distributed parameters
DESCRIPTION
Distributed Parameters. ECX 5241. Academic year 2003. Prepared by D.A.Mangala Abeysekara. ASSIGNMENT NO.1. ASSIGNMENT NO.2. ASSIGNMENT NO.3. ASSIGNMENT NO.4. ASSIGNMENT NO.1. Symmetry in the physical universe Coordinates systems Operators in vector calculus The theorem of gauss - PowerPoint PPT PresentationTRANSCRIPT
Distributed ParametersECX 5241
Academic year 2003
Prepared by
D.A.Mangala Abeysekara
ASSIGNMENT NO.1
ASSIGNMENT NO.2
ASSIGNMENT NO.3
ASSIGNMENT NO.4
ASSIGNMENT NO.1
Symmetry in the physical universe
Coordinates systems
Operators in vector calculus
The theorem of gauss
The theorem of Stokes
Symmetry in the physical universe
Most of the problems in the physical universe exhibit the spherical or cylindrical symmetry.
Examples
• the gravity field of the Earth is to the first order spherical symmetric.
• Waves excited by a stone thrown into water are usually cylindrically symmetric
•An earthquake excites a tsunami in the ocean
The Cartesian coordinates is usually not very convenient to use for study such a problems
The reason for this is that the theory is usually much simpler when one selects a coordinate system with symmetry properties that are the same as the symmetry properties of the physical system that one wants to study.
There are two coordinates systems used for study the symmetric system.
•Spherical Coordinates
•Cylindrical Coordinates
Relationship between the Cartesian coordinates and spherical coordinates
z
y
xr
ˆ
ˆ
ˆ
0cossin
sinsincoscoscos
cossinsincossin
ˆ
ˆ
ˆ
Relationship between the Cartesian coordinates and cylindrical coordinates
zz
ry
rx
sin
cos
0
sin
cos
ˆ
r
0
cos
sin
ˆ
1
0
0
z
Operators in vector calculus
There are several operators can be identified in the vector calculus. They are,
Gradient (f )
Divergent(.v )
Curl (v )
Gradient
Lets consider the particle, which moved according to the function f from point A to point B.(f Is a function of x and y)
Grad f =
yf
xff
/
/
The divergence of a vector field
dz
dxdy
vx
Outward flux through the right hand surface perpendicular through the x-axis
vx (x+dx, y, z) dydz
The flux through the left hand surface perpendicular through the x- axis
– vx (x, y, z) dydz
The the total outward flux through the two surface
vx (x+dx, y, z) dydz - vx (x, y, z) dydz =
(vx/x)dxdydz
dvvdvz
v
y
v
x
vd zyx ).(
(.v)= d/dv
The divergence of a vector field is the outward flux of the vector field per unit volume
The curl of a vector field
xyyx
zxxz
yzzy
zyx
zyx
vv
vv
vv
vvv
zyx
curlV
ˆˆˆ
Curl V=v
Physical meaning of the curl operator
The component of curl v in a certain direction is the closed line integral of v along a closed path perpendicular to this direction, normalized per unit surface area.
VS
dVvvdS ).(
The theorem of gauss
The theorem of Stokes
SC
dSvdrv ).(.
Example
The magnetic field induced by a straight current
I
B
The Maxwell equation for the curl
JB 0
)(rBB
dSrBdrBSC )(
dSJBdrSC
ˆ0
ˆ2
0
r
IB
Solving this equation,
(Using Stoke theorem)
Reference
A Guided Tour Of Mathematical Methods For The Physical Science
By
Role Sineder
Cambridge University Press
Thank you for your attention
ASSIGNMENT NO.2
Classification of PDE
Acoustic sound propagation in gas
A partial differential equation of the form
is said to be
Elliptic if 4AC-B2>0
Parabolic if 4AC-B2=0
Hyperbolic if 4AC-B2<0
y
u
x
uuyxF
y
uC
yx
uB
x
uA ,,,,
2
22
2
2
Consider the partial differential equation,
where A,B and C are constants.
Define variables,
and
where ,, and are constants.
