Armin Halilovic
Math. Exercises
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 u = (u1 , u2 , u3 ) and v = ( v1 , v2 , v3 ) then
 
(scalar or dot product)
u • v = u1v1 + u2 v2 + u3v3



i
j
k
 
(vector or cross product)
u × v = u1
u2
u3
v1
v2
v3
In some books is also considered outer product defined by
 u1 
   
u ⊗ v =  u2  ( v1
u 
 3
v2
 u1v1

v3 ) =  u2 v1
u v
 3 1
u1v2
u2 v2
u3 v 2
u1v3 

u2 v3 
u3v3 
GRADIENT, DIVERGENCE, CURL
DEL (NABLA) OPERATOR , LAPLACIAN OPERATOR
GRADIENT
Let ϕ ( x, y , z ) be a scalar field.
The gradient is the vector field defined by
∂ϕ ∂ϕ ∂ϕ
grad (ϕ ) = ( ,
, )
∂x ∂y ∂z
DIVERGENCE

Let F = ( P( x, y , z ), Q ( x, y , z ), R( x, y , z )) be a vector field, continuously differentiable with
respect to x, y and z.

Then the divergence of F is the scalar field defined by

∂P ∂Q ∂R
+
+
div ( F ) =
∂x ∂y ∂z
CURL.

The curl of F is the vector field defined by



i
j
k

∂
∂
∂
∂R ∂Q  ∂P ∂R  ∂Q ∂P 
curl ( F ) =
=( −
)i + ( − ) j + (
− )k
∂x ∂y ∂z
∂y ∂z
∂z ∂x
∂x ∂y
P Q R

∂R ∂Q ∂P ∂R ∂Q ∂P
or
−
− )
curl ( F ) = ( −
,
,
∂y ∂z ∂z ∂x ∂x ∂y
DEL (NABLA) OPERATOR
The vector differential operator
 ∂  ∂  ∂
∂ ∂ ∂
∇=i
=( , , )
+ j
+k
∂x
∂y
∂z
∂x ∂y ∂z
is called del or nabla .
1 / 27
Armin Halilovic
Math. Exercises
Using ∇ we can denote grad, div and curl as below:
grad (ϕ ) = ∇ϕ


div ( F ) = ∇ • F


curl ( F ) = ∇ × F


Note that F • ∇ is not the same as ∇ • F .

∂
∂
∂
F •∇=P
+Q
+R .
∂x
∂y
∂z
LAPLACIAN OPERATOR
∂2
∂2
∂2
The Laplacian operator, ∆ = ∇ 2 = 2 + 2 + 2 , is defined for a scalar field U(x,y,z) by
∂x
∂y
∂z
∂ 2U ∂ 2U ∂ 2U
∆U = ∇ U = 2 + 2 + 2 ,
∂y
∂x
∂z

and for a vector field F = ( P ( x, y , z ), Q ( x, y , z ), R( x, y , z )) by


∆F = ∇ 2 F = ( ∆P, ∆Q , ∆R ) .
2
Some formulas for polar and cylindrical coordinates
Polar coordinates ( 2 dim)
F
eϑ
er
j
eϑ
j
er
P
r
ϑ
ϑ
i
i
transformation: x = r cos θ ,
y = r sin θ ,
area element: dA = r dr dθ

 



standard basis: er = cos θ i + sin θ j , eθ = − sin θ i + cos θ j ,


[Remark 1 : Note that er , eθ 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 F = Fx i + Fy j in Cartesian coord. and F = Fr er + Fϑ eθ the same vector in polar
coordinates then
Fr = Fx cos θ + Fy sin θ , Fϑ = − Fx sin θ + Fy cos θ .

[Remark 2: Vi can derive these formulas by calculating the components of F in the


directions of er and eθ . Thus
2 / 27
Armin Halilovic
Math. Exercises
 




Fr = F • er = ( Fx i + Fy j ) • (cos θ i + sin θ j ) = Fx cos θ + Fy sin θ , similarly
 




Fϑ = F • eϑ = ( Fx i + Fy j ) • (− sin θ i + cos θ j ) = − Fx sin θ + Fy cos θ . ]
Cylindrical coordinates (r ,θ , z ) :
transformation: x = r cos θ ,
y = r sin θ ,
volume element: dV = r dr dθ dz
z=z
ez =k
z
F
eϑ
er
y
ϑ
r
x


 

 

standard basis: er = cos θ i + sin θ j , eθ = − sin θ i + cos θ j , e z = k








If F = Fx i + Fy j + Fz k in Cartesian coord. and F = Fr er + Fϑ eθ + Fz k the same vector in
cylindrical coordinates then we have following vector components relationship:
Fρ = Fx cos θ + Fy sin θ , Fϑ = − Fx sin θ + Fy cos θ , Fz = Fz
[Remark 3: For example, we can get Fρ = Fx cos θ + Fy sin θ in the following way:

 




Fr = F • er = ( Fx i + Fy j + Fz k ) • (cos θ i + sin θ j ) = Fx cos θ + Fy sin θ ]
scalar field: f (r ,θ , z )
∂f  1 ∂f  ∂f 
er +
eθ + e z
r ∂θ
∂r
∂z
2
1 ∂  ∂f  1 ∂ f ∂ 2 f
laplacian: ∇ 2 f = ∆f =
+
r  +
r ∂r  ∂r  r 2 ∂θ 2 ∂z 2

vector field: F = ( Fr , Fθ , Fz )

 1 ∂ (r ⋅ Fr ) 1 ∂ ( Fθ ) ∂Fz
divergence: div( F ) = ∇  F =
+
+
r ∂θ
r
∂z
∂r



er r ⋅ eθ e z

 1 ∂
∂
∂
curl: curl ( F ) = ∇ × F =
=
r ∂r
∂θ
∂z
Fr r ⋅ Fθ Fz
gradient: grad ( f ) = ∇f =
 1 ∂Fz ∂Fθ
−

∂z
 r ∂θ
   ∂F ∂F
e r +  r − z
∂r
 ∂z

  1  ∂ (rFθ ) ∂Fr
−
eθ + 
r  ∂r
∂θ

3 / 27

e k

Armin Halilovic
Math. Exercises
EXERCISES
1. Find


a) div(F ) ,
b) grad (div( F )) and

if F = ( y + x 2 , z, x 2 )

c) curl (F )


2. Find grad (div(curl ( F ))) if F = ( x + y + z , x 2 + z 2 , x + y )
3. Which one of the following functions
a) f1 ( x, y , z ) = 3x 2 + 2 y 2 + z 2
b) f 2 ( x, y , z ) = x + ln( y 2 + z 2 )
c) f 3 ( x, y , z ) = x 3 + exp( y + z )
satisfies the Laplace equation
∆f =0?
4. Find ∆f + ∇ • (∇ × (∇f
if f ( x, y, z ) = x 3 + y 2 + z .
)))
5. Write the general transport equation

∂ ( ρϕ )
+ ∇ • ( ρϕU ) = ∇ • ( Γ ⋅ ∇ϕ ) + Sφ
∂t
without using operators div, ∇ , ∆ , curl or grad.

