The Shooting Method for Boundary-value Problems Douglas Wilhelm Harder, M.Math. LEL Department of Electrical and Computer Engineering University of Waterloo.

The Shooting Method forBoundary-value Problems

Douglas Wilhelm Harder, M.Math. LELDepartment of Electrical and Computer Engineering

University of Waterloo

Waterloo, Ontario, Canada

ece.uwaterloo.ca

[email protected]

© 2012 by Douglas Wilhelm Harder. Some rights reserved.

2

Outline

This topic to approximating boundary-value problems– We will describe boundary-value problems (BVPs)– We will look at solutions with linear ordinary differential

equations (ODEs)– We will consider solutions for non-linear ODEs

• This will require successive approximations using the secant method

The Shooting Method for Boundary-value Problems

3

Outcomes Based Learning Objectives

By the end of this laboratory, you will understand:– Boundary-value problems (BVPs)– How to use an initial-value problem (IVP) solver to approximate

solutions to BVPs– The solution to linear BVPs– The application of the secant method to solve non-linear BVPs


4

Using Matlab

These methods assume that the student has written the initial-value problem solver pd45 with the signature

dp45( f, x_rng, u0, h, eps_abs ) which uses the Dormand-Prince method

If this routine is not available, you are welcome to use the built-in Matlab routine ode45 which has the similar signature

ode45( f, x_rng, u0 )


5

2nd-order ODEs

A boundary-value problem in one dimension is any 2nd-order ODE

F(x, u(x), u(1)(x), u(2)(x)) = 0

with two constraints

u(a) = ua

u(b) = ub

In general, we will look at functions of the form

u(2)(x) = f(x, u(x), u(1)(x))


6

2nd-order ODEs

Consider the corresponding initial-value problem for this 2nd-order ODE

F(x, u(x), u(1)(x), u(2)(x)) = 0

with two constraints

u(a) = ua

u(1)(a) = ua(1)

Here we specify the slope at the left-hand point x = a


7

2nd-order ODEs

Thus, a boundary-value problem could be restated as:– Given one initial condition u(a) = ua, what slope is required at that

initial point so that the solution to that initial condition passes through the point (b, ub)?

(b, ub)

(a, ua)


8

2nd-order ODEs

We will consider two possible cases:– When the ODE is linear, and– When it is not

(b, ub)

(a, ua)


9

2nd-order Linear ODEs

A 2nd-order ODE is said to be linear if it can be written as

u(2)(x) + q(x) u(1)(x) + r(x) u(x) = g(x)

or

u(2)(x) = g(x) – q(x) u(1)(x) – r(x) u(x)

We will also consider the corresponding homogenous LODE:

u(2)(x) + q(x) u(1)(x) + r(x) u(x) = 0

or

u(2)(x) = –q(x) u(1)(x) – r(x) u(x)


10


To demonstrate 2nd-order LODEs, we will consider the example

The corresponding homogeneous LODE is the

2 1sin 1

2 1.5

3 2.5

u x x u x u x

u

u

2 1sin 0u x x u x u x


11


Recall that if ug(x) is a solution to a 2nd-order LODE and u0(x) is a solution to the corresponding homogenous LODE, then

ug(x) + c·u0(x)

is also a solution the LODE


2

0 0 02 g g g

d du x cu x q x u x cu x r x u x cu x

dx dx

2 2 1 10 0 0g g gu x c u x q x u x cq x u x r x u x cr x u x

2 1 2 10 0 0g g gu x q x u x r x u x c u x q x u x r x u x

0g x c

g x

g(x) 0

12


Suppose we have redefine a boundary-value problem in terms of two IVPs, the first on the LODE with constraints

and the second on the homogenous LODE with constraints

Let ug(x) and u0(x) be the solutions, respectively

1 0

au a u

u a

1

0

1

u a

u a


13


For example, given the LODE from before, we create the two IVPs

The plots of the solutions:

