e-mail : [email protected] webpage : · without using operators div, ∇, ∆, curl or grad. here ....

Armin Halilovic Math. Exercises

E-mail : [email protected] webpage : www.sth.kth.se/armin MATH. EXERCISES. GRADIENT, DIVERGENCE, CURL DEL (NABLA) OPERATOR , LAPLACIAN OPERATOR , CONTINUITY AND NAVIER-STOKES EQUATIONS VECTOR PRODUCTS If ),,( 321 uuuu = and ),,( 321 vvvv =

then

332211 vuvuvuvu ++=• (scalar or dot product)

321

321

vvvuuukji

vu

=× (vector or cross product)

In some books is also considered outer product defined by

=⊗

3

2

1

uuu

vu )( 321 vvv =

332313

322212

312111

vuvuvuvuvuvuvuvuvu

GRADIENT, DIVERGENCE, CURL DEL (NABLA) OPERATOR , LAPLACIAN OPERATOR GRADIENT Let ),,( zyxϕ be a scalar field. The gradient is the vector field defined by

),,()(zyx

grad∂∂

∂∂

∂∂

=ϕϕϕϕ

DIVERGENCE Let )),,(),,,(),,,(( zyxRzyxQzyxPF =

be a vector field, continuously differentiable with respect to x, y and z. Then the divergence of F

is the scalar field defined by

zR

yQ

xPFdiv

∂∂

+∂∂

+∂∂

=)(

CURL. The curl of F

is the vector field defined by

kyP

xQj

xR

zPi

zQ

yR

RQPzyx

kji

Fcurl

)()()()(∂∂

−∂∂

+∂∂

−∂∂

+∂∂

−∂∂

=∂∂

∂∂

∂∂

=

or ),,()(yP

xQ

xR

zP

zQ

yRFcurl

∂∂

−∂∂

∂∂

−∂∂

∂∂

−∂∂

=

DEL (NABLA) OPERATOR The vector differential operator

=∇ ),,(zyxz

ky

jx

i∂∂

∂∂

∂∂

=∂∂

+∂∂

+∂∂

is called del or nabla .

1 / 28


Using ∇ we can denote grad, div and curl as below:

)(ϕgrad = ϕ∇ FFdiv

•∇=)( FFcurl

×∇=)( Note that ∇•F

is not the same as F

•∇ .

∇•F

=z

Ry

Qx

P∂∂

+∂∂

+∂∂ .

LAPLACIAN OPERATOR

The Laplacian operator, 2

2

2

2

2

22

zyx ∂∂

+∂∂

+∂∂

=∇=∆ , is defined for a scalar field U(x,y,z) by

2

2

2

2

2

22

zU

yU

xUUU

∂∂

+∂∂

+∂∂

=∇=∆ ,

and for a vector field )),,(),,,(),,,(( zyxRzyxQzyxPF =

by ),,(2 RQPFF ∆∆∆=∇=∆

. Some formulas for polar and cylindrical coordinates Polar coordinates ( 2 dim)

ereϑ

ϑ

i

jer

eϑ

ϑ

i

j

r P

F

transformation: θcosrx = , θsinry = , area element: θddrrdA = standard basis: ,cossin,sincos jiejier

θθθθ θ +−=+= [Remark 1 : Note that θeer

, vary ( depend on θ ) when we move from point to point, this is the reason why this basis, in some books, is called “ local basis” .] If jFiFF yx

+= in Cartesian coord. and θϑeFeFF rr

+= the same vector in polar coordinates then

θθ sincos yxr FFF += , θθϑ cossin yx FFF +−= .

[Remark 2: Vi can derive these formulas by calculating the components of F

in the directions of re and θe

. Thus

2 / 28


θθθθ sincos)sin(cos)( yxyxrr FFjijFiFeFF +=+•+=•=

, similarly

θθθθϑϑ cossin)cossin()( yxyx FFjijFiFeFF +−=+−•+=•=

. ] Cylindrical coordinates ),,( zr θ : transformation: θcosrx = , θsinry = , z=z volume element: dzddrrdV θ=

er

eϑ

ez =k

x

y

z

ϑr

F

standard basis: kejiejie zr

=+−=+= ,cossin,sincos θθθθ θ

If kFjFiFF zyx

++= in Cartesian coord. and kFeFeFF zrr

++= θϑ the same vector in cylindrical coordinates then we have following vector components relationship: θθ sincos yxr FFF += , θθϑ cossin yx FFF +−= , zz FF = [Remark 3: For example, we can get θθ sincos yxr FFF += in the following way:

θθθθ sincos)sin(cos)( yxzyxrr FFjikFjFiFeFF +=+•++=•=

] scalar field: ),,( zrf θ

gradient: zr ezfef

re

rfffgrad

∂∂

+∂∂

+∂∂

=∇= θθ1)(

laplacian: 2

2

2

2

22 11

zff

rrfr

rrff

∂∂

+∂∂

+

∂∂

∂∂

=∆=∇θ

vector field: ),,( zr FFFF θ=

divergence: z

FFrr

Frr

FFdiv zr

∂∂

+∂

∂+

∂⋅∂

=∇=θθ )(1)(1)(

curl:

zr

zr

FFrFzr

eere

rFFcurl

θ

θ

θ⋅

∂∂

∂∂

∂∂

⋅

=×∇=

1)( =

krzr

rz eF

rrF

re

rF

zFe

zFF

r

∂∂

−∂

∂+

∂∂

−∂∂

+

∂∂

−∂∂

θθθ

θθ )(11

3 / 28


EXERCISES 1. Find a) )(Fdiv

, b) ))(( Fdivgrad

and c) )(Fcurl

if ),,( 22 xzxyF +=

2. Prove the following identities a ) 0)(( =Fcurldiv

(or 0)( =×∇•∇ F

) b) 0))((

=ϕdivcurl (or 0)(

=∇×∇ ϕ ) 3. Which one of the following functions a) 222

1 23),,( zyxzyxf ++= b) )ln(),,( 22

2 zyxzyxf ++= c) )exp(),,( 3

3 zyxzyxf ++= d) zyxyxzyxf 455),,( 22

4 +++−= satisfies the Laplace equation f∆ =0 ? 4. Find )))(( ff ∇×∇•∇+∆ if zyxzyxf ++= 23),,( . 5. Write the general transport equation