Here U = (u , v, w) . Functions ρ , ϕ , Γ, S , u, v, w are real functions of t, x, y and z.
6. Which one, if any, of the following functions
a) ϕ1 ( x, y, z ) = x + y + z
4
2
ϕ 2 ( x, y , z ) = x 2 + y 2 + z
2
2
2
c) ϕ 3 ( x, y, z ) = x + y + z
b)
satisfies the equation

∇ • (ϕU ) = ∇ • ( Γ ⋅ (∇ϕ )) + S
?

Here Γ = 5 , U = (1, 2, 3) and S = 2 x + 4 y − 23 .
7. Find which one (if any) of the following functions
a) ϕ1 ( x, y, z ) = x + y + z
2
2
2
ϕ 2 ( x, y , z ) = x 2 + y 2 + 5 z
2
2
2
c) ϕ 3 ( x, y, z ) = 5 x + y + z
b)
satisfies the equation

∂ ( rϕ )
+ div( rϕU ) = div(Γgradϕ ) + S
∂t

where ρ=3, Γ = 2 , U = (2, 3, 4) and S = 12 x + 18 y + 52 .
4 / 27
Armin Halilovic
Math. Exercises
8. (exam 1, 2008)
A) Write the general transport equation

∂ ( ρϕ )
( eq 1)
+ ∇ • ( ρϕU ) = ∇ • ( Γ ⋅ (∇ϕ )) + S
∂t
without using operators div, ∇ , ∆ , curl or grad.

Here U = (u , v, w) . Functions ρ , ϕ , Γ, S , u, v, w are real functions of t, x, y and z.

B) Let ρ = 2 , Γ = 3 , U = (1, 2, 4) .
Find S in the equation (eq 1) if we now that the function
ϕ ( x, y , z ) = x + y 2 + z 3
satisfies the equation.
9 (Q6, exam 2, 2008)
Consider the following equation


∂ ( ρϕ )
( eq 1)
+ ∇ • ( ρϕU ) = ∇ • ( Γ ⋅ (∇ϕ )) + ∇ • (∇ × U ) + 6 y − 2 xyz − 4
∂t

Let ρ = 1 , Γ = constant , U = ( 2, 3, − xy ) .
Find the constant Γ in the equation (eq 1) if we now that the function
ϕ ( x , y , z ) = 2t + x + y 2 + z 2
satisfies the equation.
10. If possible, find f ( x, y ) for the given partial derivatives
∂
∂
f ( x, y ) and
f ( x, y ) .
∂x
∂y
∂
∂
f ( x, y ) = 2 xy and
f ( x, y ) = x 2 + 2 y .
∂x
∂y
∂
∂
b)
f ( x, y ) = 2 x + y and
f ( x, y ) = x .
∂x
∂y
∂
∂
c)
f ( x, y ) = ye xy and
f ( x, y ) = xe xy .
∂x
∂y
∂
∂
d)
f ( x, y ) = 2 x + y and
f ( x, y ) = 5 x .
∂x
∂y
( Hint: Necessary condition: If f ( x, y ) has continues derivatives then
the mixed derivatives of f ( x, y ) should be equal.
Thus

∂ ∂
 ∂  ∂
(*)
f ( x, y  = 
f ( x, y 

∂y  ∂x
 ∂x  ∂y

is the necessary condition for the existence of a function f ( x, y ) that has the given
derivatives.
11. Determine the value of a for which the system of partial differential equations
a)
∂
∂
f ( x, y ) = axy + y and
f ( x, y ) = x 2 + x .
∂x
∂y
has solutions. Then find f ( x, y ) corresponding to this value of a.
5 / 27
Armin Halilovic
Math. Exercises
12. If possible, find f ( x, y, z ) for the given partial derivatives
∂
∂
f ( x, y , z ) ,
f ( x, y , z )
∂x
∂y
∂
f ( x, y , z ) .
∂z
∂
∂
∂
a)
f ( x, y, z ) = xy + y + 3 z 2
f ( x, y, z ) = yz ,
f ( x, y, z ) = xz + z and
∂x
∂z
∂y
∂
∂
∂
b)
f ( x, y, z ) = xy + y + 3 z 2
f ( x, y, z ) = yz ,
f ( x, y, z ) = xz + z and
∂z
∂x
∂y
∂
∂
∂
c)
f ( x, y, z ) = yze xyz ,
f ( x, y, z ) = xye xyz
f ( x, y, z ) = xze xyz and
∂x
∂z
∂y
∂
∂
∂
d)
f ( x, y, z ) = yz ,
f ( x, y, z ) = xy + y + xz
f ( x, y, z ) = xz + z and
∂x
∂z
∂y
( Hint: Necessary condition: If f ( x, y, z ) has continuous derivatives then
the mixed derivatives of f ( x, y ) should be equal.
Thus
∂ ∂  ∂  ∂ 
Con1 :
f= 
f

∂y  ∂x  ∂x  ∂y 
and
∂ ∂  ∂ ∂ 
f= 
f

∂z  ∂x  ∂x  ∂z 
∂  ∂  ∂ ∂ 

Con3 :
f= 
f
∂z  ∂y  ∂y  ∂z 
are the necessary condition for the existence of a function f ( x, y ) that has the given
derivatives.
Con 2 :
13. Determine the values of a and b for which the system of partial differential equations
∂
∂
∂
f ( x, y, z ) = ax 2 yz + 2 x ,
f ( x, y, z ) = bx 3 y + 2 z
f ( x, y, z ) = x 3 z + 1 and
∂z
∂x
∂y
has solutions. Then find f ( x, y, z ) 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

V = (u ( x,y,z ) , v( x,y,z ) , w ( x,y,z )).
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 , g = (0,0,− g ) i.e. g x = g y =0 and
g z = −g
( where g ≈ 9.81m / s 2 )
Incompressible continuity equation:
∂u ∂v ∂w
eq1.
+
+
=0
∂x ∂y ∂z
Navier Stokes equations:
x component:
6 / 27
Armin Halilovic
Math. Exercises
 ∂u
∂u
∂u
∂u 
∂P
∂ 2u ∂ 2u ∂ 2u
+u
+v
+ w  = −
+ ρg x + µ ( 2 + 2 + 2 )
∂x
∂y
∂z 
∂x
∂x
∂y
∂z
 ∂t
y component:
 ∂v
∂v
∂v
∂v 
∂P
∂ 2v ∂ 2v ∂ 2v
ρ  + u + v + w  = −
+ ρg y + µ ( 2 + 2 + 2 )
∂x
∂y
∂z 
∂y
∂x
∂y
∂z
 ∂t
z component:
 ∂w
∂P
∂2w ∂2w ∂2w
∂w
∂w
∂w 
+ ρg z + µ ( 2 + 2 + 2 )
+u
+v
+ w  = −
ρ 
∂z
∂x
∂y
∂z 
∂x
∂y
∂z
 ∂t

a) V = (2 x + 3 y, 4 x − 2 y, 0)

b) V = (3 x + 4 y, 2 x − 3 y, − 2)

c) V = (1 + 2 y, − 4 x, 2)
ρ 
eq2.
eq3.
eq4.
15. (exam 1, 2009)
A) Consider the following equation


∂ ( ρϕ )
+ ∇ • ( ρϕU ) = ∇ • (Γ ⋅ (∇ϕ )) + ∇ • (∇ × U ) + 16 x + 16 y + 4 xz + 8 yz − 24
∂t

Let ρ = 2 , Γ = constant , U = (4, 4, x + 2 y ) .
Find the constant Γ in the equation (eq 1) if we now that the function
( eq 1)
ϕ ( x, y, z ) = 1 + 3t + x 2 + y 2 + z 2
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

