10.0 INTRODUCTION

The areas of applications of partial differential equations is too large compared to ordinary differential equations. Partial differential equations occur especially in problems involving wave phenomena and heat conduction. In practical problems we seek to obtain unique solution of ordinary and partial differential equations subject to certain specific conditions which are called boundary value conditions. The differential equation with boundary value conditions is called a boundary-value problem.

For a differential equation when conditions are prescribed at the same point, we call them as initial conditions. When conditions are prescribed at different points, we call them as boundary conditions. The initial conditions and boundary conditions together are known as boundary-value conditions.

For example: For the one-dimensional wave-equation the conditions and which are prescribed at different points are called boundary conditions.

But the conditions , which are prescribed at the same point are initial conditions.

All the four conditions , , together are called boundary-value conditions.

The partial differential equation with these four conditions is called a boundary-value problem.

Certain types of boundary value problems can be solved by replacing the derivatives or partial derivations by approximate differences and thus reducing them to a difference equation. Thereby the given boundary value problem is converted to a system of linear equations which are solved by iteration methods.

10.1 FINITE DIFFERENCE METHODS FOR SOLUTION OF SECOND ORDER ORDINARY DIFFERENTIAL EQUATIONS

Consider the boundary value problem with the boundary conditions , . We divide the interval into n equal subintervals of width h, by the points where

Let be the corresponding values of y. In this method, the derivatives in the differential equations are replaced by their finite difference approximations, namely,

Substituting these values in the given differential equation we get a difference equation, which can be solved using the boundary conditions.

Note: (1) is called forward difference approximation, where as (2) is a central difference approximation for y_i′ and (3) is called central difference approximation for y_i″.

WORKED EXAMPLES

Example 1

Using the finite difference method, find y(0.25), y(0.5) and y(0.75) satisfying the differential equation subject to the boundary conditions

Solution

Given

Here

We know

(1)

Given

If i = 1, then (1)

(2)

If i = 2, then (1)

(3)

If i = 3, then (1)

(4)

Solve (2), (3), (4) to find y₁, y₂, y₃

(5)

Substituting in (3),

Thus,

Note: The equations (2), (3), (4) could be solved by Gauss elimination method.

Example 2

Solve the equation with the boundary conditions

Solution

Given

∴

We shall divide the interval (0, 1) into 4 equal parts

We know

∴ the equation becomes

The equations are (1)

(2)

(3)

We use Gauss elimination method to find y₁, y₂ and y₃

(4)

(6)

∴

Example 3

Solve xy′ + y = 0, y (1) = 1, y (2) = 2 with by using finite difference method.

Solution

Given

∴ and

Since

We know

∴ the equation becomes

, i = 1, 2, 3

∴ the equations are

(1)

(2)

(3)

(4)

(5)

Adding, (6)

 

Subracting,

From(5),

From(6),

Thus

Exercises 10.1

1. Solve the boundary value problem for by finite difference method
2. Solve by finite difference method:
3. by taking
4. Using finite difference method, solve the differential equation with
5. Solve the equation , with taking using finite difference method.
6. Solve , given taking , using finite difference method.
7. Solve the equation and using finite difference method.

Answers 10.1

(1)Take n = 4, y(0.5) = 0.1403 (2) 0.062, 0.250, 0.562 (3) –0.5556, –0.8889, –1

(5) 0.1389 (5) y(0,5) = 0.4706 (6) 0.2, 0.4, 0.6, 0.8

10.2 NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS

We have seen numerical solution of second order ordinary differential equations by replacing the derivatives by the approximate differences. Similarly partial differential equations are also solved numerically by finite difference method. The partial derivatives appearing in the equation and the boundary conditions are replaced by their finite difference approximations. Thereby the given equation is converted into a system of a linear equations which are solved by iteration methods.

10.2.1 Classifications of Second Order Partial Differential Equations

Let u be a function of two independent variables x and y. Consider the partial differential equation of u of the form

(1)

where A, B, C are functions of x and y or constants.

The second degree terms are linear whereas the part may be non-linear. So the equation (1) is called a second order quasi-linear equation. Incase is linear then (1) is called a second order linear equation.

At a point (x, y) in the xy-plane, the equation (1) is

1. elliptic if
2. Parabolic if
3. Hyperbolic if

Note:

1. If at all points of a region R if , then the equation is elliptic or parabolic or hyperbolic in R.
2. It is quite possible an equation elliptic in one region may be parabolic in another region and hyperbolic in yet another region.

