Armin Halilovic Math. Exercises

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E-mail : armin@sth.kth.se 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)()()()(

∂∂−

∂∂+

∂∂−

∂∂+

∂∂−

∂∂=

∂∂

∂∂

∂∂=

or ),,()(yP

xQ

xR

zP

zQ

yRFcurl

∂∂−

∂∂

∂∂−

∂∂

∂∂−

∂∂=

r

DEL (NABLA) OPERATOR The vector differential operator

=∇ ),,(zyxz

ky

jx

i∂∂

∂∂

∂∂=

∂∂+

∂∂+

∂∂ rrr

is called del or nabla .

Armin Halilovic Math. Exercises

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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.

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

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

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

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

rrr+= in Cartesian coord. and θϑeFeFF rr

rrr+= the same vector in cylindrical

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

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

in the directions of rer and θer . Thus

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

, similarly

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

. ] 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

rrrrrrrr =+−=+= ,cossin,sincos θθθθ θ

If kFjFiFF zyx

rrrr++= in Cartesian coord. and kFeFeFF zrr

rrrr++= θϑ the same vector in

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

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

] scalar field: ),,( zrf θ

gradient: zr ezfef

re

rfffgrad rrr

∂∂+

∂∂+

∂∂=∇= θθ

1)(

laplacian: 2

2

2

2

22 11

zff

rrfr

rrff

∂∂+

∂∂+⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂=Δ=∇

θ

vector field: ),,( zr FFFF θ=r

divergence: z

FFrr

Frr

FFdiv zr

∂∂

+∂

∂+

∂⋅∂

=∇=θ

θ )(1)(1)(v

or

curl:

zr

zr

FFrFzr

eere

rFFcurl

θ

θ

θ⋅

∂∂

∂∂

∂∂

⋅

=×∇=

rrr

vr 1)( =

krzr

rz eF

rrF

re

rF

zFe

zFF

rrrr⎟⎠⎞

⎜⎝⎛

∂∂

−∂

∂+⎟

⎠⎞

⎜⎝⎛

∂∂

−∂

∂+⎟

⎠⎞

⎜⎝⎛

∂∂

−∂∂

θθθ

θθ )(11

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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 .

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

SUt

+∇⋅Γ•∇=•∇+∂

∂ ))(()()( ϕρϕρϕ r ( eq 1)

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.

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•∇=•∇+∂

∂ xyzyUUt

rrϕρϕρϕ ( 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 ),( yxfy∂

∂ .

a) ),( yxfx∂

∂ = xy2 and ),( yxfy∂

∂ = yx 22 + .

b) ),( yxfx∂

∂ = yx +2 and ),( yxfy∂

∂ = x .

c) ),( yxfx∂

∂ = xyye and ),( yxfy∂

∂ = xyxe .

d) ),( yxfx∂

∂ = yx +2 and ),( yxfy∂

∂ = 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

),( yxfx∂

∂ = yaxy + and ),( yxfy∂

∂ = xx +2 .

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

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

∂ , ),,( zyxfy∂

∂

and ),,( zyxfz∂

∂ .

a) yzzyxfx

=∂∂ ),,( , zxzzyxf

y+=

∂∂ ),,( and 23),,( zyxyzyxf

z++=

∂∂

b) yzzyxfx

=∂∂ ),,( , zxzzyxf

y+=

∂∂ ),,( and 23),,( zyxyzyxf

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 + , ),,( zyxfy∂

∂ = 13 +zx and ),,( zyxfz∂

∂ = 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 ))()()(( =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:

Armin Halilovic Math. Exercises

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)( 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

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

24841616)())(()()( −++++×∇•∇+∇⋅Γ•∇=•∇+∂

∂ yzxzyxUUt

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

. 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 ))()()(( =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

.

Armin Halilovic Math. Exercises

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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=∂∂

θzu

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

==∂

∂rr

uz

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

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)

Armin Halilovic Math. Exercises

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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 −+−++==r

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

rrr

∂∂+

∂∂+

∂∂=

∂∂

∂∂

∂∂= ),,()( (*) .

