MATH. EXERCISES. GRADIENT, DIVERGENCE, CURL DEL … · Armin Halilovic 2 Math. Exercises Using ∇ we can denote grad, div and curl as below: grad(ϕ)=∇ϕ div F F r r ( ) =∇•

Armin Halilovic 1 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 =r and ),,( 321 vvvv =

r then

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

321

321

vvvuuukji

vu

rrr

rr=× (vector or cross product)

In some books is also considered outer product defined by

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=⊗

3

2

1

uuu

vu rr )( 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 =

r be a vector field, continuously differentiable with

respect to x, y and z. Then the divergence of F

r is the scalar field defined by

zR

yQ

xPFdiv

∂∂

+∂∂

+∂∂

=)(r

CURL. The curl of F

r is the vector field defined by

kyP

xQj

xR

zPi

zQ

yR

RQPzyx

kji

Fcurlrrr

rrr

r)()()()(

∂∂

−∂∂

+∂∂

−∂∂

+∂∂

−∂∂

=∂∂

∂∂

∂∂

=

),,(yP

xQ

xR

zP

zQ

yR

∂∂

−∂∂

∂∂

−∂∂

∂∂

−∂∂

=

DEL (NABLA) OPERATOR The vector differential operator

=∇ ),,(zyxz

ky

jx

i∂∂

∂∂

∂∂

=∂∂

+∂∂

+∂∂ rrr

is called del or nabla .


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

)(ϕgrad = ϕ∇ FFdivrr

•∇=)( FFcurlrr

×∇=)( Note that ∇•F

r is not the same as F

r•∇ .

∇•Fr

=z

Ry

Qx

P∂∂

+∂∂

+∂∂ .

LAPLACIAN OPERATOR

The Laplacian operator, 2

2

2

2

2

22

zyx ∂∂

+∂∂

+∂∂

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

2

2

2

2

2

22

zU

yU

xUUU

∂∂

+∂∂

+∂∂

=∇=Δ ,

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

by ),,(2 RQPFF ΔΔΔ=∇=Δ

rr.

Cylindrical coordinates ),,( zr θ : transformation: θcosrx = , θsinry = , z=z volume element: dzddrrdV θ= local basis: kejiejie zr

rrrrrrrr=+−=+= ,cossin,sincos θθθθ θ

Vector components relationship: θθρ sincos yx FFF += , θθϑ cossin yx FFF +−= , zz FF = scalar field: ),,( zrf θ

gradient: zr ezfef

re

rfffgrad rrr

∂∂

+∂∂

+∂∂

=∇= θθ1)(

laplacian: 2

2

2

2

22 11

zff

rrfr

rrff

∂∂

+∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=Δ=∇θ

vector field: ),,( zr FFFF θ=r

divergence: z

FFrr

Frr

FFdiv zr

∂∂

+∂

∂+

∂⋅∂

=∇=θθ )(1)(1)(

vo

r

curl:

zr

zr

FFrFzr

eere

rFFcurl

θ

θ

θ⋅

∂∂

∂∂

∂∂

⋅

=×∇=

rrr

vr 1)( =

krzr

rz eF

rrF

re

rF

zFe

zFF

rrrr⎟⎠⎞

⎜⎝⎛

∂∂

−∂

∂+⎟

⎠⎞

⎜⎝⎛

∂∂

−∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

−∂∂

θθθ

θθ )(11


EXERCISES 1. Find a) )(Fdiv

r , b) ))(( Fdivgrad

r and c) )(Fcurl

r

if ),,( 22 xzxyF +=r

2. Find )))((( Fcurldivgrad

r if ),,( 22 yxzxzyxF ++++=

r

3. Which one of the following functions a) 222

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

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

3 zyxzyxf ++= satisfies the Laplace equation

fΔ =0? 4. Find )))(( ff ∇×∇•∇+Δ if zyxzyxf ++= 23),,( . 5. Write the general transport equation

φϕρϕρϕ SUt

+∇⋅Γ•∇=•∇+∂

∂ )()()( r

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

r. 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

r ?