WORKED EXAMPLES

Example 1

Test the nature of (i)

(ii)

(iii)

(iv)

Solution

(i) Given

Here

So, the equation is elliptic if ie. at all points on the right side of y-axis satisfying the equation. The equation is hyperbolic if ie. at all points on the left side of y-axis satisfying the equation and parabolic if ie at all points of y-axis satisfying the equation.

(ii) Given

Here

So, the equation is hyperbolic in the and quadrants, elliptic in the and quadrants and parabolic at all points in the x-axis and y -axis which satisfy the equation.

(iii) Given

Here

∴ the equation is hyperbolic for all (x, y) satisfying the equation

(iv) Given

Here

If then (as for all )

or for all

∴ the equation is hyperbolic if or and for all

If (as x² > 0 ∴ x ≠ 0)

∴ the equation is elliptic if and

If then or

∴ the equation is parabolic if

ie. on the y-axis or on the line or

Depending on the nature of the equations, methods of solving will differ.

Some well known equations, we consider for numerical solution are the following:

1. is Laplace’s equation

   ie. two dimensional heat equation in steady-state. It is elliptic type.

2. Poisson’s equation.

   It is elliptic type

3. one dimensional wave equation.

   It is hyperbolic type

4. one dimensional heat equation.

   It is parabolic type.

10.2.2 Finite Difference Approximations to Partial Derivatives

Consider a rectangular region in the xy-plane with sides parallel to the axes.

Divide this region into a network of rectangles of sides h and k units by drawing lines , , 1, 2, 3, ... The points of intersection of these family of lines are called mesh points or lattice points or grid points.

The point of intersection of is called the grid point

If u is a function of two independent variables x and y then the value of at the grid point is denoted by . Thus

Consider , where x is a variable and is fixed then is single variable function.

The Taylor’s series expansion of it in the neighbourhood of is

Put , then

If h is small, we approximate as

(1)

Replacing h by −h in (1), we get

(1) ∴ (2)

∴

This is called the forward difference approximation for

(2) ∴

This is called the backward difference approximation for

(1) − (2) ∴

∴

This is called the central difference approximation to .

The central difference approximation is better than the forward and backward difference approximations.

In the same way the approximation for 2^nd derivative is

Using backward difference approximations for and we get

and

In the same way, the first and second partial derivatives w.r.to y can be obtained.

The forward difference approximation is

The backward difference approximation is

The central difference approximation is

and

Since , the above formulae can be written as

– forward difference

– backward difference

– central difference

– forward difference

– backward difference

– central difference

10.2.3 Solution of Laplace Equation

Given the equation , the solution is satisfied by every point in a region subject to certain boundary conditions. Consider a rectangular region R for which is known at the boundary. For simplicity, take R to be a square region and divide it into small square regions of side h.

Replacing and by differences

the equation becomes

(1)

This shows that the value of u at any interior mesh point is the average value of the four adjacent mesh points on the lines through

ie. the average of the mesh point values above, below, left and right of

The formula (1) is called the standard five point formula (SFPF)

We know that Laplace equation remains invariant when the coordinate axes are rotated through . So we can also use the following formula.

(2)

This shows that the value of is the average of the values of the four mesh points through the diagonals of .

Figure 10.3

The formula (2) is called the diagonal five point formula (DFPF) as in Fig 10.3.

Now to find the initial values of u at the interior mesh points, we first find at the centre of the square by taking the mean of the 4 boundary values at the ends of the horizontal and vertical through .

Next we compute the initial values of the centers of the large squares with as a corner.

ie. we find by DFPF

The values at the remaining four mesh points are got by using SFPF

Thus, we have

Having got all the values of , the accuracy of these values are improved by any one of the following iteration methods.

1. Gauss Jacobi method:

If be the iterative value of , then an iterative procedure to SFPF is given by

2. Gauss-Seidel method

The iterative formula for SFPF is given by

Here we use the latest available iterative values and hence the rate of convergence will be faster by this method. Infact, it is nearly twice faster than Jacobi method.

The application of Gauss-Seidel iteration to solve boundary value problems was due to Liebmann. So this method is known as Liebmann’s method.

Note: It can be proved that the error in the diagonal formula is four times than in the standard formula. So, as far as possible we use the standard five point formula.

WORKED EXAMPLES

Example 1

Determine by iteration method the values at the interior lattice points of a square region of the harmonic function u whose boundary values are given as shown in the figure below

Solution

Since u is harmonic, it satisfies Laplace equation

Let u₁, u₂, u₃, … , u₉ be the values of u at the interior points.

