E-mail : [email protected] webpage : · without using operators div, ∇, ∆, curl or grad. Here ....
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Armin Halilovic Math. Exercises
E-mail : [email protected] webpage : www.sth.kth.se/armin MATH. EXERCISES. GRADIENT, DIVERGENCE, CURL DEL (NABLA) OPERATOR , LAPLACIAN OPERATOR , CONTINUITY AND NAVIER-STOKES EQUATIONS VECTOR PRODUCTS If ),,( 321 uuuu = and ),,( 321 vvvv =
then
332211 vuvuvuvu ++=• (scalar or dot product)
321
321
vvvuuukji
vu
=× (vector or cross product)
In some books is also considered outer product defined by
=⊗
3
2
1
uuu
vu )( 321 vvv =
332313
322212
312111
vuvuvuvuvuvuvuvuvu
GRADIENT, DIVERGENCE, CURL DEL (NABLA) OPERATOR , LAPLACIAN OPERATOR GRADIENT Let ),,( zyxϕ be a scalar field. The gradient is the vector field defined by
),,()(zyx
grad∂∂
∂∂
∂∂
=ϕϕϕϕ
DIVERGENCE Let )),,(),,,(),,,(( zyxRzyxQzyxPF =
be a vector field, continuously differentiable with respect to x, y and z. Then the divergence of F
is the scalar field defined by
zR
yQ
xPFdiv
∂∂
+∂∂
+∂∂
=)(
CURL. The curl of F
is the vector field defined by
kyP
xQj
xR
zPi
zQ
yR
RQPzyx
kji
Fcurl
)()()()(∂∂
−∂∂
+∂∂
−∂∂
+∂∂
−∂∂
=∂∂
∂∂
∂∂
=
or ),,()(yP
xQ
xR
zP
zQ
yRFcurl
∂∂
−∂∂
∂∂
−∂∂
∂∂
−∂∂
=
DEL (NABLA) OPERATOR The vector differential operator
=∇ ),,(zyxz
ky
jx
i∂∂
∂∂
∂∂
=∂∂
+∂∂
+∂∂
is called del or nabla .
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Armin Halilovic Math. Exercises
Using ∇ we can denote grad, div and curl as below:
)(ϕgrad = ϕ∇ FFdiv
•∇=)( FFcurl
×∇=)( Note that ∇•F
is not the same as F
•∇ .
∇•F
=z
Ry
Qx
P∂∂
+∂∂
+∂∂ .
LAPLACIAN OPERATOR
The Laplacian operator, 2
2
2
2
2
22
zyx ∂∂
+∂∂
+∂∂
=∇=∆ , is defined for a scalar field U(x,y,z) by
2
2
2
2
2
22
zU
yU
xUUU
∂∂
+∂∂
+∂∂
=∇=∆ ,
and for a vector field )),,(),,,(),,,(( zyxRzyxQzyxPF =
by ),,(2 RQPFF ∆∆∆=∇=∆
. Some formulas for polar and cylindrical coordinates Polar coordinates ( 2 dim)
ereϑ
ϑ
i
jer
eϑ
ϑ
i
j
r P
F
transformation: θcosrx = , θsinry = , area element: θddrrdA = standard basis: ,cossin,sincos jiejier
θθθθ θ +−=+= [Remark 1 : Note that θeer
, vary ( depend on θ ) when we move from point to point, this is the reason why this basis, in some books, is called “ local basis” .] If jFiFF yx
+= in Cartesian coord. and θϑeFeFF rr
+= the same vector in polar coordinates then
θθ sincos yxr FFF += , θθϑ cossin yx FFF +−= .
[Remark 2: Vi can derive these formulas by calculating the components of F
in the directions of re and θe
. Thus
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Armin Halilovic Math. Exercises
θθθθ sincos)sin(cos)( yxyxrr FFjijFiFeFF +=+•+=•=
, similarly
θθθθϑϑ cossin)cossin()( yxyx FFjijFiFeFF +−=+−•+=•=
. ] Cylindrical coordinates ),,( zr θ : transformation: θcosrx = , θsinry = , z=z volume element: dzddrrdV θ=
er
eϑ
ez =k
x
y
z
ϑr
F
standard basis: kejiejie zr
=+−=+= ,cossin,sincos θθθθ θ
If kFjFiFF zyx
++= in Cartesian coord. and kFeFeFF zrr
++= θϑ the same vector in cylindrical coordinates then we have following vector components relationship: θθ sincos yxr FFF += , θθϑ cossin yx FFF +−= , zz FF = [Remark 3: For example, we can get θθ sincos yxr FFF += in the following way:
θθθθ sincos)sin(cos)( yxzyxrr FFjikFjFiFeFF +=+•++=•=
] scalar field: ),,( zrf θ
gradient: zr ezfef
re
rfffgrad
∂∂
+∂∂
+∂∂
=∇= θθ1)(
laplacian: 2
2
2
2
22 11
zff
rrfr
rrff
∂∂
+∂∂
+
∂∂
∂∂
=∆=∇θ
vector field: ),,( zr FFFF θ=
divergence: z
FFrr
Frr
FFdiv zr
∂∂
+∂
∂+
∂⋅∂
=∇=θθ )(1)(1)(
curl:
zr
zr
FFrFzr
eere
rFFcurl
θ
θ
θ⋅
∂∂
∂∂
∂∂
⋅
=×∇=
1)( =
krzr
rz eF
rrF
re
rF
zFe
zFF
r
∂∂
−∂
∂+
∂∂
−∂∂
+
∂∂
−∂∂
θθθ
θθ )(11
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Armin Halilovic Math. Exercises
EXERCISES 1. Find a) )(Fdiv
, b) ))(( Fdivgrad
and c) )(Fcurl
if ),,( 22 xzxyF +=
2. Prove the following identities a ) 0)(( =Fcurldiv
(or 0)( =×∇•∇ F
) b) 0))((
=ϕdivcurl (or 0)(
=∇×∇ ϕ ) 3. Which one of the following functions a) 222
1 23),,( zyxzyxf ++= b) )ln(),,( 22
2 zyxzyxf ++= c) )exp(),,( 3
3 zyxzyxf ++= d) zyxyxzyxf 455),,( 22
4 +++−= satisfies the Laplace equation f∆ =0 ? 4. Find )))(( ff ∇×∇•∇+∆ if zyxzyxf ++= 23),,( . 5. Write the general transport equation
φϕρϕρϕ SUt
+∇⋅Γ•∇=•∇+∂
∂ )()()(
without using operators div, ∇ , ∆ , curl or grad. Here ),,( wvuU =
. Functions S,,, Γϕρ , u, v, w are real functions of t, x, y and z. 6. Which one, if any, of the following functions a) zyxzyx ++= 24
