MATH. EXERCISES. GRADIENT, DIVERGENCE, CURL DEL … · Armin Halilovic 2 Math. Exercises Using ∇...
Transcript of MATH. EXERCISES. GRADIENT, DIVERGENCE, CURL DEL … · Armin Halilovic 2 Math. Exercises Using ∇...
Armin Halilovic 1 Math. Exercises
E-mail : [email protected] webpage : www.sth.kth.se/armin MATH. EXERCISES. GRADIENT, DIVERGENCE, CURL DEL (NABLA) OPERATOR , LAPLACIAN OPERATOR , CONTINUITY AND NAVIER-STOKES EQUATIONS VECTOR PRODUCTS If ),,( 321 uuuu =r and ),,( 321 vvvv =
r then
332211 vuvuvuvu ++=•rr (scalar or dot product)
321
321
vvvuuukji
vu
rrr
rr=× (vector or cross product)
In some books is also considered outer product defined by
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛=⊗
3
2
1
uuu
vu rr )( 321 vvv =⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
332313
322212
312111
vuvuvuvuvuvuvuvuvu
GRADIENT, DIVERGENCE, CURL DEL (NABLA) OPERATOR , LAPLACIAN OPERATOR GRADIENT Let ),,( zyxϕ be a scalar field. The gradient is the vector field defined by
),,()(zyx
grad∂∂
∂∂
∂∂
=ϕϕϕϕ
DIVERGENCE Let )),,(),,,(),,,(( zyxRzyxQzyxPF =
r be a vector field, continuously differentiable with
respect to x, y and z. Then the divergence of F
r is the scalar field defined by
zR
yQ
xPFdiv
∂∂
+∂∂
+∂∂
=)(r
CURL. The curl of F
r is the vector field defined by
kyP
xQj
xR
zPi
zQ
yR
RQPzyx
kji
Fcurlrrr
rrr
r)()()()(
∂∂
−∂∂
+∂∂
−∂∂
+∂∂
−∂∂
=∂∂
∂∂
∂∂
=
),,(yP
xQ
xR
zP
zQ
yR
∂∂
−∂∂
∂∂
−∂∂
∂∂
−∂∂
=
DEL (NABLA) OPERATOR The vector differential operator
=∇ ),,(zyxz
ky
jx
i∂∂
∂∂
∂∂
=∂∂
+∂∂
+∂∂ rrr
is called del or nabla .
Armin Halilovic 2 Math. Exercises
Using ∇ we can denote grad, div and curl as below:
)(ϕgrad = ϕ∇ FFdivrr
•∇=)( FFcurlrr
×∇=)( Note that ∇•F
r is not the same as F
r•∇ .
∇•Fr
=z
Ry
Qx
P∂∂
+∂∂
+∂∂ .
LAPLACIAN OPERATOR
The Laplacian operator, 2
2
2
2
2
22
zyx ∂∂
+∂∂
+∂∂
=∇=Δ , is defined for a scalar field U(x,y,z) by
2
2
2
2
2
22
zU
yU
xUUU
∂∂
+∂∂
+∂∂
=∇=Δ ,
and for a vector field )),,(),,,(),,,(( zyxRzyxQzyxPF =r
by ),,(2 RQPFF ΔΔΔ=∇=Δ
rr.
Cylindrical coordinates ),,( zr θ : transformation: θcosrx = , θsinry = , z=z volume element: dzddrrdV θ= local basis: kejiejie zr
rrrrrrrr=+−=+= ,cossin,sincos θθθθ θ
Vector components relationship: θθρ sincos yx FFF += , θθϑ cossin yx FFF +−= , zz FF = scalar field: ),,( zrf θ
gradient: zr ezfef
re
rfffgrad rrr
∂∂
+∂∂
+∂∂
=∇= θθ1)(
laplacian: 2
2
2
2
22 11
zff
rrfr
rrff
∂∂
+∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
=Δ=∇θ
vector field: ),,( zr FFFF θ=r
divergence: z
FFrr
Frr
FFdiv zr
∂∂
+∂
∂+
∂⋅∂
=∇=θθ )(1)(1)(
vo
r
curl:
zr
zr
FFrFzr
eere
rFFcurl
θ
θ
θ⋅
∂∂
∂∂
∂∂
⋅
=×∇=
rrr
vr 1)( =
krzr
rz eF
rrF
re
rF
zFe
zFF
rrrr⎟⎠⎞
⎜⎝⎛
∂∂
−∂
∂+⎟
⎠⎞
⎜⎝⎛
∂∂
−∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
−∂∂
θθθ
θθ )(11
Armin Halilovic 3 Math. Exercises
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 4 Math. Exercises
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•∇=•∇+
∂∂ xyzyUU
trr
ϕρϕρϕ ( eq 1)
Let 1=ρ , Γ= constant , ),3,2( xyU −=r
. Find the constant Γ in the equation (eq 1) if we now that the function 222),,( zyxtzyx +++=ϕ satisfies the equation.
