# Applying shooting method to 4nd two point value boundary problem

1. Oct 12, 2012

### pgioun

Hi,
I was wondering if you can apply shooting method to a 4nd differential eq.
two point value boundary problem,

specifically I want to use this method to solve Euler-Bernoulli eq.

EI y(4)(x)=f(x), y(0)=0,y'(0)=0,y(L)=0,y'(L)=0.

Normally, if you have a 2nd order two point value boundary problem,
you can solve it with shooting method by making it an initial condition problem,
''guessing'' the one of the two initial conditions, and iterating (changing your guess) until you meat the 2nd boundary condition.

But when you have a 4nd differentila eq. you should ''guess'' 2 initial conditions... so, how this would work?
Thanks

2. Oct 12, 2012

### AlephZero

You have to iterate on the two unknown parameters, insteead of just one.

How to solve this type of "optimisation" problem with many unknowns is a (large) topic in numerical analysiss, but a simple method that doesn't need any mathematical analysis of the particular problem you want to solve is http://en.wikipedia.org/wiki/Nelder–Mead_method

The basic idea is similar to searching along a line. Instead of finding an interval that you includes the solution and then making it smaller, you use the two unknown parameters as coordinates to define points on a plane, find a triangular region that contains the solution, and then choose a new point on the plane for each iteration that will shrink the area of the triangle.

There are much more efficient methods (FWIW my choice of "go-to" method would be the BFGS algorithm - google for it), but whether it's worth learning about them depends on the size of the problem. It's certainly worth reducing the computing time to solve a problem from say 100 hours to 1 hour, but reducing it from 10 sec to 0.1 sec might not be worth doing, unless it takes you less than 10 seconds to write and debug the code!

3. Oct 17, 2012

### Staff: Mentor

You want to make good on the two boundary conditions at the far end. The error in satisfying each of these conditions is a function of the two guesses at the other end:

ε1= ε1 (g1 , g2)

ε2= ε2 (g1 , g2)

You want g1 and g2 to be such that ε1 and ε2 are zero. This can be regarded as two non-linear equations in two unknowns. You can solve these equations using Newton-Raphson. You need to integrate from one boundary to the other using the two present guesses and get the corresponding values of the two errors. Then you need to evaluate the partial derivatives of the errors with respect to the two guesses. This is done by perturbing each of the guesses at the near boundary and then integrating to evaluate numerically the changes in the errors at the other boundary. You then apply the Newton-Raphson formula, which involves solving two linear algebraic equations in two unknowns for the changes in the guesses to make the errors come out to zero. This procedure is repeated until the errors are close enough to zero.