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CHAPTER – 3
Application Of Spline Collocation To The Partial Differential Equation Of One As Well As Two
Space Variables
3.1 Spline Formula To Solve Parabolic Linear Partial
Differential Equation Having One Space Variable
3.2 Flow of Electricity In A Cable of Transmission Lines
3.3 Spline Solutions With Explicit Method and Implicit Method
3.4 The Heat Conduction Problem
3.5 Spline Solutions With Explicit Method and Implicit Method
3.6 Spline Formula To Solve Parabolic Partial Differential
Equation With Two Space Variables
3.7 Heat Flow In A Thin Rectangular Plate
3.8 Spline Solutions With Explicit Method and Implicit Method
3.9 Spline Formula To Solve Hyperbolic Partial Differential
Equation With One Space Variables
3.10 The Flow Of Electricity In The Transmission Lines.
3.11 Spline Solutions With Explicit Method and Implicit Method
3.12 Vibrating String Problem
3.13 Spline Solutions With Explicit Method and Implicit Method
3.14 Spline Formula To Solve Hyperbolic Partial Differential
Equation With Two Space Variables
3.15 The Problem of Vibrating Membrane
3.16 Spline Solutions With Explicit Method and Implicit Method
3.17 Conclusion
101
3.1 SPLINE FORMULA TO SOLVE PARABOLIC LINEAR
PARTIAL DIFFERENTIAL EQUATION HAVING ONE
SPACE VARIABLE :
Consider parabolic PDE having one space variable.
0 t l, x 0 u c u xx2
t ><<= …(3.1.1)
with Dirichilet boundary conditions
0 t)u(l,
0 t)u(0,
=
= …(3.1.1a)
along with initial condition,
l x 0 ; f(x) 0) u(x, ≤≤= …(3.1.1b)
We divide the region l x 0 ≤≤ into, say n-equal
subinterval each of width h. Let us denote the points of
subdivisions by n210 x....., , x, x,x . Let us have the solution on
hand at time tj∆ at the mesh points n210 x....., , x, x,x . Let j i,u
denote the value of u at the thi mesh point at time tj∆ .
We approximate the function u at time tj∆ by a cubic
spline S(X): calculate the value of )(xS i′′ by solving set
of simultaneous equations given by (2.4.15a) for i = 1, 2, 3,
…, n-1. We should note that values of u at x = 0 and x = 1
are already known. Now discretizing left side of equation
(3.1.1) by forward difference formula and replacing right
side by the second derivatives )(xS i′′ at thj level like explicit
scheme in finite difference. We get
j i,2
j i,1j i, S c t / )u - (u ′′=∆+ …(3.1.3)
where j i,S ′′ denotes )(xS i′′ at thj level.
102
Substitute values of j i,S ′′ from (3.1.3) into equation
(2.4.15a) and get
j 1,ij i,j 1,-i
1j 1,i1j i,1j 1,-i
u 6r) (1 u 12r) - (4 u 6r) (1
u u 4 u
+
++++
++++=
++
…(3.1.4)
where 22 ht / c r ∆= . This set of simultaneous equations can
now be solved equation (3.1.4) known as cubic spline
Explicit formula to solve equation (3.1.1).
The finite difference replacement of (3.1.1)
corresponding to implicit scheme is
)S S( (1/2) c t
u - u1j i,j i,
2j i,1j i,+
+ ′′+′′=∆
…(3.1.5)
where 1j i,j i, S ,S +′′′′ denote second derivatives at i xx = at the
time level j and j + 1 respectively.
However, second derivatives at th1) (j+ level cannot be
computed as the values of u are not known. We can express
(3.1.4) in terms of u as follows.
We use the relationship (2.14.5a) and rewrite it in the
following forms.
)u - u 2 - (u )(6/h S S S j 1,ij i,j1,-i2
j 1,ij i,j 1,-i ++ =′′+′′+′′ …(3.1.6)
1) -(n ...., 3, 2, 1, i
)u - u 2 - (u )(6/h S S S 1j 1,i1j i,1j1,-i2
1j 1,i1j i,1j 1,-i
=
=′′+′′+′′ ++++++++
…(3.1.7)
putting the value of 1j i,S +′′ etc from (3.1.5) into (3.1.6), we get
103
[ ]
)u u 2 - u ( (6/h2)
S ) u - u ( t)c /2(
S - ) u - u ( t)c /2(4 S - ) u - u ( t)c /2(
1j 1,i1j i,1j 1,-i
j 1,ij 1,i1j 1,i2
j i,j i,1j i,2
j 1,-ij 1,-i1j 1,-i2
++++
++++
++
+=
′′+∆+
′′+∆+′′∆
which gives
j 1,ij i,j 1,-i
j 1,ij i,j1,-ij 1,ij i,j1,-i
j 1,ij i,j1,-ij 1,ij i,j 1,-i2
1j 1,i1j i,1j 1,-i
u 3r) (1 u 6r) - (4 u 3r) (1
)u u 4 (u )u u 2 - (u3r
)u u 4 (u )S S 4 S( t/2)(c
u 3r) - (1 u 6r) (4 u 3r) - (1
+
++
++
++++
++++=
++++=
+++′′+′′+′′∆=
+++
…(3.1.8)
where 1) -(n ...., 3, 2, 1, i , ht / c r 22 =∆= . Above equation
(3.1.8) is known as cubic spline implicit formula to solve
equation (3.1.1).
Now 1j n,1j 0, u and u ++ are known due to the prescribed
boundary conditions. The set of simultaneous equations
obtained in explicit as well as implicit scheme contains only
(n – 1) unknowns. These (n – 1) equations in (n – 1)
unknown can be solved by any standard method.
Once the values of u are known at th1) (j+ level, we can
proceed to compute the next level (j + 2) by repeating the
same process. For each set of simultaneous equations in
(n – 1) unknowns give tri-diagonal matrix. It can be solved
by any standard method, thus the method can proceed by
steps.
The convergence and stability of these methods totally
depend upon the values of r. For explicit scheme, we have to
choose the value of r such that 1/2 r 0 ≤≤ . Implicit scheme
104
has advantage that there is no limitation on the value of r
for convergence and stability along with small values are
more accurate. Values much larger than unity are not
desirable. These two methods will be discussed late on by
taking its actual approximation to a problem.
3.2 FLOW OF ELECTRICITY IN A CABLE OF
TRANSMISSION LINES :
This is the study of the derivation of partial differential
equations from physical principles, we consider the flow of
electricity in a cable of transmission line. We assume the
cable to be imperfectly insulated, so that there is both
capacitance and current leakage to ground. Figure (3.2.1a)
shows such a cable with the electromotive source x = 0 and
the load x = 1.
Figure (3.2.1)
Flow of electricity in a cable
Here the current returns through the earth. Consider
an element PQ of the cable with P at a distance x from the
source e(x, t), the potential i(x, t), the current at p at any
time t. Let R denote the resistance, L inductance, G the
105
conductance (leak ness) and C the capacitance per unit
length of the cable. Let x∆ be the length of the element PQ
and i i ∆+ , e e ∆+ be the current and the electromotive force,
respectively, at Q. Figure (3.2.1b) gives the electrical circuit
for the element PQ. Since the drop in voltage from P to Q is
due to the resistance xR∆ and inductance xL∆ , we have
t / i x)(L i x)(R e - ∂∂∆+∆=∆ …(3.2.1)
Again, since the drop in the current t -∆ is due to the
conductance xG∆ and capacitance xC∆ , we have
t / e x)(C e x)(G i - ∂∂∆+∆=∆ …(3.2.2)
Dividing equations (3.2.1) and (3.2.2) by x∆ and taking the
limit as 0 x →∆ , we get
0 Ri )t / i( L x / i =+∂∂+∂∂ …(3.2.3)
and
0 Ge )t / e( C x / i =+∂∂+∂∂ …(3.2.4)
Differentiating equation (3.2.3) with respect to x and
equation (3.2.4) with respect to t, we get
0 x / iR )t x / i( L x / e 222 =∂∂+∂∂∂+∂∂
and
0 t / eG ) t / e( C t x / i 222 =∂∂+∂∂+∂∂∂
If we eliminate the term tx / i2 ∂∂∂ between these two
equations and then substitute for t / i ∂∂ from equation
(3.2.3), we obtain
RGe t / e LG) (Re ) t / e( Le x / e 2222 +∂∂++∂∂=∂∂ …(3.2.5)
Similarly we obtain
RGi t / i LG) (Re ) t / i( Le x / i 2222 +∂∂++∂∂=∂∂ …(3.2.6)
106
Equation (3.2.5) and (3.2.6) are known as the telephone
equations.
Two special case of the telephone equations are
t / e Rc x / e 22 ∂∂=∂∂ …(3.2.7)
t / i Rc x / i 22 ∂∂=∂∂ ...(3.2.8)
Equation (3.2.7) and (3.2.8) are obtained from
equations (3.2.5) and (3.2.6) by taking G = L = 0 i.e. leakage
and inductance are negligible, for example, in telegraphic
transmission through submarine cables. These equations
are known as telegraphic equations. Mathematically these
equations are similar to parabolic type linear PDE having
one space variable. Equations (3.2.7) and (3.2.8) can be
rewrite as
xxt u ) l/RC ( u =
where u(x, t) stands for either i(x, t) or e(x, t).
3.3 (A) SPLINE SOLUTIONS WITH EXPLICIT SCHEME :
Let xxt u 1/Rc u = …(3.3.1)
Let the initial current in the cable be f(x), then the
initial condition is
u(x, 0) = f(x) …(3.3.2)
where f(x) is the given function. Also the boundary
conditions are
u(0, t) = 0 and u(l, t) = 0 ∀ t …(3.3.3)
We shall determine the solution of the equation (3.3.1)
satisfying equations (3.3.2) and (3.3.3). Suppose the length
of the cable is l and the given function is
1 x 0 ; sin f(x) <<= π …(3.3.4)
107
Let length of the cable is subdivided into 20
subintervals. We get 1/20 x h =∆= . Also let 1/400 t =∆ and
RC = 1/100, we get r = 0.01 which gives
1 + 6r = 1.06 and 4 – 12r = 3.88
Substituting the values of 1 + 6r and 4 – 12r in
equation (3.1.4) and using initial conditions, we get
For j = 0
0.934523
u (1.06) u (3.88) u (1.06) u u 4 u 1, i 0 2,0 1,0 0,1 2,1 1,1 0,
=
++=++=
Since 0.934523 u u 4 0 u 1 2,1 1,1 0, =+=
1.846040
u (1.06) u (3.88) u (1.06) u u 4 u 2, i 0 3,0 2,0 1,1 3,1 2,1 1,
=
++=++=
2.712093
u (1.06) u (3.88) u (1.06) u u 4 u 3, i 0 4,0 3,0 2,1 4,1 3,1 2,
=
++=++=
3.511370 u u 4 u 4, i 1 5,1 4,1 3, =++=
4.224184 u u 4 u 5, i 1 6,1 5,1 4, =++=
4.832986 u u 4 u 6, i 1 7,1 6,1 5, =++=
5.322783 u u 4 u 7, i 1 8,1 7,1 6, =++=
5.621516 u u 4 u 8, i 1 9,1 8,1 7, =++=
5.900351 u u 4 u 9, i 1 10,1 9,1 8, =++=
5.973899 u u 4 u 10, i 1 11,1 10,1 9, =++=
Since 1 11,1 9, u and u are symmetric, we get
5.973899 u 4 u 2 i 210,1 9, =+
Here we get 10 algebraic equations in 10 unknowns
with tri-diagonal matrix. This system of equations is
solved easily by any well known method like crout’s method,
108
similarly for j = 1, we get another 10 algebraic equations.
This can be solved easily by above method. Proceeding in
this way, the results obtained by explicit method are shown
in table (3.3.1) and plotted in figure (3.3.1).