1 ),,(ϕ
b) zyxzyx ++= 222 ),,(ϕ
c) 2223 ),,( zyxzyx ++=ϕ
satisfies the equation SU +∇⋅Γ•∇=•∇ ))(()( ϕϕ
? Here 5=Γ , )3,2,1(=U
and 2342 −+= yxS . 7. Find which one (if any) of the following functions a) 222
1 ),,( zyxzyx ++=ϕ
b) zyxzyx 5),,( 222 ++=ϕ
c) 2223 5),,( zyxzyx ++=ϕ
satisfies the equation
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Armin Halilovic Math. Exercises
SgraddivUdivt
+Γ=+∂
∂ )()()( ϕρϕρϕ
where ρ=3, 2=Γ , )4,3,2(=U
and 521812 ++= yxS . 8. (exam 1, 2008) A) Write the general transport equation
SUt
+∇⋅Γ•∇=•∇+∂
∂ ))(()()( ϕρϕρϕ
( eq 1)
without using operators div, ∇ , ∆ , curl or grad. Here ),,( wvuU =
. Functions S,,, Γϕρ , u, v, w are real functions of t, x, y and z. B) Let 2=ρ , 3=Γ , )4,2,1(=U
. Find S in the equation (eq 1) if we now that the function 32),,( zyxzyx ++=ϕ satisfies the equation. 9. (Q6, exam 2, 2008) Consider the following equation
426)())(()()(−−+×∇•∇+∇⋅Γ•∇=•∇+
∂∂ xyzyUU
t
ϕρϕρϕ ( eq 1)
Let 1=ρ , Γ= constant , ),3,2( xyU −=
. Find the constant Γ in the equation (eq 1) if we now that the function 222),,( zyxtzyx +++=ϕ satisfies the equation.
10. If possible, find ),( yxf for the given partial derivatives ),( yxfx∂∂ and ),( yxf
y∂∂ .
a) ),( yxfx∂∂ = xy2 and ),( yxf
y∂∂ = yx 22 + .
b) ),( yxfx∂∂ = yx +2 and ),( yxf
y∂∂ = x .
c) ),( yxfx∂∂ = xyye and ),( yxf
y∂∂ = xyxe .
d) ),( yxfx∂∂ = yx +2 and ),( yxf
y∂∂ = x5 .
( Hint: Necessary condition: If ),( yxf has continues derivatives then the mixed derivatives of ),( yxf should be equal. Thus
∂∂
∂∂
=
∂∂
∂∂ yxf
yxyxf
xy,(,( (*)
is the necessary condition for the existence of a function ),( yxf that has the given derivatives. 11. Determine the value of a for which the system of partial differential equations
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Armin Halilovic Math. Exercises
),( yxfx∂∂ = yaxy + and ),( yxf
y∂∂ = xx +2 .
has solutions. Then find ),( yxf corresponding to this value of a.
12. If possible, find ),,( zyxf for the given partial derivatives ),,( zyxfx∂∂ , ),,( zyxf
y∂∂
and ),,( zyxfz∂∂ .
a) yzzyxfx
=∂∂ ),,( , zxzzyxf
y+=
∂∂ ),,( and 23),,( zyxyzyxf
z++=
∂∂
b) yzzyxfx
=∂∂ ),,( , zxzzyxf
y+=
∂∂ ),,( 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 + , ),,( zyxf
y∂∂ = 13 +zx and ),,( zyxf
z∂∂ = zybx 23 +
has solutions. Then find ),,( zyxf corresponding to these values of a and b. 14. We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field
.x,y,z, w x,y,z, vx,y,zuV ))()()(( =
Use the following equations ( continuity and Navier Stokes equations) to find en expression for pressure P(x,y,z) as a function of x,y and z, where ρ =constant, µ =constant , ),0,0( gg −=
i.e. xg = yg =0 and
)/81.9 where( 2smggg z ≈−=
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Armin Halilovic Math. Exercises
Incompressible continuity equation:
0=∂∂
+∂∂
+∂∂
zw
yv
xu eq1.
Navier Stokes equations: x component:
)( 2
2
2
2
2
2
zu
yu
xug
xP
zuw
yuv
xuu
tu
x ∂∂
+∂∂
+∂∂
++∂∂
−=
∂∂
+∂∂
+∂∂
+∂∂ µρρ eq2.
y component:
)( 2
2
2
2
2
2
zv
yv
xvg
yP
zvw
yvv
xvu
tv
y ∂∂
+∂∂
+∂∂
++∂∂
−=
∂∂
+∂∂
+∂∂
+∂∂ µρρ eq3.
z component:
)( 2
2
2
2
2
2
zw
yw
xwg
zP
zww
ywv
xwu
tw
z ∂∂
+∂∂
+∂∂
++∂∂
−=
∂∂
+∂∂
+∂∂
+∂∂ µρρ eq4.
a) )0,24,32( yxyxV −+=
b) )2,32,43( −−+= yxyxV
c) )2,4,21( xyV −+=
15. (exam 1, 2009) A) Consider the following equation
24841616)())(()()(−++++×∇•∇+∇⋅Γ•∇=•∇+
∂∂ yzxzyxUU
t
ϕρϕρϕ ( eq 1)
Let 2=ρ , Γ= constant , )2,4,4( yxU +=
. Find the constant Γ in the equation (eq 1) if we now that the function 22231),,( zyxtzyx ++++=ϕ satisfies the equation. B) We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field
.x,y,z, w x,y,z, vx,y,zuV ))()()(( =
Use the following equations ( continuity and Navier Stokes equations) to find en expression for pressure P(x,y,z) as a function of x,y and z, where ρ =constant, µ =constant ,
),0,0( gg −= i.e. xg = yg =0 and )/81.9 where( 2smgggz ≈−= and
)22,24,46( zyxV −−+=
. 16. (exam 2, 2009) We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field
.x,y,z, w x,y,z, vx,y,zuV ))()()(( =
Use the following equations ( continuity and Navier Stokes equations) to find first i) parameter a and then
7 / 28
Armin Halilovic Math. Exercises
ii) en expression for pressure P(x,y,z) as a function of x,y and z, where ρ =constant, µ =constant ,
),0,0( gg −= i.e. xg = yg =0 and )/81.9 where( 2smgggz ≈−= and
)1,5,32( azyxV −−+=
.