Using chain rule we can show that ,
02
22
2
2
y
uC
yx
uB
x
uA
tx tx
0)(
]2)(2[
)(
222
2
222
2
CBAu
uCBA
CBAu
If we can select ,, and such that,
and
Then
and the general solution is ,
022 CBA
022 CBA
0
u
)()( qpu
hyperbolic Equations It can be shown that, if the equation is hyperbolic if B2>4AC,
=2A=2A
Satisfies this condition. Hence hyperbolic equations have two characteristic given by,
ACBB 42 ACBB 42
tconstACBBAx tan)4(2 2
tconstACBBAx tan)4(2 2
Parabolic Equation
It can be shown that the equation is parabolic, if 4AC-B2=0
The solution of the quadric equation,
is
If =2A, and =-B, the coefficient of , that is,
02
CBA
A
B
2
u
0)4(
4)2(2
2)(2
2
BAC
ACABAB
CBA
The partial differential equation is reduce to,
The solution is,
Where p and q are arbitrary functions.
02
2
u
)()( qpu
The parabolic equations have only one characteristic given by,
2Ax-Bt = constant
Elliptic Equation
Elliptic equations have no characteristics. However, the transformation,
Reduce the partial differential equation to,
24
2
BAC
BtAx
t
02
2
2
2
uu
Let consider the cubic element of the fluid (gas).
Y
Z
X
x
zyxx
pp
)(zyp z
y
x
zyxx
pFx
net force in the positive x direction is
similarly the net force in the positive y and z directions are
zyxy
pFy
zyxz
pFz
The net vector force on the cubical element is therefore
.zyxpF
where
z
pk
y
pj
x
pip
Using Newton’s second low
2
2
0 tp
Let write the divergence of the gradient of pressure
2
2
0
).(.
tpp
The incremental pressure and the accompanying dilation are linearly related through the bulk modules B,
BBp
By eliminating .
2
2
22 1
t
p
cp
f
where 2/1
0
B
c f
The left side of equation is the three dimensional laplacian of the pressure p
Thank you for your attention
Potential of the velocity field in fluid
ASSIGNMENT NO.3
IRROTTIONAL FLOW
Lets consider the two point in the fluid with distance of dr
z
x
yA
B
dr
Les conceder the two points of A and B as a diagonal of the rectangular pipe.
A
C
B
E
dy
dz
dx
D
We can write the velocity of the point c can be given in the term of the velocity of point A.
dxx
Vvv Ac
dxx
Vvv Ac
This equation can be written in three dimensional form
dxkx
Vzdxj
x
Vdxi
x
Vvv yx
Ac
Hence the velocity at the point c can be represent in the term of velocity of the point A
dxx
Vx
dxx
Vy
dxx
Vz
A
E
D
Cy
x
z
Like in the point c, the velocities at the other two are also can be written relative to the point A
E
D
dxx
Vx
dxx
Vy
dxx
Vz
A
Cy
x
z
dyy
Vy
dyy
Vx
dyy
Vz
dzz
Vy
dzz
Vz
dzz
Vx
Now conceder particle C. it is clear that is the rate of elongation of line segment AC.
dxxVx
We can write the elongation rate per unit original length,as . This is known as the normal stain
xVx xx
.
.
.
zzz
yyy
xxx
z
V
y
Vx
V
Hence,
Lets investigate the rate of the angular change of the sides of the rectangular pipe.
The average rate of rotation about the z axis of the orthogonal line segments AC and AD is
y
V
x
Vxy
z 2
1
The rate of change of the angle CAD (a right angle at time t ) becomes
y
V
x
Vxy
Thus we know that the time rate of change of the shear angle xy so that
y
V
x
Vxy
yxxy
..
Similarly,
x
V
z
V zxzxxz
..
y
V
z
Vzy
zyyz
..
Accordingly, we have available to describe the deformation rate of the rectangular parallel piped the strain rate terms which we now set forth as follows:
...
..
.
...
22
22
22
zzyzxz
yzyy
xy
xzxyxx
=strain rate tensor
Now lets consider the rigid body rotation. Thus, the expression
y
V
x
Vxy
2
1
Is actually more than just the average rotation of line segments dx and dy about the z axis . It represents for a deformable medium what we maybe consider as the rigid_body angular velocity z about the z axis. This is,
y
V
x
Vxy
z 2
1
Similarly,
x
V
z
V zxy 2
1
z
V
y
V yzx 2
1
kx
V
z
Vj
z
V
y
Vi
y
V
x
Vzxyzxy
2
1
2
1
2
1
Thus,
VcurlV 2
1
2
1
Hence,
kx
V
z
Vj
z
V
y
Vi
y
V
x
Vcurl zxyzxy
2
1
2
1
2
1
At this time ,we define irrotational flow as those for which =0 at each point in the flow.