We find the initial values of u₁, u₂, …, u₉.

Then by iteration improve these values.

We now start the iteration to improve these values. We start with u₁ and proceed along horizontal.

Liebmann’s iteration formula is

I Iteration: SFPF

II Iteration:

III Iteration:

We notice that II and III iteration values are almost same

Remark: We have worked out elaborately. We can find the values of u’s at each grid point indicating the values at the point itself as in the table schematically.

Using the latest available values.

Example 2

Given the values of u(x, y) on the boundary of the square in the figure, evaluate the function u(x, y) satisfying the Laplace equation — ²u=0 at the pivotal points of the figure by Gauss-Seidel method.

Solution

To get the initial values of u₁, u₁, u₃, u₄ we can not use either SFPF or DFPF to find any values. So we shall assume u₄ = 0 as the initial value.

Then

I Iteration:

II Iteration:

III Iteration:

IV Iteration:

V Iteration:

We take u1 = 1208, u2 = 792, u3 = 1042, u4 = 458

Example 3

By iteration method, solve the elliptic equation over a square region of side 4, satisfying the boundary conditions

By dividing the square into 16 square mashes of side 1 and always correcting the computed values to two places of decimals, obtain the values of u at 9 interior pivotal points.

Solution

Given equation is

The boundary condition gives all boundary values zero on y-axis.

, means on x-axis the values are 0, 3, 6, 9, 12

, gives the values on x = 4 as 12, 13, 14, 15, 16.

, means on y = 4 the boundary values are 0, 1, 4, 9, 16.

Let us take the 9 interior grid points as u₁, u₂, u₃, …, u₉

First we have to find the initial values.

The other four values u₂, u₄, u₆, u₈ can be found by SFPF

I Iteration:

II Iteration:

III Iteration:

From the II and III iterations we can take the values of u correct to 2 places as

10.2.4 Poisson Equation

An equation of the form is called Poisson’s equation.(1)

Proceeding as in Laplace equations we get the standard five point formula for poisson equation as

(2)

By applying (2) at each interior mesh point, we obtain a system of linear equations in the Pivotal values i.j.

We solve by Gauss-Siedel method

method

WORKED EXAMPLES

Example 4

Solve the Poisson equation

∇²u = -10(x² + y² + 10), 0 ≤ x ≤ 3, 0 ≤ y ≤ 3, u = 0

on the boundary.

Solution

Given

Solve the Poisson equation

The region is a square bounded by x = 0, x = 3, y = 0, y = 3

Divide it into a square mesh with side h = 1.

Let u₁, u₂, u₃, u₄ be the values of u at the interior mesh points A, B, C, D

∴

(1)

(2)

(3)

(4)

(5)

(2)′

(3)′

(4)′

(5)′

Note: In the above problem which is symmetric in x and y. That is when x and y are interchanged the equation is not changed. So the function is symmetric about the line y = x. So the solution is symmetric about y = x.

The boundary conditions are also symmetric

∴

This is the reason we got u₁ = u₄

This observation will be helpful in solving, when the number of mesh points are more in a square.

Example 5

Solve in the square mesh given u = 0 on the four boundaries dividing the square into 16 subsquares of length 1 unit.

Solution

Given

Here h = 1

Let be the values of u at the interior grid points.

Choose u₅ as origin and the x, y axes along the lines u₅ u₆ and u₅ u₂

Since the equation is symmetric in x and y and the boundary values are same when x and y are interchanged.

The values of u at the mesh points symmetric about y = x and about the axes will be the same

Hence we have to find only u₁, u₂, u₅

The difference equation given by the SFPF for poisons equation is

(1)

(2)

(3)

(4)

Example 6

Solve the poisson equation Given that

Solution

Given

The equation is symmetric about x and y.

When x and y are interchanged it is unaffected and so symmetric about y = x.

The boundary conditions are also symmetric about y = x.

Let u₁, u₂, u₃, u₄ be the values of u at the mesh points.

∴

The difference equation by SFPF is

(1)

Now

(1)

(2)

(3)

Substitute for u₄ in (1) we get

(4)

(5)

10.3 ONE DIMENSIONAL HEAT EQUATION

The one dimensional heat equation is

(where c is the specific heat of the material, r is the density and k is the thermal conductivity).

It is a parabolic equation.

(1)

We shall see two methods to solve for u(x, t) subject to the boundary conditions.

