Download 14.1 Shooting Method for Solving Linear BVPs

Ordinary Differential Equations
Boundary Value Problems
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14.1 Shooting Method for Solving Linear BVPs
• We investigate the second-order, two-point boundary-value problem of
the form
y  f ( x, y, y)
,a  x  b
with Dirichlet boundary conditions
y (b)  
y (a)   ,
Or Neuman boundary conditions
y(a )   ,
y(b)  
Or mixed boundary condition
y(a)  c1 y (a)   ,
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y(b)  c 2 y (b)  
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14.1 Shooting Method for Solving Linear BVPs
• Simple Boundary Conditions
y  p( x) y  q( x) y  r ( x), a  x  b,
y (a)  ya, y (b)  yb
The approach is to solve the two IVPs
u  p ( x)u  q( x)u  r ( x),
v  p( x)v  q ( x)u,
u ( a )  ya ,
v ( a )  0,
u (a )  0,
v( a )  1.
If v (b)  0, the solution of the original two-point BVP is given by
y(x) = u(x) + Av(x) is found from the requirement that y(b) = u(b) + Av(b) = yb
A
yb  u (b)
v(b)
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
y ( x)  u ( x) 
yb  u (b)
v( x).
v(b)
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14.3 Shooting Method for Solving Linear BVPs
• EX14.1
y 
2
y, y (1)  10,
x
y (2)  0,
• convert to the pair of initial-value problems
2
u ,
x
2
v 
v
x
u  
w1  w2,
wx 
y ( x)  w1x 
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u (1)  10,
u (1)  0,
v (1)  0,
v(1)  0,
2
w 2,
x
w3  w4,
w4 
2
w 4,
x
0  w1(b)
w3( x).
w3(b)
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14.1 Shooting Method for Solving Linear BVPs
• General Boundary Condition at x = b
y  p( x) y  q( x) y  r ( x)
The condition at x=b involves a linear combination of y(b) and y’(b)
y (a)  ya, y(b)  cy (b)  yb
The approach is to solve the two IVPs
u  p ( x)u  q( x)u  r ( x),
v  p( x)v  q ( x)u,
u ( a )  ya ,
v ( a )  0,
u (a )  0,
v( a )  1.
If v(b)  cv (b)  0 there is a unique solution, given by
yb  u(b)  cu (b)
y ( x)  u ( x) 
v( x).
v(b)  cv(b)
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14.3 Shooting Method for Solving Linear BVPs
• General Boundary Condition at Both Ends of the Interval
y  p( x) y  q( x) y  r ( x)
y(a)  c1 y (a)  ya,
y(b)  c 2 y (b)  yb.
the approach is to solve two IVPs
u  p ( x)u  q( x)u  r ( x),
v  p( x)v  q ( x)u,
u (a )  0,
v( a )  1,
u ( a )  ya,
v( a )  c1.
If v(b)  c 2v(b)  0 , there is a unique solution, given by
yb  u(b)  c 2u (b)
y ( x)  u ( x) 
v( x).
v(b)  c 2v(b)
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14.3 Shooting Method for Solving Linear BVPs
• Ex 14.4
2x
2
2

