsdz

ryssbrsybds

rsdy

rxssarsxads

rsdx

oo

oo

oo

+−=→=

+−=→=

+−=→=

, (Ex. 1.5)

where )(),(),( rzryrx ooo are yet arbitrary functions of r but will be determined from the given curve χ . 3. From the given constraint, ),( rsx must contain χ at oss = . Therefore, )(),( rrso χx = ; consequently, from Eq.

(Ex. 1.5) and given )(rχ ,

)()(),(

)()(),()(),(

rzrzrsz

ryryrsyrrxrsx

oo

oo

oo

χ

χ

==

====

.

4. Thus, the integral surface ),( rsx of the vector field ),,( cba=V which contains the given curve

( ))(),(,)( rzryrr χχ=χ at os is given parametrically by

)()(),(

)()(),()(),(

rzsscrsz

ryssbrsyrssarsx

o

o

o

χ

χ

+−=

+−=+−=

.

2103-601 Advanced Engineering Mathematics Lecture Note 2.0: Integral Curve and Surface of A Vector Field ABJ

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5. If, e.g., ( ) ),0,()(),(,)( 2rrrzryrr == χχχ , then: or if, ( ) ),,()(),(,)( 2rrrrzryrr == χχχ , then:

2)(),(

)(),()(),(

rsscrsz

ssbrsyrssarsx

o

o

o

+−=

−=+−=

, 2)(),(

)(),()(),(

rsscrsz

rssbrsyrssarsx

o

o

o

+−=

+−=+−=

.

In addition, since ),(),,( rsyrsx can be solved for ),(),,( yxryxs , i.e.,

y

baxyxr

sby

yxs o

−=

+=

),(

),(,

abaybxyxr

sbayxyxs o

−−

=

+−−

=

),(

),(,

we have the spatial represention of the surface

2

),( ⎟⎠

⎞⎜⎝

⎛ −+=

baybxy

bcyxz ,

2

),( ⎟⎠

⎞⎜⎝

⎛−−

+−−

=abaybx

bayxcyxz .

ANS

z1

2. Exercise Problem 1. Given the vector field V , the point ),,( ooo zyx , and the initial curve )(rχ below:

a) Find the parametric representation of the family of integral curves of the vector field V . b) Find the parametric representation of the integral curve that passes through the point ),,( ooo zyx at

the value of the curve parameter 0=os . c) Find the parametric representation of the integral surface that contains the given initial curve )(rχ . d) Find the spatial representation of the integral curve in b), by i) using the result from b), and ii)

integrate directly from the spatial differential equation, Eq. 1.6. e) Find the spatial representation of the integral surface in c), by i) using the result from c), and ii)

integrate directly from the spatial differential equation, Eq. 1.6. f) Use Matlab, Mathcad, etc., to plot the integral surface in e).

i. )0,,( ba=V , )1,0,0(),,( =ooo zyx , )cos,0,()( rrr =χ , where ba, are constants.

ii. )0,,( yx=V , )1,0,1(),,( =ooo zyx , ),0,()( 2rrr =χ .

iii. )0,,( yx −=V , )1,0,1(),,( =ooo zyx , ),0,()( 2rrr =χ .

iv. )0,,( xy=V , )1,0,1(),,( =ooo zyx , ),0,()( 2rrr =χ .

v. )0,,( xy −=V , )1,0,1(),,( =ooo zyx , ),0,()( 2rrr =χ .

vi. ),,( zyx=V , )1,2,1(),,( =ooo zyx , ),1,()( 2rrrr +=χ .

Fig. 1.5. Integral surface of Ex. 1.3 for the case with ),,( cba = (0,1,2) and ( )2,0,)( rrr =χ . Note that the figure is not to scale and the origin does not locate at the vertex of the shown axes.