Table (3.3.1)
Current distribution in the cable through cubic spline explicit method
Current in the cable u →
X t=0.0 t=1/400 t=1/200 t=3/400 t=1/100 t = 1/80
0.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.05 0.156434 0.156396 0.156357 0.156318 0.156280 0.156241 0.10 0.309017 0.308941 0.308864 0.308788 0.308711 0.308635 0.15 0.453991 0.453878 0.453766 0.453654 0.453542 0.453429 0.20 0.587785 0.587640 0.587495 0.587349 0.587204 0.587059 0.25 0.707107 0.706932 0.706757 0.706582 0.706407 0.706233 0.30 0.809017 0.808817 0.808617 0.808417 0.808217 0.808017 0.35 0.891007 0.890786 0.890566 0.890346 0.890126 0.889906 0.40 0.951057 0.950821 0.950586 0.950351 0.950116 0.949882 0.45 0.987688 0.987444 0.987200 0.986956 0.986712 0.986468 0.50 1.000000 0.999753 0.999506 0.999258 0.999011 0.998764
0.000000
0.200000
0.400000
0.600000
0.800000
1.000000
1.200000
0.0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
X
U
t = 0.0
t = 1/400
t = 1/200
t = 3/400
t = 1/100
t = 1/80
Figure (3.3.1)
Current distribution in the cable through cubic spline explicit method
109
3.3(B) SPLINE SOLUTIONS WITH IMPLICIT METHOD :
In this section we discuss the solution of equation
(3.3.1) by implicit scheme. Substitute the values of r and
using initial conditions in (3.1.8), we get,
For j = 0 i.e. at t = 1/400
0.934639 u (0.97) u (4.06) u (0.97) 1, i 1 2,1 1,1 0, =++=
since 0 u 0,1 = , we get
0.934639 u (0.97) u (4.06) 1 2,1 1, =+
1.846264 u (0.97) u (4.06) u (0.97) 2, i 1 3,1 2,1 1, =++=
2.712429 u (0.97) u (4.06) u (0.97) 3, i 1 4,1 3,1 2, =++=
3.511804 u (0.97) u (4.06) u (0.97) 4, i 1 5,1 4,1 3, =++=
4.22407 u (0.97) u (4.06) u (0.97) 5, i 1 6,1 5,1 4, =++=
4.833583 u (0.97) u (4.06) u (0.97) 6, i 1 7,1 6,1 5, =++=
5.323441 u (0.97) u (4.06) u (0.97) 7, i 1 8,1 7,1 6, =++=
5.682218 u (0.97) u (4.06) u (0.97) 8, i 1 9,1 8,1 7, =++=
5.901080 u (0.97) u (4.06) u (0.97) 9, i 1 10,1 9,1 8, =++=
5.974638 u (0.97) u (4.06) u (0.97) 10, i 1 11,1 10,1 9, =++=
since 1 11,1 9, u and u are symmetric, we get,
5.974638 u (4.06) u (0.97) 1 10,1 9, =+
Here we get 10 algebraic equations in 10 unknowns
with tri-diagonal matrix. This can be solved by any
standard method. Similarly, applying above process are, we
get the solution at t = 1/200, 3/400, 1/100, 1/80 etc. and
they are shown in table (3.3.2) and plotted in figure (3.3.2).
110
Table (3.3.2)
Current distribution in the cable through cubic spline implicit method
Current in the cable u →
X t=0.0 t=1/400 t=1/200 t=3/400 t=1/100 t = 1/80
0.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.05 0.156434 0.156396 0.156357 0.156318 0.156280 0.156241 0.10 0.309017 0.308941 0.308866 0.308789 0.308713 0.308636 0.15 0.453991 0.453878 0.453759 0.453648 0.453536 0.453430 0.20 0.587785 0.587640 0.587496 0.587351 0.587205 0.587060 0.25 0.707107 0.706932 0.706757 0.706582 0.706408 0.706233 0.30 0.809017 0.808817 0.808617 0.808417 0.808217 0.808018 0.35 0.891007 0.890786 0.890566 0.890346 0.890126 0.889906 0.40 0.951057 0.950821 0.950586 0.950352 0.950117 0.949882 0.45 0.987688 0.987444 0.987200 0.986956 0.986712 0.986468 0.50 1.000000 0.999753 0.999506 0.999258 0.999011 0.998764
0.000000
0.200000
0.400000
0.600000
0.800000
1.000000
1.200000
0.0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
X
U
t = 0.0
t = 1/400
t = 1/200
t = 3/400
t = 1/100
t = 1/80
Figure (3.3.2)
Current distribution in the cable through cubic spline implicit method
111
3.3(C) DISCUSSION OF RESULTS :
Here table (3.3.3) gives the comparison of both the
spline solutions namely explicit and implicit with exact
solutions.
From the table (3.3.3) it is clear that, the spline
solutions are fairly agree with exact solutions up to five
digits of decimal points. Figure (3.3.3(a)) indicates the error
analysis which compares the exact solution with spline
solution obtained by both the methods at t = 1/80. The
figure (3.3.3(b)) gives good agreement of curves presenting
exact and approximate solutions obtained by cubic spline
method.
Table (3.3.3)
Error analysis
Current in the cable u → t = 1/80 X USE USI UEXT USE-UEX USI-UEX
0.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.05 0.156241 0.156241 0.156242 0.000001 0.000001 0.10 0.308635 0.308636 0.308636 0.000001 0.000000 0.15 0.453429 0.453430 0.453431 0.000002 0.000001 0.20 0.587059 0.587060 0.587060 0.000001 0.000000 0.25 0.706233 0.706233 0.706235 0.000002 0.000002 0.30 0.808017 0.808078 0.808019 0.000002 0.000001 0.35 0.889906 0.889906 0.889908 0.000002 0.000002 0.40 0.949882 0.949882 0.949884 0.000002 0.000002 0.45 0.986468 0.986468 0.986471 0.000003 0.000003 0.50 0.998764 0.998764 0.998767 0.000003 0.000003
[USE and USl are current in the cable obtained by
using explicit, implicit method respectively. UEXT
denote the exact solution.]
112
0.000000
0.000001
0.000001
0.000002
0.000002
0.000003
0.000003
0.000004
0.0
0.050.100.150.200.250.300.350.400.450.50
X
ERROR
USE-UEXT
USI-UEXT
Figure (3.3.3(a))
Error analysis (at t=1/80)
113
0.000000
0.200000
0.400000
0.600000
0.800000
1.000000
1.200000
0.0
0.05
0.10
0.15
0.20
0.250.300.35
0.40
0.45
0.50
X
U
USE
USl
UEX
Figure (3.3.3(b))
Current distribution in the cable through cubic spline method
(at t=1/80)
114
3.4 THE HEAT CONDUCTION PROBLEM :
Consider a thin long rod surrounded except at
the ends with a material impervious to heat unless all the
points of the rod are at the same temperature, heat will flow
along the rod. If the rod is homogeneous and of the same
cross section throughout, we may schematically regard the
rod as a line, since the temperature of all the points of any
cross section will sensibly the same. When heat is flowing
uniformly, it is experimentally known that the amount of
heat flow across any portion of the rod is proportional to the
difference of temperatures of the end points of the portion,
to the area of cross – section and to the end points of the
flow and inversely proportional to the length of the portion
considered. Taking the limiting case when the length of the
portion considered tends to zero, we obtain the quantity of
heat 1Q that flows across any section of the rod as,
A xu
k - Qx
1
=δδ
per sec …(3.4.1)
where, k : Coefficient of conductivity
u : The temperature at a distance x from some
fixed point
on the rod
and A : Area of cross section
The negative being attached as the heat flows from a
higher to lower temperature. If we take the section at a
point x x δ+ , then a quantity of a heat that flow out across
this section is given by
115
A xu
k - Qxx
2δδ
δ
+
= per sec …(3.4.2)
thus from (3.4.1) and (3.4.2) above the quantity of heat
gained by this section per sec is
A xu
- xu
k Q - Qxxx
21
=+ δ
δδδ
δ …(3.4.3)
The rate of rise of temperature is tu/δδ . Therefore 21 Q - Q is
also given by
tu
.A.x P. S. Q - Q 21 δδδ= …(3.4.4)
where S = Specific heat
and P = Density of the rod
equating the values of 21 Q - Q from (3.4.3) and (3.4.4) and
dividing by xδ , we have
x
xu
- xu
tu
P. S. xxx
δδδ
δδ
δδ δ
= + …(3.4.5)
taking the limit of this equation as 0 x →δ , we get
2
2
xu
k tu
P. S.δδ
δδ
=
or 2
2
xu
P. S.
k
tu
δδ
δδ
=
writing P. S.
k a = the equation of one dimension heat flow is
2
2
xu
a tu
δδ
δδ
= …(3.4.6)
In shortly, we summarize these phenomena by
considering a homogeneous rod of length L. The rod is
sufficiently thin, so that the heat is disturbed equally over
116
the cross section at time t. The surface of the rod is
insulated and therefore there is no heat loss through the
boundary. The temperature distribution of the rod is given
by the solution of initial boundary value problem
≤≤=
≤=
≥=
><<=
L x 0 f(x), 0) u(x,
0 t 0, t)u(L,
0 t 0, t)u(0,
0 t L x 0 , u a u xxt
…(3.4.7)
3.5 (A) SPLINE SOLUTIONS WITH EXPLICIT METHOD :
We illustrate the problem by considering a
homogeneous rod of length l meter and fluid having unit
viscosity (v) and the temperature is given by function f(x) =
x (1 – x), then phenomena can be written as,
0 t ; 1 x 0 u a u xxt ><<= …(3.5.1)
0 t 0 0) u(l,
0 t)u(0,≥
=
= …(3.5.2)
and
1 x 0 x)- (1 x 0) u(x, ≤≤= …(3.5.3)
We shall determine the solution of the equation (3.5.1)
satisfying equation (3.5.2) and (3.5.3).
Let us take 500
1 k t ,
101
h x ==∆==∆
a = 1 and r = 0.2 which gives,
1 + 6r = 2.2 and 4 – 12r = 1.6
Substituting the values of (1 + 6r) and (4 – 12r) in
equation (3.1.4) and using initial conditions, we get
for j = 0, i = 1,
117
0.498 u (2.2) u (1.6) u (2.2) u u 4 u 0 2,0 1,0 0,1 2,1 1,1 0, =++=++
since 0 u 1 0, =
0.498 u u 4 1 2,1 1, =+
i = 2,
0.916 u (2.2) u (1.6) u (2.2) u u 4 u 0 3,0 2,0 1,1 3,1 2,1 1, =++=++
i = 3,
1.216 u (2.2) u (1.6) u (2.2) u u 4 u 0 4,0 3,0 2,1 4,1 3,1 2, =++=++
i = 4,
1.396 u (2.2) u (1.6) u (2.2) u u 4 u 0 5,0 4,0 3,1 5,1 4,1 3, =++=++
i = 5,
1.456 u (2.2) u (1.6) u (2.2) u u 4 u 0 6,0 5,0 4,1 6,1 5,1 4, =++=++
i = 6,
1.396 u (2.2) u (1.6) u (2.2) u u 4 u 0 7,0 6,0 5,1 7,1 6,1 5, =++=++
i = 6,
1.396 u (2.2) u (1.6) u (2.2) u u 4 u 0 7,0 6,0 5,1 7,1 6,1 5, =++=++
i = 6,
1.396 u (2.2) u (1.6) u (2.2) u u 4 u 0 7,0 6,0 5,1 7,1 6,1 5, =++=++
i = 7,
1.216 u (2.2) u (1.6) u (2.2) u u 4 u 0 8,0 7,0 6,1 8,1 7,1 6, =++=++
i = 8,
0.916 u (2.2) u (1.6) u (2.2) u u 4 u 0 9,0 8,0 7,1 9,1 8,1 7, =++=++
i = 9,
0 10,0 9,0 8,1 10,1 9,1 8, u (2.2) u (1.6) u (2.2) u u 4 u ++=++
0.408 u 4 u 1 9,1 8, =+
since 0 u 1 10, =
118
Hence we get 9 algebraic equations in 9 unknowns
with tri-diagonal matrix. This system of equation is solved
by any well known method. Similarly, for j = 1, we get
another 9 algebraic equations. This can be solved by above
method. Proceeding in this way, the results obtained by
explicit method are shown in table (3.5.1) and plotted in
figure (3.5.1).