2 1

1

sin 1

2 1.5

2 0

u x x u x u x

u

u

2 1

1

sin 0

2 0

2 1

u x x u x u x

u

u

ug(x)

u0(x)

(a, ua)

(b, ub)


14


Having found these two solutions, ug(x) and u0(x), we now want to find a linear combination

of these two such that it satisfies the second boundary value– For example, here we see

ug(x), ug(x) + u0(x), and ug(x) – u0(x)

It seems ug(x) + 2u0(x) will passclose to the second boundary point

0gu x cu x

ug(x)

ug(x) + u0(x)

ug(x) – u0(x)

(a, ua)

(b, ub)


15


We want to find the appropriate linear combination

to match our second boundary condition

First, we note

We want

Solving this for c yields

0gu x cu x

0 0g a au a cu a u c u

0g bu b cu b u

0

b gu u bc

u b

1 0

g a

g

u a u

u a

0

10

0

1

u a

u a


16


Easy enough; however, recall that we must find ug(x) and u0(x) but if we use dp45, the step sizes may be different…

Solution: define a system of two uncoupled equations and unknowns and solve them simultaneously

u(2)(x) = g(x) – q(x) u(1)(x) – r(x) u(x)

u(2)(x) = 0 – q(x) u(1)(x) – r(x) u(x)


17


Using our techniques from Laboratory 5, we can write this as a system of IVPs:

w1(1)(x) = w2

(x)

w2(1)(x) = g(x) – q(x) w2(x) – r(x) w1(x)

w3(1)(x) = w4

(x)

w4(1)(x) = 0 – q(x) w4(x) – r(x) w3(x)

w1(a) = ua

w2(a) = 0

w3(a) = 0

w4(a) = 1


ug(x) = w1(x)ug

(1)(x) = w2(x) u0(x) = w3(x)u0

(1)(x) = w4(x)

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For example, consider the boundary-value problem

The Matlab function would be:function [dw] = f6a( x, w ) dw = [w(2); 1 - sin(x)*w(2) - w(1); w(4); 0 - sin(x)*w(4) - w(3)];end

with the initial conditions [u_a, 0, 0, 1]'

2 1sin 1

2 1.5

3 2.5

u x x u x u x

u

u


19


For our example, we would run the following code:a = 2; b = 3; u_a = 1.5; u_b = 2.5;[x6a, u6a] = dp45( @f6a, [a, b], [u_a 0 0 1]', 0.01, 1e-8 );

c6a = (u_b - u6a(1, end))/u6a(3, end) c6a = 1.991628911663175

u6a_soln = u6a(1, :) + c6a*u6a(3, :);plot( x6a, u6a_soln, 'r' )ylim( [0, 2.5] )u6a_soln(1) ans = 1.500000000000000

u6a_soln(end) ans = 2.500000000000000


20


Maple can find an answer, but it’s not nice…> a := 2; b := 3; u_a := 1.5; u_b := 2.5;> dsolve( {(D@@2)(u)(x) + sin(x)*D(u)(x) + u(x) = 1, u(a)=u_a, u(b)=u_b} );


21


It takes Maple about two minute to plot the same figure– Note: this is not an issue with Maple—it is giving you the exact

solution and not a numeric approximation


22

2nd-order Linear ODEs: Example

For example, consider the boundary-value problem

u(2)(x) + 4 u(1)(x) + 7x u(x) = sin(x)

u(2) = 1.5

u(3) = 2.5

The Matlab function would be:function [dw] = f6b( x, w ) dw = [w(2); sin(x) - 4*w(2) - 7*x*w(1); w(4); 0 - 4*w(4) - 7*x*w(3)];end


23


For example, considera = 2; b = 3; u_a = 1.5; u_b = 2.5;[x6b, u6b] = dp45( @f6b, [a, b], [u_a 0 0 1]', 0.01, 1e-4 );u6b_soln = u6b(1,:) + (u_b - u6b(1,end))/u6b(3,end)*u6b(3,:);plot( x6b, u6b_soln, 'o-' );grid