V = (u ( x,y,z ) , v( x,y,z ) , w ( x,y,z )).
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 ,

g = (0,0,− g ) i.e. g x = g y =0 and g z = − g ( where g ≈ 9.81m / s 2 ) and

V = (6 + 4 x, 4 − 2 y , 2 − 2 z ) .
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

V = (u ( x,y,z ) , v( x,y,z ) , w ( x,y,z )).
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 ,

g = (0,0,− g ) i.e. g x = g y =0 and g z = − g ( where g ≈ 9.81m / s 2 ) and

V = (2 + 3 x, 5 − y, 1 − az ) .
7 / 27
Armin Halilovic
Math. Exercises
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
z-axis in our notation):
∂P/∂z = –1/250,
is applied in the horizontal axis (
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,
∂u z
that is
=0
∂θ
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
∂u z
=0
∂r r = 0
--------------------------------------------------------------------------------------------The continuity and the Navier-Stokes equations for an incompressible , isothermal
Newtonian flow (density ρ =const, viscosity µ =const), with a velocity field

V = (u r , uθ , u z ) in Cylindrical coordinates (r , θ , z ) :
Incompressible continuity equation
1 ∂ (ru r ) 1 ∂ (uθ ) ∂u z
eq a)
+
+
=0
r ∂r
r ∂θ
∂z
Navier-Stokes equations in Cylindrical coordinates:
r-component:
 ∂u
∂u u ∂u u 2
∂u 
r  r + u r r + θ r − θ + u z r 
∂r
∂z 
r ∂θ
r
 ∂t
 1 ∂  ∂u r  u r 1 ∂ 2 u r 2 ∂uθ ∂ 2 u r 
∂P
=−
+ rg r + µ 
+ 2 
−
r
− 2 + 2
∂r
∂z 
r ∂θ 2 r 2 ∂θ
 r ∂r  ∂r  r
θ -component:
∂u
∂u 
u ∂u
uu
 ∂u
r θ + ur θ + θ θ + r θ + u z θ 
∂r
∂z 
r ∂θ
r
 ∂t
 1 ∂  ∂uθ  uθ
1 ∂P
1 ∂ 2 uθ
2 ∂u r ∂ 2 uθ 
=−
+ rg θ + µ 
+
+
r
− 2 + 2

r ∂θ
∂z 2 
r ∂θ 2 r 2 ∂θ
 r ∂r  ∂r  r
8 / 27
eq b)
eq c)
Armin Halilovic
Math. Exercises
z-component:
u ∂u
∂u 
∂u
 ∂u
r  z + ur z + θ z + u z z 
∂r
r ∂θ
∂z 
 ∂t
 1 ∂  ∂u z  1 ∂ 2 u z ∂ 2 u z 
∂P
=−
+ rg z + µ 
+ 2 
r
+ 2
2
∂z
∂z 
 r ∂r  ∂r  r ∂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

V = (u ( x,y,z ) , v( x,y,z ) , w ( x,y,z )) = (3 x + cy, 4 + 2 x − y + bz , 3 − az )
Use the following equations ( continuity and Navier Stokes equations) , where

ρ =constant, µ =constant , g = (0,0,− g ) i.e. g x = g y =0 and g z = − g ( g ≈ 9.81m / s 2 )
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
grad ( f ) = (
∂f ∂f ∂f
∂f  ∂f  ∂f 
j+ k
i+
, , )=
∂y
∂z
∂x ∂y ∂z
∂x
(*) .
  
If we change variables x, v, z to u, v , w and replace basis vectors i , j , k with new ( linearly
  
independent) vectors e1 , e2 , e3 then we can express the same gradient vector grad ( f ) in
  
terms of variables u, v , w and vectors e1 , e2 , e3 .
  
∂f ∂f
∂f
We simple calculate the derivatives
in new variables and express i , j , k as
and
∂x ∂y
∂z
  
a linear combinations of e1 , e2 , e3 . Then we substitute those values into (*).
(See the following example.)
17 We consider a scalar field f (r , θ , z ) given in cylindrical coordinates, where
x = r cos θ ,
y = r sin θ , z = z , and basis vectors are e1 , e2 , e3 .
Find the expression for the gradient, grad ( f (r , θ , z )) ), in cylindrical coordinates, that is in
 
 ∂f ∂f
∂f
,
and
if
∂r ∂θ
∂z
terms of r ,θ , z , e1 , e2 , e2 ,
  
  
a) e1 = i , e2 = j , e3 = k
  
( we keep the same basis i , j , k ) .
9 / 27
Armin Halilovic
Math. Exercises
 
 

e2 = 2 j , e3 = 2 j + k
b)
 
e1 = i
c)
 
e1 = i cos θ ,
 
 
e2 = j sin θ , e3 = k
(exam 1, 2012; Q5 B (2 points))

 

  

d) e1 = cos θ i + sin θ j , e2 = − sin θ i + cos θ j , e3 = k (this is often used as a local basis
for cylindrical coordinates)
ANSWERS AND SOLUTIONS:
1. Solution:

∂P ∂Q ∂R
Since div( F ) =
+
+
∂x ∂y ∂z


we have F = ( y + x 2 , z, x 2 ) ⇒ div( F ) = 2 x + 0 + 0 = 2 x .

Answer a) div ( F ) = 2 x

b) Since grad (ϕ ) = ( ∂ϕ , ∂ϕ , ∂ϕ ) we have ( for ϕ = div(F ) )
∂x ∂y ∂z

grad ( div( F )) = ( 2,0,0)

Answer b) grad ( div( F )) = ( 2,0,0)
a)




i
i
j
k
 def ∂
∂
∂
∂
c) curl ( F ) = ∂x ∂y ∂z = ∂x
P
Q
R
y + x2

 
= −1i − 2 xj − k = ( −1,−2 x,−1)

j
∂
∂y
z

k
∂
∂z
x2

Answer c) curl ( F ) = ( −1,−2 x,−1)
2. Solution:

F = ( x + y + z, x 2 + z 2 , x + y)



i
j
k




∂
∂
∂
= (1 − 2 z )i − (1 − 1) j + (2 x − 1)k
curl ( F ) =
∂x
∂y
∂z
( x + y + z) ( x 2 + z 2 ) ( x + y)
= (1 − 2 z, 0, 2 x − 1)



Thus div(curl ( F )) = 0 and therefore grad (div(curl ( F ))) = (0,0,0) = 0


Answer: grad (div(rot ( F ))) = (0,0,0) = 0
10 / 27
Armin Halilovic
Math. Exercises
3.
∆f =0 ⇒
∂2 f ∂2 f ∂2 f
+
+
=0
∂x 2 ∂y 2 ∂z 2
Answer: The function f 2 ( x, y, z ) = x + ln( y 2 + z 2 ) satisfies the Laplace equation.
4. Answer:
∆f + ∇ • (∇ × (∇f ))) = ∆f + div( curl ( gradf ))) = 6 x + 2
5. Solution:

∂ ( ρϕ )
+ div( ρϕU ) = div(Γgρadϕ ) + S φ ⇒
∂t
∂ ( ρϕ )
∂ϕ
∂ϕ
∂ϕ
,Γ
, Γ ) + Sφ ⇒
+ div( ρϕu , ρϕv, ρϕw) = div(Γ
∂t
∂x
∂y
∂z
∂ ( ρϕ ) ∂ ( ρϕu ) ∂ ( ρϕv ) ∂ ( ρϕw)
∂  ∂ϕ  ∂  ∂ϕ  ∂  ∂ϕ 
+
+
+
)=
+  Γ
Γ
 + Γ
 + Sφ
∂t
∂x
∂y
∂z
∂x  ∂x  ∂y  ∂y  ∂z  ∂z 
6.
Which one (if any) of the following functions
a) ϕ1 ( x, y, z ) = x + y + z
4
2
ϕ 2 ( x, y , z ) = x 2 + y 2 + z
2
2
2
c) ϕ 3 ( x, y, z ) = x + y + z
b)
satisfies the equation

∇ • (ϕU ) = ∇ • ( Γ ⋅ (∇ϕ )) + S ?