Table (3.5.1)
Temperature in a rod through cubic spline solution by explicit method
Temperature u(x, t)
X t=0.0 t=0.002 t=0.004 t=0.006 t=0.008 t=0.01
0.0 0.00000
00
0.00000 0.00000 0.00000 0.00000 0.00000
0 0.1 0.09 0.086 0.0828 0.08008 0.07768 0.07551
04 0.2 0.16 0.156 0.152 0.14816 0.144512 0.14104
9 0.3 0.21 0.206 0.202 0.198 0.194032 0.19012
1 0.4 0.24 0.236 0.232 0.228 0.224 0.22000
6 0.5 0.25 0.246 0.242 0.238 0.234 0.23
0.6 0.24 0.236 0.232 0.228 0.224 0.22000
6 0.7 0.21 0.206 0.202 0.198 0.194032 0.19012
1 0.8 0.16 0.156 0.152 0.14816 0.144512 0.14104
9 0.9 0.09 0.086 0.0828 0.08008 0.07768 0.07551
04 1.0 0.00000
0 0.00000 0.00000 0.00000 0.00000 0.00000
0
119
0.000000
0.050000
0.100000
0.150000
0.200000
0.250000
0.300000
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
X
U
t=0.0
t=0.002
t=0.004
t=0.006
t=0.008
t = 0.01
Figure (3.5.1)
Temperature in the rod through cubic spline explicit method
3.5 (B) SPLINE SOLUTIONS WITH IMPLICIT METHOD :
In this section we discuss the solution of (3.5.1) by
implicit scheme. Substitute the values of r and using initial
conditions in (3.1.8), we get,
For j = 0 i.e. t = 0.002
i = 1 0.508 u (0.4) u (5.2) u (0.4) 1 2,1 1,1 0, =++
Since 0 u 1 0, = , 0.508 u (0.4) u (5.2) 1 2,1 1, =+
i = 2 0.928 u (0.4) u (5.2) u (0.4) 1 3,1 2,1 1, =++
i = 3 1.228 u (0.4) u (5.2) u (0.4) 1 4,1 3,1 2, =++
i = 4 1.408 u (0.4) u (5.2) u (0.4) 1 5,1 4,1 3, =++
i = 5 1.468 u (0.4) u (5.2) u (0.4) 1 6,1 5,1 4, =++
120
i = 6 1.408 u (0.4) u (5.2) u (0.4) 1 7,1 6,1 5, =++
i = 7 1.228 u (0.4) u (5.2) u (0.4) 1 8,1 7,1 6, =++
i = 8 0.928 u (0.4) u (5.2) u (0.4) 1 9,1 8,1 7, =++
i = 9 0.508 u (0.4) u (5.2) u (0.4) 1 10,1 9,1 8, =++
Since 0 u 1 10, =
0.508 u (5.2) u (0.4) 1 9,1 8, =+
Hence we get 9 algebraic equations in 10 unknowns
with tri-diagonal matrix. This system of equations is
solved by any well known method. Similarly, for j = 1, we get
another 9 algebraic equations. This can be solved by above
method. Proceeding in this way, the results obtained by
implicit method are shown in table (3.5.2) and plotted in
figure (3.5.2).
Table (3.5.2)
Temperature in a rod through cubic spline solution by implicit method
Temperature u(x, t)
X t = 0.0 t= 0.002 t =0.004 t =0.006 t =0.008 t = 0.01
0.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.000000.1 0.09000 0.08492 0.08208 0.07914 0.07703 0.074640.2 0.16000 0.15628 0.15137 0.14777 0.14357 0.140550.3 0.21000 0.20592 0.20233 0.19868 0.19411 0.189390.4 0.24000 0.23602 0.23185 0.22802 0.22366 0.220400.5 0.25000 0.24598 0.24209 0.23771 0.23440 0.229590.6 0.24000 0.23602 0.23185 0.22802 0.22366 0.220400.7 0.21000 0.20592 0.20233 0.19868 0.19411 0.189390.8 0.16000 0.15628 0.15137 0.14777 0.14357 0.140550.9 0.09000 0.08492 0.08208 0.07914 0.07703 0.074641.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
121
0.000000
0.050000
0.100000
0.150000
0.200000
0.250000
0.300000
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
X
U
t = 0.00
t = 0.002
t = 0.004
t = 0.006
t = 0.008
t = 0.01
Figure (3.5.2)
Temperature in rod through cubic spline implicit method
3.5(C) DISCUSSION OF RESULTS :
Table (3.5.3) gives the comparison of both the spline
solutions namely explicit and implicit with exact solution.
From table (3.5.3) it is clear that, the spline solutions are
fairly agree with exact solutions up to three digits of decimal
points, figure (3.5.3(a)) indicates the error analysis which
compares the exact solution with spline solution obtained
by both the methods t = 0.008. The figure (3.5.3(b)) gives
good agreement of curves presenting exact and approximate
solutions obtained by cubic spline method.
122
Table (3.5.3)
Error analysis
Temperature u(x, t) At t = 0.006
X USE USl UEX USE-UEX USI-UEX
0.0 0.000000 0.000000 0.000000 0.000000 0.000000
0.1 0.080080 0.079149 0.079683 0.000397 0.000534
0.2 0.148160 0.147771 0.148269 0.000109 0.000498
0.3 0.198000 0.198685 0.198467 0.000467 0.000218
0.4 0.228000 0.228024 0.227979 0.000021 0.000045
0.5 0.238000 0.237713 0.237569 0.000431 0.000144
0.6 0.228000 0.228024 0.227979 0.000021 0.000045
0.7 0.198000 0.198685 0.198467 0.000467 0.000218
0.8 0.148160 0.147771 0.148269 0.000109 0.000498
0.9 0.080080 0.079149 0.079683 0.000397 0.000534
1.0 0.000000 0.000000 0.000000 0.000000 0.000000
123
0.000000
0.000100
0.000200
0.000300
0.000400
0.000500
0.000600
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
X
ERROR
USE-UEX
USI-UEX
Figure (3.5.3(a))
Error analysis (at t=0.006)
124
0.000000
0.050000
0.100000
0.150000
0.200000
0.250000
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
X
U
USE
USl
UEX
Figure (3.5.3 (b))
Tem
perature distribution in the rod through cubic spline method
125
3.6 SPLINE FORMULA TO SOLVE PARABOLIC PARTIAL
DIFFERENTIAL EQUATION WITH TWO SPACE
VARIABLES :
Consider the parabolic differential equation having two
space variables.
0 t b, y 0 a, x 0 ; )u (u c u yyxx2
t ><<<<+= …(3.6.1)
With Dirichilet conditions prescribed on the
boundaries x = 0, x = a, y = 0, y = b. We should
subdivide the region a x 0 ≤≤ into, say M intervals, each of
width x∆ and b y 0 ≤≤ into N intervals of width y∆ such that
a x M =∆ and b y N =∆ . Let us denote the points of
subdivisions by xM......, , x, x,x 210 and yN ......, ,y ,y ,y 210 . Let
k - j i,u denote the value of u at the thj) (i, mesh point at the
time tk∆ . For simplicity, let us take a square region
N M a, y x, 0 =≤≤ and h(say) y x =∆=∆ .
We approximate the function u at time t∆ by a cubic
spline S(x). Discretizing the left side of equation (3.6.1) by
forward difference formula and replacing right side by twice
the second derivative i.e. )(x S 2 ij′′ at thk level like explicit
scheme in finite difference, we get
)S(2 c t / )u - (u k ij,2
k ij,1k ij, ′′=∆+ …(3.6.2)
where is k ij,S ′′ denotes )(x S ij′′ at thk
level.
Now with the help of equation (2.4.15a) and the value
of k ij,S ′′ obtained from equation (3.6.2) we get,
126
)u u 2 - (u )(6/h
t 2c
u - u
t 2c
u - u
t 2c
u - u
k 1j, ik ij,k 1j,-i2
2k 1j,i1k 1j,i
2k ij,1k ij,
2k 1j,-i1k 1j,-i
+
+++++
+=
∆+
∆+
∆
At last we get,
1 -N ....., 2, 1, j i,
)u 12r) (1 u 24r) - (4 u 12r) (1
u u 4 u
k 1j, ik ij,k 1j,-i
1k 1j, i1k ij,1k 1j,-i
=
++++=
++
+
++++
…(3.6.3)
where 22 ht / c r ∆= these set of (N – 1) x (N – 1) equation in
(N – 1) x (N – 1) unknowns can be solved by any well –
known method. The above set of simultaneous equations
gives square matrix. The equation (3.6.3) is known as cubic
spline explicit formula to solve equation (3.6.1). For this
method, the maximum possible value of r is ¼ but the
equation (3.6.1) having two space variables and equal grid
spacing, hence r < 1/6 is required for stability and
convergence. The difficulty while using the explicit scheme
is that the restriction on t∆ requires inordinately many rows
of calculations. In such case one looks for a method in
which t∆ can be made larger without lost of stability. The
implicit method was such a method.
The implicit method, the finite difference scheme of
equation (3.6.1) is
)SS( c t / )u - (u 1k ij,k ij,2
k ij,1k ij, ++ ′′+′′=∆ …(3.6.4)
127
where 1k ij,k ij, S and S +′′′′ denote second derivatives of S(x) at
ij xx = at the time interval k and k + 1 respectively.
We can express (3.6.1) in terms of u as follows. We use
the relationship (2.4.15a) and rewrite it as
)u u 2 - (u )(6/h
S S 4 S
k j,1, ik j,i,k j,1,-i2
k j,1, ik j,i,k j,1,-i
+
+
+=
′′+′′+′′ …(3.6.5)
1 - N ......., 2, 1, j i,
)u u 2 - (u )(6/h
S S 4 S
1 k j,1, i1 k j,i,1 k j,1,-i2
1 k j,1, i1 k j,i,1 k j,1,-i
=
+=
′′+′′+′′
++++
++++
…(3.6.6)
with the help of equations (3.6.5) and (3.6.6) using the
value of 1k ij,S +′′ from equation (3.6.4) we get,
{ }{ }
)u u 2 - (u )(6/h
S - )u - (u t)c / 1( 4
S - )u - (u t)c / 1( 4 S - )u - (ut c / 1
1 k j,1, i1 k j,i,1 k j,1,-i2
k j,1, ik j,1, i1k j,1, i2
k j,i,k j,i,1k j,i,2
k j,1,-ik j,1, - i1k j,1, - i2
++++
++++
++
+=
′′∆+
′′∆+′′∆
this gives
)u 6r) (1 u 12r) - (4 u 6r) (1
u 6r) - (1 u 12r) (4 u 6r) - (1
k j,1, ik j,i,k j,1,-i
k j,1, i1k j,i,1k j,1,-i
+
+++
++++=
+++ …(3.6.7)
where 22 ht / c r ∆= and 1 -N ....., 2, 1, j i, =
Above equation (3.6.7) is known as cubic spline
implicit formula to solve equation (3.6.1). Like scheme, we
get (N – 1) x (N – 1) simultaneous equations in (N – 1) x
(N – 1) unknowns. These equations with square matrix can
be solved by any standard method.
128
In both the methods, once the values of u are known at th1) (k + level, we can proceed to compute the next level (k +
2) by same techniques as above. The coefficient matrix of
the combined equation is a square matrix, however, the
system becomes a tri-diagonal one when separate cases are
handled. These two methods will be discussed later on by
taking its actual approximation to a problem.
3.7 HEAT FLOW IN A THIN RECTANGULAR PLATE :
Consider the flow of heat in a thin rectangular plate
with sides of length y andx ∆∆ along co-ordinals x and y.
A C y
B D x
Figure (3.7.1)
Thin rectangular plate
The amount of heat entering the element through the
side AB in time t∆ is
t x)y/( y)(K - x ∆∂∂∆
and that leaving the element through the opposite side CD
is
t x)y/( y)(K - x x ∆∂∂∆ ∆+
129
where K is the thermals conductional of the material and
u(x, y, t) is the temperature function. The negative signs are
taken because the heat flows in the direction of decreasing
temperature. Hence the quantity of heat remaining in the
plate as a result of entry through the side AB and exit
through the side CD is
ty }x )xu/( {K
ty } x)u/( - x)u/( {K 22
xx x
∆∆∆∂∂=
∆∆∂∂∂∂ ∆+ …(3.7.1)
upto a first approximation.
Similarly corresponding difference in the heat entering
and leaving through the remaining pair of opposite side is
tx }y )xu/( {K 22 ∆∆∆∂∂ …(3.7.2)
Hence the total heat retained by the plate in time t∆ is
the sum of results (3.7.1) and (3.7.2), which is equal to the
heat required to raise the temperature of the element by u∆ .
Thus we have
uS y)x ( t yx } )yu/( )xu/( {K 2222 ∆∆∆=∆∆∆∂∂+∂∂ ρ …(3.7.3)
where ρ is the density and S be the specific heat of the
plate. Dividing the equation (3.7.3) by Sy x ρ∆∆ and taking
limit 0 t →∆ , we get
)yu/ xu/( c t u/ 22222 ∂∂+∂∂=∂∂
where ρK/S c2 = …(3.7.4)
which is parabolic PDE having two space variables x and y
and time variable t.
Consider the edges of thin square plate of side 1 (figure
3.7.2) are kept at temperature zero and faces are perfectly
insulated.
130
y
1
0 1 x
Figure (3.7.2)
Thin square plate
Hence the flow of heat in the plate is governed by
equation (3.7.3) with boundary with boundary conditions.
0 t)y, u(0,
1 y x, 0 0 t)0, u(x,
=
≤≤= …(3.7.5)
and let initial temperature distribution in the plate be
1 y x, 0 y sin x sin 0) y, u(x, ≤≤= ππ …(3.7.6)
The given problem with boundary and initial
conditions is solved by explicit as well as implicit method as
follows.
3.8 (A) SPLINE SOLUTIONS WITH EXPLICIT METHOD :
We shall determine the solution of equation (3.7.3)
satisfying boundary conditions (3.7.5) and initial conditions
given by the equation (3.7.6) respectively, by using the
explicit formula given in equation (3.6.3).