17. Consider steady, incompressible, isothermal, laminar stationary Newtonian flow in a long round pipe in the z-direction, with constant circular cross-section of radius R=2 m. Use the continuity and the Navier-Stokes equations in cylindrical coordinates to find the velocity field V=(ur, uθ, uz) and the pressure field P (r,θ,z) if the fluid flow satisfies the following conditions:
c0. All partial derivatives with respect to time t are 0 ( Steady flow)
c1. μ=0.001 kg/(m∙s) and ρ =1000 kg/m3
c2. A Constant pressure gradient ∂P/∂z = –1/250 Pa/m is applied in the horizontal axis ( z-axis in our notation): ∂P/∂z = –1/250, c3. The flow is parallel to the z axis, that is ur =0 and uθ =0. c4. We assume that the flow is axisymmetric . The velocity does not depend on θ,
that is 0=∂∂θ
zu
c5. Boundary cond. 1 ( No-slip boundary condition, Vfluid=Vwall ): If r=2 then uz= 0
c6. Boundary condition 2: uz has maximum at r=0 that is 00=
=∂∂
rruz
---------------------------------------------------------------------------------------------
The continuity and the Navier-Stokes equations for an incompressible , isothermal Newtonian flow (density ρ =const, viscosity µ =const), with a velocity field
),,( zr uuuV θ=
in Cylindrical coordinates ),,( zr θ : Incompressible continuity equation
0)(1)(1
=∂∂
+∂∂
+∂
∂z
uurr
rur
zr
θθ eq a)
Navier-Stokes equations in Cylindrical coordinates: r-component:
∂∂
+∂∂
−∂∂
+−
∂∂
∂∂
++∂∂
−=
∂∂
+−∂∂
+∂∂
+∂∂
2
2
22
2
22
2
211zuu
ru
rru
rur
rrg
rP
zuu
ruu
ru
ruu
tu
rrrrr
rz
rrr
r
θθµρ
θρ
θ
θθ
eq b)
θ -component:
8 / 28
Armin Halilovic Math. Exercises
∂∂
+∂∂
+∂∂
+−
∂∂
∂∂
++∂∂
−=
∂∂
++∂∂
+∂∂
+∂∂
2
2
22
2
22
2111zuu
ru
rru
ru
rrr
gPr
zu
uruuu
ru
ru
ut
u
r
zr
r
θθθθθ
θθθθθθ
θθµρ
θ
θρ
eq c)
z-component:
∂∂
+∂∂
+
∂∂
∂∂
++∂∂
−=
∂∂
+∂∂
+∂∂
+∂∂
2
2
2
2
211
zuu
rrur
rrg
zP
zuuu
ru
ruu
tu
zzzz
zz
zzr
z
θµρ
θρ θ
eq d)
18. (Exam 1 March 2012, question A , 4points.) We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field
)3,24,3())()()(( azbzyxcyxx,y,z, w x,y,z, vx,y,zuV −+−++==
Use the following equations ( continuity and Navier Stokes equations) , where ρ =constant, µ =constant , ),0,0( gg −=
i.e. xg = yg =0 and )/81.9 ( 2smgggz ≈−= to find: i) parameters a, b and c ii) en expression for pressure P(x,y,z) as a function of x,y and z. The GRADIENT VECTOR with change of variables and basis. The gradient vector for the function f(x,y,z) is defined as
kzfj
yfi
xf
zf
yf
xffgrad
∂∂
+∂∂
+∂∂
=∂∂
∂∂
∂∂
= ),,()( (*) .
If we change variables x, v, z to u, v , w and replace basis vectors kji
,, with new ( linearly independent) vectors 321 ,, eee then we can express the same gradient vector )( fgrad in terms of variables u, v , w and vectors 321 ,, eee .
We simple calculate the derivatives zf
yf
xf
∂∂
∂∂
∂∂ and in new variables and express kji
,, as
a linear combinations of 321 ,, eee . Then we substitute those values into (*). (See the following example.) 19. We consider a scalar field ),,( zrf θ given in cylindrical coordinates, where
zzryrx === , sin ,cos θθ , and basis vectors are 321 ,, eee .
9 / 28
Armin Halilovic Math. Exercises
Find the expression for the gradient, )),,(( zrfgrad θ ), in cylindrical coordinates, that is in
terms of zff
rfeeezr
∂∂
∂∂
∂∂ and ,, , ,,,, 221 θ
θ
if
a) kejeie
=== 321 ,, ( we keep the same basis kji
,, ) .
b) kjejeie
+=== 2,2 321
c) kejeie
=== 321 , sin ,cos θθ (exam 1, 2012; Q5 B (2 points)) d) kejiejie
=+−=+= 321 ,cossin,sincos θθθθ (this is often used as a local basis for cylindrical coordinates) 20. (exam 2016; Q6 A (2 points)) Derive the Cauchy momentum equation
gDt
VD
ρρ +⋅∇= σ .
21. Find the flux of the vector field kyxjyixF
)( +++= upward through the surface yxz 231 ++= , 10 ≤≤ x , 20 ≤≤ y .
