Boundary  Value  Problems  

Classical  ODE  Problems  Initial Value Problem (IVP) vs Boundary Value Problem (BVP)

(1) IVP equation

(2) BVP equation

An example of BVP is heat conduction along a long, thin rod with length L.

Boundary  Value  Problem  If the rod is not insulated along its length and the system is at a steady state, the equation is given by

where: - 𝛼 is a heat transfer coefficient (m-2) that parameterizes the is a heat transfer coefficient (m-2) that parameterizes the rate of heat dissipation to the surrounding air - Ta is the temperature of the surrounding air (0C)

(12.1)

To obtain a solution for equation (12.1), there must be appropriate boundary conditions. A simple case is where the temperatures at the ends of the rod are held at fixed values:

Boundary  Value  Problem  -  Example in (12.1) is called a “two-point boundary value

problem (BVP)” because the solution is known at two points -  A two-point BVP includes an ODE and the value of the

solutions at two different locations. -  The least order of ODE for BVP is two because (generally)

first order ODE cannot satisfy two conditions.

Suppose, for heat conducting problem along a rod in equation (12.1), the length of the rod is m and the temperatures at is and at is Also, the heat transfer coefficient m-2. We assume that the temperature of the surrounding air is a gradient of the form

Boundary  Value  Problem  Here, the heat conduction problem in equation (12.1) becomes

We can solve the equation (12.2) analytically, by taking and by applying the boundary conditions, from which we get that the particular solution of (12.2) is given by

(12.2)

(12.3)

The  Shoo:ng  Method  This method is based on converting the two-point BVP into an equivalent IVP.

For example, we use the problem in equation (12.1) where now the length of the rod is L = 10 m and

(12.4)

The second order ODE in (12.1) is now expressed as two first order ODEs, by

Fourth-­‐Order  Runge-­‐KuBa  Methods  

The classical fourth-order Runge-Kutta method

where

The  Shoo:ng  Method  To solve the system of ODEs (12.4), we require an initial value for the new variable z. We guess a value, here z(0) = 10. The numerical solution can obtained using an IVP ODE solver, such as a 4th Runge-Kutta method. We set the step size h = 2.

First, we must solve for all the slope at the beginning of the interval (at x = 0):

Then we calculate the first values of T and z at the midpoint:

The  Shoo:ng  Method  We use the values of T and z to compute the first set of midpoint slopes:

These are used to determine the second set of midpoint predictions,

which can be used to compute the second set of midpoint slopes:

The  Shoo:ng  Method  These are used to determined the predictions at the end of the interval:

which can be used to compute the endpoint slopes

The values of k’s can then be used to compute

The  Shoo:ng  Method  Proceeding in a like manner for the remaining steps yields:

x T z 0 40 10 2 60.5347 10.6033 4 82.6961 11.6322 6 107.3737 13.1297 8 135.5579 15.1505 10 168.3797 17.7810

Here the value at the end of the interval of T(10) = 168.3797 which differs from the boundary condition of T(10) = 200. Therefore we make another guess, z(0) = 20, and perform the computation again.

The  Shoo:ng  Method  Then the result of T(10) = 285.8980 is obtained.

Now, since the original ODE is linear, the values

z(0) = 10 resulting T(10) = 168.3797 and

z(0) = 20 resulting T(10) = 285.8980

to calculate the value of z(0) that yields T(10) = 200, that is:

are linearly related. We use a linear interpolation formula

The  Shoo:ng  Method  

The  Shoo:ng  Method  For nonlinear problems, such as

where β = 5×10-8, liner interpolation is no longer can be used to calculate the value of z(0) that yields T(10) = 200. Here we have to make guesses every iteration. Hence, the shooting method is not effective anymore.

(12.5)

Finite-­‐Difference  Methods  Finite-difference methods are the most common alternatives to the shooting method. In these techniques, finite divided differences are substituted for the derivatives in the original equation.

Recall second derivative of centered finite-divided formula with error O(h2):

Consider a simple BVP

(12.6)

(12.7)

which has an exact solution: (prove it!!!)

Finite-­‐Difference  Methods  Divide the interval [0, 1] into M subintervals, with step size

and the points

Let Yi be the approximation to y (xi). For each i = 2, 3, …, M we approximate

by

and the boundary conditions are:

Finite-­‐Difference  Methods  Now we have a linear system of M+1 equations in M+1 unknowns:

which can be transformed into matrix notation.

Finite-­‐Difference  Methods  In matrix notation:

or

and solve it

Finite-­‐Difference  Methods