Let 1/400 t and 0.001 c 1/20, h 2 =∆== which gives
R = 0.001
Hence 1 + 12r = 1.012
131
4 – 24r = 3.976
Substituting the values of 1 + 12r, 4 – 24r with initial
and boundary conditions in equation (3.6.3), we get
For k = 0
For j = 1
i = 1 0.146220 u 4u u 1 1, 2,1 1, 1, 1 1, 0, =++
Since 0 u 1 1, 0, = we have
0.146220 u 4u 1 1, 2,1 1, 1, =+
i = 2 0.288841 u 4u u 1 1, 3,1 1, 2, 1 1, 1, =++
i = 3 0.424349 u 4u u 1 1, 4,1 1, 3, 1 1, 2, =++
Similarly we get 19 x 19 simultaneous equations in 19
x 19 unknowns, for j = 1, 2, 3, ……, 19 where i = 1, 2, 3,
….., 19. This can be solved by any standard method. Once
the results are obtained, for results for th1) (k + level, results
for th2) (k + level are obtained in similar manner discussed
as above. Due to the symmetry of the solutions, the results
are given for 0.5 y 0 0.5, x 0 ≤≤≤≤ at t = 1/400, 3/400 and
1/80 respectively in the table (3.8.1(a)) – (3.8.1(c)). Results
are obtained by explicit method for y = 0.25 plotted in the
figures (3.8.1).
132
Table (3.8.1(a))
Tem
perature distribution in thin rectangular plate through explicit method
Tem
perature in thin rectangular plate u →
at t = 1/400
y
x 0.0
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.05
0.0
00
00
0
0.0
24
47
0
0.0
48
33
9
0.0
71
01
6
0.0
91
94
5
0.1
10
61
0
0.1
26
55
2
0.1
39
37
7
0.1
48
77
0
0.1
54
50
0
0.1
56
42
7
0.10
0.0
00
00
0
0.0
48
33
9
0.0
95
48
7
0.1
40
28
4
0.1
81
62
7
0.2
18
49
7
0.2
49
98
8
0.2
75
32
3
0.2
93
87
8
0.3
05
19
7
0.3
09
00
2
0.15
0.0
00
00
0
0.0
71
01
6
0.1
40
28
4
0.2
06
09
7
0.2
66
83
6
0.3
21
00
4
0.3
67
26
8
0.4
04
48
9
0.4
31
74
9
0.4
48
37
9
0.4
53
96
8
0.20
0.0
00
00
0
0.0
91
94
5
0.1
81
62
7
0.2
66
83
6
0.3
45
47
4
0.4
15
60
6
0.4
75
50
5
0.5
23
69
5
0.5
58
98
9
0.5
80
52
0
0.5
87
75
6
0.25
0.0
00
00
0
0.1
10
61
0
0.2
18
49
7
0.3
21
00
4
0.4
15
60
6
0.4
99
97
5
0.5
72
03
3
0.6
30
00
6
0.6
72
46
5
0.6
98
36
7
0.7
07
07
2
0.30
0.0
00
00
0
0.1
26
55
2
0.2
49
98
8
0.3
67
26
8
0.4
75
50
5
0.5
72
03
3
0.6
54
47
6
0.7
20
80
4
0.7
69
38
3
0.7
99
01
7
0.8
08
97
7
0.35
0.0
00
00
0
0.1
39
37
7
0.2
75
32
3
0.4
04
48
9
0.5
23
69
5
0.6
30
00
6
0.7
20
80
4
0.7
93
85
3
0.8
47
35
6
0.8
79
99
3
0.8
90
96
2
0.40
0.0
00
00
0
0.1
48
77
0
0.2
93
87
8
0.4
31
74
9
0.5
58
98
9
0.6
72
46
5
0.7
69
38
3
0.8
47
35
6
0.9
04
46
4
0.9
39
30
1
0.9
51
00
9
0.45
0.0
00
00
0
0.1
54
50
0
0.3
05
19
7
0.4
48
37
9
0.5
80
52
0
0.6
98
36
7
0.7
99
01
7
0.8
79
99
3
0.9
39
30
1
0.9
75
48
0
0.9
87
64
0
0.50
0.0
00
00
0
0.1
56
42
7
0.3
09
00
2
0.4
53
96
8
0.5
87
75
6
0.7
07
07
2
0.8
08
97
7
0.8
90
96
2
0.9
51
00
9
0.9
87
64
0
0.9
99
95
1
133
Table (3.8.1(b))
Tem
perature distribution in thin rectangular plate through explicit method
Tem
perature in thin rectangular plate u →
at t = 3/400
y
x 0.0
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.05
0.0
00
00
0
0.0
24
46
8
0.0
48
33
4
0.0
71
00
9
0.0
91
93
6
0.1
10
59
9
0.1
26
53
9
0.1
39
36
3
0.1
48
75
6
0.1
54
48
6
0.1
56
41
1
0.10
0.0
00
00
0
0.0
48
36
4
0.0
95
47
7
0.1
40
27
0
0.1
81
60
9
0.2
18
47
6
0.2
49
96
3
0.2
75
29
5
0.2
93
84
9
0.3
05
16
7
0.3
08
97
1
0.15
0.0
00
00
0
0.0
71
00
9
0.1
40
27
0
0.2
06
07
7
0.2
66
80
9
0.3
20
97
5
0.3
67
22
9
0.4
04
05
2
0.4
31
70
4
0.4
48
33
5
0.4
53
92
3
0.20
0.0
00
00
0
0.0
91
93
6
0.1
81
60
9
0.2
66
80
9
0.3
45
44
0
0.4
15
56
5
0.4
75
45
8
0.5
23
64
3
0.5
58
93
4
0.5
80
46
3
0.5
87
69
8
0.25
0.0
00
00
0
0.1
10
59
9
0.2
18
47
6
0.3
20
97
5
0.4
15
56
5
0.4
99
92
6
0.5
71
97
7
0.6
29
94
3
0.6
72
39
9
0.6
98
29
8
0.7
07
00
2
0.30
0.0
00
00
0
0.1
26
53
9
0.2
49
96
3
0.3
67
22
9
0.4
75
45
8
0.5
71
97
7
0.6
54
41
2
0.7
20
73
3
0.7
69
30
7
0.7
98
93
8
0.8
08
89
7
0.35
0.0
00
00
0
0.1
39
36
3
0.2
75
29
5
0.4
04
45
2
0.5
23
64
3
0.6
29
94
3
0.7
20
73
3
0.7
93
77
5
0.8
47
27
2
0.8
79
90
6
0.8
90
87
4
0.40
0.0
00
00
0
0.1
48
75
4
0.2
93
84
9
0.4
31
70
4
0.5
58
93
4
0.6
72
39
9
0.7
69
30
7
0.8
47
27
2
0.9
04
37
4
0.9
39
20
8
0.9
50
91
5
0.45
0.0
00
00
0
0.1
54
48
6
0.3
05
16
7
0.4
48
33
5
0.5
80
46
3
0.6
98
29
8
0.7
98
93
8
0.8
79
90
6
0.9
39
20
8
0.9
75
38
3
0.9
87
54
2
0.50
0.0
00
00
0
0.1
56
41
1
0.3
08
97
1
0.4
53
92
3
0.5
87
69
8
0.7
07
00
2
0.8
08
89
7
0.8
90
87
4
0.9
50
91
5
0.9
87
54
2
0.9
99
85
2
134
Table (3.8.1(c))
Tem
perature distribution in thin rectangular plate through explicit method
Tem
perature in thin rectangular plate u →
at t = 1/80
y
x 0.0
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.05
0.0
00
00
0
0.0
24
46
6
0.0
48
32
9
0.0
71
00
2
0.0
91
92
7
0.1
10
58
8
0.1
26
52
7
0.1
39
35
0
0.1
48
74
1
0.1
54
47
0
0.1
56
39
5
0.10
0.0
00
00
0
0.0
48
32
9
0.0
95
46
8
0.1
40
25
6
0.1
81
59
1
0.2
18
45
4
0.2
49
93
8
0.2
75
26
8
0.2
93
82
0
0.3
05
13
7
0.3
08
94
0
0.15
0.0
00
00
0
0.0
71
00
2
0.1
40
25
6
0.2
06
05
7
0.2
66
78
3
0.3
20
94
2
0.3
67
18
9
0.4
04
01
9
0.4
31
65
7
0.4
48
29
2
0.4
53
87
7
0.20
0.0
00
00
0
0.0
91
92
7
0.1
81
59
1
0.2
66
72
9
0.3
45
40
6
0.4
15
52
4
0.4
75
41
1
0.5
23
59
1
0.5
58
87
9
0.5
80
40
5
0.5
87
64
0
0.25
0.0
00
00
0
0.1
10
58
8
0.2
18
45
4
0.3
20
94
2
0.4
15
52
4
0.4
99
87
6
0.5
71
92
0
0.6
29
88
1
0.6
72
33
2
0.6
98
22
9
0.7
06
93
2
0.30
0.0
00
00
0
0.1
26
52
7
0.2
49
93
8
0.3
67
18
9
0.4
75
41
1
0.5
71
92
0
0.6
54
34
7
0.7
20
66
2
0.7
69
23
1
0.7
98
86
0
0.8
08
81
6
0.35
0.0
00
00
0
0.1
39
35
0
0.2
75
26
8
0.4
04
40
9
0.5
23
59
1
0.6
29
88
1
0.7
20
66
2
0.7
93
69
7
0.8
47
18
8
0.8
79
82
0
0.8
90
78
6
0.40
0.0
00
00
0
0.1
48
75
1
0.2
93
82
0
0.4
31
65
7
0.5
58
87
9
0.6
72
33
2
0.7
69
23
1
0.8
47
18
8
0.9
04
28
5
0.9
39
11
6
0.9
50
82
1
0.45
0.0
00
00
0
0.1
54
47
0
0.3
05
13
7
0.4
48
29
2
0.5
80
40
5
0.6
98
22
9
0.7
98
86
0
0.8
79
82
0
0.9
39
11
6
0.9
75
28
7
0.9
87
44
4
0.50
0.0
00
00
0
0.1
56
39
5
0.3
08
94
0
0.4
53
87
7
0.5
87
64
0
0.7
06
93
2
0.8
08
81
6
0.8
90
78
6
0.9
50
82
1
0.9
87
44
4
0.9
99
75
3
135
0.000000
0.100000
0.200000
0.300000
0.400000
0.500000
0.600000
0.700000
0.800000
0.0
0.050.100.150.200.250.300.350.400.450.50
X
U
t=1/400
t=3/400
t=1/80
Figure (3.8.1)
Tem
perature distribution in thin rectangular plate through cubic spline explicit method
(at y=0.25)
136
3.8(B) SPLINE SOLUTIONS WITH IMPLICIT METHOD :
Using implicit formula given by equation (3.6.7), the
solution of equation (3.7.3) satisfying boundary and initial
conditions, those are given in section (3.7) is obtained as
follows :
For k = 0
For j = 1
i = 1, 0.146224 u (0.994) u (4.012) u (0.994) 1 1, 2,1 1, 1, 1 1, 0, =++
i = 2, 0.288848 u (0.994) u (4.012) u (0.994) 1 1, 3,1 1, 2, 1 1, 1, =++
i = 3, 0.424359 u (0.994) u (4.012) u (0.994) 1 1, 4,1 1, 3, 1 1, 2, =++
Proceeding in this way, we get 19 x 19 simultaneous
equations in 19 x 19 unknowns, for j = 1, 2, 3, ……, 19
where i = 1, 2, 3, ….., 19. Solving the set of equations by
any well known method, the temperature distribution in the
plate is obtained. Once the results are obtained, for results
for th1) (k + level, the results for th2) (k + level are obtained in
similar manner discussed as above. Due to the symmetry of
the solutions, the results are given for 0.5 y 0 0.5, x 0 ≤≤≤≤ at
t = 1/400, 3/400 and 1/80 respectively in the tables
(3.8.2(a)) – (3.8.2(c)). Results are obtained by implicit
method for y = 0.25 plotted in the figures (3.8.2(a)).