22. Use the Divergence Theorem to find the flux of the vector field
kzjyzixF
2212 −+= out of the sphere S with equation 9222 =++ zyx ANSWERS AND SOLUTIONS: 1. Solution:
a) Since zR
yQ
xPFdiv
∂∂
+∂∂
+∂∂
=)(
we have ),,( 22 xzxyF +=
⇒ xxFdiv 2002)( =++=
. Answer a) xFdiv 2)( =
b) Since ),,()(zyx
grad∂∂
∂∂
∂∂
=ϕϕϕϕ we have ( for )(Fdiv
=ϕ )
)0,0,2())(( =Fdivgrad
Answer b) )0,0,2())(( =Fdivgrad
c)
)1,2,1(21
)(22
−−−=−−−=
+∂∂
∂∂
∂∂
=∂∂
∂∂
∂∂
=
xkjxi
xzxyzyx
kji
RQPzyx
kji
Fcurldef
10 / 28
Armin Halilovic Math. Exercises
Answer c) )1,2,1()( −−−= xFcurl
2. Hint. Use the definitions of div and curl. 3.
f∆ =0 ⇔ 02
2
2
2
2
2
=∂∂
+∂∂
+∂∂
zf
yf
xf
Answer: The functions )ln(),,( 22
2 zyxzyxf ++= and zyxyxzyxf 455),,( 22
4 +++−= satisfies the Laplace equation. 4. Answer: )))(( ff ∇×∇•∇+∆ = )))(( gradfcurldivf +∆ = 26 +x 5. Solution:
⇒+Γ=+∂
∂φϕρϕρϕ SgraddivUdiv
t)()()(
⇒+∂∂
Γ∂∂
Γ∂∂
Γ=+∂
∂φ
ϕϕϕρϕρϕρϕρϕ Szyx
divwvudivt
),,(),,()(
φϕϕϕρϕρϕρϕρϕ Szzyyxxz
wy
vx
ut
+
∂∂
Γ∂∂
+
∂∂
Γ∂∂
+
∂∂
Γ∂∂
=∂
∂+
∂∂
+∂
∂+
∂∂ ))()()()(
6. Which one (if any) of the following functions a) zyxzyx ++= 24
1 ),,(ϕ
b) zyxzyx ++= 222 ),,(ϕ
c) 2223 ),,( zyxzyx ++=ϕ
satisfies the equation SU +∇⋅Γ•∇=•∇ ))(()( ϕϕ
? Here 5=Γ , )3,2,1(=U
and 2342 −+= yxS . Solution : The equation
SU +∇⋅Γ•∇=•∇ ))(()( ϕϕ
can be written as
11 / 28
Armin Halilovic Math. Exercises
1.) (eq 234255532
2342)5,5,5()3,2,(
)()(
2
2
2
2
2
2
−++∂∂
+∂∂
+∂∂
=∂∂
+∂∂
+∂∂
⇒−++∂∂
∂∂
∂∂
=
⇒+Γ=
yxzyxzyx
yxzyx
divdiv
SgraddivUdiv
ϕϕϕϕϕϕ
ϕϕϕϕϕϕ
ϕϕ
a) zyxzyx ++== 241 ),,(Let ϕϕ
Vi calculate the derivatives of 1ϕ and substitute in the left hand side (LHS) and right hand side of the equation (eq1).
LHS: 34432 3 ++=∂∂
+∂∂
+∂∂ yx
zyxϕϕϕ
RHS= 1342602342555 22
2
2
2
2
2
−++=−++∂∂
+∂∂
+∂∂ yxx yx
zyxϕϕϕ
Whence RHSLHS ≠ Thus the function zyxzyx ++= 24
1 ),,(ϕ is not a solution to the equation b) zyxzyx ++== 22
2 ),,(ϕϕ LHS= yx 423 ++ , RHS= yx 423 ++−
Whence RHSLHS ≠ , and the function zyxzyx ++= 222 ),,(ϕ is not a solution to the
equation c) Let 222
3 ),,( zyxzyx ++==ϕϕ Then LHS= zyx 642 ++ , RHS= yx 427 ++
Thus RHSLHS ≠ , and the function 2223 ),,( zyxzyx ++=ϕ is not a solution to the
equation. Answer: None of the functions satisfies the equation 7. Answer: Function zyxzyx 5),,( 22
2 ++=ϕ satisfies the equation. 8. (exam 1, 98) A) Write the general transport equation
SUt
+∇⋅Γ•∇=•∇+∂
∂ ))(()()( ϕρϕρϕ
( eq 1)
without using operators div, ∇ , ∆ , curl or grad. Here ),,( wvuU =
. Functions S,,, Γϕρ , u, v, w are real functions of t, x, y and z.
12 / 28
Armin Halilovic Math. Exercises
B) Let 2=ρ , 3=Γ , )4,2,1(=U
. Find S in the equation (eq 1) if we now that the function 32),,( zyxzyx ++=ϕ satisfies the equation. Solution: A)
SUt
+∇⋅Γ•∇=•∇+∂
∂ ))(()()( ϕρϕρϕ
⇒
⇒+Γ=+∂
∂ SgraddivUdivt
)()()( ϕρϕρϕ
⇒+∂∂
Γ∂∂
Γ∂∂
Γ=+∂
∂ Szyx
divwvudivt
),,(),,()( ϕϕϕρϕρϕρϕρϕ
Szzyyxxz
wy
vx
ut
+
∂∂
Γ∂∂
+
∂∂
Γ∂∂
+
∂∂
Γ∂∂
=∂
∂+
∂∂
+∂
∂+
∂∂ ϕϕϕρϕρϕρϕρϕ ))()()()( (eq2)
B) We substitute 2=ρ , 3=Γ , )4,2,1(=U
and 32),,( zyxzyx ++=ϕ in the equation (eq2) and get
Szzyyxxzyx
+
∂∂
∂∂
+
∂∂
∂∂
+
∂∂
∂∂
=∂
∂+
∂∂
+∂
∂+
ϕϕϕφϕϕ 333))8()4()2(0
Szzy +++=+++ 186024820 2 . Consequently
2241884 zzyS +−+−= 9. (Q6, exam 2, 2008) Consider the following equation
426)())(()()(−−+×∇•∇+∇⋅Γ•∇=•∇+
∂∂ xyzyUU
t
ϕρϕρϕ ( eq 1)
Let 1=ρ , Γ= constant , ),3,2( xyU −=
. Find the constant Γ in the equation (eq 1) if we now that the function 222),,( zyxtzyx +++=ϕ satisfies the equation. Solution:
426)())(()()(−−+×∇•∇+∇⋅Γ•∇=•∇+
∂∂ xyzyUU
t
ϕρϕρϕ ⇒
⇒−−++Γ=+∂
∂ 426))(()()()( xyzyUcurldivgraddivUdivt
ϕρϕρϕ
(since )0,,()( yxUcurl −=
we have 011))(( =+−=Ucurldiv
)
13 / 28
Armin Halilovic Math. Exercises
⇒−−++∂∂
Γ∂∂
Γ∂∂
Γ=+∂
∂ 4260),,(),,()( xyzyzyx
divwvudivt
ϕϕϕρϕρϕρϕρϕ
4260))()()()(−−++
∂∂
Γ∂∂
+
∂∂
Γ∂∂
+
∂∂
Γ∂∂
=∂
∂+
∂∂
+∂
∂+
∂∂ xyzy
zzyyxxzw
yv
xu
tϕϕϕρϕρϕρϕρϕ (eq2)
We substitute 1=ρ , , ),3,2( xyU −=
and 222),,( zyxtzyx +++=ϕ in the equation (eq2) and get
426))()3()2()1(−−+
∂∂
Γ∂∂
+
∂∂
Γ∂∂
+
∂∂
Γ∂∂
=∂−∂
+∂
∂+
∂∂
+∂
∂ xyzyzzyyxxz
xyyxt
ϕϕϕφϕϕϕ
( Note that Γ is a constant)
248
4262202622
=Γ⇒Γ=
⇒−−+Γ+Γ+=−++ xyzyxyzy
Answer: 2=Γ
10. If possible, find ),( yxf for the given partial derivatives ),( yxfx∂∂ and ),( yxf
y∂∂ .