24


Maple can find an answer, but it’s not nice…> a := 2; b := 3; u_a := 1.5; u_b := 2.5;> dsolve( {(D@@2)(u)(x) + 4*D(u)(x) + 7*x*u(x) = sin(x), u(a) = u_a, u(b) = u_b}, u(x) );


25


Comparing the results with Maple’s plot, we see they are similar


26


If we repeat using b = 4, we get significantly more varying, but still similar, graphs


27

2nd-order Non-linear ODEs

What happens if the ODE is not linear?– In this case, there is no guarantee that we can find a solution

using this simple technique– Instead, we will use

u(2)(x) = f(x, u(x), u(1)(x))

and let us(x) be the solution to the IVP

u(a) = ua

u(1) (a) = s

Now, define a new function

err(s) = us(b) – ub


28


When this function has a zero, we have a solution to the boundary-value problem


What is err(s) in Matlab?function [du_b] = err_shot( s ) [t_out, u_out] = ... dp45( @f, [a, b], [u_a, s], h, eps_abs );

du_b = u_out(1, end) - u_b;end


29


When this function has a zero, we have a solution tothe boundary-value problem


In reality, however, it will require a number of other parameters:

function [du_b] = err_shot(s, f, x_rng, u_bndry, h, eps_abs)

u_a = u_bndry(1); u_b = u_bndry(2);

[~, u_out] = dp45( f, x_rng, [u_a, s]', h, eps_abs );

du_b = u_out(1, end) - u_b;end


30

Secant Method

We will use the secant method for fining this root:– Suppose we are trying to find a root of a real-valued function of a

real variable err(s)

– Suppose s1 and s2 are two initial approximations of the root

s1s2

err(s)


31

Secant Method

We can find a better approximation of the root by finding the interpolating straight line that passes through these points

s1s2

err(s)


32

Secant Method

The formula for this new point is

s1s2

1 2 2 13

2 1

err err

err err

s s s ss

s s

err(s)


s3

33

Secant Method

In the first case, we considered linear ODEs– Essentially, what we are doing here is asking:

Given these two approximations,

what would our next approximation be if the ODE was linear?

s1s2

err(s)


s3

34

Secant Method

Now, suppose we have s2 and s3, the next step is to use these approximations to find the next approximation, s4, and so on...

s1s2

2 3 3 24

3 2

err err

err err

s s s ss

s s

err(s)


s3

35

Secant Method

Thus, we will continue to iterate, using the secant method at each step...

With any iterative numerical method, we must know under what conditions we will:– Halt with success, and– Indicate a failure to find a solution

After all, not all iterative numerical methods converge...


36

Secant Method

For the secant method, we must have two approximations, s1 and s2

First, if |err(s1)| < |err(s2)|, we will swap s1 and s2

– We’re assuming that s2 is more accurate


37

Secant Method

We will iterate at most Nmax times:– With each iteration, we will approximate a new point s

– If |s2 – s| < estep and |err(s)| < eabs, we will assume we are finished and we will return the value s (shooting, however, will take one more step)

– Otherwise, we will set s1 = s2 and s2 = s

If we have iterated Nmax times and not found a solution, we will throw an exception with an appropriate message


38

The Shooting Method

Normally, the secant method simply returns the slope

Your function, shooting, however, will have to return the x- and u-values corresponding to that optimal slope– Thus, your last steps before you successfully return will be:

[x_out, u_out] = dp45( f, x_rng, [u_bndry(1), s]', h, eps_abs );

return;


39

The Shooting Method

The arguments offunction [x_out, u_out] = shooting( s1, s2, f, x_rng, u_bndry, ... h, eps_abs, eps_step, N_max )

are as follows:s1 and s2 are the two approximations of the slopes

f is a function handle for the differential equation: u(2)(x) = f(x, u(x), u(1)(x))

[a, b] is a row vector defining the range on which we are approximating the boundary-value problem

[ua, ub] is a row vector defining the boundary conditions:

u(a) = ua and u(b) = ub

h is the initial step size for the function dp45eabs is the parameter passed to dp45 and is also the parameter

used by the secant method

estep is the parameter used by the secant method for the step size

Nmax is the parameter used by the secant method to define themaximum number of iterations


40

The Shooting Method

Question: what do we use as initial conditions?