(2)

(3)

the initial conditions u(x, 0) = f(x), 0 < x < l(4)

10.3.1 Schmidt’s Method [Explicit Method]

Consider a rectangular mesh in the x-t plane with side lengths h in the x-direction and k in the t-direction. Let us denote the mesh point (x, y) = (ih, jk) as simply i, j.

Then we have the approximate relations of differences

Substituting in the equations (1), we get

∴

where is the mesh ratio parameter.

(5)

The boundary conditions can be written as

and the initial condition can be written as

u_x,0 = f(ih), i = 1,2,3,...

We choose the time interval k in such a way that the coefficient of ui_,j in (5) is zero.

∴

Then the equation becomes

∴ (6)

This equation is called the Bender-Schmidt recurrence equation.

Note:

1. The equation (6) gives the value of u at the (i, j + 1)^th interior mesh point as the average of the values of u at the surrounding points x_{i + 1} and x _{i − 1} at previous time t_j
   
2. Schmidt method is called the explicit method because each subsequent computation u at t = t_{j + 1} is explicitly given from the previous u-value at t = t_j
3. Bender-Schmidt method is valid only for . To obtain more accurate results h should be small and hence k should be very small. This makes the computation pretty long because many time-intervals would be required to cover the prescribed region. The solution will become unstable if λ exceeds . The reason for this in the explicit method is the difference in orders of the finite-difference approximations for the spatial derivative and the time derivative .

The next method, proposed by Crank and Nicolson in 1947 is a technique that makes these finite difference approximations of the same order. This method does not restrict λ and also reduces the amount of computations.

10.3.2 Crank-Nicolson Method [Implicit Method]

One dimensional heat equations is (1)

where with boundary conditions and the initial condition

In this method is replaced by the average of its central-difference approximations on the j^th and (j + 1)^th time rows.

∴

and

∴ the equation (1) becomes

Put

∴ (2)

This is called the Crank-Nicolson difference formula

Note:

• The left hand side of the formula contains three unknown values of u at the (j + 1)^th time level while all the three values on the right hand side are known values of u at the j^th level.
• Hence Crank-Nicolson formula is called a 2-level implicit relation because it does not give the value of u at t = t_{j + 1} directly in terms of u at t = t_j
• To simplify the scheme, we choose k such that l = 1
• The Crank-Nicolson formula becomes

(3)

If there are n internal mesh points on each row then the formula (2) gives a system of n simultaneous equations for the n unknown values in terms of the known boundary values. These equations can be soled by Gauss Seidel method. Similarly the values at the other internal mesh points on each row can be found.

WORKED EXAMPLES

Example 1

Solve and and, choosing h = k = 1 and using Bender-Schmidt formula find the values upto t = 5.

Solution

We have to choose

Bender-Schmidt recurrence formula for is

(1)

x varies from 0 to 4 ∴ i = 0, 1, 2, 3, 4, j ≥ 0

The boundary value conditions become

We shall tabulate the meshpoint values of u

If j = 0, then

If i = 1, then

If i = 2, then

If i = 3, then

Similarly for j = 1, 2, 3, 4, 5. The values can be found and they are tabulated.

Now we shall tabulate the mesh point value of u

Example 2

Find the values of the function u(x, t) satisfying the differential equation and the boundary conditions

u(0, t) = 0 = u(8, t) and at the points x = i, i = 0, 1, 2, 3, 4,...8 and j = 0, 1, 2, 3, 4, 5.

Solution

Given

Given i = 0, 1, 2, …, 8 and j = 0, 1, 2, 3, 4, 5

, So,

Bender-Schmidt recurrence formula for is

The boundary conditions are

and

For i = 0, 1, 2, …, 8, we get

These are the entries in the first row. Similarly we can find the other rows mesh points.

Now we shall tabulate the mesh point values of u

Example 3

Solve and u(x, 0) = 4(4 - x) choosing h = k = 1 using Bender-schmidt formula.

Solution

Given

Here a =1 ∴ a² = 1

The Bender-Schmidt formula for is

Since l = 1, the equation becomes

(1)

Now x = ih = i, y = jk = j,

which are the first row mesh points. Similarly the other rows mesh points are computed.

Now we shall tabulate the mesh point values of u

Example 4

Solve with u(x, 0) = x (1 − x), 0 < x < 1,

u(0, t) = u(1, t) = 0 t > 0. Using explicit method with for 3 time steps.

Solution

Given

Here

Here k is not given. We shall choose k such that

But

Then Bender-Schmidt formula is

The boundary condition are

∴

These values are the first row mesh points.