a) ),( yxfx∂∂ = xy2 and ),( yxf
y∂∂ = yx 22 + .
b) ),( yxfx∂∂ = yx +2 and ),( yxf
y∂∂ = x .
c) ),( yxfx∂∂ = xyye and ),( yxf
y∂∂ = xyxe .
d) ),( yxfx∂∂ = yx +2 and ),( yxf
y∂∂ = x5 .
( Hint: Necessary condition: If ),( yxf has continuous derivatives then the mixed derivatives of ),( yxf should be equal, i.e.
∂∂
∂∂
=
∂∂
∂∂ ),(),( yxf
yxyxf
xy (*)
is the necessary condition for the existence of a function ),( yxf that has the given derivatives. Answer: a) ),( yxf = Cyyx ++ 22 b) ),( yxf = Cxyx ++2 c) ),( yxf = Ce xy + d) No solution since the condition (*) is not fulfilled,
5),(),(1 =
∂∂
∂∂
≠
∂∂
∂∂
= yxfyx
yxfxy
.
Solution a)
Since
∂∂
∂∂
=
∂∂
∂∂ ),(),( yxf
yxyxf
xy =2x and the derivatives are continuous the
condition (*) is fulfilled and we can find ),( yxf for the given derivatives. In order to find ),( yxf we integrate with respect to x the first of the equations
),( yxfx∂∂ = xy2 (eq1)
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Armin Halilovic Math. Exercises
),( yxfy∂∂ = yx 22 + (eq2).
and get
∫ +== )(2),( 1́2 yCyxxydxyxf
Thus )(),( 1́
2 yCyxyxf += ( i ) We have integrated with respect to x, therefore the constant still depend on y. Now, to find )(1́ yC we differentiate and substitute (i) in (eq2) and get:
( ))(1́2 yCyx
y+
∂∂ = yx 22 + ⇒ ( ))(1́
2 yCy
x∂∂
+ = yx 22 + ⇒
( ))(1́ yCy∂∂ = y2 ⇒ )(1́ yC = Cy +2 .
Finally, substituting )(1́ yC = Cy +2 in (i) we have Cyyxyxf ++= 22),( (where C is a constant).
11.
Answer: From 2121),(),( =⇒+=+⇒∂∂
∂∂
=∂∂
∂∂ axayyxf
yxyxf
xy
Then for 2=a we have Cxyyxyxf ++= 2),( 12. Answer: a) Czyzxyzf +++= 3 b) Cyzxyf ++= c) Cef xyz += d) No solution since the condition 2Con is not fulfilled,
zyfzx
fxz
y +=
∂∂
∂∂
≠
∂∂
∂∂
=
Solution a)
a) yzzyxfx
=∂∂ ),,( , zxzzyxf
y+=
∂∂ ),,( 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
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Armin Halilovic Math. Exercises
),(),,( 1́ zyCxyzzyxf += ( i ) We have integrated with respect to x, therefore the constant still depend on y and z. Now, to find ),(1́ zyC we differentiate and substitute (i) in (eq2) and get:
( )),(1́ zyCxyzy
+∂∂ = zxz + ⇒ ( )),(1́ zyC
yxz
∂∂
+ = zxz + ⇒
( )),(1́ zyCy∂∂ = z ⇒ ),(1́ zyC = )(2 zCyz + .
(We have integrated with respect to y , therefore the constant still depend on and z. Thus )(),,( 2 zCyzxyzzyxf ++= (ii) Now , substituting (ii) in (eq3) we have
CzzC
zzCz
zyxyzCz
yxy
zyxyzCyzxyzz
+=
=∂∂
⇒++=∂∂
++
⇒++=++∂∂
32
22
22
22
)(
3))((
3))((
3))((
Finally, substituting CzzC += 3
2 )( in (ii) we have Czyzxyzzyxf +++= 3),,( (where C is a constant).
13. Answer: From
33 22 =⇒=⇒
∂∂
∂∂
=
∂∂
∂∂ azxzaxf
yxf
xy
133 =⇒=⇒
∂∂
∂∂
=
∂∂
∂∂ bbxxf
zxf
xz
133 =⇒=⇒
∂∂
∂∂
=
∂∂
∂∂ bbxxf
zyf
yz.