137
Table (3.8.2(a))
Tem
perature distribution in thin rectangular plate through implicit method
Tem
perature in thin rectangular plate u →
at t = 1/400
y
x 0.0
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.05
0.0
00
00
0
0.0
24
47
0
0.0
48
33
9
0.0
71
01
6
0.0
91
94
5
0.1
10
61
0
0.1
26
55
1
0.1
39
37
7
0.1
48
77
1
0.1
54
50
1
0.1
56
42
7
0.10
0.0
00
00
0
0.0
48
33
9
0.0
95
48
7
0.1
40
28
4
0.1
81
62
7
0.2
18
49
7
0.2
49
98
8
0.2
75
32
3
0.2
93
87
8
0.3
05
19
7
0.3
09
00
2
0.15
0.0
00
00
0
0.0
71
01
6
0.1
40
28
4
0.2
06
09
7
0.2
66
83
6
0.3
21
00
4
0.3
67
26
8
0.4
04
48
9
0.4
31
74
9
0.4
48
37
9
0.4
53
96
8
0.20
0.0
00
00
0
0.0
91
94
5
0.1
81
62
7
0.2
66
83
6
0.3
45
47
4
0.4
15
60
6
0.4
75
50
5
0.5
23
69
5
0.5
58
98
9
0.5
80
52
0
0.5
87
75
6
0.25
0.0
00
00
0
0.1
10
61
0
0.2
18
49
7
0.3
21
00
4
0.4
15
60
6
0.4
99
97
5
0.5
72
03
3
0.6
30
00
6
0.6
72
46
5
0.6
98
36
7
0.7
07
07
2
0.30
0.0
00
00
0
0.1
26
55
2
0.2
49
98
8
0.3
67
26
8
0.4
75
50
5
0.5
72
03
3
0.6
54
47
6
0.7
20
80
4
0.7
69
38
3
0.7
99
01
7
0.8
08
97
7
0.35
0.0
00
00
0
0.1
39
37
7
0.2
75
32
3
0.4
04
48
9
0.5
23
69
5
0.6
30
00
6
0.7
20
80
4
0.7
93
85
3
0.8
47
35
6
0.8
79
99
3
0.8
90
96
2
0.40
0.0
00
00
0
0.1
48
77
1
0.2
93
87
8
0.4
31
74
9
0.5
58
98
9
0.6
72
46
5
0.7
69
38
3
0.8
47
35
6
0.9
04
46
4
0.9
39
30
1
0.9
51
00
9
0.45
0.0
00
00
0
0.1
54
50
0
0.3
05
19
7
0.4
48
37
9
0.5
80
52
0
0.6
98
36
7
0.7
99
11
7
0.8
79
99
3
0.9
39
30
1
0.9
75
48
0
0.9
87
64
0
0.50
0.0
00
00
0
0.1
56
42
7
0.3
09
00
2
0.4
53
96
8
0.5
87
75
6
0.7
07
07
2
0.8
08
97
7
0.8
90
96
2
0.9
51
00
9
0.9
87
64
0
0.9
99
95
1
138
Table (3.8.2(b))
Tem
perature distribution in thin rectangular plate through implicit method
Tem
perature in thin rectangular plate u →
at t = 3/400
y
x 0.0
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.05
0.0
00
00
0
0.0
24
46
8
0.0
48
33
4
0.0
71
00
9
0.0
91
93
6
0.1
10
59
9
0.1
26
53
9
0.1
39
36
3
0.1
48
75
6
0.1
54
48
6
0.1
56
41
1
0.10
0.0
00
00
0
0.0
48
36
4
0.0
95
47
7
0.1
40
27
0
0.1
81
60
0
0.2
18
47
6
0.2
49
96
3
0.2
75
29
5
0.2
93
84
9
0.3
05
16
7
0.3
08
97
1
0.15
0.0
00
00
0
0.0
71
00
9
0.1
40
27
0
0.2
06
07
7
0.2
66
80
9
0.3
20
97
2
0.3
67
23
1
0.4
04
44
9
0.4
31
70
7
0.4
48
33
5
0.4
53
92
3
0.20
0.0
00
00
0
0.0
91
93
6
0.1
81
60
0
0.2
66
80
9
0.3
45
44
0
0.4
15
56
5
0.4
75
45
8
0.5
23
64
3
0.5
58
93
4
0.5
80
46
3
0.5
87
69
8
0.25
0.0
00
00
0
0.1
10
59
9
0.2
18
47
6
0.3
20
97
2
0.4
15
56
5
0.4
99
92
6
0.5
71
97
7
0.6
29
94
3
0.6
72
39
9
0.6
98
29
8
0.7
07
00
2
0.30
0.0
00
00
0
0.1
26
53
9
0.2
49
96
3
0.3
67
23
1
0.4
75
45
8
0.5
71
97
7
0.6
54
41
1
0.7
20
73
3
0.7
69
30
7
0.7
98
93
8
0.8
08
89
7
0.35
0.0
00
00
0
0.1
39
36
3
0.2
75
29
5
0.4
04
44
9
0.5
23
64
3
0.6
29
94
3
0.7
20
73
3
0.7
93
77
5
0.8
47
27
2
0.8
79
90
6
0.8
90
87
4
0.40
0.0
00
00
0
0.1
48
75
4
0.2
93
84
9
0.4
31
70
7
0.5
58
93
4
0.6
72
39
9
0.7
69
30
7
0.8
47
27
2
0.9
04
37
4
0.9
39
20
8
0.9
50
91
5
0.45
0.0
00
00
0
0.1
54
48
6
0.3
05
16
7
0.4
48
33
5
0.5
80
46
3
0.6
98
29
8
0.7
98
93
8
0.8
79
90
6
0.9
39
20
8
0.9
75
38
3
0.9
87
54
2
0.50
0.0
00
00
0
0.1
56
41
1
0.3
08
97
1
0.4
53
92
3
0.5
87
69
8
0.7
07
00
2
0.8
08
89
7
0.8
90
87
4
0.9
50
91
5
0.9
87
54
2
0.9
99
85
2
139
Table (3.8.2(c))
Tem
perature distribution in thin rectangular plate through implicit method
Tem
perature in thin rectangular plate u →
at t = 1/80
y
x 0.0
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.0
00
00
0
0.05
0.0
00
00
0
0.0
24
46
5
0.0
48
32
9
0.0
71
00
2
0.0
91
92
7
0.1
10
58
8
0.1
26
52
7
0.1
39
35
0
0.1
48
74
1
0.1
54
47
0
0.1
56
39
5
0.10
0.0
00
00
0
0.0
48
32
9
0.0
95
46
8
0.1
40
25
6
0.1
81
59
1
0.2
18
45
4
0.2
49
93
8
0.2
75
26
8
0.2
93
82
0
0.3
05
13
7
0.3
08
94
0
0.15
0.0
00
00
0
0.0
71
00
2
0.1
40
25
6
0.2
06
05
7
0.2
66
78
3
0.3
20
94
1
0.3
67
19
5
0.4
04
40
9
0.4
31
66
4
0.4
48
29
0
0.4
53
87
8
0.20
0.0
00
00
0
0.0
91
92
7
0.1
81
59
1
0.2
66
72
9
0.3
45
40
6
0.4
15
52
4
0.4
75
41
0
0.5
23
59
1
0.5
58
87
9
0.5
80
40
5
0.5
87
64
0
0.25
0.0
00
00
0
0.1
10
58
8
0.2
18
45
4
0.3
20
94
1
0.4
15
52
4
0.4
99
87
7
0.5
71
92
0
0.6
29
88
1
0.6
72
33
2
0.6
98
22
9
0.7
06
93
2
0.30
0.0
00
00
0
0.1
26
52
7
0.2
49
93
8
0.3
67
19
5
0.4
75
41
0
0.5
71
92
0
0.6
54
34
7
0.7
20
66
1
0.7
69
23
1
0.7
98
85
9
0.8
08
81
7
0.35
0.0
00
00
0
0.1
39
35
0
0.2
75
26
8
0.4
04
40
9
0.5
23
59
1
0.6
29
88
1
0.7
20
66
1
0.7
93
69
9
0.8
47
18
8
0.8
79
82
0
0.8
90
78
6
0.40
0.0
00
00
0
0.1
48
75
1
0.2
93
82
0
0.4
31
66
4
0.5
58
87
9
0.6
72
33
2
0.7
69
23
1
0.8
47
18
8
0.9
04
28
5
0.9
39
11
6
0.9
50
82
1
0.45
0.0
00
00
0
0.1
54
47
0
0.3
05
13
7
0.4
48
29
0
0.5
80
40
5
0.6
98
22
9
0.7
98
85
9
0.8
79
82
0
0.9
39
11
6
0.9
75
28
7
0.9
87
44
4
0.50
0.0
00
00
0
0.1
56
39
5
0.3
08
94
0
0.4
53
87
8
0.5
87
64
0
0.7
06
93
2
0.8
08
81
7
0.8
90
78
6
0.9
50
82
1
0.9
87
44
4
0.9
99
75
3
140
0.000000
0.100000
0.200000
0.300000
0.400000
0.500000
0.600000
0.700000
0.800000
0.0
0.050.100.150.200.250.300.350.400.450.50
X
U
t=1/400
t=3/400
t=1/80
Figure (3.8.2)
Tem
perature distribution in thin rectangular plate through cubic spline im
plicit method
(at y=0.25)
141
3.8 (C) DISCUSSION OF RESULTS :
The solutions of equation (3.7.3) obtained by implicit
as well as explicit solutions are comparing with exact
solutions as follows in table (3.8.3). Clearly the results are
accurate upto five digits of decimal places. Figure (3.8.3(a))
indicates the error analysis which compares the solutions
with spline solutions and figure (3.8.3(b)) gives the single
curve which shows that spline solutions are quite accurate
and reliable.
Table (3.8.3)
Error analysis
Temperature in thin rectangular plate u → at y = 0.25 & t = 1/80
x USE USI UEXT UEXT-USE UEXT-USI
0.0 0.000000 0.000000 0.000000 0.000000 0.000000
0.05 0.110588 0.110588 0.110589 0.000001 0.000001
0.10 0.218454 0.218454 0.218454 0.000000 0.000000
0.15 0.320942 0.320941 0.320941 0.000001 0.000000
0.20 0.415524 0.415524 0.415524 0.000000 0.000000
0.25 0.499876 0.499877 0.499877 0.000001 0.000000
0.30 0.571920 0.571920 0.571920 0.000000 0.000000
0.35 0.629881 0.629881 0.629881 0.000000 0.000000
0.40 0.672332 0.672332 0.672333 0.000001 0.000001
0.45 0.698229 0.698229 0.698228 0.000001 0.000001
0.50 0.706932 0.706932 0.706932 0.000000 0.000000
142
0.000000
0.000000
0.000000
0.000001
0.000001
0.000001
0.000001
0.0
0.050.100.150.200.250.300.350.400.450.50
X
ERROR
USE-UEXT
USI-UEXT
Figure (3.8.3(a))
Error Analysis (at y = 0.25 & t = 1/80)
143
0.000000
0.100000
0.200000
0.300000
0.400000
0.500000
0.600000
0.700000
0.800000
0.0
0.050.100.150.200.250.300.350.400.450.50
X
U
USE
USI
UEXT
Figure (3.8.3(b))
Tem
perature distribution in thin rectangular plate through cubic spline im
plicit method
(at t=1/80 & y=0.25)
144
3.9 SPLINE FORMULA TO SOLVE HYPERBOLIC
PARTIAL DIFFERENTIAL EQUATION WITH ONE
SPACE VARIABLES :
The general form of hyperbolic PDE with one space
variable x and time variable t is given by
0 t , L x 0 ; 2xu / 2 / 2c t 2u / 2 ><<∂∂=∂∂ …(3.9.1)
with a Dirichilet boundary conditions, namely
u(0, t) = 0
u(L, t) = 0 …(3.9.2)
and two initial conditions at t = 0 (Cauchy conditions)
u(x, 0) = f(x)
)()0,(u t xgx = …(3.9.3)
In equation (3.9.1), 2c is a constant term, it depends
upon some physical quantities in case of different problems.
Divide the region L x 0 ≤≤ into say n sub – intervals
each of width h) (x =∆ such that L x n =∆ . The subscript j
denotes time and i for the positions. The points of
subdivisions are n (1) 0 i ; x i = . Let j i,u denote the solution
of equation (3.9.1) at thj) (i, mesh point. Discretize the left
hand side of PDE (3.9.1) by the central difference formula
like finite difference and right side by second derivative of
cubic spline S(x) i.e. )(xS i′′ at the thj) (i, mesh point, one can
get
)S( c t)( / ) u 2u - u ( j i,22
1- j i,j i,1j i, ′′=∆++ …(3.9.4)
where j i,S ′′ = second derivative of cubic spline S(x) i. e. )(xS i′′
145
at time tj∆ substituting the values of j i,S ′′ from equation
(3.9.4) into (2.4.15a), the following relation is obtained.
] u 4u u [ -
u )6r (2 u )12r - (8 u )6r (2 )u 4u (u
1-j 1,i1-j i,1-j 1,-i
j 1,i2
j i,2
j 1,-i2
1j 1,i1j i,1j 1,-i
+
+++++
++
++++=++
…(3.9.5)
where ht / c r ∆= i = 1(1) n – 1
Above formula is known as cubic spline explicit
formula at to solve hyperbolic PDE of the form (3.9.1). It is
clear that above formula is applied for all values of 1 j≥ .