Here Γ = 5 , U = (1, 2, 3) and S = 2 x + 4 y − 23 .
Solution :
The equation

∇ • (ϕU ) = ∇ • ( Γ ⋅ (∇ϕ )) + S
can be written as

div(ϕU ) = div(Γgradϕ ) + S ⇒
5∂ϕ 5∂ϕ 5∂ϕ
div(ϕ , 2ϕ , 3ϕ ) = div(
,
,
) + 2 x + 4 y − 23 ⇒
∂x ∂y ∂z
∂ϕ 2∂ϕ 3∂ϕ 5∂ 2ϕ 5∂ 2ϕ 5∂ 2ϕ
+
+
=
+
+
+ 2 x + 4 y − 23
2
z
∂x
∂y
∂z
∂x 2
∂y 2
∂
a) Let ϕ = ϕ1 ( x, y, z ) = x + y + z
4
(eq 1.)
2
Vi calculate the derivatives of ϕ1 and substitute in the left hand side (LHS) and right hand
side of the equation (eq1).
11 / 27
Armin Halilovic
LHS:
RHS=
Math. Exercises
∂ϕ 2∂ϕ 3∂ϕ
+
+
= 4 x3 + 4 y + 3
∂x
∂y
∂z
5∂ 2ϕ 5∂ 2ϕ 5∂ 2ϕ
+
+
+ 2 x + 4 y − 23 = 60 x 2 + 2 x + 4 y − 13
2
2
2
∂y
∂x
∂z
Whence LHS ≠ RHS
Thus the function ϕ1 ( x, y, z ) = x + y + z is not a solution to the equation
4
2
b) ϕ = ϕ 2 ( x, y, z ) = x + y + z
LHS= 3 + 2 x + 4 y , RHS= − 3 + 2 x + 4 y
2
2
Whence LHS ≠ RHS , and the function
equation
ϕ 2 ( x, y, z ) = x 2 + y 2 + z is not a solution to the
c) Let ϕ = ϕ 3 ( x, y, z ) = x + y + z
Then
LHS= 2 x + 4 y + 6 z , RHS= 7 + 2 x + 4 y
2
2
2
Thus LHS ≠ RHS , and the function ϕ 3 ( x, y, z ) = x + y + z
equation.
Answer: None of the functions satisfies the equation
2
7. Answer: Function
2
2
is not a solution to the
ϕ 2 ( x, y, z ) = x 2 + y 2 + 5 z satisfies the equation.
8. (exam 1, 98)
A) Write the general transport equation

∂ ( ρϕ )
( eq 1)
+ ∇ • ( ρϕU ) = ∇ • ( Γ ⋅ (∇ϕ )) + S
∂t
without using operators div, ∇ , ∆ , curl or grad.

Here U = (u , v, w) . Functions ρ , ϕ , Γ, S , u, v, w are real functions of t, x, y and z.

B) Let ρ = 2 , Γ = 3 , U = (1, 2, 4) .
Find S in the equation (eq 1) if we now that the function
ϕ ( x, y , z ) = x + y 2 + z 3
satisfies the equation.
Solution:
A)

∂ ( ρϕ )
+ ∇ • ( ρϕU ) = ∇ • ( Γ ⋅ (∇ϕ )) + S ⇒
∂t

∂ ( rϕ )
+ div ( rϕU ) = div( Γgradϕ ) + S ⇒
∂t
∂ ( ρϕ )
∂ϕ
∂ϕ
∂ϕ
+ div ( ρϕu, ρϕv, ρϕw) = div ( Γ
,Γ
,Γ
)+S ⇒
∂t
∂x
∂y
∂z
12 / 27
Armin Halilovic
Math. Exercises
∂  ∂ϕ  ∂  ∂ϕ  ∂  ∂ϕ 
∂ ( ρϕ ) ∂ ( ρϕu ) ∂ ( ρϕv ) ∂ ( ρϕw)
) = Γ
+
+
+
+  Γ
+S
 + Γ
∂t
∂x
∂y
∂z
∂x  ∂x  ∂y  ∂y  ∂z  ∂z 
(eq2)
B)

We substitute ρ = 2 , Γ = 3 , U = (1, 2, 4) and
in the equation (eq2) and get
0+
ϕ ( x, y , z ) = x + y 2 + z 3
∂ ( 2ϕ ) ∂ ( 4ϕ ) ∂ (8φ )
∂  ∂ϕ  ∂  ∂ϕ  ∂  ∂ϕ 
+
+
) =  3  +  3  +  3  + S
∂x
∂y
∂z
∂x  ∂x  ∂y  ∂y  ∂z  ∂z 
0 + 2 + 8 y + 24 z 2 = 0 + 6 + 18 z + S .
Consequently
S = −4 + 8 y − 18 z + 24 z 2
9. (Q6, exam 2, 2008)
Consider the following equation


∂ ( ρϕ )
( eq 1)
+ ∇ • ( ρϕU ) = ∇ • ( Γ ⋅ (∇ϕ )) + ∇ • (∇ × U ) + 6 y − 2 xyz − 4
∂t

Let ρ = 1 , Γ = constant , U = ( 2, 3, − xy ) .
Find the constant Γ in the equation (eq 1) if we now that the function
ϕ ( x , y , z ) = 2t + x + y 2 + z 2
satisfies the equation.
Solution:


∂ ( ρϕ )
+ ∇ • ( ρϕU ) = ∇ • ( Γ ⋅ (∇ϕ )) + ∇ • (∇ × U ) + 6 y − 2 xyz − 4 ⇒
∂t


∂ ( rϕ )
+ div ( rϕU ) = div ( Γgradϕ ) + div ( curl (U )) + 6 y − 2 xyz − 4 ⇒
∂t


(since curl (U ) = ( − x, y ,0) we have div( curl (U )) = −1 + 1 = 0 )
∂ ( ρϕ )
∂ϕ
∂ϕ
∂ϕ
+ div( ρϕu, ρϕv, ρϕw) = div( Γ
,Γ
,Γ
) + 0 + 6 y − 2 xyz − 4 ⇒
∂t
∂x
∂y
∂z
∂ ( ρϕ ) ∂ ( ρϕu ) ∂ ( ρϕv ) ∂ ( ρϕw)
∂  ∂ϕ  ∂  ∂ϕ  ∂  ∂ϕ 
) = Γ
+
+
+
+  Γ
 + 0 + 6 y − 2 xyz − 4
 + Γ
∂t
∂x
∂y
∂z
∂x  ∂x  ∂y  ∂y  ∂z  ∂z 

