ME502-Cprogram-main Solved

A. Solving Initial value problem (Ordinary differential equation with initial condition):

We want to solve following ODE (Ordinary differential equation)

y′ = f(x,y) where y(x0) = y0;

i.e. we want to get values of y(x) at different x points (0 < x < L) such that it satisfy the above ODE.

Different Numerical Methods:

1.     Euler Method:

yn+1 = y(xn+1) = yn + hf(xn,yn),

where h = xn+1       xn. −

2.     Mid Point Method:

yn+1 = y(xn+1) = yn−1 + 2hf(xn,yn),

where h = xn+1       xn. −

Detailed Algorithm for Mid Point Method: Input:

(a)     Initial value (x0,y0),

(b)    End point L,

(c)     No. of steps N,

(d)    Function f(x,y): Should be provided as different function,

(e)     Analytical Solution (if available), y = g(x): Should be provided as another function for error calculation

Algorithm:

 ;

(ii)      Initialize a one dimensional array yval[N] with yval[0] = y0;

(iii)    esum = 0, (To calculate L2 norm of error);

(iv)    y1 = y0 + hf(x0,y0); (First step by Euler method)

(v)       esum = esum + (y1                   g(x0 + h))2; −

(vi)    yn−1 = y0;xn−1 = x0; (vii) yn = y1;xn = x0 + h; (viii) For i = 2 : N

yn+1 = yn−1 + 2hf(xn,yn); yval[i] = yn+1;

esum = esum + (yn+1      g(xn + h))2; −

xn = xn + h;

yn−1 = yn,yn = yn+1; End For(i)

 (ix) eL2 = N1 √esum;

3.    Runge-Kutta method of second order (RK2):

 ,

where h = xn+1       xn. −

4.    Runge-Kutta method of fourth order (RK4):

 ,

where h = xn+1       xn. −

Detailed Algorithm for RK4 Method: Input:

(a)     Initial value (x0,y0),

(b)    End point L,

(c)     No. of steps N,

(d)    Function f(x,y): Should be provided as different function,

(e)     Analytical Solution (if available), y = g(x): Should be provided as another function for error calculation

Algorithm:

 ;

(ii)      Initialize a one dimensional array yval[N] with yval[0] = y0;

(iii)    esum = 0, (To calculate L2 norm of error);

(iv)    yn = y0;xn = x0;

(v)       For i = 1 : N

k1 = hf(xn,yn);

 );

 );

 );

                                              y             y

esum = esum + (yn+1      g(xn + h))2; −

xn = xn + h; yn = yn+1; End For(i)

 (vi) eL2 = N1 √esum;

Problem 1: Let us consider following initial value problem

                                                                    xy′               2y = x3, where y(1) = 6.

−

We want to find distribution of y in 1 < x < 6 by above four numerical methods. Analytical solution of above ODE is given as

y(x) = x3 + 5x2.

•    Write a function which calculate f(x,y) in y′ = f(x,y).

•    Write another function which calculate analytical solution g(x) = x3 + 5x2.

•    Write four different functions Euler, Midpt, RK2, RK4 where for each function INPUTS: x0,y0,L,N and

OUTPUTS: yval[N],eL2.

In each of these functions, functions f(x,y) and g(x) will be called for required calculation.

(a)     Find yval[N] for N = 5,10,20 for all four methods. You should represent your output in tabular form. Table 1 represent one such table for N = 5. There will be similar

x
Analytical,y(x)
yn (Euler)
yn (MidPoint)
yn (RK2)
yn (RK4)
1.0
 
 
 
 
 
2.0
 
 
 
 
 
3.0
 
 
 
 
 
4.0
 
 
 
 
 
5.0
 
 
 
 
 
6.0
 
 
 
 
 
Table 1: Values of yn for different methods for N = 5.

             tables for N = 10 and N = 20.            [10+4+4=18]

(b)    Find eL2 for N = 2,5,10,15,20,25 for all four methods. Represent your results in tabular form as shown in Table 2 [12]

N
Euler
MidPoint
RK2
RK4
2
 
 
 
 
5
 
 
 
 
10
 
 
 
 
15
 
 
 
 
20
 
 
 
 
25
 
 
 
 
Table 2: L2 error norms for different methods.

B. Solving system of initial value problems: Consider following system of initial value problems

u′(x) = f(x,u,v),
u(a) = u0,
v′(x) = g(x,u,v),
v(a) = v0,
where we want to find u(x) and v(x) in a < x < b such that both u and v satisfy above coupled ODEs.