10. If possible, find ),( yxf for the given partial derivatives ),( yxfx∂∂ and ),( yxf
y∂∂ .
a) ),( yxfx∂∂ = xy2 and ),( yxf
y∂∂ = yx 22 + .
b) ),( yxfx∂∂ = yx +2 and ),( yxf
y∂∂ = x .
c) ),( yxfx∂∂ = xyye and ),( yxf
y∂∂ = xyxe .
d) ),( yxfx∂∂ = yx +2 and ),( yxf
y∂∂ = x5 .
( Hint: Necessary condition: If ),( yxf has continues derivatives then the mixed derivatives of ),( yxf should be equal. Thus
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
∂∂
=⎟⎠⎞
⎜⎝⎛∂∂
∂∂ yxf
yxyxf
xy,(,( (*)
is the necessary condition for the existence of a function ),( yxf that has the given derivatives.
Armin Halilovic 5 Math. Exercises
11. If possible, find ),,( zyxf for the given partial derivatives ),,( zyxfx∂∂ , ),,( zyxf
y∂∂
and ),,( zyxfz∂∂ .
a) yzzyxfx
=∂∂ ),,( , zxzzyxf
y+=
∂∂ ),,( and 23),,( zyxyzyxf
z++=
∂∂
b) yzzyxfx
=∂∂ ),,( , zxzzyxf
y+=
∂∂ ),,( 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. 12. We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field
.x,y,z, w x,y,z, vx,y,zuV ))()()(( =r
Use the following equations ( continuity and Navier Stokes equations) to find en expression for pressure P(x,y,z) as a function of x,y and z, where ρ =constant, μ =constant , ),0,0( gg −=
r i.e. xg = yg =0 and
)/81.9 where( 2smggg z ≈−= Incompressible continuity equation:
0=∂∂
+∂∂
+∂∂
zw
yv
xu eq1.
Navier Stokes equations: x component:
)( 2
2
2
2
2
2
zu
yu
xug
xP
zuw
yuv
xuu
tu
x ∂∂
+∂∂
+∂∂
++∂∂
−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂ μρρ eq2.
y component:
)( 2
2
2
2
2
2
zv
yv
xvg
yP
zvw
yvv
xvu
tv
y ∂∂
+∂∂
+∂∂
++∂∂
−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂ μρρ eq3.
Armin Halilovic 6 Math. Exercises
z component:
)( 2
2
2
2
2
2
zw
yw
xwg
zP
zww
ywv
xwu
tw
z ∂∂
+∂∂
+∂∂
++∂∂
−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂ μρρ eq4.
a) )0,24,32( yxyxV −+=r
b) )2,32,43( −−+= yxyxV
r
c) )2,4,21( xyV −+=r
13. (exam 1, 2009) A) Consider the following equation
24841616)())(()()(−++++×∇•∇+∇⋅Γ•∇=•∇+
∂∂ yzxzyxUU
t
rrϕρϕρϕ ( eq 1)
Let 2=ρ , Γ= constant , )2,4,4( yxU +=r
. Find the constant Γ in the equation (eq 1) if we now that the function 22231),,( zyxtzyx ++++=ϕ satisfies the equation. B) We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field
.x,y,z, w x,y,z, vx,y,zuV ))()()(( =r
Use the following equations ( continuity and Navier Stokes equations) to find en expression for pressure P(x,y,z) as a function of x,y and z, where ρ =constant, μ =constant ,
),0,0( gg −=r i.e. xg = yg =0 and )/81.9 where( 2smgggz ≈−= and
)22,24,46( zyxV −−+=r
. 14. (exam 2, 2009) We consider an incompressible ( density ρ =const), steady state ( variables do not depend on time), isothermal Newtonian flow with a given velocity field
.x,y,z, w x,y,z, vx,y,zuV ))()()(( =r
Use the following equations ( continuity and Navier Stokes equations) to find first i) parameter a and then ii) en expression for pressure P(x,y,z) as a function of x,y and z, where ρ =constant, μ =constant ,
),0,0( gg −=r i.e. xg = yg =0 and )/81.9 where( 2smgggz ≈−= and
)1,5,32( azyxV −−+=r
.