Thus, all three conditions are fulfilled if 3=a and 1=b . For these values of a and b we get Czyxyzxzyxf +++= 223),,( Calculation of the pressure field for a known velocity field for an incompressible, steady state, isothermal Newtonian flow. 14. Answer: a) CyxgzP +−−−= 22 88 ρρρ
16 / 28
Armin Halilovic Math. Exercises
b) CyxgzP +−−−= 22
217
217 ρρρ
c) CyyxgzP ++++−= ρρρρ 444 22 Solution a) We substitute 0,24,32 =−=+= wyxvyxu in eq1,2,3,4 and get ( note that al derivatives with respect to t are 0): Continuity equation: 00 = eq1i. ( identically fulfilled) Navier Stokes equations: x component:
xPx∂∂
−=ρ16 eq2i.
y component:
yPy∂∂
−=ρ16 eq3i.
z component:
gzP ρ−∂∂
−=0 eq4i.
Now eq2i. gives ),(8),,( 1́2 zyCxzyxP +−= ρ (*) .
Substitution in eq3i. implies
)(8),(
),(16
2´2
1́
1́
zCyzyCy
zyCy
+−=
⇒∂
∂−=
ρ
ρ
Hence, from (*) we have )(88),,( 2´
22 zCyxzyxP +−−= ρρ (**) Now we substitute (**) in eq4i. and get
gzCz
gzP ρρ −
∂∂
−=⇒−∂∂
−= ))((00 2´
CgzzC +−=⇒ ρ)(2´ ( where C is a constant) Finally, substituting CgzzC +−= ρ)(2´ in (**) we have
CgzyxzyxP +−−−= ρρρ 22 88),,( (where C is a constant). 15. Solution A:
24841616)())(()()(−++++×∇•∇+∇⋅Γ•∇=•∇+
∂∂ yzxzyxUU
t
ϕρϕρϕ ⇒
⇒−+++++Γ=+∂
∂ 24841616))(()()()( yzxzyxUcurldivgraddivUdivt
ϕρϕρϕ
(since )0,1,2()( −=Ucurl
we have 0))(( =Ucurldiv
)
⇒−++++∂∂
Γ∂∂
Γ∂∂
Γ=+∂
∂ 24841616),,(),,()( yzxzyxzyx
divwvudivt
ϕϕϕρϕρϕρϕρϕ
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Armin Halilovic Math. Exercises
24841616))()()()(−++++
∂∂
Γ∂∂
+
∂∂
Γ∂∂
+
∂∂
Γ∂∂
=∂
∂+
∂∂
+∂
∂+
∂∂ yzxzyx
zzyyxxzw
yv
xu
tϕϕϕρϕρϕρϕρϕ
(eq2) We substitute 2=ρ , , )2,4,4( yxU +=
and 22231),,( zyxtzyx ++++=ϕ in the equation (eq2) and get
24841616)))2(2()8()8()2(−++++
∂∂
Γ∂∂
+
∂∂
Γ∂∂
+
∂∂
Γ∂∂
=∂+∂
+∂
∂+
∂∂
+∂
∂ yzxzyxzzyyxxz
yxyxt
ϕϕϕφϕϕϕ
( Note that Γ is a constant)
5630
2484161668461166
=Γ⇒Γ=
⇒−++++Γ=++++ yzxzyxyzxzyx
Answer A: 5=Γ Solution B: We substitute zwyvxu 22,24,46 −=−=+= in eq1,2,3,4 and get ( note that al derivatives with respect to t are 0): Continuity equation: 00 = eq1i. ( identically fulfilled) Navier Stokes equations: x component:
xPx∂∂
−=+ )2416(ρ eq2i.
y component:
yPy∂∂
−=− )84(ρ eq3i.
z component:
gzPz ρρ −∂∂
−=− )44( eq4i.
Now eq2i. gives ),()248(),,( 1́2 zyCxxzyxP +−−= ρ (*) .
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:
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Armin Halilovic Math. Exercises
CgzzzyyxxzyxP +−+−+−−−= )4282248(),,( 222ρ (where C is a constant). 16. Solution )1,5,32( azyxV −−+=
First we substitute azwyvxu −=−=+= 1,5,32 in eq1 and get ( note that al derivatives with respect to t are 0): Continuity equation: 2013 =⇒=−− aa No we have )21,5,32( zyxV −−+=
Using the Navier Stokes equations we get: x component:
xPx∂∂
−=+ )69(ρ eq2i.
y component:
yPy∂∂
−=− )5(ρ eq3i.
z component:
gzPz ρρ −∂∂
−=− )24( eq4i.
Now eq2i. gives ),()629(),,( 1́
2
zyCxxzyxP +−−
= ρ (*) .
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
19 / 28
Armin Halilovic Math. Exercises
the continuity and the Navier-Stokes equations in cylindrical coordinates to find the velocity field V=(ur, uθ, uz) and the pressure field P (r,θ,z) if the fluid flow satisfies the following conditions:
c0. All partial derivatives with respect to time t are 0 ( Steady flow)
c1. μ=0.001 kg/(m∙s) and ρ =1000 kg/m3
c2. A Constant pressure gradient ∂P/∂z = –1/250 Pa/m is applied in the horizontal axis ( z-axis in our notation): ∂P/∂z = –1/250, c3. The flow is parallel to the z axis, that is ur =0 and uθ =0. c4. We assume that the flow is axisymmetric . The velocity does not depend on θ,
that is 0=∂∂θ
zu
c5. Boundary cond. 1 ( No-slip boundary condition, Vfluid=Vwall ): If r=2 then uz= 0
c6. Boundary condition 2: uz has maximum at r=0 that is 00=
=∂∂
rruz
The continuity and the Navier-Stokes equations for an incompressible , isothermal Newtonian flow (density ρ =const, viscosity µ =const), with a velocity field
),,( zr uuuV θ=
in Cylindrical coordinates ),,( zr θ : ---------------------------------------------------------------- SOLUTION Incompressible continuity equation
0)(1)(1
=∂∂
+∂∂
+∂
∂z
uurr
rur
zr
θθ eq a)
Navier-Stokes equations in Cylindrical coordinates: r-component:
∂∂
+∂∂
−∂∂
+−
∂∂
∂∂
++∂∂
−=
∂∂
+−∂∂
+∂∂
+∂∂
2
2
22
2
22
2
211zuu
ru
rru
rur
rrg
rP
zuu
ruu
ru
ruu
tu
rrrrr
rz
rrr
r
θθµρ
θρ
θ
θθ
eq b)
θ -component:
∂∂
+∂∂
+∂∂
+−
∂∂
∂∂
++∂∂
−=
∂∂
++∂∂
+∂∂
+∂∂
2
2
22
2
22
2111zuu
ru
rru
ru
rrr
gPr
zu
uruuu
ru
ru
ut
u
r
zr
r
θθθθθ
θθθθθθ
θθµρ
θ
θρ
eq c)
z-component:
20 / 28
Armin Halilovic Math. Exercises
∂∂
+∂∂
+
∂∂
∂∂
++∂∂
−=
∂∂
+∂∂
+∂∂
+∂∂
2
2
2
2
211
zuu
rrur
rrg
zP
zuuu
ru
ruu
tu
zzzz
zz
zzr
z
θµρ
θρ θ
eq d)
We choose x as a vertical axis, y an z are in a horizontal plane and the flow is parallel with the z-axis. We denote velocity vector V=(ur, uθ, uz) where ur, uθ and uz are r-component, θ-component and z-component in cylindrical coordinates. According to the assumptions we have ur =0, uθ = 0, and uz does not depend on θ. Since x is the vertical axis we have that vector g=(-g, 0,0) where g=9,81 m/s2 which in cylindrical coordinates gives
θcosgg r −= , θθ singg = and 0=zg
Now we substitute ∂P/∂z = –1/250 Pa/m, μ=0.001kg /(ms) in the continuity and Navier-Stokes equations:
Since ur =0 and uθ =0 (according to c3), continuity equation in cylindrical coordinates
0)(1)(1=
∂∂
+∂∂
+∂
∂zuu
rrru
rzr
θθ
gives
0=∂∂
zuz .