y

y

x
 1,
x2 1
x2 1
y 

y (0)  1, y(1)  y (1)  0.
y  ( x 4  3x 2  x  6) / 6
• Ex 14.5
y 

2x
2
2

y

y

x
 1,
2
2
x 1
x 1
y(0)  y (0)  0, y(1)  y (1)  3.
y  x 4 / 6  3x 2 / 2  x  1
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14.2 Shooting Method for Solving NonLinear BVPs
• Nonlinear Shooting Based on the Secant Method
use an iterative process based on the secant method presented in
Chapter 2
y  f ( x, y, y),
y(a)  ya , h( y (b), y(b))  0,
the initial slope ‘t’, begin with u’(a) = t(1) = 0, error is m(1)
Unless the absolute value of m(1) is less than the tolerance, we continue
by solving eq.
t (i)  t (i  1) 
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t (i  1)  t (i  2)
m(i  1).
m(i  1)  m(i  2)
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14.2 Shooting Method for Solving NonLinear BVPs
• EX 14.6
y  2 yy,
y (0)  1,
y (1)  y(1)  0.25  0.
 y = 1/(x+1)
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14.2 Shooting Method for Solving NonLinear BVPs
• Nonlinear Shooting Using Newton’s Method
y  f ( x, y, y),
y(a)  ya , y (b)  yb .
begin by solving the initial-value problem
u  f ( x, u, u ),
v' '  vfu ( x, u, u )  vf u ( x, u, u),
u (a )  ya ,
v(a)  0,
u (a)  tk ,
v(a)  1.
Check for convergence:
m  u (b, tk )  yb ;
if |m| < tol, stop;
Otherwise, update t:
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tk 1  tk  m / v(b, tk );
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14.2 Shooting Method for Solving NonLinear BVPs
• EX 14.8
[ y]2
y  
,
y
u  [u]2 u 1 ,
y (1)  2,
y (0)  1,
u (0)  1,
u(0)  tk .
f u ( x, u, u)  [u]2 (1)u 2 ,
f u ( x, u, u)  2[u]u1.
v  v[u]2 u 2  v2[u]u1 ,
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v (0)  0,
v(0)  1.
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14.3 Finite Difference Method for Solving Linear BVPs
• Replace the derivatives in the differential equation by finite-difference
approximations (discussed in Chapter 11).
• We now consider the general linear two-point boundary-value problem
y  p( x) y  q( x) y  r ( x), a  x  b,
• with boundary conditions
y (a)   , y (b)   ,
• To solve this problem using finite-differences, we divide the interval [a, b]
into n subintervals, so that h=(b-a)/n. To approximate the function y(x) at
the points x1  a  h, xn1  a  (n  1)h, we use the central difference
formulas from Chapter 11:
y( xi ) 
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yi 1  2 yi  yi 1
yi 1  yi 1

,
y
(
x
)

i
h2
2h
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Finite Difference Method for Solving Linear BVPs
• Substituting these expressions into the BVP and writing p( xi ) as
pi , q( xi ) as qi , and r ( xi ) as ri gives
____
yi 1  2 yi  yi 1
yi 1  yi 1
 pi
 qi yi  ri
2
h
2h
• Further algebraic simplification leads to a tridiagonal system for
the unknowns y1 ,, yn1 , viz.
h
h


2
2
1

p
y

(
2

h
q
)
y

1

p
y

h
ri , i  1, , n  1,




i
i 1
i
i
i
i 1
2
2


• where y0  y(a)   and yn  y(b)   .
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Finite Difference Method for Solving Linear BVPs
• Expanding this expression into the full system gives
h