15. Consider steady, incompressible, isothermal, laminar stationary Newtonian flow in a long round pipe in the z-direction, with constant circular cross-section of radius R=2 m. Use the continuity and the Navier-Stokes equations in cylindrical coordinates to find the velocity field V=(ur, uθ, uz) and the pressure field P (r,θ,z) if the fluid flow satisfies the following conditions:
Armin Halilovic 7 Math. Exercises
c0. All partial derivatives with respect to time t are 0 ( Steady flow)
c1. μ=0.001 kg/(m·s) and ρ =1000 kg/m3
c2. A Constant pressure gradient ∂P/∂z = –1/250 Pa/m is applied in the horizontal axis ( z-axis in our notation): ∂P/∂z = –1/250, c3. The flow is parallel to the z axis, that is ur =0 and uθ =0. c4. We assume that the flow is axisymmetric . The velocity does not depend on θ,
that is 0=∂∂θ
zu
c5. Boundary cond. 1 ( No-slip boundary condition, Vfluid=Vwall ): If r=2 then uz= 0
c6. Boundary condition 2: uz has maximum at r=0 that is 00=
=∂∂
rruz
---------------------------------------------------------------------------------------------
The continuity and the Navier-Stokes equations for an incompressible , isothermal Newtonian flow (density ρ =const, viscosity μ =const), with a velocity field
),,( zr uuuV θ=r
in Cylindrical coordinates ),,( zr θ : Incompressible continuity equation
0)(1)(1
=∂∂
+∂
∂+
∂∂
zuu
rrru
rzr
θθ eq a)
Navier-Stokes equations in Cylindrical coordinates: r-component:
⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
−∂∂
+−⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
++∂∂
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+−∂∂
+∂∂
+∂∂
2
2
22
2
22
2
211zuu
ru
rru
rur
rrg
rP
zuu
ruu
ru
ruu
tu
rrrrr
rz
rrr
r
θθμρ
θρ
θ
θθ
eq b)
θ -component:
⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
+∂∂
+−⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
++∂∂
−=
⎟⎠⎞
⎜⎝⎛
∂∂
++∂∂
+∂∂
+∂∂
2
2
22
2
22
2111zuu
ru
rru
ru
rrr
gPr
zu
uruuu
ru
ru
ut
u
r
zr
r
θθθθθ
θθθθθθ
θθμρ
θ
θρ
eq c)
z-component:
Armin Halilovic 8 Math. Exercises
⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
++∂∂
−=
⎟⎠⎞
⎜⎝⎛
∂∂
+∂∂
+∂∂
+∂∂
2
2
2
2
211
zuu
rrur
rrg
zP
zuuu
ru
ruu
tu
zzzz
zz
zzr
z
θμρ
θρ θ
eq d)
Armin Halilovic 9 Math. Exercises
ANSWERS AND SOLUTIONS: 1. Solution:
a) Since zR
yQ
xPFdiv
∂∂
+∂∂
+∂∂
=)(r
we have ),,( 22 xzxyF +=r
⇒ xxFdiv 2002)( =++=r
. Answer a) xFdiv 2)( =
r
b) Since ),,()(zyx
grad∂∂
∂∂
∂∂
=ϕϕϕϕ we have ( for )(Fdiv
r=ϕ )
)0,0,2())(( =Fdivgradr
Answer b) )0,0,2())(( =Fdivgrad
r
c)
)1,2,1(21
)(22
−−−=−−−=
+∂∂
∂∂
∂∂
=∂∂
∂∂
∂∂
=
xkjxi
xzxyzyx
kji
RQPzyx
kji
Fcurldef
rrr
rrrrrr
r
Answer c) )1,2,1()( −−−= xFcurlr
2. Solution:
),,( 22 yxzxzyxF ++++=r
)()()(
)(22 yxzxzyx
zyx
kji
Fcurl
++++∂∂
∂∂
∂∂
=
rrr
r= kxjiz
rrr)12()11()21( −+−−−
)12,0,21( −−= xz Thus 0))(( =Fcurldiv
r and therefore )))((( Fcurldivgrad
r= 0)0,0,0(
r=
Answer: )))((( Frotdivgradr
= 0)0,0,0(r
= 3.