We substitute ρ = 1 , , U = ( 2, 3, − xy ) and
in the equation (eq2) and get
(eq2)
ϕ ( x , y , z ) = 2t + x + y 2 + z 2
∂ (1ϕ ) ∂ ( 2ϕ ) ∂ (3ϕ ) ∂ ( − xyφ )
∂  ∂ϕ  ∂  ∂ϕ  ∂  ∂ϕ 
+
+
+
) = Γ
+  Γ
 + 6 y − 2 xyz − 4
 + Γ
∂t
∂x
∂y
∂z
∂x  ∂x  ∂y  ∂y  ∂z  ∂z 
( Note that Γ is a constant)
2 + 2 + 6 y − 2 xyz = 0 + 2Γ + 2Γ + 6 y − 2 xyz − 4 ⇒
8 = 4Γ ⇒
Γ=2
Answer: Γ = 2
10. If possible, find f ( x, y ) for the given partial derivatives
13 / 27
∂
∂
f ( x, y ) and
f ( x, y ) .
∂x
∂y
Armin Halilovic
Math. Exercises
∂
∂
f ( x, y ) = 2 xy and
f ( x, y ) = x 2 + 2 y .
∂x
∂y
∂
∂
b)
f ( x, y ) = 2 x + y and
f ( x, y ) = x .
∂x
∂y
∂
∂
c)
f ( x, y ) = ye xy and
f ( x, y ) = xe xy .
∂x
∂y
∂
∂
d)
f ( x, y ) = 2 x + y and
f ( x, y ) = 5 x .
∂x
∂y
( Hint: Necessary condition: If f ( x, y ) has continuous derivatives then
the mixed derivatives of f ( x, y ) should be equal, i.e.
a)

∂ ∂
 ∂  ∂
(*)
f ( x, y ) 
f ( x, y )  = 

∂y  ∂x
 ∂x  ∂y

is the necessary condition for the existence of a function f ( x, y ) that has the given
derivatives.
Answer:
a) f ( x, y ) = x 2 y + y 2 + C b) f ( x, y ) = x 2 + xy + C c) f ( x, y ) = e xy + C
d) No solution since the condition (*) is not fulfilled,

∂ ∂
 ∂  ∂
f ( x, y )  ≠ 
f ( x, y )  = 5 .
1= 
∂y  ∂x
 ∂x  ∂y

Solution a)

∂ ∂
 ∂  ∂
Since
and the derivatives are continuous the
f ( x, y )  = 
f ( x, y )  =2x

∂y  ∂x
 ∂x  ∂y

condition (*) is fulfilled and we can find f ( x, y ) for the given derivatives.
In order to find f ( x, y ) we integrate with respect to x the first of the equations
∂
(eq1)
f ( x, y ) = 2 xy
∂x
∂
(eq2).
f ( x, y ) = x 2 + 2 y
∂y
and get
f ( x, y ) = ∫ 2 xydx = x 2 y + C´1 ( y )
Thus
(i)
f ( x, y ) = x 2 y + C´1 ( y )
We have integrated with respect to x, therefore the constant still depend on y.
Now, to find C´1 ( y ) we differentiate and substitute (i) in (eq2) and get:
∂
∂ 2
x y + C´1 ( y ) = x 2 + 2 y ⇒ x 2 + (C´1 ( y ) ) = x 2 + 2 y ⇒
∂y
∂y
(
)
∂
(C´1 ( y ) ) = 2 y ⇒ C´1 ( y ) = y 2 + C .
∂y
Finally, substituting C´1 ( y ) = y 2 + C in (i) we have
f ( x, y ) = x 2 y + y 2 + C (where C is a constant).
14 / 27
Armin Halilovic
Math. Exercises
11.
∂ ∂
∂ ∂
f ( x, y ) =
f ( x, y ) ⇒ ay + 1 = 2 x + 1 ⇒ a = 2
∂y ∂x
∂x ∂y
Then for a = 2 we have f ( x, y ) = x 2 y + xy + C
Answer: From
12.
Answer:
a) f = xyz + yz + z 3 + C b) f = xy + yz + C c) f = e xyz + C
d) No solution since the condition Con 2 is not fulfilled,
∂ ∂  ∂ ∂ 
f = y+z
f≠ 
y=

∂z  ∂x  ∂x  ∂z 
Solution a)
∂
∂
∂
a)
f ( x, y, z ) = xy + y + 3 z 2
f ( x, y, z ) = yz ,
f ( x, y, z ) = xz + z and
∂z
∂x
∂y
Since the conditions Con1,2,3 are fulfilled and we can find f ( x, y, z ) for the given
derivatives.
In order to find f ( x, y, y ) we integrate with respect to x the first of the equations
∂
(eq1)
f ( x, y, z ) = yz
∂x
∂
(eq2)
f ( x, y, z ) = xz + z
∂y
∂
(eq3)
f ( x, y, z ) = xy + y + 3 z 2
∂z
and get
f ( x, yz ) = ∫ yzdx = xyz + C´1 ( y, z )
Thus
(i)
f ( x, y, z ) = xyz + C´1 ( y, z )
We have integrated with respect to x, therefore the constant still depend on y and z.
Now, to find C´1 ( y, z ) we differentiate and substitute (i) in (eq2) and get:
∂
(xyz + C´1 ( y, z ) ) = xz + z ⇒ xz + ∂ (C´1 ( y, z ) ) = xz + z ⇒
∂y
∂y
∂
(C´1 ( y, z ) ) = z ⇒ C´1 ( y, z ) = yz + C 2 ( z ) .
∂y
(We have integrated with respect to y , therefore the constant still depend on and z.
Thus f ( x, y, z ) = xyz + yz + C 2 ( z )
(ii)
Now , substituting (ii) in (eq3) we have
15 / 27
Armin Halilovic
Math. Exercises
∂
( xyz + yz + C 2 ( z )) = xy + y + 3 z 2 ⇒
∂z
∂
xy + y + (C 2 ( z )) = xy + y + 3 z 2 ⇒
∂z
∂
(C 2 ( z )) = 3 z 2
∂z
C2 ( z) = z 3 + C
Finally, substituting C 2 ( z ) = z 3 + C in (ii) we have
f ( x, y, z ) = xyz + yz + z 3 + C (where C is a constant).
13.
Answer:
From
∂ ∂  ∂  ∂ 
f  ⇒ ax 2 z = 3x 2 z ⇒ a = 3
f  = 