21 / 28
Armin Halilovic Math. Exercises
This tells us that uz is not a function of z. Furthermore, since uz velocity does not depend on θ (assumption c4) we conclude that uz depends only on r. To simplify notation we denote
)(rwuz = (*)
Now we substitute
θcosggr −= , θθ singg = and 0=zg
∂P/∂z = –1/250 Pa/m, μ=0.001kg /(ms)
in the Navier-Stokes equations:
The r-component of the Navier-Stokes equation gives:
θρ cos0 grP−
∂∂
−= ( eq r-c)
The θ-component of the Navier-Stokes equation:
θρθ
sin10 gPr
+∂∂
−= ( eq θ-c)
The Z-component of the Navier-Stokes equation (where )(rwuz = and 2501
−=∂∂
zP ) givs:
∂∂
∂∂
+=rwr
rr1
10001
25010 ( eq z-c)
Step 1. We find the pressure ),,( zrPP θ= .
In order to find the pressure P we solve ( eq r-c), ( eq θ-c) and the equation 2501
−=∂∂
zP that
is
θρ cosgrP
−=∂∂
θρθ
singrP+=
∂∂
2501
−=∂∂
zP
From these equations we get
CgrzP +−−= θρ cos2501
22 / 28
Armin Halilovic Math. Exercises
Step 2. We find the velocity component )(rwuz = .
We solve ( eq z-c) with boundaries c5 and c6:
∂∂
∂∂
+=rwr
rr1
10001
25010 ( eq z-c)
0)2( =w (c5)
00=
=∂∂
rrw (c6)
( Remark: Technically, we can write drdw instead
rw∂∂ since w is now a function of only one
variable)
From ( eq z-c) we have
∂∂
∂∂
+=rwr
rr1
10001
25010
⇒−=
∂∂
∂∂ r
rwr
r4
⇒+−=∂∂
122 Cr
rwr (substitution 0=r and (c6) ⇒ 01 =C )
⇒−=∂∂ 22r
rwr
⇒−=∂∂ r
rw 2
⇒+−= 22 Crw (substitution 2=r and (c5) ⇒ 42 =C )
⇒+−= 42rw
Thus 4)( 2 +−== rrwuz and
V )4 0, (0,),,( 2 +−== ruuu zr θ .
Answer :
CgrzP +−−= θρ cos2501 ,
V = )4 0, (0, 2 +− r
23 / 28
Armin Halilovic Math. Exercises
18. We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field
)3,24,3())()()(( azbzyxcyxx,y,z, w x,y,z, vx,y,zuV −+−++==
Use the following equations ( continuity and Navier Stokes equations) , where ρ =constant, µ =constant , ),0,0( gg −=
i.e. xg = yg =0 and )/81.9 ( 2smgggz ≈−= to find: i) parameters a, b and c ii) en expression for pressure P(x,y,z) as a function of x,y and z. Incompressible continuity equation:
0=∂∂
+∂∂
+∂∂
zw
yv
xu eq1.
Navier Stokes equations: x component:
)( 2
2
2
2
2
2
zu
yu
xug
xP
zuw
yuv
xuu
tu
x ∂∂
+∂∂
+∂∂
++∂∂
−=
∂∂
+∂∂
+∂∂
+∂∂ µρρ eq2.
y component:
)( 2
2
2
2
2
2
zv
yv
xvg
yP
zvw
yvv
xvu
tv
y ∂∂
+∂∂
+∂∂
++∂∂
−=
∂∂
+∂∂
+∂∂
+∂∂ µρρ eq3.
z component:
)( 2
2
2
2
2
2
zw
yw
xwg
zP
zww
ywv
xwu
tw
z ∂∂
+∂∂
+∂∂
++∂∂
−=
∂∂
+∂∂
+∂∂
+∂∂ µρρ eq4.
------------------------------------------------------------- We substitute ,3,24,3 azwbzyxvcyxu −=+−+=+= in eq1,2,3,4 and get ( note that al derivatives with respect to t are 0): Continuity equation: 202 =⇒=− aa eq1i. Thus ,23,24,3 zwbzyxvcyxu −=+−+=+= Navier Stokes equations: x component:
eq2i.
y component: eq3i.
z component:
24 / 28
Armin Halilovic Math. Exercises
eq4i.