Answer: as an engineer, you will understand your problem; you should know the expected behaviour around the point x = a– In these laboratories, you will be given the initial points– In reality, you will understand what they should be when you get

there


41

The Shooting Method

Consider the following non-linear boundary-value problem:

The corresponding function isfunction [dw] = f6c( x, w ) dw = [w(2); sin(x) - 4*w(2)*w(1) + 2*x*w(1)^2];end

22 14 2 sin

2 1.5

3 2.5

u x u x u x x u x x

u

u


42

The Shooting Method

When I run the function [x6c, u6c] = shooting( -3, -3.1, @f6c, [2, 3], [1.5, 2.5]', ... 0.01, 1e-6, 1e-6, 20 );

I get a solution after four iterations where the approximations of the initial slopes are:

–3.326745004558498

–3.323136496135223

–3.323176635129875

–3.323176642445732

These last two slopes are sufficiently close enough and the u6c(1, end) = 2.499999999999991 close enough to 2.5


43

The Shooting Method

We can look at the plot and the result:plot( xs, us(1,:) )

us(1,end) ans = 2.499999999999991

length( xs ) ans = 36


44

The Shooting Method

You might get slightly different values depending on your implementation of the secant method; however, the final result should bereasonably close


45

The Shooting Method

Plotting approximations with the various slopes:slopes = [-3 -3.1 -3.326745004558498 -3.323136496135223 ... -3.323176635129875 -3.323176642445732];for s = slopes [x6s, u6s] = dp45( @f6c, [2, 3], [1.5, s]', 0.01, 1e-6 ); plot( x6s, u6s(1,:) )

u6s(1,end)end

s1 = 3

s2 = 3.1

All others...


46

The Shooting Method

If you plot the relative errors of the approximation of u6s(1,end) as an approximation of ub = 2.5, we observe the fast (i.e., O(hf)) convergence of the secant method

s u6s(1,end) AbsoluteError

–3 2.668638013935938 1.686 × 10–1

–3.1 2.617026509067240 1.170 × 10–1

–3.326745004558498 2.498107476136670 1.893 × 10–3

–3.323136496135223 2.500021288156047 2.129 × 10–5

–3.323176635129875 2.500000003879338 3.879 × 10–9

–3.323176642445732 2.499999999999991 7.105 × 10–15


47

Hints

You will work with various values of s and err(s) – Instead of continually recalculating value err(s), just calculate

err1 = err_shot( s1 ); err2 = err_shot( s2 );

and when you calculate a new value of s, just use a simpler expression with the variables s1, s2, err1 and err2 and then immediately calculate

errs = err( s );

– When you update s1 and s2, update err1 and err2


48

Summary

We have looked at using the function dp45 to approximate a boundary-value problem– The shooting method converts a BVP into an IVP– For linear IVPs, we use the ODE and the homogeneous ODE

and find a linear combination of the two solutions– For non-linear IVPs, we use dp45 and the secant method to

guide us to an approximation


49

References

[1] Glyn James, Modern Engineering Mathematics, 4th Ed., Prentice Hall, 2007.

[2] Glyn James, Advanced Modern Engineering Mathematics, 4th Ed., Prentice Hall, 2011.

[3] John H. Mathews and Kurtis D. Fink, Numerical Methods using Matlab, 4th Ed., Prentice Hall, 2004, pp. 529-534.

[4] Shooting Method, http://en.wikipedia.org/wiki/Shooting_method.


The Shooting Method for Boundary-value Problems Douglas Wilhelm Harder, M.Math. LEL Department of Electrical and Computer Engineering University of Waterloo.

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