For irrotational flow, we require that
0
y
V
x
Vxy
0
z
V
y
V yz
0
x
V
z
V zx
Thus it become clear that another criterion for irrotationality, and the one we will use,is
Curl V = 0
The velocity potential
If velocity components at all points in a region of flow can be expressed as continuous partial derivatives of a scalar function (x,y,z,t) thusly
z
tzyxV
y
tzyxV
x
tzyxV
z
y
x
),,,(
),,,(
),,,(
In general we can write
gragV
0
0
0
22
22
22
xyyx
zxxz
yzzy
Then the flow must be irrotational. Hence,
Relationship between the stream function and the velocity field.
It will now be demonstrated that a relativity simple relation exists between the stream function (x,y,t) and the velocity field v(x,y,t).
y
x
X0,y0
X,y
dy
dxVx
The flow q associated with the path from x0y0 to the extremity of dy may be expressed in two ways. These are equated as
dyVtyxdyy
tyxtyx xyxyx
),,(),,(
),,(0000
After cancellation of equal terms in both sides,
yVx
Performing the same computation for the extreme point of the dx segment leads to the result.
xVy
Relation between the steam function and velocity potential for flows which are irrotational as well as two dement ional and incompressible.
We will now add the restriction of irrotationality to the previous restrictions. This means the existence of a velocity potential . For two dimensional flow, must be a function of x,y, and t. By equating the corresponding expressions of velocity involving the stream function and velocity potential,the following relationship can be established.
yx
xy
In complex variable theory, this equation is known as Cauchy_Riemann equation.
ReferenceMechanics of Fluids
By
Irving H Shames
THANK YOU FOR YOUR ATTENTION
MATLAB SOLUTIONS FOR PDE
There are two methods for solve the PDEs in Matlab.
•Using Command Line window
•Using GUI(Graphical User Interface) window
ASSIGNMENT NO.4
•Finite element method
Using the MATLAB PDE TOOLBOX
The MATLAB PDE Toolbox is a tool for solving two-dimensional linear partial differential equations by finite element methods. This presentation is intended to guide you through the use of the graphical user interface for interactive use. It assumes that the reader is running MATLAB and the PDE toolbox in a graphical environment.At the MATLAB prompt, type pdetool. You should see a window like the one below pop up.
This is an example which have been done in second assignment , and which you might solve by Fourier series methods, by hand; you will find that you obtain the solution faster with the PDE Toolbox, and that you also obtain a visual representation, which may contribute to the more important goal of understanding.
1. Draw the region of interest: suppose we want the rectangle (x,y) = [0, 3] X [0, 1]. o In the Options menu, select Grid and Snap; then set the Axes' Limits to [-1.5 1.5] and [-1.5 1.5]. When finished, click Apply and Close the Axes' Limits window.
1.
o In the Draw menu, select Rectangle/Square (not centered). o In the drawing region, click and hold with the cursor at the origin, and drag it to the point with coordinates (3,1). The region will be shaded and marked R1.
2. Set up the boundary conditions: suppose we want the unknown solution function u to be -1 on the long horizontal sides, and +1 on the short vertical sides. o In the Boundary menu, select Boundary Mode. The window changes to this:
o In the drawing region, double click on the left edge of the rectangle. The boundary condition window pops up. The default Dirichlet condition h*u=r is OK, the default h=1 is OK; change the value of r to +1.
3.
o Do the same thing with the right edge: double-click, set the value. On the top and bottom edges, set r to -1. Be patient and careful. When the boundary conditions are correctly set, you should be able to click on any edge, and see the correct value for r. 4. Specify the particular PDE of interest: o In the PDE menu, select PDE specification. The pde specification window pops up; we are doing the default type of problem, hyperbolic, of the form -div(c*grad(u)) + a*u = f, where a=0, f=0,d=1 and c=1. Change the values as needed, and click on OK.
4. Generate the mesh: o In the Mesh menu, select Initialize Mesh. This time we will use the default mesh.
5. Solve the Finite Element problem: o In the Solve menu, select Solve PDE. The figure in the main drawing area changes.
6. Generate other plots: o In the Plot menu, select Parameters. The plot parameter window pops up; click on the check boxes for color, contour, and height, then on the Plot button.
A new figure window pops up, with the appropriate plot.