φϕρϕρϕ SUt

+∇⋅Γ•∇=•∇+∂

∂ )()()(

without using operators div, ∇ , ∆ , curl or grad. Here ),,( wvuU =

. Functions S,,, Γϕρ , u, v, w are real functions of t, x, y and z. 6. Which one, if any, of the following functions a) zyxzyx ++= 24

1 ),,(ϕ

b) zyxzyx ++= 222 ),,(ϕ

c) 2223 ),,( zyxzyx ++=ϕ

satisfies the equation SU +∇⋅Γ•∇=•∇ ))(()( ϕϕ

? Here 5=Γ , )3,2,1(=U

and 2342 −+= yxS . 7. Find which one (if any) of the following functions a) 222

1 ),,( zyxzyx ++=ϕ

b) zyxzyx 5),,( 222 ++=ϕ

c) 2223 5),,( zyxzyx ++=ϕ

satisfies the equation

4 / 28


SgraddivUdivt

+Γ=+∂

∂ )()()( ϕρϕρϕ

where ρ=3, 2=Γ , )4,3,2(=U

and 521812 ++= yxS . 8. (exam 1, 2008) A) Write the general transport equation

SUt

+∇⋅Γ•∇=•∇+∂

∂ ))(()()( ϕρϕρϕ

( eq 1)


. Functions S,,, Γϕρ , u, v, w are real functions of t, x, y and z. B) Let 2=ρ , 3=Γ , )4,2,1(=U

. Find S in the equation (eq 1) if we now that the function 32),,( zyxzyx ++=ϕ satisfies the equation. 9. (Q6, exam 2, 2008) Consider the following equation

426)())(()()(−−+×∇•∇+∇⋅Γ•∇=•∇+

∂∂ xyzyUU

t

ϕρϕρϕ ( eq 1)

Let 1=ρ , Γ= constant , ),3,2( xyU −=

. Find the constant Γ in the equation (eq 1) if we now that the function 222),,( zyxtzyx +++=ϕ satisfies the equation.

10. If possible, find ),( yxf for the given partial derivatives ),( yxfx∂∂ and ),( yxf

y∂∂ .

a) ),( yxfx∂∂ = xy2 and ),( yxf

y∂∂ = yx 22 + .

b) ),( yxfx∂∂ = yx +2 and ),( yxf

y∂∂ = x .

c) ),( yxfx∂∂ = xyye and ),( yxf

y∂∂ = xyxe .

d) ),( yxfx∂∂ = yx +2 and ),( yxf

y∂∂ = x5 .

( Hint: Necessary condition: If ),( yxf has continues derivatives then the mixed derivatives of ),( yxf should be equal. Thus

∂∂

∂∂

=

∂∂

∂∂ yxf

yxyxf

xy,(,( (*)

is the necessary condition for the existence of a function ),( yxf that has the given derivatives. 11. Determine the value of a for which the system of partial differential equations

5 / 28


),( yxfx∂∂ = yaxy + and ),( yxf

y∂∂ = xx +2 .

has solutions. Then find ),( yxf corresponding to this value of a.

12. If possible, find ),,( zyxf for the given partial derivatives ),,( zyxfx∂∂ , ),,( zyxf

y∂∂

and ),,( zyxfz∂∂ .

a) yzzyxfx

=∂∂ ),,( , zxzzyxf

y+=

∂∂ ),,( and 23),,( zyxyzyxf

z++=

∂∂

b) yzzyxfx

=∂∂ ),,( , zxzzyxf

y+=


z++=

∂∂

c) xyzyzezyxfx

=∂∂ ),,( , xyzxzezyxf

y=

∂∂ ),,( and xyzxyezyxf

z=

∂∂ ),,(

d) yzzyxfx

=∂∂ ),,( , zxzzyxf

y+=

∂∂ ),,( and xzyxyzyxf

z++=

∂∂ ),,(

( Hint: Necessary condition: If ),,( zyxf has continuous derivatives then the mixed derivatives of ),( yxf should be equal. Thus

∂∂

∂∂

=

∂∂

∂∂ f

yxf

xyCon :1

∂∂

∂∂

=

∂∂

∂∂ f

zxf

xzCon :2

∂∂

∂∂

=

∂∂

∂∂ f

zyf

yzCon :3

are the necessary condition for the existence of a function ),( yxf that has the given derivatives. 13. Determine the values of a and b for which the system of partial differential equations

),,( zyxfx∂∂ = xyzax 22 + , ),,( zyxf

y∂∂ = 13 +zx and ),,( zyxf

z∂∂ = zybx 23 +

has solutions. Then find ),,( zyxf corresponding to these values of a and b. 14. We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field

.x,y,z, w x,y,z, vx,y,zuV ))()()(( =

Use the following equations ( continuity and Navier Stokes equations) to find en expression for pressure P(x,y,z) as a function of x,y and z, where ρ =constant, µ =constant , ),0,0( gg −=

i.e. xg = yg =0 and

)/81.9 where( 2smggg z ≈−=

6 / 28


Incompressible continuity equation:

0=∂∂

+∂∂

+∂∂

zw

yv

xu eq1.

Navier Stokes equations: x component:

)( 2

2

2

2

2

2

zu

yu

xug

xP

zuw

yuv

xuu

tu

x ∂∂

+∂∂

+∂∂

++∂∂

−=

∂∂

+∂∂

+∂∂

+∂∂ µρρ eq2.

y component:

)( 2

2

2

2

2

2

zv

yv

xvg

yP

zvw

yvv

xvu

tv

y ∂∂

+∂∂

+∂∂

++∂∂

−=

∂∂

+∂∂

+∂∂

+∂∂ µρρ eq3.

z component:

)( 2

2

2

2

2

2

zw

yw

xwg

zP

zww

ywv

xwu

tw

z ∂∂

+∂∂

+∂∂

++∂∂

−=

∂∂

+∂∂

+∂∂

+∂∂ µρρ eq4.

a) )0,24,32( yxyxV −+=

b) )2,32,43( −−+= yxyxV

c) )2,4,21( xyV −+=

15. (exam 1, 2009) A) Consider the following equation

24841616)())(()()(−++++×∇•∇+∇⋅Γ•∇=•∇+

∂∂ yzxzyxUU

t

ϕρϕρϕ ( eq 1)

Let 2=ρ , Γ= constant , )2,4,4( yxU +=

. Find the constant Γ in the equation (eq 1) if we now that the function 22231),,( zyxtzyx ++++=ϕ satisfies the equation. B) We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field

.x,y,z, w x,y,z, vx,y,zuV ))()()(( =

Use the following equations ( continuity and Navier Stokes equations) to find en expression for pressure P(x,y,z) as a function of x,y and z, where ρ =constant, µ =constant ,

),0,0( gg −= i.e. xg = yg =0 and )/81.9 where( 2smgggz ≈−= and

)22,24,46( zyxV −−+=

. 16. (exam 2, 2009) We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field

.x,y,z, w x,y,z, vx,y,zuV ))()()(( =

Use the following equations ( continuity and Navier Stokes equations) to find first i) parameter a and then

7 / 28


ii) en expression for pressure P(x,y,z) as a function of x,y and z, where ρ =constant, µ =constant ,

),0,0( gg −= i.e. xg = yg =0 and )/81.9 where( 2smgggz ≈−= and

)1,5,32( azyxV −−+=

.