∂y  ∂x  ∂x  ∂y 
∂ ∂  ∂ ∂ 
f  ⇒ x 3 = bx 3 ⇒ b = 3
f = 

∂z  ∂x  ∂x  ∂z 
∂  ∂  ∂ ∂ 

f  = 
f  ⇒ x 3 = bx 3 ⇒ b = 3 .
∂z  ∂y  ∂y  ∂z 
Thus, all three conditions are fulfilled if a = 3 and b = 3 .
For these values of a and b we get f ( x, y, z ) = x 3 yz + y + z 2 + C
Calculation of the pressure field for a known velocity field for an incompressible, steady
state, isothermal Newtonian flow.
14.
Answer:
a) P = − ρgz − 8ρx 2 − 8ρy 2 + C
17 2 17 2
ρx − ρy + C
2
2
2
c) P = − ρgz + 4 ρx + 4 ρy 2 + C
b) P = − ρgz −
Solution a)
We substitute u = 2 x + 3 y, v = 4 x − 2 y, w = 0 in eq1,2,3,4
and get ( note that al derivatives with respect to t are 0):
Continuity equation:
eq1i.
0=0
( identically fulfilled)
Navier Stokes equations:
x component:
16 / 27
Armin Halilovic
Math. Exercises
∂P
eq2i.
∂x
y component:
∂P
eq3i.
16 ρy = −
∂y
z component:
∂P
eq4i.
− ρg
0=−
∂z
Now eq2i. gives P( x, y, z ) = −8 ρx 2 + C´1 ( y, z )
Substitution in eq3i. implies
∂C ( y, z )
⇒
16 ρy = − ´1
∂y
16 ρx = −
(*) .
C´1 ( y, z ) = −8 ρy 2 + C´2 ( z )
Hence, from (*) we have
(**)
P ( x, y, z ) = −8 ρx 2 − 8 ρy 2 + C´2 ( z )
Now we substitute (**) in eq4i. and get
∂
∂P
− ρg ⇒ 0 = − (C´2 ( z )) − ρg
0=−
∂z
∂z
( where C is a constant)
⇒ C´2 ( z ) = − ρgz + C
Finally, substituting C´2 ( z ) = − ρgz + C in (**) we have
P ( x, y, z ) = −8 ρx 2 − 8 ρy 2 − ρgz + C (where C is a constant).
15.
Solution A:


∂ ( ρϕ )
+ ∇ • ( ρϕU ) = ∇ • (Γ ⋅ (∇ϕ )) + ∇ • (∇ × U ) + 16 x + 16 y + 4 xz + 8 yz − 24 ⇒
∂t


∂ ( rϕ )
+ div( rϕU ) = div(Γgradϕ ) + div(curl (U )) + 16 x + 16 y + 4 xz + 8 yz − 24 ⇒
∂t


(since curl (U ) = (2,−1, 0) we have div(curl (U )) = 0 )
∂ ( ρϕ )
∂ϕ
∂ϕ
∂ϕ
,Γ
, Γ ) + 16 x + 16 y + 4 xz + 8 yz − 24 ⇒
+ div( ρϕu , ρϕv, ρϕw) = div(Γ
∂t
∂x
∂y
∂z
∂ ( ρϕ ) ∂ ( ρϕu ) ∂ ( ρϕv) ∂ ( ρϕw)
∂  ∂ϕ  ∂  ∂ϕ  ∂  ∂ϕ 
+  Γ
) = Γ
+
+
+
 + 16 x + 16 y + 4 xz + 8 yz − 24
 + Γ
∂t
∂x
∂y
∂z
∂x  ∂x  ∂y  ∂y  ∂z  ∂z 
(eq2)

We substitute ρ = 2 , , U = (4, 4, x + 2 y ) and
in the equation (eq2) and get
ϕ ( x, y, z ) = 1 + 3t + x 2 + y 2 + z 2
∂ (2ϕ ) ∂ (8ϕ ) ∂ (8ϕ ) ∂ (2( x + 2 y )φ )
∂  ∂ϕ  ∂  ∂ϕ  ∂  ∂ϕ 
+  Γ
) = Γ
+
+
+
 + 16 x + 16 y + 4 xz + 8 yz − 24
 + Γ
∂t
∂x
∂y
∂z
∂x  ∂x  ∂y  ∂y  ∂z  ∂z 
( Note that Γ is a constant)
6 + 16 x + 1 6 y + 4 xz + 8 yz = 6Γ + 16 x + 16 y + 4 xz + 8 yz − 24 ⇒
30 = 6Γ ⇒
Γ=5
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Armin Halilovic
Math. Exercises
Answer A: Γ = 5
Solution B:
We substitute u = 6 + 4 x, v = 4 − 2 y, w = 2 − 2 z in eq1,2,3,4
and get ( note that al derivatives with respect to t are 0):
Continuity equation:
eq1i.
0=0
( identically fulfilled)
Navier Stokes equations:
x component:
∂P
eq2i.
ρ (16 x + 24) = −
∂x
y component:
∂P
eq3i.
ρ (4 y − 8) = −
∂y
z component:
∂P
eq4i.
ρ (4 z − 4) = −
− ρg
∂z
Now eq2i. gives P ( x, y, z ) = ρ (−8 x 2 − 24 x) + C´1 ( y, z )
Substitution in eq3i. implies
∂C ( y, z )
ρ (4 y − 8) = − ´1
⇒
∂y
(*) .
C´1 ( y, z ) = ρ (−2 y 2 + 8 y ) + C´2 ( z )
Hence, from (*) we have
(**)
P( x, y, z ) = ρ (−8 x 2 − 24 x) + ρ (−2 y 2 + 8 y ) + C´2 ( z )
Now we substitute (**) in eq4i. and get
∂
∂P
ρ (4 z − 4) = −
− ρg ⇒ ρ (4 z − 4) = − (C´2 ( z )) − ρg
∂z
∂z
2
( where C is a constant)
⇒ C´2 ( z ) = − ρgz + ρ (−2 z + 4 z ) + C
Finally, substituting C´2 ( z ) = − ρgz + C in (**) we have
P( x, y, z ) = ρ (−8 x 2 − 24 x) + ρ (−2 y 2 + 8 y ) + ρ (−2 z 2 + 4 z ) − ρgz + C
Answer B:
P ( x, y, z ) = ρ (−8 x 2 − 24 x − 2 y 2 + 8 y − 2 z 2 + 4 z − gz ) + C
(where C is a constant).
16. Solution

V = (2 + 3 x, 5 − y, 1 − az )
First we substitute u = 2 + 3 x, v = 5 − y, w = 1 − az in eq1
and get ( note that al derivatives with respect to t are 0):
Continuity equation:
3 −1− a = 0 ⇒ a = 2

No we have V = (2 + 3 x, 5 − y, 1 − 2 z )
Using the Navier Stokes equations we get:
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Armin Halilovic
x component:
ρ (9 x + 6) = −
∂P
∂x
y component:
∂P
ρ ( y − 5) = −
∂y
z component:
∂P
ρ (4 z − 2) = −
− ρg
∂z
Math. Exercises
eq2i.
eq3i.
eq4i.
− 9x 2
Now eq2i. gives P( x, y, z ) = ρ (
− 6 x) + C´1 ( y, z )
2
Substitution in eq3i. implies
∂C ( y, z )
ρ ( y − 5) = − ´1
⇒
∂y
(*) .
− y2
+ 5 y ) + C´2 ( z )
2
Hence, from (*) we have
− 9x 2
− y2
(**)
P ( x, y , z ) = ρ (
− 6 x) + ρ (
+ 5 y ) + C´2 ( z )
2
2
We substitute (**) in eq4i. and get
∂
∂P
ρ (4 z − 2) = −
− ρg ⇒ ρ (4 z − 2) = − (C´2 ( z )) − ρg
∂z
∂z
2
( where C is a constant)
⇒ C´2 ( z ) = − ρgz + ρ (−2 z + 2 z ) + C
Finally, substituting C´2 ( z ) in (**) we have
C´1 ( y, z ) = ρ (
P ( x, y , z ) = ρ (
− 9x 2
− y2
+ 6 x) + ρ (
+ 5 y ) + − ρgz + ρ (−2 z 2 + 2 z ) + C
2
2
Answer :
9x 2
y2
P ( x, y , z ) = ρ ( −
+ 6x −
+ 5 y − gz − 2 z 2 + 2 z ) + C
2
2
(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
z-axis in our notation):
∂P/∂z = –1/250,
is applied in the horizontal axis (
c3. The flow is parallel to the z axis, that is ur =0 and uθ =0.
19 / 27
Armin Halilovic
Math. Exercises
c4. We assume that the flow is axisymmetric . The velocity does not depend on θ,
∂u z
that is
=0
∂θ
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
∂u z
=0
∂r r = 0
The continuity and the Navier-Stokes equations for an incompressible , isothermal
Newtonian flow (density ρ =const, viscosity µ =const), with a velocity field