For 3 time steps the mesh point values are given in the table

Example 5

Solve the equation with the conditions u(0, t) = 0, u(4, t) = 8 and with the conditions u(0, t) = 0, u(4, t) = 8 and , taking and upto 5 time units.

Solution

Given

Since we have to use the general Schmidt’s formula

The boundary value conditions are u(0, t) = 0 ∴ u_0,j = 0 j = 0, 1, 2, 3, 4, 5

When x = 4, t = 8,

Now we shall tabulate the mesh point value of u

Example 6

Using Crank-Nicolson’s implicit scheme, solve 16 ut = uxx, 0 < x < 1, t > 0, given that u(x, 0) = 0, u(0, t) = 0, u(1, t) = 100t, compute u for one-time step.

Solution

Given uxx = 16 ut

Here

Here (0, 1) is divided, choose .

Choose k such that

Since l = 1, we use the simplified form of Crank-Nicolson formula

Boundary value conditions are

We have to compute the values of u for only one time step.

Put j = 0, i = 0, 1, 2, 3, 4.

Substituting for u₁ and u₃ in (2), we get

Example 7

Solve given that u(x, 0) = 20, u(0, t) = 0, u(5, t) = 100. Compute u for one time step with h=1 by Crank-Nicolson method.

Solution

Given

Here

k is not given, so choose k such that l = 1

Since l = 1, the simplified Crank-Nicolson formula is used to compute only one time step.

(1)

where x = i h = i, t = k j = j

On the line x = 0, u = 0 for all t. That is for all j

Using formula (1)

Substituting (5) in (4),

(6)

Substituting (2) in (3),

(7)

∴

Example 8

Solve subject to the conditions u(x, 0) = sin p x, 0 ≤ x ≤ 1 u(0, t) = u(1, t) = 0, using Crank-Nicolson method taking

Solution

Given uxx = ut

Crank-Nicolson formula is

Since , we get

Multiply by 4,

(1)

The boundary conditions are

We shall find one time step values.

Putting j = 0, i = 1 in (1), we get

∴ (1)

Putting j = 0, i= 2, in (1) we get,

∴

∴

Substituting in (1), we get

Thus we have computed the value of u for one time step.

Example 9

Using Crank-Nicolson scheme Solve given u(x, 0) = u(0, t) = 0, u(1, t) = t, choosing h = 0.5,

Solution

Given

Here

So we have to use Crank-Nicolson general formula

Putting

Multiply by 2, (1)

The boundary value conditions are

∴

We shall find one-step values.

Putting j = 0, i = 1 in (1), we get

Example 10

Solve by Crank-Nicolson’s method for 0 < x < 1, t > 0 given that u(0, t) = 0, u (1, t) = 0 and u(x, 0) = 100x(1 - x). Compute u for one time step with

Solution

Given u_xx= u_t

Here

So, we choose k such that

Since l = 1, we use Crank-Nicolson simplified formula

(1)

The boundary value conditions are

We have to compute on time step values of u

Putting i = 1, j = 0 in (1), we get

(2)

Putting i = 2, j = 0 in (1), we get

(3)

Putting i = 3, j = 0 in (1), we get

(4)

From (2) and (4), we get u_1,1 = u_3,1

∴ (3) ∴ u_2,1

∴ 4u_2,1

∴ 4u_2,1 [Using (2)]

∴

∴ u_2,1 = 14.29

∴

and u_3,1 = 9.82

Thus

Example 11

Obtain the Crank-Nicholson finite difference method by taking Hence, find u(x, t) in the rod for two time steps for the heat equation given u(x, 0) = sin px, u(0, t) = 0, u(1, t) = 0. Take h = 0.2.

Solution

Given

For derivation of Crank-Nicholson formula for one dimensional heat equation where refer page [In this question c is a]

Here and h = 0.2

Now l = 1

Since l = 1, we use the simplified Crank-Nicolson formula

(1)

where x = i h = 0.2 i, t = k j = 0.04 j and 0 ≤ x ≤ 1

When x = 0, i = 0 and when x = 1, ∴ i = 0, 1, 2, 3, 4, 5

The boundary value conditions are

and u(x, 0) = sin px ∴ u(0.2i, 0) = sinp (0.2i) = sin

On the line x = 0, u = 0t ie. when i = 0, u = 0 j

and on the line x = 1, u = 0j ie. when i = 5, u = 0 j

For two time steps we have to compute for j = 0, 1, 2

∴

∴

We shall tabulate the values

We shall now compute the unknown u’s.