The system (eq2i, eq3i, eq4i) is solvable only if mixed derivatives are equal:
242:1 =⇒−=−⇒
∂∂
∂∂
=
∂∂
∂∂ ccP
yxP
xyCon ρρ
00:2 =⇒=⇒
∂∂
∂∂
=
∂∂
∂∂ bbcP
zxP
xzCon ρ
003:3 =⇒=−⇒
∂∂
∂∂
=
∂∂
∂∂ bbP
zyP
yzCon ρ
Thus c=2 and b=0 We solve simplified equations
and get
CzgzyxxyzyxzyxP ++−+−−−−−= )648422
52
13(),,( 222
ρ
Answer. CzgzyxxyzyxzyxP ++−+−−−−−= )64842
25
213(),,( 2
22
ρ
19. We consider a scalar field ),,( zrf θ given in cylindrical coordinates, where
zzryrx === , sin ,cos θθ , and basis vectors are 321 ,, eee . Find the expression for the gradient, )),,(( zrfgrad θ ), in cylindrical coordinates, that is in
terms of zff
rfeeezr
∂∂
∂∂
∂∂ and ,, , ,,,, 321 θ
θ
if
a) kejeie
=== 321 ,, ( we keep the same basis kji
,, ) .
b) kjejeie
+=== 2,2 321
c) kejeie
=== 321 , sin ,cos θθ (exam 1, 2012; Q5 B (2 points)) d) kejiejie
=+−=+= 321 ,cossin,sincos θθθθ (this is often used as a local basis for cylindrical coordinates)
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Armin Halilovic Math. Exercises
Solution: In x,y, z variables we have
kzfj
yfi
xf
zf
yf
xffgrad
∂∂
+∂∂
+∂∂
=∂∂
∂∂
∂∂
= ),,()( ( eq1)
For cylindrical coordinates we have θcosrx = , θsinry = , z=z
First we write the derivatives ,xf∂∂ and
yf∂∂ in θ,r coordinates (the variable z is in both
coord systems) .
Solving the following system for ,xf∂∂ and
yf∂∂ ,
)cos()sin(
sincos
θθθ
θθ
ryfr
xff
yf
xf
rf
⋅∂∂
+−⋅∂∂
=∂∂
⋅∂∂
+⋅∂∂
=∂∂
we get
θθθ∂∂
−∂∂
=∂∂ f
rrf
xf sincos
θθθ∂∂
+∂∂
=∂∂ f
rrf
yf cossin (**)
We substitute the derivatives (**) in ( eq1) and get
*)*(* )cos(sin)sin(cos)( kzfjf
rrfif
rrffgrad
∂∂
+∂∂
+∂∂
+∂∂
−∂∂
=θ
θθθ
θθ
To solve problems a) , b) c) and d) we must express kji
,, as a linear combinations of
321 ,, eee and substitute them into (***) . a) From (***), since 321 ,, ekejei
=== , we have immediately
321 )cos(sin)sin(cos)( ezfef
rrfef
rrffgrad
∂∂
+∂∂
+∂∂
+∂∂
−∂∂
=θ
θθθ
θθ
b) From kjejeie
+=== 2,2 321 we find
232
1 ,2
, eekejei
−=== (eq b)
Then we put kji
,, from ( eq b) into (***) and get
)(2
)cos(sin)sin(cos)( 232
1 eezfef
rrfef
rrffgrad
−∂∂
+∂∂
+∂∂
+∂∂
−∂∂
=θ
θθθ
θθ
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Armin Halilovic Math. Exercises
and, after collecting components for 321 , eee
321 )cos21sin
21()sin(cos)( e
zfe
zff
rrfef
rrffgrad
∂∂
+∂∂
−∂∂
+∂∂
+∂∂
−∂∂
=θ
θθθ
θθ
c) From
kejeie
=== 321 , sin ,cos θθ we have
321 ,
sin,
cosekejei
===θθ
( eq c)
Putting kji
,, from ( eq c) into (***) gives
sin
)cos(sincos
)sin(cos)( 321 e
zfef
rrfef
rrffgrad
∂∂
+∂∂
+∂∂
+∂∂
−∂∂
=θθ
θθθθ
θθ
d) kejiejie
=+−=+= 321 ,cossin,sincos θθθθ .
We can solve d) in the same manner as in a,b,c but this time we can just collect terms rf∂∂ ,
θ∂∂f ,
zf∂∂ and get the result:
kzfjf
rrfif
rrfk
zfj
yfi
xffgrad
∂∂
+∂∂
+∂∂
+∂∂
−∂∂
=∂∂
+∂∂
+∂∂
= )cos(sin)sin(cos)(θ
θθθ
θθ
kzfjif
rji
rf
∂∂
++−⋅∂∂
++⋅∂∂
= )cossin(1)sin(cos θθθ
θθ
3211 e
zfef
re
rf
∂∂
+⋅∂∂
+⋅∂∂
=θ
Answer:
a) 321 )cos(sin)sin(cos)( ezfef
rrfef
rrffgrad
∂∂
+∂∂
+∂∂
+∂∂
−∂∂
=θ
θθθ
θθ
b)
321 )cos21sin
21()sin(cos)( e
zfe
zff
rrfef
rrffgrad
∂∂
+∂∂
−∂∂
+∂∂
+∂∂
−∂∂
=θ
θθθ
θθ
c) sin
)cos(sincos
)sin(cos)( 321 e
zfef
rrfef
rrffgrad
∂∂
+∂∂
+∂∂
+∂∂
−∂∂
=θθ
θθθθ
θθ
d) 3211),,(( e
zfef
re
rfzrfgrad
∂∂
+⋅∂∂
+⋅∂∂
=θ
θ
20. See http://ingforum.haninge.kth.se/armin/AR_2000/HL2008/DERIV_NAVIER_STOKES.pdf 21. Solution:
yxz 231 ++= )1,2,3()1,,( −−=′−′−= yx zzN
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Armin Halilovic Math. Exercises
),,( yxyxF +=
yxyxyxNF −−=++−−= 223
4)24()2(1
0
2
0
1
0
−=−−=−−==Φ ∫ ∫ ∫∫∫ dxxdyyxdxdxdyNFD
Answer: 4−=Φ 22. Solution:
122212)( =−+= zzFdiv
∫∫∫∫∫∫ ===ΦKK
dxdydzdVFdiv 12
12× Volume(K)= ππ 43233412 3 =××
Answer: π432=Φ
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