However, for j = 0, it becomes
] u 4u u [ -
u )6r (2 u )12r - (8 u )6r (2 )u 4u u
1- 1,i1- i,1- 1,-i
1,0i2
0 i,2
0 1,-i2
1 1,i1 i,1 1,-i
+
++
++
++++=++
…(3.9.6)
where i = 1(1) n – 1
it involves the term u 1- 1,i+ , 1- i,u and 1- 1,-iu which are
unknowns. To deal with them, we consider the function u =
u(x, t) to be extended backward in time, the term 1- tt = make
a good sense. Most of the time, we get periodic functions for
u versus at a given point, we can consider zero time as an
arbitrary point at which we know the value of u. So, to get
the values for the fictitious points at 1- tt = we use the initial
condition (initial velocity)
0 at t g(x) t u / ==∂∂
By central difference approximation, we have
)g(x t 2 / )u - (u 0) x,(u i1- i,1 i,t =∆=
giving 0 at t t )2g(x - u u i1 i,1- i, =∆= only
i = 1(1) n – 1 …(3.9.7)
146
The values of 0(1)n i ; u 1- i, = and using initial and
boundary conditions with equation (3.9.6) gives system of
(n – 1) simultaneous linear equations in (n – 1) unknowns.
The system has a tri-diagonal matrix, which can be solved
by any well-known method. After calculating the values of u
for j = 0, we apply equation (3.9.5) for 1 j≥ , we get (n – 1)
simultaneous linear equations in (n – 1) unknowns with
tri-diagonal matrix, again 1 r ≤ is the required condition for
convergence and stability of this cubic spline explicit
method.
Just like the implicit scheme for obtaining solution of
parabolic partial differential equations, we have implicit
scheme for hyperbolic differential equations. Implicit
scheme is unconditionally stable i.e. stable for all the values
for r. In implicit method the discretization of the differential
equation at any mesh point (i, j) is done by replacing time
derivative by the central difference formula as done in
explicit scheme and the space derivative is replaced by
average of second derivatives of cubic spline S(x) at the
th1) - (j and th1) (j+ level i.e. /2)S S( 1j i,1-j i, +′′+′′ , equation (3.9.1)
becomes,
/2)S S( c )t ( / )u 2u - (u 1j i,1-j i,22
1-j i,j i,1j i, ++ ′′+′′=∆+ …(3.9.8)
The values of 1j i,S +′′ obtained from equation (3.11.8) and
with the help of equation (2.4.15a) the following relation is
obtained.
147
)u 4u 2(u
u )1 - (3r u )4 (6r - u )1 - (3r
u )3r - (1 u )6r (4 u )3r - (1
j 1,-ij i,j 1,i
1-j 1,-i2
1-j i,2
j 1,i2
1j 1,-i2
1j i,2
1j 1,i2
+++
++=
+++
+
+
++++
…(3.9.9)
where 1-n 1(1) i ; h t / c r =∆=
The above equation (3.9.9) is known as cubic spline
implicit formula to solve hyperbolic PDE of the form (3.9.1).
Like explicit scheme, described as above, the equation
(3.9.9) gives (n – 1) simultaneous linear equations in (n – 1)
unknowns with the coefficient matrix of tri-diagonal form.
For j = 0, here we can also use the initial condition in
similar manner described as above.
3.10 THE FLOW OF ELECTRICITY IN THE
TRANSMISSION LINES :
The problem is already discussed in chapter – 3 in
section 3.2. In addition to that, for higher frequency the
effect of R and G are negligible and equations (3.2.4) and
(3.2.5) reduce to
2222
2222
t / i (LC) x / i
t / e (LC) x / e
∂∂=∂∂
∂∂=∂∂
i.e. 2222 tu / (LC) xu / ∂∂=∂∂ …(3.10.1)
where u stands for either i(x, t). L is the inductance and C is
the capacitance per unit length of the transmission line i.e.
cable. This equation is known as hyperbolic PDE with one
space variable x and time variable t.
148
Let l = length of the transmission line = 1 and using
following initial and boundary conditions, the solution of
above equation (3.10.1) i.e. the current distribution is
obtained as follows.
Dirichilet boundary conditions :
u(0, t) = 0
u(l, t) = 0 …(3.10.2)
Initial conditions (cauchy conditions at t = 0)
1 x 0 ; x sin 0) u(x, ≤≤= π
U(x, 0) = 0 …(3.10.3)
3.11(A) SPLINE SOLUTIONS WITH EXPLICIT METHOD :
Using initial and boundary conditions described by
equations (3.10.2) and (3.10.3), the solutions of equation
(3.10.1) are obtained by explicit formula given in the
equation (3.9.5).
Let the length of the transmission line i.e. l = 1 divide
the region 1 x 0 ≤≤ into 10 sub – intervals each of width h =
0.1. Let L.C = 0.01, and 0.01 t =∆ . These gives r = 0.01.
Hence, 2.0006 r62 2 =+ and 7.9988 r128 2 =−
Having calculated the values of 1(1)n i ; u 1- i, = from
equation (3.9.7) with initial conditions, substitute the values
of 1- i,u and 2r62 + , 2r128 − into the equation (3.9.6) we get,
For j = 0
i = 1
149
1.823844
2/}u (2.0006) u (7.9988) u (2.0006) {
u 4u u
0 2,0 1,0 0,
1 2,1 1,1 0,
=
++=
++
Since 0 u 1 0, =
1.823844 u 4u 1 2,1 1, =+
i = 2 3.469157 u 4u u 1 3,1 2,1 1, =++
i = 3 4.774886 u 4u u 1 4,1 3,1 2, =++
i = 4 5.613215 u 4u u 1 5,1 4,1 3, =++
i = 5 5.902084 u 4u u 1 6,1 5,1 4, =++
But 1 6,1 4, u andu are same due to symmetry of the
results
5.902084 4u 2u 1 5,1 4, =+
The above five simultaneous equations in five
unknowns with in diagonal coefficient matrix can be easily
solved. Once the values of u are known at the first level of
time, the process can be repeated for second time level and
so on. It should be noted that the number of unknowns are
reduced because of the symmetry of the problem otherwise
we would have obtained nine simultaneous equations for
nine internal mesh points. The results obtained by the
explicit method are given in the table (3.11.1) and are
plotted in the figures (3.11.1). Due to the symmetry of the
problem, results are given for 0.5 x 0 ≤≤ only.
150
Table (3.11.1)
Current distribution in the cable through cubic spline explicit method
Current Distribution in the cable →
X t = 0.0 t = 0.01 t = 0.02 t = 0.03 t = 0.04 t = 0.05
0.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.1 0.309017 0.309015 0.309011 0.309003 0.308992 0.308978 0.2 0.587785 0.587782 0.587773 0.587759 0.587738 0.587712 0.3 0.809017 0.809013 0.809000 0.808981 0.808953 0.808916 0.4 0.951057 0.951052 0.951038 0.951015 0.950982 0.950940 0.5 1.000000 0.999995 0.999978 0.999955 0.999920 0.999875
0.000000
0.200000
0.400000
0.600000
0.800000
1.000000
1.200000
0.0 0.1 0.2 0.3 0.4 0.5
X
U
t = 0.0
t = 0.01
t = 0.02
t = 0.03
t = 0.04
t = 0.05
Figure (3.11.1)
Current distribution in the cable through cubic spline explicit method
151
3.11(B) SPLINE SOLUTIONS WITH IMPLICIT METHOD:
Using the same initial and boundary conditions and
the values of other parameters discussed in section 3.11(A),
the solution of equation (3.11.1) is obtained by implicit
formula that is given by the equation (3.9.9), derived as
follows.
For j = 0,
i = 1 1.823853 u (0.9997) u (4.0006) u (0.9997) 1 2,1 1,1 0, =++
Since 0 u 1 0, =
1.823853 u (0.9997) u (4.0006) 1 2,1 1, =+
i = 3 4.774909 u (0.9997) u (4.0006) u (0.9997) 1 4,1 3,1 2, =++
i = 4 5.613243 u (0.9997) u (4.0006) u (0.9997) 1 5,1 4,1 3, =++
i = 5 5.902113 u (0.9997) u (4.0006) u (0.9997) 1 6,1 5,1 4, =++
Since 1 6,1 4, u andu are same,
5.902113 u (4.0006) u (1.9994) 1 5,1 4, =+
The above five simultaneous equations in five
unknowns with coefficient matrix tri-diagonal one, is solved
by any well-known method. Like the explicit scheme, once
the values of u are known at the first level of time, the
process can be repeated for second time level and so on.
Due to symmetry of the problem the numbers of unknowns
are reduced otherwise we get nine equations in nine
unknowns. The results obtained by the implicit method are
given in the table (3.11.2) and are plotted in the figures
(3.11.2). Also due to the symmetry of the process results are
given for 0.5 x 0 ≤≤ only.
152
Table (3.11.2)
Current distribution in the cable through cubic spline implicit method
Current distribution in the cable →
X t =0.0 t = 0.01 t = 0.02 t = 0.03 t = 0.04 t = 0.05
0.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.1 0.309017 0.309015 0.309011 0.309003 0.308992 0.308978 0.2 0.587785 0.587782 0.587773 0.587759 0.587738 0.587711 0.3 0.809017 0.809013 0.809000 0.808981 0.808953 0.808917 0.4 0.951057 0.951052 0.951038 0.951014 0.950981 0.950939 0.5 1.000000 0.999995 0.999978 0.999955 0.999920 0.999875
0.000000
0.200000
0.400000
0.600000
0.800000
1.000000
1.200000
0.0 0.1 0.2 0.3 0.4 0.5
X
U
t = 0.0
t = 0.01
t = 0.02
t = 0.03
t = 0.04
t = 0.05
Figure (3.11.2)
Current distribution in the cable through cubic spline implicit method
153
3.11 (C) DISCUSSION OF RESULTS :
The current distributions obtained by the solution of
equation (3.10.1) by explicit as well as implicit methods are
compared with the exact solutions at t = 0.04 in the table
(3.11.3). Figure (3.11.3) indicates the error analysis, which
compares the exact solutions with spline solutions, obtained
by both the methods, Explicit as well as Implicit, at t =
0.04. From the table (3.11.3) and figure (3.11.3), it is clear
that solutions obtained by both the methods are fairly good
and correct up to five decimal places. Also implicit scheme
gives better results than the explicit one.
Table (3.11.3)
Error analysis
Current distribution u →
X USE USl UEXT USE-UEXT USI-UEXT
0.0 0.000000 0.000000 0.000000 0.000000 0.000000
0.1 0.308992 0.308992 0.308992 0.000000 0.000000
0.2 0.587738 0.587738 0.587739 0.000001 0.000001
0.3 0.808953 0.808953 0.808953 0.000000 0.000000
0.4 0.950982 0.950981 0.950981 0.000001
587
0.000000
0.5 0.999920 0.999920 0.999921 0.000001
0.706706
0.000001
[USE, USl and UEXT denotes cubic spline explicit, cubic
spline implicit and exact solutions]
154
0.0000000
0.0000002
0.0000004
0.0000006
0.0000008
0.0000010
0.0000012
0.0
0.1
0.2
0.3
0.4
0.5
X
ERROR
USE-UEXT
USI-UEXT
Figure (3.11.3(a))
Error analysis (at t = 0.04)
155
0.000000
0.200000
0.400000
0.600000
0.800000
1.000000
1.200000
0.0
0.1
0.2
0.3
0.4
0.5
X
U
USE
USl
UEXT
Figure (3.11.2(b)
Current Distribution in the cable through cubic spline method
(at t = 0.04)
156
3.12 VIBRATING STRING PROBLEM :
Many problems in mathematical physics reduce to the
solving of partial differential equations. The partial
differential equations play an important role in the theory of
elasticity, hydraulics and other some branches of
engineering. One of the most important problems in
mathematical physics is the vibration of a stretched string.
Simplicity and frequent occurrence in many branches of
mathematical physics make it classical example in the
theory of partial differential equations.
Let us consider a stretched string of length L fixed at
the end points. The problem here is to determine the
equation of motion which characterizes the position u(x, t)
of the string at time t after an initial disturbance is given.
In order to obtain a simple equation we make the
following assumptions.
(1) The string is flexible and elastic that is the string
cannot resist bending moments and thus the tension
in the string is always in the direction of the tangent to
the existing profile of the string.
(2) There is no elongation of a single segment of the string
and hence by Hook’s law the tension is constant.
(3) The weight of the string is small compared with the
tension in the string.
(4) The deflection is small compared with the length of the
string.
(5) The slope of the displaced string at any point is small
compared with unity.
157
(6) There is only pure transverse vibration.
u u
T
b T
a
c h x x x δ+ x
Figure (3.12.1)
String position and tension
Let T be the tension at the end points as shown in
figure (3.12.1). Consider the vibration of an elastic string of
length L in the vertical plane, the ends of which are fixed.
Take the origin at one fixed end, the x-axis along the length
of the string (when undisturbed) and the y-axis
perpendicular to it.
The displacement u of any point of the string is a
function of two variables, x its distance from o, and the time
t. To obtain the relation between u, x and t, take a small
element xδ of the string at a distance x from the origin of
which the displacement by y at time t. Let m be the mass
per unit length of the string. Then considering vertical
displacement of these elements, its acceleration is 22 tu / δδ .