Putting j = 0 in (1) we get the second row values of u’s

(2)

Put i = 2, then

(3)

Put i = 3, then

(4)

Put i = 4, then

(5)

Substituting (5) in (4) we get

∴

∴(6)

Now substituting (2) in (3), we get

∴(7)

(6) × 15 ∴ 225u_3,1 106.5945

(7) × 4 ∴ 16u_3,1 − 60u_2,1 = −28.4264

Subtracting we get, 209u_3,1 = 135.0209 ∴ u_3,1 = 0.6460

Substituting in (6), we get

15(0.6460) − 4u_2,1 = 7.1063

∴ 4u_2,1 = 15(0.6460) − 7.1063 = 2.5837

∴

To find the third row values of u, put j = 1 in (1)

∴

Put i = 1, then

⇒ (8)

Put i = 2, then

(9)

Put i = 3, then

(10)

Put i = 4, then

(11)

Substituting (11) in (10) we get,

∴ (12)

Substituting (8) in (9), we get,

∴

∴ (13)

(12) × 15 ∴ 225u_3,2 − 60u_2,2 = 72.402

(13) × 4 ∴ 16u_3,2 − 60u_2,2 = −19.3068

Subtracting we get, 209 u_3,2 = 91.7088 ∴ u_3,2 = 0.4388

∴ 15u_2,2 = 4(0.4388) + 4.8267 = 6.5819 ∴ u_2,2 = 0.4388

∴

and

Thus the two time step values are given in the table

 

10.4 ONE-DIMENSIONAL WAVE EQUATION

One dimensional wave equation (of vibrating string) is

(1)

This is hyperbolic type.

We shall solve this for u(x, t) using finite difference method subject to the boundary conditions.

u(0, t) = 0, u(l, t) = 0(2)

and the initial conditions

u(x, 0) = f (x), ut(x, 0) = 0(3)

Assume equal interval h for x variable and equal interval k for t variable and x = ih, t = jk and u(x, t) = u(ih, jk) = ui,j

We know the partial derivatives in terms of differences

∴ the equation (1) becomes

∴

Let then the equation becomes

∴

∴ (4)

This equation is an explicit formula for the solution of the wave equation because it gives the value of u at t = t_{j + 1}, explicitly in terms of the values of u at t = t_j and t = t_{j − 1}

The simplest form of the equation (4) is obtained by putting

∴ the equation (4) becomes,

(5)

The boundary conditions are

and if l = nh

The initial conditions are u(x, 0) = f(x) ∴ u(ih, 0) = f(ih)

∴ ui _,0 = f(ih), i = 0, 1, 2, 3 ,…

We shall use central difference approximation for

When t =0, u_t(x,0) = 0 ∴ ⇒

When j = 0,

∴ for all i

Putting j = 0 in (5), we get

∴−_1,0

∴

For i = 1, 2, 3, …, we get the values of u in the second row.

Note: The period of vibration of the string of length l given by is we have to find the values of u at the mesh points for one period, unless otherwise specified.

WORKED EXAMPLES

Example 1

Solve 16uxx = utt, u(0, t) = u(5, t) = 0, u(x, 0) = x2(5 - x), ut(x, 0) = 0 taking h = 1 and upto one half of the period of vibration.

Solution

Given 16uxx − utt = 0

Here a² = 16 ∴ a = 4.

Now

Choose l = 1 and h = 1, then

Since l = 1, the simplest difference equation is

(1)

Now x = ih =i, t = jk = u(x, y) = u = ui,j

From the boundary conditions u(0, t) = u(5, t) = 0, it is obvious 0 ≤ x ≤ 5 and length of the string l = 5

∴ period of vibration

Half the period = 1.25 secs.

So, we have to find values of t upto t = 1.25 ∴ = 1.25 ∴ j = 5

(2)

(3)

∴ (4)

When t = 0, ∴

Since t = 0 ∴ j = 0

When j = 0, (5)

(2) ∴

(3) ∴

(4) ∴

Putting j = 0 in (1), we get

[Using (5)]

∴−_1,0

∴ (6)

Put i = 1,

To fill up the 3^rd row put j = 1 in (1)

∴

Putting i = 1, 2, 3, 4, we get

To fill up the 4^th row put j = 2 in (1)

∴

Putting i = 1, 2, 3, 4 we get

To fill up the 5^th row put j = 3 in (1)

Putting i = 1, 2, 3, 4, we get

To fill up the last row put j = 4 in (1)

Putting i = 1, 2, 3, 4, we get

Example 2

Solve the wave equation

and using h = k = 0.1 for 3 time steps.