fΔ =0⇒
02
2
2
2
2
2
=∂∂
+∂∂
+∂∂
zf
yf
xf
Answer: The function )ln(),,( 222 zyxzyxf ++= satisfies the Laplace equation.
4. Answer: )))(( ff ∇×∇•∇+Δ = )))(( gradfcurldivf +Δ = 26 +x 5. Solution:
Armin Halilovic 10 Math. Exercises
⇒+Γ=+∂
∂φϕρϕρϕ SgraddivUdiv
t)()()( r
⇒+∂∂
Γ∂∂
Γ∂∂
Γ=+∂
∂φ
ϕϕϕρϕρϕρϕρϕ Szyx
divwvudivt
),,(),,()(
φϕϕϕρϕρϕρϕρϕ Szzyyxxz
wy
vx
ut
+⎟⎠⎞
⎜⎝⎛
∂∂
Γ∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
Γ∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
Γ∂∂
=∂
∂+
∂∂
+∂
∂+
∂∂ ))()()()(
6. Which one (if any) of the following functions a) zyxzyx ++= 24
1 ),,(ϕ
b) zyxzyx ++= 222 ),,(ϕ
c) 2223 ),,( zyxzyx ++=ϕ
satisfies the equation SU +∇⋅Γ•∇=•∇ ))(()( ϕϕ
r ?
Here 5=Γ , )3,2,1(=Ur
and 2342 −+= yxS . Solution : The equation
SU +∇⋅Γ•∇=•∇ ))(()( ϕϕr
can be written as
1.) (eq 234255532
2342)5,5,5()3,2,(
)()(
2
2
2
2
2
2
−++∂∂
+∂∂
+∂∂
=∂∂
+∂∂
+∂∂
⇒−++∂∂
∂∂
∂∂
=
⇒+Γ=
yxzyxzyx
yxzyx
divdiv
SgraddivUdiv
ϕϕϕϕϕϕ
ϕϕϕϕϕϕ
ϕϕr
a) zyxzyx ++== 241 ),,(Let ϕϕ
Vi calculate the derivatives of 1ϕ and substitute in the left hand side (LHS) and right hand side of the equation (eq1).
LHS: 34432 3 ++=∂∂
+∂∂
+∂∂ yx
zyxϕϕϕ
RHS= 1342602342555 22
2
2
2
2
2
−++=−++∂∂
+∂∂
+∂∂ yxx yx
zyxϕϕϕ
Whence RHSLHS ≠ Thus the function zyxzyx ++= 24
1 ),,(ϕ is not a solution to the equation b) zyxzyx ++== 22
2 ),,(ϕϕ
Armin Halilovic 11 Math. Exercises
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=+∂
∂ Szyx
divwvudivt
),,(),,()( ϕϕϕρϕρϕρϕρϕ
Szzyyxxz
wy
vx
ut
+⎟⎠⎞
⎜⎝⎛
∂∂
Γ∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
Γ∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
Γ∂∂
=∂
∂+
∂∂
+∂
∂+
∂∂ ϕϕϕρϕρϕρϕρϕ ))()()()( (eq2)
B) We substitute 2=ρ , 3=Γ , )4,2,1(=U
r and 32),,( zyxzyx ++=ϕ
in the equation (eq2) and get
Szzyyxxzyx
+⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
=∂
∂+
∂∂
+∂
∂+
ϕϕϕφϕϕ 333))8()4()2(0
Szzy +++=+++ 186024820 2 . Consequently
Armin Halilovic 12 Math. Exercises
2241884 zzyS +−+−= 9. (Q6, exam 2, 2008) Consider the following equation
426)())(()()(−−+×∇•∇+∇⋅Γ•∇=•∇+
∂∂ xyzyUU
trr
ϕρϕρϕ ( eq 1)
Let 1=ρ , Γ= constant , ),3,2( xyU −=r
. Find the constant Γ in the equation (eq 1) if we now that the function 222),,( zyxtzyx +++=ϕ satisfies the equation. Solution:
426)())(()()(−−+×∇•∇+∇⋅Γ•∇=•∇+
∂∂ xyzyUU
trr
ϕρϕρϕ ⇒
⇒−−++Γ=+∂
∂ 426))(()()()( xyzyUcurldivgraddivUdivt
rrϕρϕρϕ