17. Consider steady, incompressible, isothermal, laminar stationary Newtonian flow in a long round pipe in the z-direction, with constant circular cross-section of radius R=2 m. Use the continuity and the Navier-Stokes equations in cylindrical coordinates to find the velocity field V=(ur, uθ, uz) and the pressure field P (r,θ,z) if the fluid flow satisfies the following conditions:

c0. All partial derivatives with respect to time t are 0 ( Steady flow)

c1. μ=0.001 kg/(m∙s) and ρ =1000 kg/m3

c2. A Constant pressure gradient ∂P/∂z = –1/250 Pa/m is applied in the horizontal axis ( z-axis in our notation): ∂P/∂z = –1/250, c3. The flow is parallel to the z axis, that is ur =0 and uθ =0. c4. We assume that the flow is axisymmetric . The velocity does not depend on θ,

that is 0=∂∂θ

zu

c5. Boundary cond. 1 ( No-slip boundary condition, Vfluid=Vwall ): If r=2 then uz= 0

c6. Boundary condition 2: uz has maximum at r=0 that is 00=

=∂∂

rruz

---------------------------------------------------------------------------------------------

The continuity and the Navier-Stokes equations for an incompressible , isothermal Newtonian flow (density ρ =const, viscosity µ =const), with a velocity field

),,( zr uuuV θ=

in Cylindrical coordinates ),,( zr θ : Incompressible continuity equation

0)(1)(1

=∂∂

+∂∂

+∂

∂z

uurr

rur

zr

θθ eq a)

Navier-Stokes equations in Cylindrical coordinates: r-component:

∂∂

+∂∂

−∂∂

+−

∂∂

∂∂

++∂∂

−=

∂∂

+−∂∂

+∂∂

+∂∂

2

2

22

2

22

2

211zuu

ru

rru

rur

rrg

rP

zuu

ruu

ru

ruu

tu

rrrrr

rz

rrr

r

θθµρ

θρ

θ

θθ

eq b)

θ -component:

8 / 28


∂∂

+∂∂

+∂∂

+−

∂∂

∂∂

++∂∂

−=

∂∂

++∂∂

+∂∂

+∂∂

2

2

22

2

22

2111zuu

ru

rru

ru

rrr

gPr

zu

uruuu

ru

ru

ut

u

r

zr

r

θθθθθ

θθθθθθ

θθµρ

θ

θρ

eq c)

z-component:

∂∂

+∂∂

+

∂∂

∂∂

++∂∂

−=

∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

211

zuu

rrur

rrg

zP

zuuu

ru

ruu

tu

zzzz

zz

zzr

z

θµρ

θρ θ

eq d)

18. (Exam 1 March 2012, question A , 4points.) We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field

)3,24,3())()()(( azbzyxcyxx,y,z, w x,y,z, vx,y,zuV −+−++==

Use the following equations ( continuity and Navier Stokes equations) , where ρ =constant, µ =constant , ),0,0( gg −=

i.e. xg = yg =0 and )/81.9 ( 2smgggz ≈−= to find: i) parameters a, b and c ii) en expression for pressure P(x,y,z) as a function of x,y and z. The GRADIENT VECTOR with change of variables and basis. The gradient vector for the function f(x,y,z) is defined as

kzfj

yfi

xf

zf

yf

xffgrad

∂∂

+∂∂

+∂∂

=∂∂

∂∂

∂∂

= ),,()( (*) .

If we change variables x, v, z to u, v , w and replace basis vectors kji

,, with new ( linearly independent) vectors 321 ,, eee then we can express the same gradient vector )( fgrad in terms of variables u, v , w and vectors 321 ,, eee .

We simple calculate the derivatives zf

yf

xf

∂∂

∂∂

∂∂ and in new variables and express kji

,, as

a linear combinations of 321 ,, eee . Then we substitute those values into (*). (See the following example.) 19. We consider a scalar field ),,( zrf θ given in cylindrical coordinates, where

zzryrx === , sin ,cos θθ , and basis vectors are 321 ,, eee .

9 / 28


Find the expression for the gradient, )),,(( zrfgrad θ ), in cylindrical coordinates, that is in

terms of zff

rfeeezr

∂∂

∂∂

∂∂ and ,, , ,,,, 221 θ

θ

if

a) kejeie

=== 321 ,, ( we keep the same basis kji

,, ) .

b) kjejeie

+=== 2,2 321

c) kejeie

=== 321 , sin ,cos θθ (exam 1, 2012; Q5 B (2 points)) d) kejiejie

=+−=+= 321 ,cossin,sincos θθθθ (this is often used as a local basis for cylindrical coordinates) 20. (exam 2016; Q6 A (2 points)) Derive the Cauchy momentum equation

gDt

VD

ρρ +⋅∇= σ .

21. Find the flux of the vector field kyxjyixF

)( +++= upward through the surface yxz 231 ++= , 10 ≤≤ x , 20 ≤≤ y .

