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Boundary Value Problems in ODE & PDE
1 Solution of Boundary Value Problems in ODE
2 Solution of Laplace Equation and Poisson Equation
Solution of Laplace Equation – Leibmann`s iteration process
Solution of Poisson Equation
3 Solution of One Dimensional Heat Equation
Bender-Schmidt Method
Crank- Nicholson Method
4 Solution of One Dimensional Wave Equation

**Boundary value Problems in ODE&PDE**

**BOUNDARY
VALUE PROBLEMS IN ODE & PDE**

**1
Solution of Boundary Value Problems in ODE**

**2
Solution of Laplace Equation and Poisson Equation**

Solution of Laplace Equation – Leibmann`s iteration
process

Solution of Poisson Equation

**3
Solution of One Dimensional Heat Equation**

Bender-Schmidt Method

Crank- Nicholson Method

**4
Solution of One Dimensional Wave Equation**

**1 Solution of Boundary
value problems in ODE **

**Introduction**

The solution of a
differential equation of second order of the form *F *(*x*,* y*,*
y’ *,* y’’ *) 0* *contains
two arbitrary constants. These constants are determined by means of two
conditions. The conditions on *y* and *y*’or their combination are prescribed at two different values of x
are called ** boundary conditions**.

The differential
equation together with the boundary conditions is called a ** boundary value
problem**.

In this chapter ,we
consider the finite difference method of solving linear boundary value problems
of the form.

**Finite difference approximations to derivatives**

**First derivative approximations**

**Second derivative approximations**

**Third derivative approximations**

**Fourth derivative
approximations**

**Solution of ordinary differential
equations of Second order**

**Problems**

**Sol:**

(i)
Divide the interval[1,2] into two
sub-intervals with h=(2-1)/2=0.5

**2 Solution of
Laplace Equation and Poisson equation **

Partial differential
equations with boundary conditions can be solved in a region by replacing the
partial derivative by their finite difference approximations. The finite
difference approximations to partial derivatives at a point (x_{i},y_{i})
are given below. The xy-plane is divided
into a network of rectangle of lengh ∆*x= h* and breadth ∆*y=k*
by drawing the lines x=ih and y=jk, parallel to x and y axes.The points
of intersection of these lines are called grid points or mesh points or lattics
points.The grid points ( *x _{i}* ,

*Note*

The most general linear
P.D.E of second order can be written as

Where A,B,C,D,E,F are in general functions of x and
y.

The equation (1) is said
to be

Elliptic if B^{2}-4AC<0

Parabolic if B^{2}-4AC=0

Hyperbolic if B^{2}-4AC>0

**Solution of
Laplace equation u _{xx}+u_{yy}=0**

This formula is called *Standard
five point formula*

**Leibmann’s**
**Iteration Process**

We compute *u*_{1}
,*u*_{3} ,*u*_{7} .*u*_{9} by using diagonal five point formula (DFPF)

Finally we compute *u*_{2}
,*u*_{4} ,*u*_{6} ,*u*_{8} by using standard five point formula.

The use of Gauss-seidel iteration method to solve
the system of equations obtained by finite difference method is called *Leibmann’s
method.*

**Problems**

**Sol:**

**Let ***u*_{1}**
**,**
***u*_{2}**
**……*u*_{9}**
**be the
values of u at the interior mesh points of the given region.By symmetry about**
**the lines
AB and the line CD,we observe

Hence it is enough to find *u*_{1} , *u*_{2} , *u*_{4} , *u*_{5}

*Calculation of
rough values*

*u*_{5}*
*=1500

*u*_{1}*
*=1125

*u*_{2}*
*=1187.5

*u*_{4}*
*=1437.5

Gauss-seidel scheme

2.When steady state
condition prevail,the temperature distribution of the plate is represented by
Laplace equation u_{xx}+u_{yy}=0.The temperature along the
edges of the square plate of side 4 are given by along x=y=0,u=x^{3}
along y=4 and u=16y along x=4,divide the square plate into 16 square meshes of
side h=1 ,compute the temperature a iteration process.

**Solution of
Poisson equation**

This expression is
called the replacement formula.applying this equation at each internal mesh
point ,we get a system of linear equations in u_{i},where u_{i}
are the values of u at the internal mesh points.Solving the equations,the
values u_{i} are known.

