Numerical Analysis
Module XVII
Numerical solutions for Ordinary Differential Equations-III
(Boundary Value Problems and Eigen Value Problems)
Objectives
From this session a learner is expected to achieve the following:
- Familiarize with boundary value problems.
- Study to solve boundary value problems by exact and numerical methods.
- Learn that finite difference method is helpful in solving boundary value problems.
- Study Richardson’s Extrapolation Formula
- Familiarise with eigen value problems
- Familiarise with Sturm-Liouville’s Problemsand numerical solution of the same.
Introduction
A second-order differential equation must have two conditions for its numerical solution. If these conditions are given at the start, then the problem is an initial value problem. In this session we consider the case when the conditions may be at different points, usually at the endpoints of the region of interest, and such problems are called boundary value problems. We first discuss exact solution and then discuss numerical solution of boundary value problems. We will see that finite difference method is very useful in finding the solution of boundary value problems. Richardson’s Extrapolation Formula also will be discussed. Eigen Value Problems, in particular Sturm-Liouville’s Problem, will be discussed. Both exact and numerical solutions of such problems will be discussed.
Boundary Value Problems
Since the general solution to a second order differential equation …(1)
contains two arbitrary constants, we need two conditions for obtaining a particular solution. Some times the two conditions are of the type
… (2)
The conditions in (2) are called boundary conditions since they refer to the boundary points of an interval Equation (1) and conditions in (2) together constitute what is known as a boundary value problem.
If the function g has the value zero for each x, and if the boundary values and are also zero, then the problem (1) with conditions (2) is called homogeneous. Otherwise, the problem is nonhomogeneous.
Remark: Some times the boundary conditions have the form
and
Problem 1: Solve the boundary value problem
Solution:
The general solution to the given differential equation is
Using the first of boundary
conditions, we obtain
The second boundary condition implies that so Thus the solution of the given boundary value problem is
Problem 2:Solve the boundary value problem
Solution:
The general solution is given by
The first boundary condition requires that The second boundary condition gives Since we must have
Consequently, for all is the only solution of the given boundary value problem. This example illustrates that a homogeneous boundary value problem may have only the trivial solution.
Problem 3: Solve the boundary value problem
Solution:
The general solution is given by
The first boundary condition requires that The second boundary condition is also satisfied when regardless the value of Thus the solution of the given boundary value problem is
where is still arbitrary. This example illustrates that a homogeneous boundary value problem may have infinitely many solutions.
Problem 4: Solve
Solution:
The general solution is given by
Using the first of the boundary conditions we have
The second boundary condition gives
These two conditions on are incompatible if so the problem has no solution in that case. However, if then both
boundary conditions are satisfied provided that where is still arbitrary.
In this case a solution to the given boundary value problem is
where is still arbitrary. This example illustrates that a homogeneous boundary value problem may have no solution, and also that under special circumstances it may have infinitely many solutions.
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Numerical Solution of Boundary Value Problems
We consider the boundary value problem
…(3)
with the boundary conditions
… (4)
If is any solution of the equation (3), which satisfies the condition and is any nontrivial solution of the equation
which satisfies the conditionthen the function
satisfies (3) and the condition for any constant value of Further, if and are continuous on then a solution to the given boundary value problem is found if can be determined such that
Problem 5: Solve the boundary value problem
Solution:
A particular solution to the given differential equation is
Hence we put
where at the origin. Using Picard’s method,
Now using the boundary condition,
gives
We now describe finite-difference method.
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Finite Difference Method
We consider the boundary value problem
with the boundary conditions
The finite differnce method for the solution of a boundary value problem consists in replacing the derivativs occuring in the differential equation (and the boundary conditions) by means of their finite difference approximations and then solving the resulting linear system of equations. The finite difference approximations to the derivatives are obtained as follows:
The Taylor series representation of is given by
which gives
.
The above is the forward
difference approximation for .
Similarly, which gives
.
The above is the backward difference approximation for . Also
which gives
.…(5)
The above is the central difference approximation for .