(since )0,,()( yxUcurl −=r
we have 011))(( =+−=Ucurldivr
)
⇒−−++∂∂
Γ∂∂
Γ∂∂
Γ=+∂
∂ 4260),,(),,()( xyzyzyx
divwvudivt
ϕϕϕρϕρϕρϕρϕ
4260))()()()(−−++⎟
⎠⎞
⎜⎝⎛
∂∂
Γ∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
Γ∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
Γ∂∂
=∂
∂+
∂∂
+∂
∂+
∂∂ xyzy
zzyyxxzw
yv
xu
tϕϕϕρϕρϕρϕρϕ (eq2)
We substitute 1=ρ , , ),3,2( xyU −=r
and 222),,( zyxtzyx +++=ϕ in the equation (eq2) and get
426))()3()2()1(−−+⎟
⎠⎞
⎜⎝⎛
∂∂
Γ∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
Γ∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
Γ∂∂
=∂−∂
+∂
∂+
∂∂
+∂
∂ xyzyzzyyxxz
xyyxt
ϕϕϕφϕϕϕ
( Note that Γ is a constant)
248
4262202622
=Γ⇒Γ=
⇒−−+Γ+Γ+=−++ xyzyxyzy
Answer: 2=Γ
10. If possible, find ),( yxf for the given partial derivatives ),( yxfx∂∂ and ),( yxf
y∂∂ .
a) ),( yxfx∂∂ = xy2 and ),( yxf
y∂∂ = yx 22 + .
b) ),( yxfx∂∂ = yx +2 and ),( yxf
y∂∂ = x .
c) ),( yxfx∂∂ = xyye and ),( yxf
y∂∂ = xyxe .
d) ),( yxfx∂∂ = yx +2 and ),( yxf
y∂∂ = x5 .
( Hint: Necessary condition: If ),( yxf has continuous derivatives then
Armin Halilovic 13 Math. Exercises
the mixed derivatives of ),( yxf should be equal, i.e.
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
∂∂
=⎟⎠⎞
⎜⎝⎛∂∂
∂∂ ),(),( yxf
yxyxf
xy (*)
is the necessary condition for the existence of a function ),( yxf that has the given derivatives. Answer: a) ),( yxf = Cyyx ++ 22 b) ),( yxf = Cxyx ++2 c) ),( yxf = Ce xy + d) No solution since the condition (*) is not fulfilled,
5),(),(1 =⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
∂∂
≠⎟⎠⎞
⎜⎝⎛∂∂
∂∂
= yxfyx
yxfxy
.
Solution a)
Since ⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
∂∂
=⎟⎠⎞
⎜⎝⎛∂∂
∂∂ ),(),( yxf
yxyxf
xy =2x and the derivatives are continuous the
condition (*) is fulfilled and we can find ),( yxf for the given derivatives. In order to find ),( yxf we integrate with respect to x the first of the equations
),( yxfx∂∂ = xy2 (eq1)
),( yxfy∂∂ = yx 22 + (eq2).
and get
∫ +== )(2),( 1́2 yCyxxydxyxf
Thus )(),( 1́
2 yCyxyxf += ( i ) We have integrated with respect to x, therefore the constant still depend on y. Now, to find )(1́ yC we differentiate and substitute (i) in (eq2) and get:
( ))(1́2 yCyx
y+
∂∂ = yx 22 + ⇒ ( ))(1́
2 yCy
x∂∂
+ = yx 22 + ⇒
( ))(1́ yCy∂∂ = y2 ⇒ )(1́ yC = Cy +2 .
Finally, substituting )(1́ yC = Cy +2 in (i) we have Cyyxyxf ++= 22),( (where C is a constant).