 (2  h 2 q1 ) y1  1  p1  y2
2

h


2
1  p1  y1  (2  h q2 ) y2  1  p2
2



h

 h 2 r1  1  p1  ,
2

h
 y3
2
h
h


2
1  pi  yi 1  (2  h qi ) yi  1  pi  yi 1
2
2



 h 2 r2 ,

 h 2 ri ,

h
h


2
2
1  p1n  2  yn 3  (2  h qn  2 ) yn  2  1  pn  2  yn 1  h rn  2 ,
2
2


h
h


2
2
1  pn 1  yn  2  (2  h qn 1 ) yn 1  h rn 1  1  pn 1   ,
2
2


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Example 14.9 A Finite-Difference Problem
• Use the finite-difference method to solve the problem
y  y  x( x  4),
0  x  4,
• with y(0)=y(4)=0 and n=4 subintervals.
• Using the central difference formula for the second derivative, we
find that the differential equation becomes the system.
y( xi ) 
yi 1  2 yi  yi 1
 yi  xi ( xi  4),
2
h
i  1,2,3.
• For this example, h = 1 ,i =1, y = 0, and i = 3, y = 0. Substituting
this values, we obtain
y2  2 y1  0  y1  1(1  4),
y3  2 y2  y1  y2  2(2  4),
0  2 y3  y2  y3  3(3  4).
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Example 14.9 A Finite-Difference Problem
• Combining like terms and simplifying gives
 3 y1  y2
 3,
y1  3 y2  y3  4,
y2  3 y3  3,
• Solving, we find that y_1=13/7, y_2 = 18/7, and y_3 = 13/7.
• We note for comparison that the exact solution of this problem is
2(1  e 4 )  x 2(1  e 4 ) x
y  4 4 e  4 4 e  x 2  4 x  2.
e e
e e
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Example 14.9 A Finite-Difference Problem
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Example 14.10 A Matlab Script for a Linear FDP.
• The Matlab script that follows solves the BVP.
y  2 y  2 y,
y(0)  0.1,
y(3)  0.1e3cos(3).
function S_linear_FD
aa = 0; bb = 3; n = 300;
p = 2*ones(1, n-1); q = -2*ones(1, n-1); r = zeros(1, n-1);
ya = 0.1; yb = 0.1*exp(3)*cos(3);
h = (bb-aa)/n; h2 = h/2; hh= h*h;
x = linspace(aa+h, bb, n);
a = zeros(1, n-1); b = a;
a(1:n-2) = 1 - p(1, 1:n-2)*h2; d = -(2 + hh*q);
b(2:n-1) = 1 + p(1, 2:n-1)*h2;
c(1) = hh*r(1) - (1+p(1)*h2)*ya;
c(2:n-2) = hh*r(2:n-2);
c(n-1) = hh*r(n-1) - (1 - p(n-1)*h2)*yb;
y = Thomas(a, d, b, c)
xx = [aa
x]; yy = [ya y yb];
out = [xx'
yy']; disp(out)
plot(xx, yy), grid on, hold on
plot(xx, 0.1*exp(xx).*cos(xx))
hold off
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Example 14.10 A Matlab Script for a Linear FDP.
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14.4 FDM for Solving Nonlinear BVPs
• We consider the nonlinear ODE-BVP of the form
y  2 y  2 y, y(0)  0.1, y(3)  0.1e3cos(3).
• Assume that there are constants Q* , Q* and P * such that
0  Q*  f y ( x, y, y)  Q* and | f y ( x, y, y) | P*
• Use a finite-difference grid with spacing h  2 / P* ,and let f i denote
the result of evaluating f at xi using ( yi 1  yi 1 ) /( 2h) for yi .
• The ODE then becomes the system
yi 1  2 yi  yi 1
 f i  0.
2
h
• An explicit iteration scheme, analogous to the SOR method
1
yi 
yi 1  2yi  yi 1  h 2 f i ,
2(1   )


• where y0   and yn   . The process will converge for w  h 2Q* / 2.
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Example 14.12 Solving a Nonlinear BVP by Using FDM
• Consider again the nonlinear BVP
2

y
y  
,
y(0)  1,
y (1)  2.
y
• We illustrate the use of the iterative procedure just outlined by
taking a grid with h = ¼. The general form of the difference equation
is
yi 


1
yi 1  2yi  yi 1  h 2 f i ,
2(1   )
• where
fi
2

( yi 1  yi 1 ) /( 2h)

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yi
yi21  2 yi 1 yi 1  yi21

.
2
4h yi (1   )
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Example 14.12 Solving a Nonlinear BVP by Using FDM
• Substituting the rightmost expression for f_i into the equation for
y_i, we obtain

yi21  2 yi 1 yi 1  yi21 
1
yi 
 yi 1  2yi  yi 1 
.
2(1   ) 
4 yi

• The computed solution after 10 iterations agrees very closely with
the exact solution.
function S_nonlinear_FD
ya = 1; yb = 2; a = 0; b = 1;
max_it = 10; n = 4; w = 0.1; ww = 1/(2*(1+w)); h = (b-a)/n
y(1:n-1) = 1
for k = 1:max_it
y(1) = ww*(ya+2*w*y(1)+y(2)+(ya^2-2*ya*y(2)+y(2)^2)/(4*y(1)));
y(2) = ww*(y(1)+2*w*y(2)+y(3)+(y(1)^2-2*y(1)*y(3)+y(3)^2)/(4*y(2)));
y(3) = ww*(y(2)+2*w*y(3)+yb+(y(2)^2-2*y(2)*yb+yb^2)/(4*y(3)));
end
x = [ a
a+h
a+2*h
a+3*h
b ]; z = [ ya y yb ];
plot(x, z), hold on, zz = sqrt(3*x+1); plot(x, zz), hold off
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Example 14.12 Solving a Nonlinear BVP by Using FDM
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