V = (u r , uθ , u z ) in Cylindrical coordinates (r , θ , z ) :
---------------------------------------------------------------SOLUTION
Incompressible continuity equation
1 ∂ (ru r ) 1 ∂ (uθ ) ∂u z
eq a)
+
+
=0
r ∂r
r ∂θ
∂z
Navier-Stokes equations in Cylindrical coordinates:
r-component:
 ∂u
∂u u ∂u u 2
∂u 
r  r + u r r + θ r − θ + u z r 
r ∂θ
r
∂r
∂z 
 ∂t
 1 ∂  ∂u r  u r 1 ∂ 2 u r 2 ∂uθ ∂ 2 u r 
∂P
=−
+ rg r + µ 
−
+ 2 
r
− 2 + 2
r ∂θ 2 r 2 ∂θ
∂r
∂z 
 r ∂r  ∂r  r
θ -component:
∂u
∂u 
u ∂u
uu
 ∂u
r θ + ur θ + θ θ + r θ + u z θ 
∂r
∂z 
r ∂θ
r
 ∂t
 1 ∂  ∂uθ  uθ
1 ∂P
1 ∂ 2 uθ
2 ∂u r ∂ 2 uθ 
=−
+ rg θ + µ 
+
+
r
− 2 + 2

r ∂θ
∂z 2 
r ∂θ 2 r 2 ∂θ
 r ∂r  ∂r  r
z-component:
u ∂u
∂u
∂u 
 ∂u
r  z + ur z + θ z + u z z 
∂r
r ∂θ
∂z 
 ∂t
 1 ∂  ∂u z  1 ∂ 2 u z ∂ 2 u z 
∂P
=−
+ rg z + µ 
+ 2 
r
+ 2
2
∂z 
∂z
 r ∂r  ∂r  r ∂θ
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eq d)
eq b)
eq c)
Armin Halilovic
Math. Exercises
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, θcomponent 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
g r = − g cos θ , gθ = g sin θ and g z = 0
Now we substitute ∂P/∂z = –1/250 Pa/m, μ=0.001kg /(ms) in the continuity and NavierStokes equations:
Since ur =0 and uθ =0 (according to c3), continuity equation in cylindrical coordinates
1 ∂ ( ru r ) 1 ∂ (uθ ) ∂u z
+
+
=0
r ∂r
r ∂θ
∂z
gives
∂u z
=0.
∂z
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
u z = w(r )
(*)
Now we substitute
g r = − g cos θ , gθ = g sin θ and g z = 0
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Armin Halilovic
Math. Exercises
∂P/∂z = –1/250 Pa/m, μ=0.001kg /(ms)
in the Navier-Stokes equations:
The r-component of the Navier-Stokes equation gives:
∂P
( eq r-c)
− ρg cos θ
∂ρ
The θ-component of the Navier-Stokes equation:
1 ∂P
( eq θ-c)
0=−
+ ρg sin θ
ρ ∂θ
0=−
The Z-component of the Navier-Stokes equation (where u z = w(r ) and
0=
1
1  1 ∂  ∂w 
+
r

250 1000  r ∂r  ∂r 
1
∂P
) givs:
=−
250
∂z
( eq z-c)
Step 1. We find the pressure P = P( r, θ , z ) .
In order to find the pressure P we solve ( eq r-c), ( eq θ-c) and the equation
is
∂P
= − ρg cos θ
∂ρ
∂P
= + ρρg sin θ
∂θ
1
∂P
=−
∂z
250
From these equations we get
P=−
1
z − rgr cos θ + C
250
Step 2. We find the velocity component u z = w(r ) .
We solve ( eq z-c) with boundaries c5 and c6:
0=
1
1  1 ∂  ∂w 
+
r

250 1000  r ∂r  ∂r 
w( 2) = 0
( eq z-c)
(c5)
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1
∂P
that
=−
250
∂z
Armin Halilovic
∂w
=0
∂r r = 0
( Remark: Technically, we can write
Math. Exercises
(c6)
∂w
dw
instead
since w is now a function of only one
∂r
dr
variable)
From ( eq z-c) we have
0=
1
1  1 ∂  ∂w 
+
r

250 1000  r ∂r  ∂r 
∂  ∂w 
r
 = −4 r ⇒
∂r  ∂r 
r
r
∂w
= −2r 2 + C1 ⇒ (substitution r = 0 and (c6) ⇒ C1 = 0 )
∂r
∂w
= −2r 2 ⇒
∂r
∂w
= −2r ⇒
∂r
w = − r 2 + C 2 ⇒ (substitution r = 2 and (c5) ⇒ C 2 = 4 )
w = −r 2 + 4 ⇒
Thus u z = w( r ) = − r 2 + 4 and
V = (u r , uθ , u z ) =(0, 0, − r 2 + 4) .
Answer :
1
P=−
z − rgr cos θ + C ,
250
V = (0, 0, − r 2 + 4)
18.
We consider an incompressible ( density ρ =const), steady state ( variables do not depend on
time), isothermal Newtonian flow with a given velocity field

V = (u ( x,y,z ) , v( x,y,z ) , w ( x,y,z )) = (3 x + cy, 4 + 2 x − y + bz , 3 − az )
Use the following equations ( continuity and Navier Stokes equations) , where