% Animation for wave propagation problem for ass2
%echo on
clc
g='squareg'; % The unit square
b='Boundary'; % (boundary condition) boundary is the function defined by the user and should be created before run the program
c=1;
a=0; coefficient of the hyperbolic function
f=0;
d=1;
[p,e,t]=initmesh('squareg'); %Mesh(p and e are mesh parameters)
clc
x=p(1,:)';
y=p(1,:)';
u0=atan(cos(pi/2*x)); % The initial conditions:
ut0=3*sin(pi*x).*exp(sin(pi/2*y)); % The initial conditions:
clc
n=20;
tlist=linspace(0,5,n); % We want the solution at 20 points in time between 0 and 5.
uu=hyperbolic(u0,ut0,tlist,b,p,e,t,c,a,f,d); %Solve hyperbolic problem
Clc % To speed up the plotting, we interpolate to a rectangular grid.
delta=-1:0.1:1;
[uxy,tn,a2,a3]=tri2grid(p,t,uu(:,1),delta,delta); NOT ESSENTIAL FOR RUN THE PROGRAM
gp=[tn;a2;a3];
newplot; % prepare the new figure window with axis
M=moviein(n); %initialized movie frame memory
umax=max(max(uu)); maximum and minimum values for Z axis
umin=min(min(uu));
for i=1:n,...
if rem(i,10)==0,...
fprintf('%d ',i);...
end,...
pdeplot(p,e,t,'xydata',uu(:,i),'zdata',uu(:,i),'zstyle','continuous',...
'colormap','gray','mesh','on','xygrid','off','gridparam',gp,'colorbar','on');...
axis([-1 1 -1 1 umin umax]); caxis([umin umax]);...
title('Acoustic Wave in Gas');
zlabel('Pressure level');
M(:,i)=getframe;...
if i==n,...
fprintf('done\n');...
end,...
end
nfps=5; used for speed up animation
movie(M,10,nfps); not essential for run the program
echo off
Program for function “boundary”function [q,g,h,r]=Boundary(p,e,u,time)
% Boundary Boundary condition data
bl=[
1 1 1 1
0 1 0 1
1 1 1 1
1 1 1 1
48 1 48 1
48 1 48 1
48 48 42 48
48 48 120 48
49 49 49 49
48 48 48 48
];
if any(size(u))
[q,g,h,r]=pdeexpd(p,e,u,time,bl);
else
[q,g,h,r]=pdeexpd(p,e,time,bl);
end
This program is as same as in the programs in MATLAB demos.Refer MATLAB demo files for details
When function u and its derivatives are single_valued,finite and continuous function of x,then by Taylor’s theorem,
...)('''6
1)(''
2
1)(')()( 32 xuhxuhxhuxuhxu
and
...)('''6
1)(''
2
1)(')()( 32 xuhxuhxhuxuhxu
1
2
Finite element methods
Addition of these expansions gives
)()('')(2)()( 42 hOxuhxuhxuhxu
)}()(2)({1
)(''22
2
hxuxuhxuhdx
xdxu
xx
By subtraction eqn2 from eqn1
)}()({2
1)(' hxuhxu
hdx
duxu
xx
3
4
5
A P
B
The slope of the chord PB is given by the forward difference formula.
)}()({1
)(' xuhxuh
xu 6
The slope of the chord AP is given by the backward difference formula.
)}()({1
)(' hxuxuh
xu 7
Notation for function of the independent variables x and t. Subdivide the x_t plane into sets of equal rectangles of sides x=h, t=k, as shown in Figer,and let the co_ordinate (x,t) of the representative mesh point P be
X=ih; t=jk
Where I and j are integers.
P(ih,jk)
i,j
i,j-1
i+1,j
i,j+1
i,j+1
t
x
k
h
Denote the value of u at P by
jip ujkihuu ,),(
Then by equation 4
2
,
2
2
2
2 },)1{(},{2},)1{(
h
jkhiujkihujkhiu
u
u
u
u
jip
2
,1,,1
,
2
2 }2
h
uuu
u
u jijiji
ji
With leading error of order h2.
Similarly,
2
,,,
,
2
2 }121
k
uuu
u
u jijiji
ji
With leading error of order k2.
With this notation the forward difference approximation for u/t at P is
k
uu
t
u jiji ,1,
With leading error of k
THANK YOU FOR YOUR ATTENTION