Solution

Given

Here a² = 1 ∴ a = 1. h = k = 0.1

∴

Since l = 1, the simplest equation is (1)

x = ih = (0.1)i, t = j k = (0.1)j, u(x, y) =

Since 0 ≤ x ≤ 1, x = 1 ∴ 0.1 i = 1 ∴ i = 10. So i = 0, 1, 2, ..., 10

Since t takes 3 steps, j = 0, 1, 2, 3

Given

∴

When t = 0, u(x, 0) = 0 ∴

∴

Now t = 0 ∴ j = 0

When j = 0,

(1) ∴

 

∴

∴

Putting i =1, 2, ..., 9, we get

To find the values of u in the third row put j = 1 in (1)

Putting i = 1, 2, 3, …, 9, we get

To find the values of u in the 4^th row put j = 2 in (1)

Putting i = 1, 2, ..., 9, we get

Example 3

Approximate the solution to the wave equation u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = sin2p x, 0 £ x £ 1 and 0 £ x £ 1 with Δx = 0.25 and Δt = 0.25 for 3 time steps.

Solution

Given

Here a² = 1 ∴ a =1; Δx = h = 0.25, Δt = k = 0.25

∴

Since l = 1, the difference equation is

(1)

Since i varies from 0 to 1 and

We have to compute for 3 time steps. ∴ j = 0, 1, 2, 3

∴ (2)

u(0, t) = 0 for t > 0 ∴ u_0,j = 0, j = 1, 2, 3 (3)

∴ (4)

∴

So we have the first row values of u.

To find second row mesh point values of u.

When t = 0 u_t (x,0) = 0 ∴ ∴

When j = 0, j = 0]

When j = 0, (1) ∴

∴

∴ (5)

Putting i = 1, 2, 3, we get

To find the third row values of u, put j = 1 in (1)

∴

Putting i = 1, 2, 3, we get

To find the fourth row values of u, put j = 2 in (1)

∴

Putting i = 1, 2, 3, we get,

Example 4

Solve numerically 4uxx = utt with the boundary conditions u(0, t) = 0, u(4, t) = 0, t > 0 and the initial conditions ut(x, 0) = 0, u(x, 0) = x(4 − x), taking h = 1 for four time steps.

Solution

Given 4uxx = utt

Here a² = 4 ∴ a = 2; h = 1

We shall choose k such that l = 1 ∴

Since l = 1, the difference equation is

(1)

From boundary condition, 0 ≤ x ≤ 4 and so i = 0, 1, 2, 3, 4

(2)

Since we have to compute for 4 time steps j = 0, 1, 2, 3, 4

(3)

∴ (4)

∴ [From (2)]

and [From (3)]

From (4),

So we have the first row values of u

Now to find other rows mesh point values of u

When t = 0, ∴ u_i,j −₁= 0

When j = 0 j = 0]

∴ (5)

When j = 0, (1) ∴

= [Using (5)]

∴

∴

When i = 1,

When i = 2,

When i = 3,

To find the values of u in the third row put j = 1 in (1)

∴

Putting i = 1, 2, 3 we get

To find the values of u in the 4th row, put j = 2 in (1)

∴

Putting i = 1, 2, 3, we get

To find the values of u in the 5th row, put j = 3 in (1)

∴

Putting i = 1, 2, 3, we get

Exercises 10.2

(I) Elliptic equations:

1. Solve uxx + uyy = 0 for the following mesh with boundary values as shown in the figure. Iterate until the maximum difference between the successive values at any point is less than 0.005.
   
2. Solve uxx + uyy = 0 numerically for the following mesh with boundary conditions as shown below.
   
3. Solve in the square mesh given u = 0 on the four boundaries dividing the square into 16 subsquares of length 1 unit.
   
4. 0 over the square region of side n, satisfying the boundary condition
5. 0, 0 ≤ y ≤ 4 (ii) u(4,y) = 12 y ≤ 4
6. 3x, 0 ≤ y ≤ 4 (iv) u(x,4) = x², 0 ≤ y ≤ 4.
7. 8 + 2y, u(x, 0) = , u(x, 4) = x² with
8. 