22. Use the Divergence Theorem to find the flux of the vector field

kzjyzixF

2212 −+= out of the sphere S with equation 9222 =++ zyx ANSWERS AND SOLUTIONS: 1. Solution:

a) Since zR

yQ

xPFdiv

∂∂

+∂∂

+∂∂

=)(

we have ),,( 22 xzxyF +=

⇒ xxFdiv 2002)( =++=

. Answer a) xFdiv 2)( =

b) Since ),,()(zyx

grad∂∂

∂∂

∂∂

=ϕϕϕϕ we have ( for )(Fdiv

=ϕ )

)0,0,2())(( =Fdivgrad

Answer b) )0,0,2())(( =Fdivgrad

c)

)1,2,1(21

)(22

−−−=−−−=

+∂∂

∂∂

∂∂

=∂∂

∂∂

∂∂

=

xkjxi

xzxyzyx

kji

RQPzyx

kji

Fcurldef

10 / 28


Answer c) )1,2,1()( −−−= xFcurl

2. Hint. Use the definitions of div and curl. 3.

f∆ =0 ⇔ 02

2

2

2

2

2

=∂∂

+∂∂

+∂∂

zf

yf

xf

Answer: The functions )ln(),,( 22

2 zyxzyxf ++= and zyxyxzyxf 455),,( 22

4 +++−= satisfies the Laplace equation. 4. Answer: )))(( ff ∇×∇•∇+∆ = )))(( gradfcurldivf +∆ = 26 +x 5. Solution:

⇒+Γ=+∂

∂φϕρϕρϕ SgraddivUdiv

t)()()(

⇒+∂∂

Γ∂∂

Γ∂∂

Γ=+∂

∂φ

ϕϕϕρϕρϕρϕρϕ Szyx

divwvudivt

),,(),,()(

φϕϕϕρϕρϕρϕρϕ Szzyyxxz

wy

vx

ut

+

∂∂

Γ∂∂

+

∂∂

Γ∂∂

+

∂∂

Γ∂∂

=∂

∂+

∂∂

+∂

∂+

∂∂ ))()()()(

6. Which one (if any) of the following functions a) zyxzyx ++= 24

1 ),,(ϕ

b) zyxzyx ++= 222 ),,(ϕ

c) 2223 ),,( zyxzyx ++=ϕ

satisfies the equation SU +∇⋅Γ•∇=•∇ ))(()( ϕϕ

? Here 5=Γ , )3,2,1(=U

and 2342 −+= yxS . Solution : The equation

SU +∇⋅Γ•∇=•∇ ))(()( ϕϕ

can be written as

11 / 28


1.) (eq 234255532

2342)5,5,5()3,2,(

)()(

2

2

2

2

2

2

−++∂∂

+∂∂

+∂∂

=∂∂

+∂∂

+∂∂

⇒−++∂∂

∂∂

∂∂

=

⇒+Γ=

yxzyxzyx

yxzyx

divdiv

SgraddivUdiv

ϕϕϕϕϕϕ

ϕϕϕϕϕϕ

ϕϕ

a) zyxzyx ++== 241 ),,(Let ϕϕ

Vi calculate the derivatives of 1ϕ and substitute in the left hand side (LHS) and right hand side of the equation (eq1).

LHS: 34432 3 ++=∂∂

+∂∂

+∂∂ yx

zyxϕϕϕ

RHS= 1342602342555 22

2

2

2

2

2

−++=−++∂∂

+∂∂

+∂∂ yxx yx

zyxϕϕϕ

Whence RHSLHS ≠ Thus the function zyxzyx ++= 24

1 ),,(ϕ is not a solution to the equation b) zyxzyx ++== 22

2 ),,(ϕϕ LHS= yx 423 ++ , RHS= yx 423 ++−

Whence RHSLHS ≠ , and the function zyxzyx ++= 222 ),,(ϕ is not a solution to the

equation c) Let 222

3 ),,( zyxzyx ++==ϕϕ Then LHS= zyx 642 ++ , RHS= yx 427 ++

Thus RHSLHS ≠ , and the function 2223 ),,( zyxzyx ++=ϕ is not a solution to the

equation. Answer: None of the functions satisfies the equation 7. Answer: Function zyxzyx 5),,( 22

2 ++=ϕ satisfies the equation. 8. (exam 1, 98) A) Write the general transport equation

SUt

+∇⋅Γ•∇=•∇+∂

∂ ))(()()( ϕρϕρϕ

( eq 1)


. Functions S,,, Γϕρ , u, v, w are real functions of t, x, y and z.

12 / 28


B) Let 2=ρ , 3=Γ , )4,2,1(=U

. Find S in the equation (eq 1) if we now that the function 32),,( zyxzyx ++=ϕ satisfies the equation. Solution: A)

SUt

+∇⋅Γ•∇=•∇+∂

∂ ))(()()( ϕρϕρϕ

⇒

⇒+Γ=+∂

∂ SgraddivUdivt

)()()( ϕρϕρϕ

⇒+∂∂

Γ∂∂

Γ∂∂

Γ=+∂

∂ Szyx

divwvudivt

),,(),,()( ϕϕϕρϕρϕρϕρϕ

Szzyyxxz

wy

vx

ut

+

∂∂

Γ∂∂

+

∂∂

Γ∂∂

+

∂∂

Γ∂∂

=∂

∂+

∂∂

+∂

∂+

∂∂ ϕϕϕρϕρϕρϕρϕ ))()()()( (eq2)

B) We substitute 2=ρ , 3=Γ , )4,2,1(=U

and 32),,( zyxzyx ++=ϕ in the equation (eq2) and get

Szzyyxxzyx

+

∂∂

∂∂

+

∂∂

∂∂

+

∂∂

∂∂

=∂

∂+

∂∂

+∂

∂+

ϕϕϕφϕϕ 333))8()4()2(0

Szzy +++=+++ 186024820 2 . Consequently

2241884 zzyS +−+−= 9. (Q6, exam 2, 2008) Consider the following equation

426)())(()()(−−+×∇•∇+∇⋅Γ•∇=•∇+

∂∂ xyzyUU

t

ϕρϕρϕ ( eq 1)

Let 1=ρ , Γ= constant , ),3,2( xyU −=

. Find the constant Γ in the equation (eq 1) if we now that the function 222),,( zyxtzyx +++=ϕ satisfies the equation. Solution:

426)())(()()(−−+×∇•∇+∇⋅Γ•∇=•∇+

∂∂ xyzyUU

t

ϕρϕρϕ ⇒

⇒−−++Γ=+∂

∂ 426))(()()()( xyzyUcurldivgraddivUdivt

ϕρϕρϕ

(since )0,,()( yxUcurl −=

we have 011))(( =+−=Ucurldiv

)

13 / 28


⇒−−++∂∂

Γ∂∂

Γ∂∂

Γ=+∂

∂ 4260),,(),,()( xyzyzyx

divwvudivt

ϕϕϕρϕρϕρϕρϕ

4260))()()()(−−++

∂∂

Γ∂∂

+

∂∂

Γ∂∂

+

∂∂

Γ∂∂

=∂

∂+

∂∂

+∂

∂+

∂∂ xyzy

zzyyxxzw

yv

xu

tϕϕϕρϕρϕρϕρϕ (eq2)