Here 5=Γ , )3,2,1(=Ur

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

SgraddivUdivt

+Γ=+∂

∂ )()()( ϕρϕρϕ r

where ρ=3, 2=Γ , )4,3,2(=Ur

and 521812 ++= yxS .


8. (exam 1, 2008) A) Write the general transport equation

SUt

+∇⋅Γ•∇=•∇+∂

∂ ))(()()( ϕρϕρϕ r ( eq 1)



B) Let 2=ρ , 3=Γ , )4,2,1(=U

r.

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

trr

ϕρϕρϕ ( eq 1)

Let 1=ρ , Γ= constant , ),3,2( xyU −=r

. 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. 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. 12. 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 ))()()(( =r

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 −=

r i.e. xg = yg =0 and

)/81.9 where( 2smggg 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.

a) )0,24,32( yxyxV −+=r

b) )2,32,43( −−+= yxyxV

r

c) )2,4,21( xyV −+=r

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

24841616)())(()()(−++++×∇•∇+∇⋅Γ•∇=•∇+

∂∂ yzxzyxUU

t

rrϕρϕρϕ ( eq 1)

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

. 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 ))()()(( =r

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

)22,24,46( zyxV −−+=r

. 14. (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 ))()()(( =r

Use the following equations ( continuity and Navier Stokes equations) to find first i) parameter a and then 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 −=r i.e. xg = yg =0 and )/81.9 where( 2smgggz ≈−= and

)1,5,32( azyxV −−+=r

.

15. 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=r

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

0)(1)(1

=∂∂

+∂

∂+

∂∂

zuu

rrru

rzr

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:


⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

++∂∂

−=

⎟⎠⎞

⎜⎝⎛

∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

211

zuu

rrur

rrg

zP

zuuu

ru

ruu

tu

zzzz

zz

zzr

z

θμρ

θρ θ

eq d)


ANSWERS AND SOLUTIONS: 1. Solution:

a) Since zR

yQ

xPFdiv

∂∂

+∂∂

+∂∂

=)(r

we have ),,( 22 xzxyF +=r

⇒ xxFdiv 2002)( =++=r

. Answer a) xFdiv 2)( =

r

b) Since ),,()(zyx

grad∂∂

∂∂

∂∂

=ϕϕϕϕ we have ( for )(Fdiv

r=ϕ )

)0,0,2())(( =Fdivgradr

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

r

c)

)1,2,1(21

)(22

−−−=−−−=

+∂∂

∂∂

∂∂

=∂∂

∂∂

∂∂

=

xkjxi

xzxyzyx

kji

RQPzyx

kji

Fcurldef

rrr

rrrrrr

r

Answer c) )1,2,1()( −−−= xFcurlr

2. Solution:

),,( 22 yxzxzyxF ++++=r

)()()(

)(22 yxzxzyx

zyx

kji

Fcurl

++++∂∂

∂∂

∂∂

=

rrr

r= kxjiz

rrr)12()11()21( −+−−−

)12,0,21( −−= xz Thus 0))(( =Fcurldiv

r and therefore )))((( Fcurldivgrad

r= 0)0,0,0(

r=

Answer: )))((( Frotdivgradr

= 0)0,0,0(r

= 3.

fΔ =0⇒

02

2

2

2

2

2

=∂∂

+∂∂

+∂∂

zf

yf

xf

Answer: The function )ln(),,( 222 zyxzyxf ++= satisfies the Laplace equation.

4. Answer: )))(( ff ∇×∇•∇+Δ = )))(( gradfcurldivf +Δ = 26 +x 5. Solution:


⇒+Γ=+∂

∂φϕρϕρϕ SgraddivUdiv

t)()()( r

⇒+∂∂

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

r ?