ρ =constant, µ =constant , g = (0,0,− g ) i.e. g x = g y =0 and g z = − g ( g ≈ 9.81m / s 2 )
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.
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Armin Halilovic
Math. Exercises
Incompressible continuity equation:
∂u ∂v ∂w
eq1.
+
+
=0
∂x ∂y ∂z
Navier Stokes equations:
x component:
 ∂u
∂u
∂u 
∂u
∂P
∂ 2u ∂ 2u ∂ 2u
ρ  + u + v + w  = −
+ ρg x + µ ( 2 + 2 + 2 )
∂x
∂y
∂z 
∂x
∂x
∂y
∂z
 ∂t
y component:
 ∂v
∂P
∂ 2v ∂ 2v ∂ 2v
∂v
∂v
∂v 
+ ρg y + µ ( 2 + 2 + 2 )
ρ  + u + v + w  = −
∂y
∂x
∂y
∂z 
∂x
∂y
∂z
 ∂t
z component:
 ∂w
∂w
∂w
∂w 
∂P
∂2w ∂2w ∂2w
+ w  = −
ρ 
+u
+v
+ ρg z + µ ( 2 + 2 + 2 )
∂x
∂y
∂z 
∂z
∂x
∂y
∂z
 ∂t
eq2.
eq3.
eq4.
------------------------------------------------------------We substitute u = 3 x + cy, v = 4 + 2 x − y + bz , w = 3 − az ,
and get ( note that al derivatives with respect to t are 0):
Continuity equation:
2−a = 0⇒ a = 2
in eq1,2,3,4
eq1i.
Thus
u = 3 x + cy, v = 4 + 2 x − y + bz , w = 3 − 2 z ,
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:
Con1 :
∂ ∂  ∂ ∂ 
 P  =  P  ⇒ −2 ρc = −4 ρ ⇒ c = 2
∂y  ∂x  ∂x  ∂y 
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Armin Halilovic
Con 2 :
Con3 :
Math. Exercises
∂ ∂  ∂ ∂ 
 P  =  P  ⇒ ρbc = 0 ⇒ b = 0
∂z  ∂x  ∂x  ∂z 
∂ ∂  ∂ ∂ 
 P  =  P  ⇒ −3ρb = 0 ⇒ b = 0
∂z  ∂y  ∂y  ∂z 
Thus c=2 and b=0
We solve simplified equations
and get
13 x 2 5 y 2
−
− 2 z 2 − 4 xy − 8 x + 4 y − gz + 6 z ) + C
2
2
13 x 2 5 y 2
Answer. P( x, y, z ) = ρ (−
−
− 2 z 2 − 4 xy − 8 x + 4 y − gz + 6 z ) + C
2
2
P ( x, y , z ) = ρ ( −
19. We consider a scalar field f (r , θ , z ) given in cylindrical coordinates, where
x = r cos θ ,
y = r sin θ , z = z , and basis vectors are e1 , e2 , e3 .
Find the expression for the gradient, grad ( f (r , θ , z )) ), in cylindrical coordinates, that is in
 
 ∂f ∂f
∂f
,
and
if
∂r ∂θ
∂z
terms of r,θ , z, e1 , e2 , e3 ,
  
  
  
a) e1 = i , e2 = j , e3 = k
( we keep the same basis i , j , k ) .
 
 
  
b)
e1 = i e2 = 2 j , e3 = 2 j + k
c)
 
e1 = i cos θ ,
 
 
e2 = j sin θ , e3 = k
(exam 1, 2012; Q5 B (2 points))

 

  

d) e1 = cos θ i + sin θ j , e2 = − sin θ i + cos θ j , e3 = k (this is often used as a local basis
for cylindrical coordinates)
Solution:
In x,y, z variables we have
∂f ∂f ∂f
∂f  ∂f  ∂f 
grad ( f ) = ( , , ) =
i+
j+ k
∂x ∂y ∂z
∂x
∂y
∂z
For cylindrical coordinates we have x = r cos θ ,
25 / 27
( eq1)
y = r sin θ ,
z=z
Armin Halilovic
Math. Exercises
First we write the derivatives
∂f
∂f
, and
in r ,θ coordinates (the variable z is in both
∂x
∂y
coord systems) .
Solving the following system for
∂f
∂f
, and
,
∂x
∂y
∂f ∂f
∂f
=
⋅ cos θ +
⋅ sin θ
∂r ∂x
∂y
∂f ∂f
∂f
=
⋅ (−r sin θ ) +
⋅ (r cos θ )
∂θ ∂x
∂y
we get
∂f sin θ
∂f
−
= cos θ
∂r
∂x
r
∂f
∂f cos θ
= sin θ
+
r
∂y
∂r
∂f
∂θ
∂f
∂θ
(**)
We substitute the derivatives (**) in ( eq1) and get
∂f sin θ ∂f 
∂f cos θ ∂f  ∂f 
−
)i + (sin θ
+
) j + k (* * *)
∂r
r ∂θ
∂r
r ∂θ  ∂z
 
To solve problems a) , b) c) and d) we must express i , j , k as a linear combinations of
  
e1 , e2 , e3 and substitute them into (***) .
grad ( f ) = (cos θ
a)
 
 
 
From (***), since i = e1 , j = e2 , k = e3 , we have immediately
∂f sin θ ∂f 
∂f cos θ ∂f  ∂f 
−
+
grad ( f ) = (cos θ
)e1 + (sin θ
)e2 + e3
∂r
∂r
r ∂θ
∂z
r ∂θ
 
b) From e1 = i
 
 

e2 = 2 j , e3 = 2 j + k we find
  
 e2
(eq b)
j = , k = e3 − e2
2 
 
Then we put i , j , k from ( eq b) into (***) and get
 
i = e1 ,

∂f sin θ ∂f 
∂f cos θ ∂f e2 ∂f  
−
+
grad ( f ) = (cos θ
)e1 + (sin θ
) + (e3 − e2 )
∂r
∂r
r ∂θ
r ∂θ 2 ∂z
 

and, after collecting components for e1 e2 , e3
∂f sin θ ∂f 
∂f 1 cos θ ∂f ∂f  ∂f 
1
−
+
− )e2 + e3
grad ( f ) = (cos θ
)e1 + ( sin θ
∂r
∂r 2 r ∂θ ∂z
∂z
r ∂θ
2
c) From
 
e1 = i cosθ ,
 
 
e2 = j sin θ , e3 = k
we have
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Armin Halilovic
Math. Exercises


 


e2
e1
i =
, j=
, k = e3 ( eq c)
cos θ
sin θ
  
Putting i , j , k from ( eq c) into (***) gives


∂f sin θ ∂f
e1
∂f cos θ ∂f e2
∂f 
)
)
grad ( f ) = (cos θ
−
+ (sin θ
+
+ e3
∂r
r ∂θ cos θ
∂r
r ∂θ sin θ ∂z








d) e1 = cos θ i + sin θ j , e2 = − sin θ i + cos θ j , e3 = k .
We can solve d) in the same manner as in a,b,c but this time we can just collect terms
∂f
∂f
,
and get the result:
∂θ ∂z
∂f  ∂f  ∂f 
∂f sin θ ∂f 
∂f cos θ ∂f  ∂f 
grad ( f ) =
i+
j + k = (cos θ
−
)i + (sin θ
+
)j + k
∂x
∂y
∂z
∂r
r ∂θ
∂r
r ∂θ
∂z


 1 ∂f

 ∂f
∂f
=
⋅ (cos θ i + sin θ j ) +
⋅ (− sin θ i + cos θ j ) + k
r ∂θ
∂r
∂z
∂f  1 ∂f  ∂f 
=
⋅ e1 +
⋅ e2 + e3
∂r
r ∂θ
∂z
Answer:
∂f cos θ ∂f  ∂f 
∂f sin θ ∂f 
+
)e1 + (sin θ
)e2 + e3
−
a) grad ( f ) = (cos θ
∂r
∂r
∂z
r ∂θ
r ∂θ
∂f sin θ ∂f 
1
∂f 1 cos θ ∂f ∂f  ∂f 
−
)e1 + ( sin θ
+
− )e2 + e3
b) grad ( f ) = (cos θ
∂r
r ∂θ
2
∂r 2 r ∂θ ∂z
∂z


∂f sin θ ∂f
e
∂f cos θ ∂f e2
∂f 
) 1 + (sin θ
)
−
+
+ e3
c) grad ( f ) = (cos θ
∂r
r ∂θ cos θ
∂r
r ∂θ sin θ ∂z
∂f  1 ∂f  ∂f 
⋅ e2 + e3
⋅ e1 +
d) grad ( f (r , θ , z ) =
r ∂θ
∂r
∂z
27 / 27
∂f
,
∂r