Runge-Kutta of Fourth order (RK4) for system of ODEs:

 ,

Detailed Algorithm for RK4 Method for system of ODEs:

Input:

1.    Initial value (x0,u0,v0),

2.    End point L,

3.    No. of steps N,

4.    Function f(x,u,v) and g(x,u,v): Should be provided as different function,

5.    Analytical Solution (if available), u = l(x),v = m(x): Provided for error calculation.

Algorithm:

(i)        h = L−Nx0;

(ii)      Initialize two one dimensional arrays uval[N] with uval[0] = u0 and vval[N] with vval[0] = v0;

(iii)    esumu = 0, esumv = 0 (To calculate L2 norm of errors for both u and v);

(iv)    un = u0;vn = v0;xn = x0;

(v)       For i = 1 : N

k1 = hf(xn,un,vn);

l1 = hg(xn,un,vn);

 );

 );

 );

 );

k4 = hf(xn+1,un + k3,vn + l3);

l4 = hg(xn+1,un + k3,vn + l3);

 );

 );

vval[i] = vn+1;

esumu = esumu + (un+1     l(xn + h))2; −

esumv = esumv + (vn+1 m(xn + h))2; xn = xn + h; −

un = un+1; vn = vn+1; End For(i)

(vi)       euL2 = N1 √esumu;

(vii)    evL2 = N1 √esumv;

Problem 2: Let us consider following system of initial value problem

            where y(1) = e,     where z(1) = 2.

We want to find distribution of y and z in 1 < x < 5 by RK4 method. Analytical solution of above set of ODEs is given as

 .

•    Write two functions which calculate f(x,y,z) in y′ = f(x,y,z) and g(x,y,z) in z′ = g(x,y,z) .

•    Write two functions l(x) and m(x) which calculate two analytical solutions.

•    Write the function RK4system where

INPUTS: x0,y0,z0,L,N and

OUTPUTS: yval[N],zval[N],eyL2,ezL2.

In RK4system, functions f(x,y,z),g(x,y,z),l(x),m(x) will be called for required calculation.

x
Analytical,y(x)
yn (RK4)
Analytical,z(x)
zn (RK4)
1.0
 
 
 
 
1.8
 
 
 
 
2.6
 
 
 
 
3.4
 
 
 
 
4.2
 
 
 
 
5.0
 
 
 
 
Table 3: Values of yn and zn for RK4 method for N = 5.

(a)     Find yval[N] and zval[N] for N = 5,10,20 with RK4 method. You should represent your output in tabular form. Table 3 represent one such table for N = 5. There will be similar tables for N = 10 and N = 20. [10+4+4=18]

(b)    Find eyL2 and ezL2 for N = 2,5,10,15,20,25. Represent your results in tabular form as shown in Table 4   [12]

N
eyL2
ezL2
 
 
 
Table 4: Error norms (eyL2 and ezL2).

C. Expressing higher order ODE as system of first order ODEs:

Let us consider following higher order ODE

                          y′′′             xy′′ + y′                     8y4 = y tanx, where y(1) = 5,y′(1) = 0,y′′(1) = 10

                                  −                −

We can express above equation as system of 3 first order ODEs as below

y′ = u,        y(1) = 5 u′ = v, u(1) = 0

                                            v′ = y tanx + 8y4                u + xv                                v(1) = 10

−

Problem 3: Let us consider following higher order ODE 3y′′ + xy′ 3y + 20x + 5x2 = 0 subjected to y(0) = 10,y′(0) = 19. −

We want to find distribution of y and y′ in 0 < x < 5 by RK4 method. Analytical solution of above ODE is given as

y = x3 + 5x2 + 19x + 10

x
Analytical,y(x)
yn (RK4)
Analytical,y′(x)
 
0.0
 
 
 
 
1.0
 
 
 
 
2.0
 
 
 
 
3.0
 
 
 
 
4.0
 
 
 
 
5.0
 
 
 
 
Table 5: Values of yn and  for RK4 method for N = 5.

(a)     Find yval[N] and  ] for N = 5,10,20 with RK4 method. You should represent your output in tabular form. Table 5 represent one such table for N = 5. There will be similar tables for N = 10 and N = 20. [10+4+4=18]

(b)    Find eyL2 and ey′L2 for N = 2,5,10,15,20,25. Represent your results in tabular form as shown in Table 6 [12]

N
eyL2
ey′L2
 
 
 
Table 6: Error norms (eyL2 and ey′L2).