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

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

We simple calculate the derivatives zf

yf

xf

∂∂

∂∂

∂∂ and in new variables and express kji

rrr,, as

a linear combinations of 321 ,, eee rrr . Then we substitute those values into (*). (See the following example.) 17 We consider a scalar field ),,( zrf θ given in cylindrical coordinates, where

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

terms of zff

rfeeezr

∂∂

∂∂

∂∂ and ,, , ,,,, 221 θ

θ rrr if

a) kejeie

rrrrrr === 321 ,, ( we keep the same basis kjirrr

,, ) .

Armin Halilovic Math. Exercises

10 / 27

b) kjejeierrrrrrr +=== 2,2 321

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

d) kejiejie

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

=

Armin Halilovic Math. Exercises

11 / 27

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=+

∂∂

φϕϕϕρϕρϕρϕρϕ 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).

Armin Halilovic Math. Exercises

12 / 27

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)

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.

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=+

∂∂ S

zyxdivwvudiv

t),,(),,()( ϕϕϕρϕρϕρϕρϕ

Armin Halilovic Math. Exercises

13 / 27

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•∇=•∇+∂

∂ xyzyUUt

rrϕρϕρϕ ( 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•∇=•∇+∂

∂ xyzyUUt

rrϕρϕρϕ ⇒

⇒−−++Γ=+∂

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

rrϕρϕρϕ

(since )0,,()( yxUcurl −=r

we have 011))(( =+−=Ucurldivr

)

⇒−−++∂∂Γ

∂∂Γ

∂∂Γ=+

∂∂ 4260),,(),,()( xyzy

zyxdivwvudiv

tϕϕϕρϕρϕρϕρϕ

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

∂∂=

∂−∂+

∂∂+

∂∂+

∂∂ xyzy

zzyyxxzxy

yxtϕϕϕφϕϕϕ

( 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 ),( yxfy∂

∂ .

Armin Halilovic Math. Exercises

14 / 27

a) ),( yxfx∂

∂ = xy2 and ),( yxfy∂

∂ = yx 22 + .

b) ),( yxfx∂

∂ = yx +2 and ),( yxfy∂

∂ = x .

c) ),( yxfx∂

∂ = xyye and ),( yxfy∂

∂ = xyxe .

d) ),( yxfx∂

∂ = yx +2 and ),( yxfy∂

∂ = 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 =⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂≠⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂= yxf

yxyxf

xy.

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).

Armin Halilovic Math. Exercises

15 / 27

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+=

∂∂ ),,( and 23),,( zyxyzyxf

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

Armin Halilovic Math. Exercises

16 / 27

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

333 =⇒=⇒⎟⎠⎞

⎜⎝⎛

∂∂

∂∂=⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂ bbxxf

zxf

xz

333 =⇒=⇒⎟⎠⎞

⎜⎝⎛

∂∂

∂∂=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

∂∂ bbxxf

zyf

yz.

Thus, all three conditions are fulfilled if 3=a and 3=b . For these values of a and b we get Czyyzxzyxf +++= 23),,( 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

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:

Armin Halilovic Math. Exercises

17 / 27

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•∇=•∇+∂

∂ yzxzyxUUt

rrϕρϕρϕ ⇒

⇒−+++++Γ=+∂

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

rrϕρϕρϕ

(since )0,1,2()( −=Ucurlr

we have 0))(( =Ucurldivr

)

⇒−++++∂∂Γ

∂∂Γ

∂∂Γ=+

∂∂ 24841616),,(),,()( yzxzyx

zyxdivwvudiv

tϕϕϕρϕρϕρϕρϕ

24841616))()()()(−++++⎟

⎠⎞

⎜⎝⎛

∂∂Γ

∂∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂Γ

∂∂+⎟

⎠⎞

⎜⎝⎛

∂∂Γ

∂∂=

∂∂

+∂

∂+

∂∂

+∂

∂ yzxzyxzzyyxxz

wy

vx

ut

ϕϕϕρϕρϕρϕρϕ

(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

∂∂=

∂+∂

+∂

∂+

∂∂

+∂

∂ yzxzyxzzyyxxz

yxyxt

ϕϕϕφϕϕϕ

( Note that Γ is a constant)

5630

2484161668461166

=Γ⇒Γ=

⇒−++++Γ=++++ yzxzyxyzxzyx

Armin Halilovic Math. Exercises

18 / 27

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 (*) .

Substitution in eq3i. implies

)()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). 16. 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:

Armin Halilovic Math. Exercises

19 / 27

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 (*) .

