{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Ch E 310 - Fall 10 - Lecture 26

# Ch E 310 - Fall 10 - Lecture 26 - Lecture 26 December 7...

This preview shows pages 1–9. Sign up to view the full content.

Lecture 26 – December 7, 2010 Agenda: Boundary Value Problems (solving ODEs) Finite Difference Methods Derivative (Neumann) Boundary Conditions Solving Non-linear ODEs HW 6 questions and help Course Evaluations (if you haven’t done this already)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Finite Difference Methods An alternative to the shooting method is to use finite difference approaches We looked at these previously for numerically approximating derivatives This method entails converting a linear ODE into a system of simultaneous algebraic equations Once formulated, we can use methods for linear systems to provide a solution The key difference in boundary value problems is that we divide the domain (space, time, etc.) into nodes and specify conditions at each node simultaneously This leads to a solution at each node and thus overall
Finite Difference Methods   T T h dx T d 2 2 0 Let’s revisit the heated rod example; the ODE is: The spatial domain is discretized into a series of nodes: The second derivative is depicted by a finite difference: 2 1 1 2 2 2 x T T T T d i i i K 200 T K 400 T K 300 T

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Forward Finite Difference Formulas
Backward Finite Difference Formulas

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Centered Finite Difference Formulas
Substituting the finite difference into the ODE, we get: Collecting terms yields: This can be written for any of the interior nodes along the length of the rod The first and last nodes ( T 0 and T n ) are subject to chosen boundary conditions (here, 300 and 400 K ) This leads to a system of n –1 linear algebraic equations and n –1 unknowns – we know how to solve this! Finite Difference Methods   0 2 2 1 1 i i i i T T h x T T T   T x h T T x h T i i i 2 1 2 1 2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Use the finite difference method to solve the first heated rod example from Lecture 25 where x = 2 m From the finite difference form of the ODE, we can write an equation at each interior node i For node 1 we have: T 0 + (2 + 0.05×2 2 ) T 1 T 2 = 0.05×2 2 ×200 T 0 + 2.2 T 1 T 2 = 40 ( where T 0 = 300 K) 2.2 T 1 T 2 = 340 ( two unknowns )
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 22

Ch E 310 - Fall 10 - Lecture 26 - Lecture 26 December 7...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online