9. (II) Parabolic equations:

   • (6) Solve subject to the conditions u(x, 0) = , u(0, t) = u(1, t) = 0

     Using Schmidt method taking

   • (7) Solve uxx = 32 ut = taking h = 0.25 for t > 0, 0 < x < 1 and u(x, 0) = 0, u(0, t) = 0, u(1, t) = t using Schmidt method.
   • (8) Using Crank-Nicolson method solve uxx = 16 ut , 0 < x < 1, t > 0 given u(x, 0) = 0, u(0, t) = 0 and u(1, t) = 100 t, choosing .
   • − Nicoloson method 0 < x < 2, t > 0, u (0, t) = u (2, t) = 0, t > 0 and u(x, 0) using h = 0.5, k = 0.25 for two time steps.
   • (10) Using Crank-Nicolson method solve ut = uxx in , t > 0, given u(0, t) = 0 u(l, t)
   • and u(x, 0) =
   • Take h = 0.1 and find u for one step in t.

   (III) Hyperbolic equations:

   • 0.5, with spacing of 0.1 subject to y (0, t) = 0, y(1, t) = 0, yt (x, 0) = 10 − x).
   • − x²), (x, 0) = 0, u(0, t) = u(1, t) = 0, t > 0 by finite difference method for one time step with h = 0.25.
   • (13) Solve 0 < x < 1, t > 0, given u(x, 0) = 0, ut (x, 0) = u(0, t) = 0 and u(1, t) = 100 sin p t. Compute u for 4 time steps with h = 0.25.
   • − utt = 0 for u at the pivotal points, given u(0, t) = u(5, t) = 0, ut(x, 0) = 0 and
   • u(x,0) = for one half period of vibration.
   • (15) Solve the boundary value problem utt = uxx = 0 with the conditions u(0, t) = u(1, t) = 0, u(x, 0) and ut(x, 0) = 0 taking h = k = 0.1 for . Compare your solution with the exact solution at x = 0.5 and t = 0.3.
   • Answers 10.2
   • 1.999, 2.999, 3.999 (2) u₁ = u₄ = 1.333, u₂ = u₃= 1.6667
   • − 3 (4) 2.37, 5.59, 9.87, 2.88, 6.13,
   • − 2, u₅ = − 2 9.88, 3.01, 6.16, 9.51
   • 4.9, u₃ = 9, u₄ = 2.07, u₅ = 4.69,
   • u₆ = 8.07, u₇ = 1.57, u₈ = 3.71, u₉ = 6.57]
   • (6)
   
   • (7)
   
   • 1.7857, u ₂ = 7.1429, u₃ = 26.7857
   • u₁ = 0.2857, u₂ = 0.1429, u₃ = 0.2857, u₄ = 0.0816, u₅ = 0.1837, u₆ = 0.0816
   • (10)
   
   • (11)
   
   • (12)
   
   • (13)
   
   • (14)
   
   • Find the exact solution when t = 0.3 or j = 3
   

Short Answer Questions

• 0.
• 2. Identify the type of the equation f_xx + 2f_xy + f_yy = 0.
• 3. Classify the partial differential equation u_xx + 2u_xy + 4u_yy = 0, x > 0, y > 0.
• 4. For what values of x and y, the equation xf_xx + yf_yy = 0, x > 0, y > 0 is elliptic?
• 5. Obtain the finite difference approximation for the differential equation
• 6. Name at least two numerical methods that are used to solve one dimensional diffusion equation.
• 7. Write down Laplace equation and its finite difference analogue and the standard five point formula.
• 8. State the explicit finite difference scheme for one dimensional wave equation
• 9. Find the explicit finite differences scheme for one dimensional wave equation
• 10. Write down the Crank-Nicolson formula to solve u_t = u_xx.
• 11. Write down the implicit formula to solve one dimensional heat flow equation
• 12. Why is Crank - Nicolson’s scheme is called an implicit scheme?
• 13. What type of equations can be solved by using Crank-Nicolson’s formula?
• 0.
• 15. State Leibmann’s iteration process formula.
• 16. Write down the finite difference form of the Poisson equation ∴²u = f(x, y).
• 17. Write down the Schemidt’s explicit formula for solving heat flow equation.
• − au_t = 0.
• 19. Write down the Crank-Nicolson difference scheme to solve u_xx = au_t with u(0, t) = T₀, (l, t) = T₁ and the initial condition as u(x, 0) = f(x).
• Write down the Forward difference approximation to u_x(x₀, y₀).
• 21. Write down the backward difference approximation of u_x(x₀, y₀).
• 22. Write down the finite difference approximation to u_xx, where u = u(x, y) taking, h, k as the step sizes at the point (x, y).