We substitute 1=ρ , , ),3,2( xyU −=

and 222),,( zyxtzyx +++=ϕ in the equation (eq2) and get

426))()3()2()1(−−+

∂∂

Γ∂∂

+

∂∂

Γ∂∂

+

∂∂

Γ∂∂

=∂−∂

+∂

∂+

∂∂

+∂

∂ xyzyzzyyxxz

xyyxt

ϕϕϕφϕϕϕ

( Note that Γ is a constant)

248

4262202622

=Γ⇒Γ=

⇒−−+Γ+Γ+=−++ xyzyxyzy

Answer: 2=Γ

10. If possible, find ),( yxf for the given partial derivatives ),( yxfx∂∂ and ),( yxf

y∂∂ .

a) ),( yxfx∂∂ = xy2 and ),( yxf

y∂∂ = yx 22 + .

b) ),( yxfx∂∂ = yx +2 and ),( yxf

y∂∂ = x .

c) ),( yxfx∂∂ = xyye and ),( yxf

y∂∂ = xyxe .

d) ),( yxfx∂∂ = yx +2 and ),( yxf

y∂∂ = x5 .

( Hint: Necessary condition: If ),( yxf has continuous derivatives then the mixed derivatives of ),( yxf should be equal, i.e.

∂∂

∂∂

=

∂∂

∂∂ ),(),( yxf

yxyxf

xy (*)

is the necessary condition for the existence of a function ),( yxf that has the given derivatives. Answer: a) ),( yxf = Cyyx ++ 22 b) ),( yxf = Cxyx ++2 c) ),( yxf = Ce xy + d) No solution since the condition (*) is not fulfilled,

5),(),(1 =

∂∂

∂∂

≠

∂∂

∂∂

= yxfyx

yxfxy

.

Solution a)

Since

∂∂

∂∂

=

∂∂

∂∂ ),(),( yxf

yxyxf

xy =2x and the derivatives are continuous the

condition (*) is fulfilled and we can find ),( yxf for the given derivatives. In order to find ),( yxf we integrate with respect to x the first of the equations

),( yxfx∂∂ = xy2 (eq1)

14 / 28


),( yxfy∂∂ = yx 22 + (eq2).

and get

∫ +== )(2),( 1́2 yCyxxydxyxf

Thus )(),( 1́

2 yCyxyxf += ( i ) We have integrated with respect to x, therefore the constant still depend on y. Now, to find )(1́ yC we differentiate and substitute (i) in (eq2) and get:

( ))(1́2 yCyx

y+

∂∂ = yx 22 + ⇒ ( ))(1́

2 yCy

x∂∂

+ = yx 22 + ⇒

( ))(1́ yCy∂∂ = y2 ⇒ )(1́ yC = Cy +2 .

Finally, substituting )(1́ yC = Cy +2 in (i) we have Cyyxyxf ++= 22),( (where C is a constant).

11.

Answer: From 2121),(),( =⇒+=+⇒∂∂

∂∂

=∂∂

∂∂ axayyxf

yxyxf

xy

Then for 2=a we have Cxyyxyxf ++= 2),( 12. Answer: a) Czyzxyzf +++= 3 b) Cyzxyf ++= c) Cef xyz += d) No solution since the condition 2Con is not fulfilled,

zyfzx

fxz

y +=

∂∂

∂∂

≠

∂∂

∂∂

=

Solution a)

a) yzzyxfx

=∂∂ ),,( , zxzzyxf

y+=


z++=

∂∂

Since the conditions Con1,2,3 are fulfilled and we can find ),,( zyxf for the given derivatives. In order to find ),,( yyxf we integrate with respect to x the first of the equations

yzzyxfx

=∂∂ ),,( (eq1)

zxzzyxfy

+=∂∂ ),,( (eq2)

23),,( zyxyzyxfz

++=∂∂ (eq3)

and get

∫ +== ),(),( 1́ zyCxyzyzdxyzxf Thus

15 / 28


),(),,( 1́ zyCxyzzyxf += ( i ) We have integrated with respect to x, therefore the constant still depend on y and z. Now, to find ),(1́ zyC we differentiate and substitute (i) in (eq2) and get:

( )),(1́ zyCxyzy

+∂∂ = zxz + ⇒ ( )),(1́ zyC

yxz

∂∂

+ = zxz + ⇒

( )),(1́ zyCy∂∂ = z ⇒ ),(1́ zyC = )(2 zCyz + .

(We have integrated with respect to y , therefore the constant still depend on and z. Thus )(),,( 2 zCyzxyzzyxf ++= (ii) Now , substituting (ii) in (eq3) we have

CzzC

zzCz

zyxyzCz

yxy

zyxyzCyzxyzz

+=

=∂∂

⇒++=∂∂

++

⇒++=++∂∂

32

22

22

22

)(

3))((

3))((

3))((

Finally, substituting CzzC += 3

2 )( in (ii) we have Czyzxyzzyxf +++= 3),,( (where C is a constant).

13. Answer: From

33 22 =⇒=⇒

∂∂

∂∂

=

∂∂

∂∂ azxzaxf

yxf

xy

133 =⇒=⇒

∂∂

∂∂

=

∂∂

∂∂ bbxxf

zxf

xz

133 =⇒=⇒

∂∂

∂∂

=

∂∂

∂∂ bbxxf

zyf

yz.

Thus, all three conditions are fulfilled if 3=a and 1=b . For these values of a and b we get Czyxyzxzyxf +++= 223),,( Calculation of the pressure field for a known velocity field for an incompressible, steady state, isothermal Newtonian flow. 14. Answer: a) CyxgzP +−−−= 22 88 ρρρ

16 / 28


b) CyxgzP +−−−= 22

217

217 ρρρ

c) CyyxgzP ++++−= ρρρρ 444 22 Solution a) We substitute 0,24,32 =−=+= wyxvyxu in eq1,2,3,4 and get ( note that al derivatives with respect to t are 0): Continuity equation: 00 = eq1i. ( identically fulfilled) Navier Stokes equations: x component:

xPx∂∂

−=ρ16 eq2i.

y component:

yPy∂∂

−=ρ16 eq3i.

z component:

gzP ρ−∂∂

−=0 eq4i.

Now eq2i. gives ),(8),,( 1́2 zyCxzyxP +−= ρ (*) .