If δφφφ and + be the angle made by the tangents at the
extremities of this elements (along which the tension t at
158
each point acts), with the x-axis then the vertical component
of the force to which this element is subjected is
φδφφ sin T - ) (sin T +
and as φ is small, we can write
φφ sin tan = , so these becomes
φδφφ tan T - ) ( tan T +
+ xx x xu
- xu
Tδδ
δδ
δ
or
x x
xu
- xu
Txx x
δδ
δδ
δδ
δ
+ …(3.12.1)
Thus for the vertical motion of this element, we have
the equation of motion as
x x
xu
- xu
T
t
u )x . m (
xx x
2
2
δδ
δδ
δδ
δδδ
δ
=+
…(3.12.2)
canceling x δ on both sides and taking the limit of the
equation (3.12.2), we have
2
2
2
2
xu
T t
u m
δδ
δδ
=
or
2
22
2
2
xu
c t
u
δδ
δδ
= where m / T c2 = …(3.12.3)
T : Tension in a string
m : Mass per unit length
159
This is the PDE having the vibrations of the string and
is called the WAVE equations, which is of HYPERBOLIC
TYPE.
Suppose the string of length 2 meter fixed at its both
ends, the initial velocity g(x) is to be taken 0 and from the
based on mathematical physics it can be known that the
initial displacement f(x) is ) Lx / (sin π . We also assume that
constant C takes value 1, then these problem becomes a
initial boundary value problem defined as
0 t ; 2 x 0 ; u u xxtt ><<= …(3.12.3)
2 x 0 2x
sin 0) u(x, ≤≤
=π
…(3.12.3a)
0 t 0 t)u(2,
0 t0 t)u(0,
2 x 0 ; 0 0) u t(x,
≥=
≥=
≤≤=
…(3.12.3b)
3.13(A) SPLINE SOLUTION WITH EXPLICIT METHOD :
Using initial and boundary conditions described by
equation (3.12.3a) and (3.12.3b), the solution of equation
(3.12.3) is obtained by explicit formula given in equation
(3.9.5).
Let the string of length 2 meter fixed at its both ends
and divide the region 2 x 0 ≤≤ with 0.05 201
t 0.2 51
h ==∆== .
These gives r = 0.25.
Hence 7.25 12r - 8 and 2.375 6r 2 22 ==+ .
160
Having calculated the values of 1(1)n i ; u 1- i, = from
equation (3.9.7) with initial condition, substitute the values
1- i,u and 22 12r - 8 ,6r 2 + into equation (3.9.6), we get
For j = 0 and i = 1
{ }1.818177
u (2.375) u (7.25) u (2.375) 21
u u 4 u 0 2,0 1,0 0,1 2,1 1,1 0,
=
++=++
Since 0 u 1 0, =
1.818177 u u 4 1 2,1 1, =+
i = 2 3.458383 u u 4 u 1 3,1 2,1 1, =++
i = 3 4.760060 u u 4 u 1 4,1 3,1 2, =++
i = 4 5.595787 u u 4 u 1 5,1 4,1 3, =++
Hence we get nine equations in nine unknowns for
i=1,2,…,10 and j=1,2,3,4,5. The above nine simultaneous
equations in nine unknowns with in diagonal co-efficient
matrix can be easily solved. Once the value of u are known
at first level of time, the process can be repeated for second
time level and so on. The results obtained by explicit
method are given in table (3.13.1) and are plotted in figures
(3.13.1). Due to the symmetry of the problem results are
given for 1 x 0 ≤≤ only.
161
Table (3.13.1)
Displacement of vibrating string through cubic spline explicit method
Displacement u(x, t)
X t= 0.0 t= 0.05 t= 0.1 t= 0.15 t= 0.20 t= 0.25
0.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.2 0.309075 0.308058 0.305190 0.300430 0.2938 0.285341 0.4 0.587786 0.585962 0.580507 0.571453 0.558897 0.542938 0.6 0.809016 0.806508 0.799000 0.786838 0.769760 0.747802 0.8 0.951056 0.948108 0.939282 0.924632 0.904292 0.878408 1.0 1.000000 0.996899 0.987618 0.972214 0.950766 0.923410 1.2 0.951056 0.948108 0.939282 0.924632 0.904292 0.878408 1.4 0.809016 0.806508 0.799000 0.786838 0.769760 0.747802 1.6 0.587786 0.585962 0.580507 0.571453 0.558897 0.542938 1.8 0.309075 0.308058 0.305190 0.300430 0.2938 0.285341 2.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
0.000000
0.200000
0.400000
0.600000
0.800000
1.000000
1.200000
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
X
U
t= 0.0
t= 0.05
t= 0.1
t= 0.15
t= 0.20
t= 0.25
Figure (3.13.1)
Displacement of vibrating string through cubic spline explicit method
162
3.13 (B) SPLINE SOLUTIONS WITH IMPLICIT METHOD :
Using the same initial and boundary conditions and
the values of other parameters discussed in section (3.13A)
the solution of equation (3.12.3) is obtained by implicit
formula that is given by the equation (3.9.9), derived as
follows.
For j = 0
i = 1
{ }1.823849
)u u 4 (u 2 21
u (0.8125) u (4.375) u (0.8125)
0 0,0 1,0 2,
1 2,1 1,1 0,
=
++=
++
Since 0 u 1 0, =
1.823849 u (0.8125) u (4.375) 1 2,1 1, =+
i = 2 3.469172 u (0.8125) u (4.375) u (0.8125) 1 3,1 2,1 1, =++
i = 3 4.774909 u (0.8125) u (4.375) u (0.8125) 1 4,1 3,1 2, =++
i = 4 5.613242 u (0.8125) u (4.375) u (0.8125) 1 5,1 4,1 3, =++
i = 5 5.902113 u (0.8125) u (4.375) u (0.8125) 1 6,1 5,1 4, =++
i = 6 5.613242 u (0.8125) u (4.375) u (0.8125) 1 7,1 6,1 5, =++
i = 7 4.774909 u (0.8125) u (4.375) u (0.8125) 1 8,1 7,1 6, =++
i = 8 3.469172 u (0.8125) u (4.375) u (0.8125) 1 9,1 8,1 7, =++
i = 9 1.823849 u (0.8125) u (4.375) u (0.8125) 1 10,1 9,1 8, =++
Since 0 u 1 10, =
1.823849 u (4.375) u (0.8125) 1 9,1 8, =+
The above system of nine equations in nine unknowns
with co-efficient matrix tri-diagonal one, is solved by
163
any well known method. Like explicit scheme, once the
value of u are known at first level of time, the process can
be repeated for second time level and so on. The results
obtained by implicit method are given in the table (3.13.2)
and are plotted in figures (3.13.2). Due to the symmetry of
the process results are given for 1 x 0 ≤≤ only.
Table (3.13.2)
Displacement of vibrating string through cubic spline implicit method
Displacement u(x, t)
X t= 0.0 t= 0.05 t= 0.1 t= 0.15 t= 0.20 t= 0.25
0.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
0.2 0.309016 0.308058 0.305190 0.300430 0.2938 0.285341
0.4 0.587785 0.585962 0.580507 0.571453 0.558897 0.542938
0.6 0.809016 0.806508 0.799000 0.786838 0.769760 0.747802
0.8 0.951056 0.948108 0.939282 0.924632
0.7
0.904292 0.878408
1.0 1.000000 0.996899 0.987618 0.972214
0.706706
0.950766 0.923410
1.2 0.951056 0.948108 0.939282 0.924632
0.7
0.904292 0.878408
1.4 0.809016 0.806508 0.799000 0.786838 0.769760 0.747802
1.6 0.587785 0.585962 0.580507 0.571453 0.558897 0.542938
1.8 0.309016 0.308058 0.305190 0.300430 0.2938 0.285341
2.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
164
0.000000
0.200000
0.400000
0.600000
0.800000
1.000000
1.200000
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
X
Ut= 0.0
t= 0.05
t= 0.1
t= 0.15
t= 0.20
t= 0.25
Figure (3.13.2)
Displacement of vibrating string through cubic spline implicit method
3.13 (C) DISCUSSION OF RESULTS :
The result presented in table (3.13.1), (3.13.2) is
obtained by explicit and implicit methods are compared with
the exact solutions at t = 0.15 in the table (3.13.3).
Figure (3.13.3(a)) indicates the error analysis, which
compares the exact solutions with spline solutions, obtained
by both the methods, Explicit as well as Implicit, at
t = 0.15. From the table (3.13.3) and figure (3.13.3(a)), it is
clear that solutions obtained by both the methods are fairly
good and correct up to three decimal places. Figure
(3.13.3(b)) shows the single curve which gives accuracy of
spline solutions. Also implicit method gives better results
than the explicit one.
165
Table 3.13.3
Error analysis
x USE USI UEX UEX-USE
UEX-USI
0.0 0.000000 0.000000 0.000000 0.000000 0.000000
0.2 0.300404 0.300430 0.300477 0.000073 0.000047
0.4 0.571402 0.571453 0.571538 0.000136 0.000085
0.6 0.786468 0.786838 0.786659 0.000191 0.000179
0.8 0.924548 0.924632 0.904506 0.020042 0.020126
1.0 0.972129 0.972214 0.951056 0.021073 0.021158
1.2 0.924548 0.924632 0.904506 0.020042 0.020126
1.4 0.786468 0.786838 0.786659 0.000191 0.000179
1.6 0.571402 0.571453 0.571538 0.000136 0.000085
1.8 0.300404 0.300430 0.300477 0.000073 0.000047
2.0 0.000000 0.000000 0.000000 0.000000 0.000000
166
0.000000
0.005000
0.010000
0.015000
0.020000
0.025000
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
X
ERROR
UEX-USE
UEX-USI
Figure (3.13.3(a))
Error Analysis (at t = 0.15
)
167
0.000000
0.200000
0.400000
0.600000
0.800000
1.000000
1.200000
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
X
ERROR
USE
UEXT
USI
Figure (3.13.3(b))
Displacement of vibrating string through cubic spline method
(at t = 0.15)
168
3.14 SPLINE FORMULA TO SOLVE HYPERBOLIC
PARTIAL DIFFERENTIAL EQUATION WITH TWO
SPACE VARIABLES :
The general form of hyperbolic PDE with two space
variables is as follows:
)yu / xu / ( c t u / 2222222 ∂∂+∂∂=∂∂ …(3.14.1)
with Dirichilet boundary conditions namely,
a y 0 ; 0 t)y, u(0, ≤≤=
b x 0 ; 0 t)0, u(x, ≤≤= …(3.14.2)
and initial conditions at t = 0 (Cauchy conditions)
u(x, y, 0) = f(x, y)
y) g(x, 0) y, (x, t u / ==∂∂ …(3.14.3)
In equation (3.14.1), 2c is a constant term, it depends
upon some physical quantities in case of different types of
problems.
Divide the region a x 0 ≤≤ into say n sub – intervals
each of width x∆ such that a x n =∆ , b y 0 ≤≤ into m sub –
intervals each of width y∆ such that b x n =∆ .
The subscript k denotes the time and (i, j) denote the
position. For simplicity consider square region i.e. a = b = L
the length of the region, m = n and (say)h y x =∆=∆ . The
points of subdivisions are (1)n 0 j i, , )y ,(x ji = .
For explicit scheme, the formula is obtained in similar
manner as discussed in section (3.9), by replacing the right
side by twice the second derivative of cubic spline S(x) at at
the thj) (i, mesh point i.e. )(xS 2 j i,′′ Let j i,u , we get
169
] u 4u u [ -
u )12r (2 u )24r - (8 u )12r (2
)u 4u (u
1-k j,1,-i1-k j,i,1-k j,1,i
k j,1,-i2
k j,i,2
k j,1,i2
1k j,1,-i1k j,i,1k j,1,i
++
++++=
++
+
+
++++
…(3.14.4)
where ht / c r ∆= i, j = 1(1) n – 1
The values of 1-n 1(1) j i s'u 1- j, i = are obtained from
following relation
t )y ,2g(x - u u ji1- j, i1- j,, i ∆= …(3.14.5)
The equation (3.14.4) is known as cubic spline explicit
formula to solve hyperbolic type PDE of the form given by
the equation (3.14.1). The values obtained from equation
(3.14.5) using initial and boundary conditions, the equation
(3.14.4) gives system of (n-1) x (n-1) simultaneous
equations in (n-1) x (n-1) unknowns. After calculating the
results for k = 0; the results for k = 1, 2, ….. are obtained
in similar manner. 21
r ≤ is the required condition for
convergence and stability of this explicit method.
