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
Crank- Nicholson Method
4 Solution of One Dimensional Wave Equation
1 Solution of Boundary value problems in ODE
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
(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 (xi,yi) 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 ( xi , y j ) is denoted by (i,j) and is surrounded by the neighbouring grid points (i-1,j),(i+1,j),(i,j-1),(i,j+1) etc.,
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 B2-4AC<0
Parabolic if B2-4AC=0
Hyperbolic if B2-4AC>0
Solution of Laplace equation uxx+uyy=0
This formula is called Standard five point formula
This expression is called diagonal five point formula.
This expression is called diagonal five point formula.
Leibmann’s Iteration Process
We compute u1 ,u3 ,u7 .u9 by using diagonal five point formula (DFPF)
Finally we compute u2 ,u4 ,u6 ,u8 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.
Let u1 , u2 ……u9 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 u1 , u2 , u4 , u5
Calculation of rough values
2.When steady state condition prevail,the temperature distribution of the plate is represented by Laplace equation uxx+uyy=0.The temperature along the edges of the square plate of side 4 are given by along x=y=0,u=x3 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 ui,where ui are the values of u at the internal mesh points.Solving the equations,the values ui are known.
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
In this scheme, the values of u at a time step are obtained by solving a system of linear equations in the unknowns ui.
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 = 2ut assuming ∆x=h=1
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
The one dimensional wave equation is of hyperbolic type. In this session, we discuss the finite difference solution of the one dimensional wave equation
1.What is the error for solving Laplace and Poisson’s equations by finite difference method?
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.
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?
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 j th level. Hence it is an implicit method.
4. What are the methods to solve second order boundary-value problems?
(i)Finite difference method (ii)Shooting method.
5. What is the classification of one dimensional heat flow equation.
One dimensional heat flow equation is
B2 −4AC = 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)
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
The equation is elliptic if (2B2 ) −4AC < 0
(i.e) B2 −AC < 0. It is parabolic if B2 −AC = 0 and hyperbolic if B2−4AC > 0
9. Write a note on the stability and convergence of the solution of the difference
equation corresponding to the hyperbolic equation .
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.
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.