Substitution in eq3i. implies

)(8),(

),(16

2´2

1́

1́

zCyzyCy

zyCy

+−=

⇒∂

∂−=

ρ

ρ

Hence, from (*) we have )(88),,( 2´

22 zCyxzyxP +−−= ρρ (**) Now we substitute (**) in eq4i. and get

gzCz

gzP ρρ −

∂∂

−=⇒−∂∂

−= ))((00 2´

CgzzC +−=⇒ ρ)(2´ ( where C is a constant) Finally, substituting CgzzC +−= ρ)(2´ in (**) we have

CgzyxzyxP +−−−= ρρρ 22 88),,( (where C is a constant). 15. Solution A:

24841616)())(()()(−++++×∇•∇+∇⋅Γ•∇=•∇+

∂∂ yzxzyxUU

t

ϕρϕρϕ ⇒

⇒−+++++Γ=+∂

∂ 24841616))(()()()( yzxzyxUcurldivgraddivUdivt

ϕρϕρϕ

(since )0,1,2()( −=Ucurl

we have 0))(( =Ucurldiv

)

⇒−++++∂∂

Γ∂∂

Γ∂∂

Γ=+∂

∂ 24841616),,(),,()( yzxzyxzyx

divwvudivt

ϕϕϕρϕρϕρϕρϕ

17 / 28


24841616))()()()(−++++

∂∂

Γ∂∂

+

∂∂

Γ∂∂

+

∂∂

Γ∂∂

=∂

∂+

∂∂

+∂

∂+

∂∂ yzxzyx

zzyyxxzw

yv

xu

tϕϕϕρϕρϕρϕρϕ

(eq2) We substitute 2=ρ , , )2,4,4( yxU +=

and 22231),,( zyxtzyx ++++=ϕ in the equation (eq2) and get

24841616)))2(2()8()8()2(−++++

∂∂

Γ∂∂

+

∂∂

Γ∂∂

+

∂∂

Γ∂∂

=∂+∂

+∂

∂+

∂∂

+∂

∂ yzxzyxzzyyxxz

yxyxt

ϕϕϕφϕϕϕ

( Note that Γ is a constant)

5630

2484161668461166

=Γ⇒Γ=

⇒−++++Γ=++++ yzxzyxyzxzyx

Answer A: 5=Γ Solution B: We substitute zwyvxu 22,24,46 −=−=+= in eq1,2,3,4 and get ( note that al derivatives with respect to t are 0): Continuity equation: 00 = eq1i. ( identically fulfilled) Navier Stokes equations: x component:

xPx∂∂

−=+ )2416(ρ eq2i.

y component:

yPy∂∂

−=− )84(ρ eq3i.

z component:

gzPz ρρ −∂∂

−=− )44( eq4i.

Now eq2i. gives ),()248(),,( 1́2 zyCxxzyxP +−−= ρ (*) .


)()82(),(

),()84(

2´2

1́

1́

zCyyzyCy

zyCy

++−=

⇒∂

∂−=−

ρ

ρ

Hence, from (*) we have )()82()248(),,( 2´

22 zCyyxxzyxP ++−+−−= ρρ (**) Now we substitute (**) in eq4i. and get

gzCz

zgzPz ρρρρ −

∂∂

−=−⇒−∂∂

−=− ))(()44()44( 2´

CzzgzzC ++−+−=⇒ )42()( 22´ ρρ ( where C is a constant)

Finally, substituting CgzzC +−= ρ)(2´ in (**) we have CgzzzyyxxzyxP +−+−++−+−−= ρρρρ )42()82()248(),,( 222

Answer B:

18 / 28


CgzzzyyxxzyxP +−+−+−−−= )4282248(),,( 222ρ (where C is a constant). 16. Solution )1,5,32( azyxV −−+=

First we substitute azwyvxu −=−=+= 1,5,32 in eq1 and get ( note that al derivatives with respect to t are 0): Continuity equation: 2013 =⇒=−− aa No we have )21,5,32( zyxV −−+=

Using the Navier Stokes equations we get: x component:

xPx∂∂

−=+ )69(ρ eq2i.

y component:

yPy∂∂

−=− )5(ρ eq3i.

z component:

gzPz ρρ −∂∂

−=− )24( eq4i.

Now eq2i. gives ),()629(),,( 1́

2

zyCxxzyxP +−−

= ρ (*) .


)()52

(),(

),()5(

2´

2

1́

1́

zCyyzyC

yzyCy

++−

=

⇒∂

∂−=−

ρ

ρ

Hence, from (*) we have

)()52

()629(),,( 2´

22

zCyyxxzyxP ++−

+−−

= ρρ (**)

We substitute (**) in eq4i. and get

gzCz

zgzPz ρρρρ −

∂∂

−=−⇒−∂∂

−=− ))(()24()24( 2´

CzzgzzC ++−+−=⇒ )22()( 22´ ρρ ( where C is a constant)

Finally, substituting )(2´ zC in (**) we have

CzzgzyyxxzyxP ++−+−++−

++−

= )22()52

()629(),,( 2

22

ρρρρ

Answer :

CzzgzyyxxzyxP ++−−+−+−= )2252

62

9(),,( 222

ρ

(where C is a constant).

Q17. Consider steady, incompressible, isothermal, laminar stationary Newtonian flow in a long round pipe in the z-direction, with constant circular cross-section of radius R=2 m. Use

19 / 28


the continuity and the Navier-Stokes equations in cylindrical coordinates to find the velocity field V=(ur, uθ, uz) and the pressure field P (r,θ,z) if the fluid flow satisfies the following conditions:

c0. All partial derivatives with respect to time t are 0 ( Steady flow)

c1. μ=0.001 kg/(m∙s) and ρ =1000 kg/m3

c2. A Constant pressure gradient ∂P/∂z = –1/250 Pa/m is applied in the horizontal axis ( z-axis in our notation): ∂P/∂z = –1/250, c3. The flow is parallel to the z axis, that is ur =0 and uθ =0. c4. We assume that the flow is axisymmetric . The velocity does not depend on θ,

that is 0=∂∂θ

zu

c5. Boundary cond. 1 ( No-slip boundary condition, Vfluid=Vwall ): If r=2 then uz= 0

c6. Boundary condition 2: uz has maximum at r=0 that is 00=

=∂∂

rruz

The continuity and the Navier-Stokes equations for an incompressible , isothermal Newtonian flow (density ρ =const, viscosity µ =const), with a velocity field