Likewise implicit scheme for obtaining solution of
parabolic PDE with one space variable, we have implicit
scheme to solve hyperbolic differential equation with two
space variables. This implicit scheme is unconditionally
stable. As similar technique discussed in section (3.9) the
cubic spline implicit formula to solve equation (3.14.1) is as
follows, by replacing the right side of equation (3.14.1) i.e.
space derivatives by )S S( 1j i,1-j i, +′′+′′
170
)u 4u 2[u
u )1 - (6r u )4 (12r u )1 - (6r
u )6r - (1 u )12r (4 u )6r - (1
k j, 1,-ik j,, ik j,1,i
1-k j, 1,-i2
1-k j,, i2
1-k j, 1,i2
1k j, ,1,-i2
1k j, i,2
1k j, 1,i2
+++
+++=
+++
+
+
++++
…(3.14.6)
where 1-n 1(1) j i, ; h t / c r =∆=
Due to similar discussion as above also equation
(3.14.6) give (n – 1) x (n – 1) simultaneous equations in
(n – 1) x (n – 1) unknowns. After calculating results for
th1) (k + time level, the results for th2) (k + time level are
obtained in similar manner by repeating the process.
3.15 THE PROBLEM OF VIBRATING MEMBRANE :
The problem of vibrating membrane is well-known to
researchers. Assume that membrane is tightly stretched and
homogeneous i.e. its mass per unit area is constant, it is
perfectly flexible and is so thin that it offers no resistance
for bending. The membrane is stretched and then fixed
along its entire boundary in the xy - plane and the tension T
caused by stretching the membrane is the same at every
point in all directions and does not change during the
motion. The deflection u(x, y, t) of the membrane during the
motion is small compared to the size of the membrane.
Consider the forces acting on a small portion of the
membrane as shown in figure (3.15.1). Since the deflection
of the membrane and the angles of inclination are small the
171
sides of the portion may be taken approximately equal to
y andx ∆∆ .
xT∆
yT∆ yT∆
u
xT∆
y x ∆+
y
0 x x x x ∆+
Figure (3.15.1)
Vibrating membrane
The tension T is the force per unit length. Forces
acting on the edges are xT∆ and yT∆ which could be taken
tangent to the membrane since the membrane is perfectly
elastic. Let the force yT∆ make angles βα and with the
horizontal on the opposite edges of the membrane since the
membrane is perfectly elastic.
The resultant vertical component of force due to yT∆ is
therefore,
x )x / u(y T
}x)u / ( - x)u / ( {y T
tan y)(T - tan y)T(
sin y)(T - sin y)T(
22
xx x
∆∂∂∆=
∂∂∂∂∆=
∆∆
∆∆
∆+
αβαβ
up to a first order approximation. Note that sine have
replaced by tangent because the angles βα and are small.
172
Similarly, the forces xT∆ acting on the edges of length
x∆ can be shown to have the vertical component.
y )yu / (x T 22 ∆∂∂∆
If m be the mass per unit area of the membrane, by
Newton’s second law of motion
( )
( ) m / T c whereyu / xu / c tu /
yx yu / xu / T tu / y)x(m22222222
222222
=∂∂+∂∂=∂∂⇒
∆∆∂∂+∂∂=∂∂∆∆
…(3.15.1)
The above equation (3.15.1) is hyperbolic PDE in two
space variables x and y, and time variable t. The solution fro
the case of a rectangular membrane is given below by cubic
spline explicit as well as implicit method using following
initial and boundary conditions.
Dirichilet boundary conditions :
t x 0 0 t), 0 u(x,
ty 0 0 t)y, u(0,
≤≤=
≤≤= …(3.15.2)
Initial conditions at t = 0 (Cauchy condition)
0 ,0)y u(x,
1 y x, 0 ;y sin x sin y) f(x, 0) y, u(x,
=
≤≤== ππ …(3.15.3)
3.16(A) SPLINE SOLUTIONS WITH EXPLICIT METHOD :
Using initial and boundary conditions are described by
equations (3.15.2) and (3.15.3), the solution of equation
(3.15.1) is obtained through explicit formula given by the
equation (3.14.4) as follows.
Let the region 1 y 0 and 1 x 0 ≤≤≤≤ be divided into 10
sub-intervals, each of length h = 0.1. Let c = 0.1 and
0.01 t =∆ , these gives r = 0.01. After calculating the values
173
of 1(1)n j i, ,u 1-j i, = form equation (3.14.5), using initial
condition, substitute the values of 1-j i,u , 212r 2 + and 224r - 8
into equation (3.14.4) one gets,
For k = 0
For j = 1
i = 1 0.563599 u 4u u 1 1, 2,1 1, 1,1 1, 0, =++
Since 0 u 1 1, 0, =
0.563599 u 4u 1 1, 2,1 1, 1, =+
i = 2 1.072029 u 4u u 1 1, 3,1 1, 2,1 1, 1, =++
i = 3 1.475521 u 4u u 1 1, 4,1 1, 3,1 1, 2, =++
i = 4 1.734579 u 4u u 1 1, 5,1 1, 4,1 1, 3, =++
i = 5 1.873844 u 4u u 1 1, 6,1 1, 5,1 1, 4, =++
Since 1 1, 6,1 1, 4, u u = (Due to symmetry of the
problem)
1.823844 4u 2u 1 1, 5,1 1, 4, =+
In similar manner calculate the equation j = 1(1)9 one
gets 9 x 9 simultaneous linear equations in 9 x 9
unknowns. This system of equations can be solved by any
well-known method. Once the values of u are known at the
first level of time, the process can be repeated for second
time level and so on. The results obtained by explicit
method are given as follows. Due to the symmetry of the
problem, results are given for 0.5 x 0 0.5, y 0 ≤≤≤≤ in the table
(3.16.1(a)) and (3.16.1b). The results obtained at y = 0.1 are
plotted in the figures (3.16.1).
174
Table (3.16.1(a))
Velocity distribution in rectangular vibrating membrane through cubic spline explicit method u→ at t=0.01
y x 0.0 0.1 0.2 0.3 0.4 0.5
0.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
0.1 0.000000 0.095491 0.181634 0.249998 0.293890 0.309014
0.2 0.000000 0.181634 0.345488 0.475523 0.559012 0.587779
0.3 0.000000 0.249998 0.475523 0.654502 0.769413 0.809009
0.4 0.000000 0.293890 0.559012 0.769413 0.904499 0.951047
0.5 0.000000 0.309014 0.587779 0.809009 0.951047 0.999990
Table (3.16.1(b))
Velocity distribution in rectangular vibrating membrane through cubic spline explicit method u → at t=0.03
y x 0.0 0.1 0.2 0.3 0.4 0.5
0.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
0.1 0.000000 0.095483 0.181619 0.249978 0.293867 0.308989
0.2 0.000000 0.181619 0.345461 0.475485 0.558967 0.587732
0.3 0.000000 0.249978 0.475485 0.654450 0.769352 0.808944
0.4 0.000000 0.293867 0.558967 0.769352 0.904427 0.950917
0.5 0.000000 0.308989 0.587732 0.808944 0.950971 0.999910
175
0.000000
0.050000
0.100000
0.150000
0.200000
0.250000
0.300000
0.350000
0.0 0.1 0.2 0.3 0.4 0.5
X
U
t=0.01
t=0.03
Figure (3.16.1)
Velocity distribution in rectangular membrane through cubic spline explicit method (at y = 0.1)
3.16(B) SPLINE SOLUTIONS WITH IMPLICIT METHOD :
In similar manner discussed as earlier, using initial
and boundary condition, at 0.01 t =∆ , r = 0.01, the solution
of equation (3.14.1) is obtained by using cubic spline
implicit formula given by equation (3.14.6) as follows.
For k = 0
For j = 1
i = 1 0.563602 u (0.9994) u (4.0012) 1 1, 2,1 1, 1, =+
176
i= 2 1.072034 u (0.9994) u (4.0012) u (0.9994) 1 1, 3,1 1, 2,1 1, 1, =++
i = 3 1.475528 u (0.9994) u (4.0012) u (0.9994) 1 1, 4,1 1, 3,1 1, 2, =++
i=4 1.734587 u (0.9994) u (4.0012) u (0.9994) 1 1, 5,1 1, 4,1 1, 3, =++
i = 5 1.823853 u (4.0012) u (1.9988) 1 1, 5,1 1, 4, =+
Like explicit scheme discussed in section 3.17(A), one
gets system of 9 x 9 simultaneous linear equations in 9 x 9
unknowns. Once the values of u are obtained at the first
time level, the values for second time level are obtained in
similar manner by repeating the process. Results obtained
by cubic spline implicit method are given at different times
in the tables (3.16.2(a)) and (3.16.2(b)). The results are
plotted in the figures (3.17.2) at y = 0.1.
Table (3.16.2(a))
Velocity distribution in rectangular vibrating membrane through cubic spline implicit method u → at t = 0.01
y x 0.0 0.1 0.2 0.3 0.4 0.5
0.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
0.1 0.000000 0.095491 0.181634 0.249998 0.293890 0.309014
0.2 0.000000 0.181634 0.345488 0.475523 0.559011 0.587779
0.3 0.000000 0.249998 0.475523 0.654502 0.769413 0.809009
0.4 0.000000 0.293890 0.559011 0.769413 0.904499 0.951047
0.5 0.000000 0.309014 0.587779 0.809009 0.951047 0.999990
177
Table (3.16.2(b))
Velocity distribution in rectangular vibrating membrane through cubic spline implicit method u →at t = 0.03
y x 0.0 0.10 0.20 0.30 0.40 0.50
0.0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
0.1 0.000000 0.095483 0.181619 0.249978 0.293866 0.308989
0.2 0.000000 0.181619 0.345461 0.475485 0.558967 0.587732
0.3 0.000000 0.249978 0.475485 0.654450 0.769352 0.808944
0.4 0.000000 0.293866 0.558967 0.769352 0.904427 0.950917
0.5 0.000000 0.308989 0.587732 0.808944 0.950971 0.999910
0.000000
0.050000
0.100000
0.150000
0.200000
0.250000
0.300000
0.350000
0.0 0.1 0.2 0.3 0.4 0.5
X
U
t=0.01
t=0.03
Figure (3.16.2)
Velocity distribution in rectangular membrane through cubic spline implicit method (at y = 0.1)
178
3.16 (C) DISCUSSIONS OF RESULTS :
The velocity distribution obtained by the solution of
equation (3.14.1) by cubic spline explicit as well as implicit
methods are compared with exact solutions at y = 0.1 for
t = 0.03 in the table (3.16.3). From the table it is clear that
the solutions obtained by both these methods are accurate
up to five digits of decimal places. Figure (3.16.3(a))
indicates the error analysis which compares the exact
solutions with spline solutions obtained by both the
methods explicit as well as implicit at t = 0.03 clearly from
the figure (3.16.3(b)), the spline solutions are quite accurate
and reliable.
Table (3.16.3)
Error Analysis
Velocity distribution u → y = 0.1 & t = 0.03
X USE USI UEXT USE-UEXT USI-UEXT
0.0 0.000000 0.000000 0.000000 0.000000 0.000000
0.1 0.095483 0.095483 0.095483 0.000000 0.000000
0.2 0.181619 0.181619 0.181619 0.000000 0.000000
0.3 0.249978 0.249978 0.249978 0.000000 0.000000
0.4 0.293867 0.293866 0.293866 0.000001
587
0.000000
0.5 0.308989 0.308989 0.308989 0.000000
0.706706
0.000000
[USE, USI and UEXT denotes cubic spline explicit, cubic
spline implicit and exact solutions]
179
0.0000000
0.0000002
0.0000004
0.0000006
0.0000008
0.0000010
0.0000012
0.0
0.1
0.2
0.3
0.4
0.5
X
ERROR
USE-UEXT
USI-UEXT
Figure (3.16.3(a))
Error Analysis (at t = 0.03, y = 0.1)
180
0.000000
0.050000
0.100000
0.150000
0.200000
0.250000
0.300000
0.350000
0.0
0.1
0.2
0.3
0.4
0.5
X
U
USE
USI
UEXT
Figure (3.16.3(b))
Velocity distribution in rectangular mem
brane through cubic spline im
plicit method
(at y = 0.1, t = 0.03)
181
3.17 CONCLUSION :
In this chapter the method of spline collocation is
applied to linear PDE of parabolic and hyperbolic type with
one and two space variable. The problem describing flow of
electricity in a cable of transmission line, heat conduction
in a rod ,heat flow in thin rectangular plate, vibrating string,
vibrating membrane are discussed briefly. The governing
equations are parabolic PDE in one and two space variables
as well as hyperbolic PDE in one and two space variables.
All the equations are solved by spline explicit and spline
implicit scheme. The results are given in tabular form.
Comparison of the results shows that spline solutions are
quite reliable. The figure provides a pictorial evidence of a
good agreement of the curves presenting actual as well as
approximate solutions. Results are accurate up to five digits
of decimal place in most of the case.
In all problems, reducing the length of sub-interval
gives more closed results. An ultimate conclusion is drawn
from this work is that, both the method of spline collocation
gives accurate results with compact computations. From the
comparison of results and figure it is conclude that, implicit
scheme provide better convergence than explicit scheme.