Problems

u(0,y)=u(x,0)=0,u(x,1)=u(1,y)=100
with the square meshes ,each of length h=1/3.

**3 Solution of One dimensional heat equation**

In this session, we
will discuss the finite difference solution of one dimensional heat flow equation
by Explicit and implicit method

**Explicit
Method(Bender-Schmidt method**

This formula is called
Bender-Schmidt formula.

**Implicit method
(Crank-Nicholson method)**

This expression is called Crank-Nicholson’s implicit scheme.

This expression is called Crank-Nicholson’s implicit scheme. We note that Crank Nicholson’s scheme converges for all values of λ

The use of the above
simplest scheme is given below.

The value of u at A=Average of the values of u at B,
C, D, E

**Note**

In this scheme, the
values of u at a time step are obtained by solving a system of linear equations
in the unknowns u_{i}.

**Solved Examples when u(0,t)=0,u(4,t)=0
and with initial condition u(x,0)=x(4-x) upto t=sec**

**1.Solve ***u*** **_{xx }*= _{ }*2

**Sol:**

By Bender-Schmidt
recurrence relation ,

For applying eqn(1) ,we
choose

Here a=2,h=1.Then k=1

By initial conditions,
u(x,0)=x(4-x) ,we have

**4 Solution of One dimensional wave
equation**

**Introduction**

The one dimensional
wave equation is of hyperbolic type. In this session, we discuss the finite difference solution of the one dimensional wave equation

__Part
A__

**1.What
is the error for solving Laplace and Poisson’s equations by finite difference
method?**

__Sol:__

The error in replacing
by the difference expression is of the order . Since h=k, the error in replaing
by the difference expression is of the order .

**2**.** Define a
difference quotient.**

__Sol:__

A difference quotient
is the quotient obtained by dividing the difference between two values of a
function by the difference between two corresponding values of the independent
variable.

**3.
Why is Crank Nicholson’s scheme called
an implicit scheme?**

__Sol:__

The Schematic
representation of crank Nicholson method is shown below.

The solution value at
any point (i,j+1) on the (*j* +1)* ^{th}* level is dependent
on the solution values at the neighboring points on the same level and on three
values on the

**4. What are the methods to solve second
order boundary-value problems?**

__Sol:__

(i)Finite difference method (ii)Shooting method.

**5. What is the classification of one
dimensional heat flow equation.**

__Sol:__

One dimensional heat
flow equation is

Here
A=1,B=0,C=0

*B*^{2}*
*−4*AC
*= 0

Hence
the one dimensional heat flow equation is parabolic.

6. 6. State Schmidt’s explicit formula for solving heat flow equation

__Sol: ---- ---- __

**7. **** Write an explicit formula to solve numerically the heat equation (parabolic equation) **

__Sol:__

--------- ----------

x and k is the space in the time direction).

The above formula is a
relation between the function values at the two levels j+1 and j and is called
a two level formula. The solution value at any point (i,j+1) on the (j+1)^{th}
level is expressed in terms of the solution values at the points (i-1,j),(i,j)
and (i+1,j) on the j th level.Such a method is called explicit formula. the
formula is geometrically represented below.

8. **State the
condition for the equation** **to be**

**(i)
elliptic,(ii)parabolic(iii)hyperbolic when A,B,C are functions of ***x***
and ***y*

__Sol:__

The
equation is elliptic if (2*B ^{2}* ) −4

(i.e) *B ^{2}* −

9. **Write a note on the stability and convergence
of the solution of the difference**

**equation corresponding to the hyperbolic
equation** .

__Sol:__

For ,λ= the solution of
the difference equation is stable and coincides with the solution of the
differential equation. For λ> ,the solution is unstable.

For λ<
,the solution is stable but not convergent.

10. **State
the explicit scheme formula for the solution of the wave equation.**

__Sol:__

The formula to solve
numerically the wave equation =0 is

The schematic representation is shown below.

The solution value at
any point (i,j+1) on the ( *j* +1)^{th} level is expressed in
terms of solution values on the previous j and (j-1) levels (and not interms of
values on the same level).Hence this is an explicit difference formula.

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