11. Answer: a) Czyzxyzf +++= 3 b) Cyzxyf ++= c) Cef xyz += d) No solution since the condition 2Con is not fulfilled,
zyfzx
fxz
y +=⎟⎠⎞
⎜⎝⎛∂∂
∂∂
≠⎟⎠⎞
⎜⎝⎛∂∂
∂∂
=
Solution a)
Armin Halilovic 14 Math. Exercises
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
CzzC
zzCz
zyxyzCz
yxy
zyxyzCyzxyzz
+=
=∂∂
⇒++=∂∂
++
⇒++=++∂∂
32
22
22
22
)(
3))((
3))((
3))((
Finally, substituting CzzC += 3
2 )( in (ii) we have Czyzxyzzyxf +++= 3),,( (where C is a constant).
Calculation of the pressure field for a known velocity field for an incompressible, steady state, isothermal Newtonian flow. 12. Answer:
Armin Halilovic 15 Math. Exercises
a) CyxgzP +−−−= 22 88 ρρρ
b) CyxgzP +−−−= 22
217
217 ρρρ
c) CyxgzP +++−= 22 44 ρρρ Solution a) We substitute 0,24,32 =−=+= wyxvyxu in eq1,2,3,4 and get ( note that al derivatives with respect to t are 0): Continuity equation: 00 = eq1i. ( identically fulfilled) Navier Stokes equations: x component:
xPx∂∂
−=ρ16 eq2i.
y component:
yPy∂∂
−=ρ16 eq3i.
z component:
gzP ρ−∂∂
−=0 eq4i.
Now eq2i. gives ),(8),,( 1́2 zyCxzyxP +−= ρ (*) .
Substitution in eq3i. implies
)(8),(
),(16
2´2
1́
1́
zCyzyCy
zyCy
+−=
⇒∂
∂−=
ρ
ρ
Hence, from (*) we have )(88),,( 2´
22 zCyxzyxP +−−= ρρ (**) Now we substitute (**) in eq4i. and get
gzCz
gzP ρρ −
∂∂
−=⇒−∂∂
−= ))((00 2´
CgzzC +−=⇒ ρ)(2´ ( where C is a constant) Finally, substituting CgzzC +−= ρ)(2´ in (**) we have
CgzyxzyxP +−−−= ρρρ 22 88),,( (where C is a constant). 13. Solution A:
24841616)())(()()(−++++×∇•∇+∇⋅Γ•∇=•∇+
∂∂ yzxzyxUU
t
rrϕρϕρϕ ⇒
⇒−+++++Γ=+∂
∂ 24841616))(()()()( yzxzyxUcurldivgraddivUdivt
rrϕρϕρϕ
(since )0,1,2()( −=Ucurlr
we have 0))(( =Ucurldivr
)
Armin Halilovic 16 Math. Exercises
⇒−++++∂∂
Γ∂∂
Γ∂∂
Γ=+∂
∂ 24841616),,(),,()( yzxzyxzyx
divwvudivt
ϕϕϕρϕρϕρϕρϕ
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∂∂
=∂+∂
+∂
∂+
∂∂
+∂
∂yzxzyx
zzyyxxzyx
yxtϕϕϕφϕϕϕ
( Note that Γ is a constant)
5630
2484161668461166
=Γ⇒Γ=
⇒−++++Γ=++++ yzxzyxyzxzyx
Answer A: 5=Γ Solution B: We substitute zwyvxu 22,24,46 −=−=+= in eq1,2,3,4 and get ( note that al derivatives with respect to t are 0): Continuity equation: 00 = eq1i. ( identically fulfilled) Navier Stokes equations: x component:
xPx∂∂
−=+ )2416(ρ eq2i.
y component:
yPy∂∂
−=− )84(ρ eq3i.
z component:
gzPz ρρ −∂∂
−=− )44( eq4i.
Now eq2i. gives ),()248(),,( 1́2 zyCxxzyxP +−−= ρ (*) .
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
Armin Halilovic 17 Math. Exercises
CgzzzyyxxzyxP +−+−++−+−−= ρρρρ )42()82()248(),,( 222 Answer B:
CgzzzyyxxzyxP +−+−+−−−= )4282248(),,( 222ρ (where C is a constant). 14. Solution )1,5,32( azyxV −−+=r
First we substitute azwyvxu −=−=+= 1,5,32 in eq1 and get ( note that al derivatives with respect to t are 0): Continuity equation: 2013 =⇒=−− aa No we have )21,5,32( zyxV −−+=
r
Using the Navier Stokes equations we get: x component:
xPx∂∂
−=+ )69(ρ eq2i.
y component:
yPy∂∂
−=− )5(ρ eq3i.
z component:
gzPz ρρ −∂∂
−=− )24( eq4i.