),,( zr uuuV θ=

in Cylindrical coordinates ),,( zr θ : ---------------------------------------------------------------- SOLUTION Incompressible continuity equation

0)(1)(1

=∂∂

+∂∂

+∂

∂z

uurr

rur

zr

θθ eq a)

Navier-Stokes equations in Cylindrical coordinates: r-component:

∂∂

+∂∂

−∂∂

+−

∂∂

∂∂

++∂∂

−=

∂∂

+−∂∂

+∂∂

+∂∂

2

2

22

2

22

2

211zuu

ru

rru

rur

rrg

rP

zuu

ruu

ru

ruu

tu

rrrrr

rz

rrr

r

θθµρ

θρ

θ

θθ

eq b)

θ -component:

∂∂

+∂∂

+∂∂

+−

∂∂

∂∂

++∂∂

−=

∂∂

++∂∂

+∂∂

+∂∂

2

2

22

2

22

2111zuu

ru

rru

ru

rrr

gPr

zu

uruuu

ru

ru

ut

u

r

zr

r

θθθθθ

θθθθθθ

θθµρ

θ

θρ

eq c)

z-component:

20 / 28


∂∂

+∂∂

+

∂∂

∂∂

++∂∂

−=

∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

211

zuu

rrur

rrg

zP

zuuu

ru

ruu

tu

zzzz

zz

zzr

z

θµρ

θρ θ

eq d)

We choose x as a vertical axis, y an z are in a horizontal plane and the flow is parallel with the z-axis. We denote velocity vector V=(ur, uθ, uz) where ur, uθ and uz are r-component, θ-component and z-component in cylindrical coordinates. According to the assumptions we have ur =0, uθ = 0, and uz does not depend on θ. Since x is the vertical axis we have that vector g=(-g, 0,0) where g=9,81 m/s2 which in cylindrical coordinates gives

θcosgg r −= , θθ singg = and 0=zg

Now we substitute ∂P/∂z = –1/250 Pa/m, μ=0.001kg /(ms) in the continuity and Navier-Stokes equations:

Since ur =0 and uθ =0 (according to c3), continuity equation in cylindrical coordinates

0)(1)(1=

∂∂

+∂∂

+∂

∂zuu

rrru

rzr

θθ

gives

0=∂∂

zuz .

21 / 28


This tells us that uz is not a function of z. Furthermore, since uz velocity does not depend on θ (assumption c4) we conclude that uz depends only on r. To simplify notation we denote

)(rwuz = (*)

Now we substitute

θcosggr −= , θθ singg = and 0=zg

∂P/∂z = –1/250 Pa/m, μ=0.001kg /(ms)

in the Navier-Stokes equations:

The r-component of the Navier-Stokes equation gives:

θρ cos0 grP−

∂∂

−= ( eq r-c)

The θ-component of the Navier-Stokes equation:

θρθ

sin10 gPr

+∂∂

−= ( eq θ-c)

The Z-component of the Navier-Stokes equation (where )(rwuz = and 2501

−=∂∂

zP ) givs:

∂∂

∂∂

+=rwr

rr1

10001

25010 ( eq z-c)

Step 1. We find the pressure ),,( zrPP θ= .

In order to find the pressure P we solve ( eq r-c), ( eq θ-c) and the equation 2501

−=∂∂

zP that

is

θρ cosgrP

−=∂∂

θρθ

singrP+=

∂∂

2501

−=∂∂

zP

From these equations we get

CgrzP +−−= θρ cos2501

22 / 28


Step 2. We find the velocity component )(rwuz = .

We solve ( eq z-c) with boundaries c5 and c6:

∂∂

∂∂

+=rwr

rr1

10001

25010 ( eq z-c)

0)2( =w (c5)

00=

=∂∂

rrw (c6)

( Remark: Technically, we can write drdw instead

rw∂∂ since w is now a function of only one

variable)

From ( eq z-c) we have

∂∂

∂∂

+=rwr

rr1

10001

25010

⇒−=

∂∂

∂∂ r

rwr

r4

⇒+−=∂∂

122 Cr

rwr (substitution 0=r and (c6) ⇒ 01 =C )

⇒−=∂∂ 22r

rwr

⇒−=∂∂ r

rw 2

⇒+−= 22 Crw (substitution 2=r and (c5) ⇒ 42 =C )

⇒+−= 42rw

Thus 4)( 2 +−== rrwuz and

V )4 0, (0,),,( 2 +−== ruuu zr θ .

Answer :

CgrzP +−−= θρ cos2501 ,

V = )4 0, (0, 2 +− r

23 / 28


18. We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field

)3,24,3())()()(( azbzyxcyxx,y,z, w x,y,z, vx,y,zuV −+−++==

Use the following equations ( continuity and Navier Stokes equations) , where ρ =constant, µ =constant , ),0,0( gg −=

i.e. xg = yg =0 and )/81.9 ( 2smgggz ≈−= to find: i) parameters a, b and c ii) en expression for pressure P(x,y,z) as a function of x,y and z. Incompressible continuity equation:

0=∂∂

+∂∂

+∂∂

zw

yv

xu eq1.

Navier Stokes equations: x component:

)( 2

2

2

2

2

2

zu

yu

xug

xP

zuw

yuv

xuu

tu

x ∂∂

+∂∂

+∂∂

++∂∂

−=

∂∂

+∂∂

+∂∂

+∂∂ µρρ eq2.

y component:

)( 2

2

2

2

2

2

zv

yv

xvg

yP

zvw

yvv

xvu

tv

y ∂∂

+∂∂

+∂∂

++∂∂

−=

∂∂

+∂∂

+∂∂

+∂∂ µρρ eq3.

z component:

)( 2

2

2

2

2

2

zw

yw

xwg

zP

zww

ywv

xwu

tw

z ∂∂

+∂∂

+∂∂

++∂∂

−=

∂∂

+∂∂

+∂∂

+∂∂ µρρ eq4.

------------------------------------------------------------- We substitute ,3,24,3 azwbzyxvcyxu −=+−+=+= in eq1,2,3,4 and get ( note that al derivatives with respect to t are 0): Continuity equation: 202 =⇒=− aa eq1i. Thus ,23,24,3 zwbzyxvcyxu −=+−+=+= Navier Stokes equations: x component:

eq2i.

y component: eq3i.

z component:

24 / 28


eq4i.

