math. exercises. gradient, divergence, curl del...
TRANSCRIPT
Armin Halilovic Math. Exercises
1 / 27
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)()()()(
∂∂−
∂∂+
∂∂−
∂∂+
∂∂−
∂∂=
∂∂
∂∂
∂∂=
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
2 / 27
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
Armin Halilovic Math. Exercises
3 / 27
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
Armin Halilovic Math. Exercises
4 / 27
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 .
Armin Halilovic Math. Exercises
5 / 27
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.
Armin Halilovic Math. Exercises
6 / 27
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
7 / 27
)( 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
8 / 27
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
9 / 27
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