Also
which gives
..…(6)
To solve the boundary-value problem given by Equations (3) and (4) , we divide the range into n equal sub-intervals of width h so that
The corresponding values of
y at these points are denoted byUsing Equations (5) and (6), values of and at the point are given by
and
At the point the differential equation (3) gives where
Substituting the expressions for and the above equation gives
…(7)
where and
The above can be simplified to the following form
The above gives the folowing algebraic system of equations:
Since and are already known, the above system of equations may be solved for the unknowns
Problem6 (i) using numerical methods with step sizes and
and
(ii) Using exact method,
find the value of to the following boundary value problem: with the boundary conditions
Solution:
Comparing with the standard equation (3), we have
and hence using Eq. (7), the finite difference approximation for the given second order differential equation is
On simplification,
The boundary conditions are
The above equations give a system of equations in unknowns, viz.,
If we set then Then, the corresponding system of 3 equations in 3 unknowns is
with
Hence
or
Hence
If we set then The corresponding system of 5 equations in 5 unknowns is
With Solving the above system, we obtain
Hence
Now to find the exact value, we proceed as follows:
The characteristic equation associated with the given differential equation
is
which gives
Hence the solution to the homogeneous equation
Is
Also is a particular solution to the differential equation
.
Hence the general solution to the given differential equation
Using the boundary conditions we obtain
and
Hence the solution to the given boundary value problem is
Hence the exact value of is
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Richardson’s Extrapolation Formula
In order to improve the accuracy of the solution of a boundary value problem we use Richardson’s Extrapolation Formula. Let and be two approximate values of y obtained using Finite Difference Method with step sizes and respectively. Then
and
Eliminating A between the above two equations, we obtain
If we take and , then the above takes the form
The above is the Richardson’s Extrapolation Formula.
Problem7: Using Richardson’s formula, find to the boundary value problem
with the boundary conditions
Solution
Using Eq.(7), the finite difference approximation of the given differential equation is
On simplification the above gives
…(8)
Let ; then and in this case we have
Substituting these values in Eq.(8), with we obtain
which gives
Next let ; then and in this case we have the following system of equations:
with
Solving the above system, we obtain
Taking and and using the Richardson’s formula,
17.6. Eigen Value Problems:
Consider the boundary value problem
…(9)
with the boundary conditions
…(10)
We define eigen values and eigen functions as follows:
Definition: The values of for which nontrivial solutions of (9) and (10) occur are called eigen values, and the corresponding nontrivial solutions are called eigen functions.
Problem 8: Discuss whether is an eigen value of the boundary value problem (9) and (10). Give some eigen functions. Is an eigen value?
Solution :
Observe that the above boundary value problem (9) and (10) is the same as that in Problems2 and 3 if and respectively.
- Referring to Problem 3, we note that for Eqs. (9) and (10) have the solution
where is an arbitrary constant. This shows that for Eqs. (9) and (10) have nontrivial solutions. Hence is an eigen value. Any nonzero multiple of is an eigen function corresponding the eigen value
- Referring to Problem 2, we note that for Eqs.(9) and (10) have only the trivial solution Hence is not an eigen value.
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Sturm-Liouville’s Problem
Some times a second order differential equation is written in the form
…(11)
The problem of solving this equation with respect to the boundary conditions
and
or is called the Sturm-Liouville’s problem.
If the interval between a and b is divided into equal parts, and the derivatives are approximated by difference expressions, we obtain
This can be written in the form
where is a band matrix and is a column vector. Nontrivial solutions exist if and hence the eigen value problem discussed now can be transformed into an algebraic eigen value problem.
In the Sturm-Liouville case, we compute iteratively from
When integrating, we obtain two integration constants which can be determined from the boundary conditions; also, is obtained from the condition or from similar condition. This is illustrated in the following problem.
Problem 9: Using usual method and using numerical method, find the lowest value of and the corresponding solution to the boundary value problem
Solution:
The general solution to the differential equation
is
and the first boundary condition gives Hence
and the second boundary condition gives
.
Now for the nontrivial solution, so
gives
Hence the lowest value of is (obtained by putting) and the corresponding solution to the boundary value problem is .
To obtain numerical solution, we proceed as follows:
The iterative formula is
where we find from the condition
We start with that fulfills
Then
Integrating,
Again, integrating we obtain .
gives gives and gives and hence
i.e.,.
Repeating the procedure we get the values as in the following Table:
0 / / -1 / / 2.4
2 / / 2.459
3 / / 2.4664
4 / / 2.46729