Substitution in eq3i. implies

)()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 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.

Armin Halilovic Math. Exercises

20 / 27

c4. We assume that the flow is axisymmetric . The velocity does not depend on θ,

that is 0=∂∂

θzu

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

==∂

∂rr

uz

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)

Armin Halilovic Math. Exercises

21 / 27

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

Armin Halilovic Math. Exercises

22 / 27

∂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

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)

Armin Halilovic Math. Exercises

23 / 27

00

==∂

∂rr

w (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 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 −+−++==r

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

Armin Halilovic Math. Exercises

24 / 27

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: eq4i.

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

242:1 =⇒−=−⇒⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂=⎟

⎠⎞

⎜⎝⎛

∂∂

∂∂ ccP

yxP

xyCon ρρ

Armin Halilovic Math. Exercises

25 / 27

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

terms of zff

rfeeezr

∂∂

∂∂

∂∂ and ,, , ,,,, 221 θ

θ rrr if

a) kejeie

rrrrrr === 321 ,, ( we keep the same basis kjirrr

,, ) .

b) kjejeierrrrrrr +=== 2,2 321

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

d) kejiejie

rrrrrrrr =+−=+= 321 ,cossin,sincos θθθθ (this is often used as a local basis for cylindrical coordinates) Solution: In x,y, z variables we have

kzfj

yfi

xf

zf

yf

xffgrad

rrr

∂∂+

∂∂+

∂∂=

∂∂

∂∂

∂∂= ),,()( ( eq1)

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

Armin Halilovic Math. Exercises

26 / 27

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

rrr

∂∂+

∂∂+

∂∂+

∂∂−

∂∂=

θθθ

θθθ

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

rrr,, as a linear combinations of

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

=== , we have immediately

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

rrfef

rrffgrad rrr

∂∂+

∂∂+

∂∂+

∂∂−

∂∂=

θθθ

θθθ

b) From kjejeie

rrrrrrr +=== 2,2 321 we find

232

1 ,2

, eekejei rrrrrrr−=== (eq b)

Then we put kjirrr

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

)(2

)cos(sin)sin(cos)( 232

1 eezfef

rrfef

rrffgrad rr

rr −

∂∂+

∂∂+

∂∂+

∂∂−

∂∂=

θθθ

θθθ

and, after collecting components for 321 , eee rrr

321 )cos21sin

21()sin(cos)( e

zfe

zff

rrfef

rrffgrad rrr

∂∂+

∂∂−

∂∂+

∂∂+

∂∂−

∂∂=

θθθ

θθθ

c) From

kejeierrrrvr === 321 , sin ,cos θθ

we have

Armin Halilovic Math. Exercises

27 / 27

321 ,

sin,

cosekejei rrrrrr

===θθ

( eq c)

Putting kjirrr

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

sin

)cos(sincos

)sin(cos)( 321 e

zfef

rrfef

rrffgrad r

rr

∂∂+

∂∂+

∂∂+

∂∂−

∂∂=

θθθθ

θθθθ

d) kejiejie

rrrrrrrr =+−=+= 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

rrrrrr

∂∂+

∂∂+

∂∂+

∂∂−

∂∂=

∂∂+

∂∂+

∂∂= )cos(sin)sin(cos)(

θθθ

θθθ

kzfjif

rji

rf rrrrr

∂∂++−⋅

∂∂++⋅

∂∂= )cossin(1)sin(cos θθ

θθθ

3211 e

zfef

re

rf rrr

∂∂+⋅

∂∂+⋅

∂∂=

θ

Answer:

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

rrfef

rrffgrad rrr

∂∂+

∂∂+

∂∂+

∂∂−

∂∂=

θθθ

θθθ

b)

321 )cos21sin

21()sin(cos)( e

zfe

zff

rrfef

rrffgrad rrr

∂∂+

∂∂−

∂∂+

∂∂+

∂∂−

∂∂=

θθθ

θθθ

c) sin

)cos(sincos

)sin(cos)( 321 e

zfef

rrfef

rrffgrad r

rr

∂∂+

∂∂+

∂∂+

∂∂−

∂∂=

θθθθ

θθθθ

d) 3211),,(( e

zfef

re

rfzrfgrad rrr

∂∂+⋅

∂∂+⋅

∂∂=

θθ