The system (eq2i, eq3i, eq4i) is solvable only if mixed derivatives are equal:

242:1 =⇒−=−⇒

∂∂

∂∂

=

∂∂

∂∂ ccP

yxP

xyCon ρρ

00:2 =⇒=⇒

∂∂

∂∂

=

∂∂

∂∂ bbcP

zxP

xzCon ρ

003:3 =⇒=−⇒

∂∂

∂∂

=

∂∂

∂∂ bbP

zyP

yzCon ρ

Thus c=2 and b=0 We solve simplified equations

and get

CzgzyxxyzyxzyxP ++−+−−−−−= )648422

52

13(),,( 222

ρ

Answer. CzgzyxxyzyxzyxP ++−+−−−−−= )64842

25

213(),,( 2

22

ρ

19. We consider a scalar field ),,( zrf θ given in cylindrical coordinates, where

zzryrx === , sin ,cos θθ , and basis vectors are 321 ,, eee . Find the expression for the gradient, )),,(( zrfgrad θ ), in cylindrical coordinates, that is in

terms of zff

rfeeezr

∂∂

∂∂

∂∂ and ,, , ,,,, 321 θ

θ

if

a) kejeie

=== 321 ,, ( we keep the same basis kji

,, ) .

b) kjejeie

+=== 2,2 321

c) kejeie

=== 321 , sin ,cos θθ (exam 1, 2012; Q5 B (2 points)) d) kejiejie

=+−=+= 321 ,cossin,sincos θθθθ (this is often used as a local basis for cylindrical coordinates)

25 / 28


Solution: In x,y, z variables we have

kzfj

yfi

xf

zf

yf

xffgrad

∂∂

+∂∂

+∂∂

=∂∂

∂∂

∂∂

= ),,()( ( eq1)

For cylindrical coordinates we have θcosrx = , θsinry = , z=z

First we write the derivatives ,xf∂∂ and

yf∂∂ in θ,r coordinates (the variable z is in both

coord systems) .

Solving the following system for ,xf∂∂ and

yf∂∂ ,

)cos()sin(

sincos

θθθ

θθ

ryfr

xff

yf

xf

rf

⋅∂∂

+−⋅∂∂

=∂∂

⋅∂∂

+⋅∂∂

=∂∂

we get

θθθ∂∂

−∂∂

=∂∂ f

rrf

xf sincos

θθθ∂∂

+∂∂

=∂∂ f

rrf

yf cossin (**)

We substitute the derivatives (**) in ( eq1) and get

*)*(* )cos(sin)sin(cos)( kzfjf

rrfif

rrffgrad

∂∂

+∂∂

+∂∂

+∂∂

−∂∂

=θ

θθθ

θθ

To solve problems a) , b) c) and d) we must express kji

,, as a linear combinations of

321 ,, eee and substitute them into (***) . a) From (***), since 321 ,, ekejei

=== , we have immediately

321 )cos(sin)sin(cos)( ezfef

rrfef

rrffgrad

∂∂

+∂∂

+∂∂

+∂∂

−∂∂

=θ

θθθ

θθ

b) From kjejeie

+=== 2,2 321 we find

232

1 ,2

, eekejei

−=== (eq b)

Then we put kji

,, from ( eq b) into (***) and get

)(2

)cos(sin)sin(cos)( 232

1 eezfef

rrfef

rrffgrad

−∂∂

+∂∂

+∂∂

+∂∂

−∂∂

=θ

θθθ

θθ

26 / 28


and, after collecting components for 321 , eee

321 )cos21sin

21()sin(cos)( e

zfe

zff

rrfef

rrffgrad

∂∂

+∂∂

−∂∂

+∂∂

+∂∂

−∂∂

=θ

θθθ

θθ

c) From

kejeie

=== 321 , sin ,cos θθ we have

321 ,

sin,

cosekejei

===θθ

( eq c)

Putting kji

,, from ( eq c) into (***) gives

sin

)cos(sincos

)sin(cos)( 321 e

zfef

rrfef

rrffgrad

∂∂

+∂∂

+∂∂

+∂∂

−∂∂

=θθ

θθθθ

θθ

d) kejiejie

=+−=+= 321 ,cossin,sincos θθθθ .

We can solve d) in the same manner as in a,b,c but this time we can just collect terms rf∂∂ ,

θ∂∂f ,

zf∂∂ and get the result:

kzfjf

rrfif

rrfk

zfj

yfi

xffgrad

∂∂

+∂∂

+∂∂

+∂∂

−∂∂

=∂∂

+∂∂

+∂∂

= )cos(sin)sin(cos)(θ

θθθ

θθ

kzfjif

rji

rf

∂∂

++−⋅∂∂

++⋅∂∂

= )cossin(1)sin(cos θθθ

θθ

3211 e

zfef

re

rf

∂∂

+⋅∂∂

+⋅∂∂

=θ

Answer:

a) 321 )cos(sin)sin(cos)( ezfef

rrfef

rrffgrad

∂∂

+∂∂

+∂∂

+∂∂

−∂∂

=θ

θθθ

θθ

b)

321 )cos21sin

21()sin(cos)( e

zfe

zff

rrfef

rrffgrad

∂∂

+∂∂

−∂∂

+∂∂

+∂∂

−∂∂

=θ

θθθ

θθ

c) sin

)cos(sincos

)sin(cos)( 321 e

zfef

rrfef

rrffgrad

∂∂

+∂∂

+∂∂

+∂∂

−∂∂

=θθ

θθθθ

θθ

d) 3211),,(( e

zfef

re

rfzrfgrad

∂∂

+⋅∂∂

+⋅∂∂

=θ

θ

20. See http://ingforum.haninge.kth.se/armin/AR_2000/HL2008/DERIV_NAVIER_STOKES.pdf 21. Solution:

yxz 231 ++= )1,2,3()1,,( −−=′−′−= yx zzN

27 / 28


),,( yxyxF +=

yxyxyxNF −−=++−−= 223

4)24()2(1

0

2

0

1

0

−=−−=−−==Φ ∫ ∫ ∫∫∫ dxxdyyxdxdxdyNFD

Answer: 4−=Φ 22. Solution:

122212)( =−+= zzFdiv

∫∫∫∫∫∫ ===ΦKK

dxdydzdVFdiv 12

12× Volume(K)= ππ 43233412 3 =××

Answer: π432=Φ

28 / 28

e-mail : [email protected] webpage : · without using operators div, ∇, ∆, curl or grad. here ....

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