Now eq2i. gives ),()629(),,( 1́
2
zyCxxzyxP +−−
= ρ (*) .
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).
Armin Halilovic 18 Math. Exercises
Q15. Consider steady, incompressible, isothermal, laminar stationary Newtonian flow in a long round pipe in the z-direction, with constant circular cross-section of radius R=2 m. Use the continuity and the Navier-Stokes equations in cylindrical coordinates to find the velocity field V=(ur, uθ, uz) and the pressure field P (r,θ,z) if the fluid flow satisfies the following conditions:
c0. All partial derivatives with respect to time t are 0 ( Steady flow)
c1. μ=0.001 kg/(m·s) and ρ =1000 kg/m3
c2. A Constant pressure gradient ∂P/∂z = –1/250 Pa/m is applied in the horizontal axis ( z-axis in our notation): ∂P/∂z = –1/250, c3. The flow is parallel to the z axis, that is ur =0 and uθ =0. c4. We assume that the flow is axisymmetric . The velocity does not depend on θ,
that is 0=∂∂θ
zu
c5. Boundary cond. 1 ( No-slip boundary condition, Vfluid=Vwall ): If r=2 then uz= 0
c6. Boundary condition 2: uz has maximum at r=0 that is 00=
=∂∂
rruz
The continuity and the Navier-Stokes equations for an incompressible , isothermal Newtonian flow (density ρ =const, viscosity μ =const), with a velocity field
),,( zr uuuV θ=r
in Cylindrical coordinates ),,( zr θ : ---------------------------------------------------------------- SOLUTION Incompressible continuity equation
0)(1)(1
=∂∂
+∂
∂+
∂∂
zuu
rrru
rzr
θθ eq a)
Navier-Stokes equations in Cylindrical coordinates: r-component:
⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
−∂∂
+−⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
++∂∂
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+−∂∂
+∂∂
+∂∂
2
2
22
2
22
2
211zuu
ru
rru
rur
rrg
rP
zuu
ruu
ru
ruu
tu
rrrrr
rz
rrr
r
θθμρ
θρ
θ
θθ
eq b)
θ -component:
⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
+∂∂
+−⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
++∂∂
−=
⎟⎠⎞
⎜⎝⎛
∂∂
++∂∂
+∂∂
+∂∂
2
2
22
2
22
2111zuu
ru
rru
ru
rrr
gPr
zu
uruuu
ru
ru
ut
u
r
zr
r
θθθθθ
θθθθθθ
θθμρ
θ
θρ
eq c)
Armin Halilovic 19 Math. Exercises
z-component:
⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
++∂∂
−=
⎟⎠⎞
⎜⎝⎛
∂∂
+∂∂
+∂∂
+∂∂
2
2
2
2
211
zuu
rrur
rrg
zP
zuuu
ru
ruu
tu
zzzz
zz
zzr
z
θμρ
θρ θ
eq d)
We choose x as a vertical axis, y an z are in a horizontal plane and the flow is parallel with the z-axis. We denote velocity vector V=(ur, uθ, uz) where ur, uθ and uz are r-component, θ-component and z-component in cylindrical coordinates. According to the assumptions we have ur =0, uθ = 0, and uz does not depend on θ. Since x is the vertical axis we have that vector g=(-g, 0,0) where g=9,81 m/s2 which in cylindrical coordinates gives
θcosgg r −= , θθ singg = and 0=zg
Now we substitute ∂P/∂z = –1/250 Pa/m, μ=0.001kg /(ms) in the continuity and Navier-Stokes equations:
Since ur =0 and uθ =0 (according to c3), continuity equation in cylindrical coordinates
0)(1)(1=
∂∂
+∂∂
+∂
∂zuu
rrru
rzr
θθ
gives
0=∂∂
zuz .
Armin Halilovic 20 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 those equations we get
CgrzP +−−= θρ cos2501
Armin Halilovic 21 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