Here 5=Γ , )3,2,1(=Ur

and 2342 −+= yxS . Solution : The equation

SU +∇⋅Γ•∇=•∇ ))(()( ϕϕr

can be written as

1.) (eq 234255532

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

)()(

2

2

2

2

2

2

−++∂∂

+∂∂

+∂∂

=∂∂

+∂∂

+∂∂

⇒−++∂∂

∂∂

∂∂

=

⇒+Γ=

yxzyxzyx

yxzyx

divdiv

SgraddivUdiv

ϕϕϕϕϕϕ

ϕϕϕϕϕϕ

ϕϕr

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 r ( eq 1)



B) Let 2=ρ , 3=Γ , )4,2,1(=U

r.

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

SUt

+∇⋅Γ•∇=•∇+∂

∂ ))(()()( ϕρϕρϕ r⇒

⇒+Γ=+∂

∂ SgraddivUdivt

)()()( ϕρϕρϕ r

⇒+∂∂

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

r 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

trr

ϕρϕρϕ ( eq 1)

Let 1=ρ , Γ= constant , ),3,2( xyU −=r

. 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

trr

ϕρϕρϕ ⇒

⇒−−++Γ=+∂

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

rrϕρϕρϕ

(since )0,,()( yxUcurl −=r

we have 011))(( =+−=Ucurldivr

)

⇒−−++∂∂

Γ∂∂

Γ∂∂

Γ=+∂

∂ 4260),,(),,()( xyzyzyx

divwvudivt

ϕϕϕρϕρϕρϕρϕ

4260))()()()(−−++⎟

⎠⎞

⎜⎝⎛

∂∂

Γ∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

Γ∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

Γ∂∂

=∂

∂+

∂∂

+∂

∂+

∂∂ xyzy

zzyyxxzw

yv

xu

tϕϕϕρϕρϕρϕρϕ (eq2)

We substitute 1=ρ , , ),3,2( xyU −=r

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)

),( 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: 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

),(),,( 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).

Calculation of the pressure field for a known velocity field for an incompressible, steady state, isothermal Newtonian flow. 12. Answer:


a) CyxgzP +−−−= 22 88 ρρρ

b) CyxgzP +−−−= 22

217

217 ρρρ

c) CyxgzP +++−= 22 44 ρρρ 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). 13. Solution A:

24841616)())(()()(−++++×∇•∇+∇⋅Γ•∇=•∇+

∂∂ yzxzyxUU

t

rrϕρϕρϕ ⇒

⇒−+++++Γ=+∂

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

rrϕρϕρϕ

(since )0,1,2()( −=Ucurlr

we have 0))(( =Ucurldivr

)


⇒−++++∂∂

Γ∂∂

Γ∂∂

Γ=+∂

∂ 24841616),,(),,()( yzxzyxzyx

divwvudivt


24841616))()()()(

−++++⎟⎠⎞

⎜⎝⎛

∂∂

Γ∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

Γ∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

Γ∂∂

=∂

∂+

∂∂

+∂

∂+

∂∂

yzxzyxzzyyxxz

wy

vx

ut


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

r and 22231),,( zyxtzyx ++++=ϕ

in the equation (eq2) and get

24841616)))2(2()8()8()2(

−++++⎟⎠⎞

⎜⎝⎛

∂∂

Γ∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

Γ∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

Γ∂∂

=∂+∂

+∂

∂+

∂∂

+∂

∂yzxzyx

zzyyxxzyx

yxtϕϕϕφϕϕϕ

( 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:

CgzzzyyxxzyxP +−+−+−−−= )4282248(),,( 222ρ (where C is a constant). 14. Solution )1,5,32( azyxV −−+=r

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 −−+=

r

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).


Q15. 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=r

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

0)(1)(1

=∂∂

+∂

∂+

∂∂

zuu

rrru

rzr

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:

⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

++∂∂

−=

⎟⎠⎞

⎜⎝⎛

∂∂

+∂∂

+∂∂

+∂∂

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 .


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 those equations we get

CgrzP +−−= θρ cos2501


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

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MATH. EXERCISES. GRADIENT, DIVERGENCE, CURL DEL … · Armin Halilovic 2 Math. Exercises Using ∇ we can denote grad, div and curl as below: grad(ϕ)=∇ϕ div F F r r ( ) =∇•