{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Perhaps the simplest strategy for solving 221 is to

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: t u +1 ; uj n ; 2 ( tt)n+ + ( n) j j uj au n x ; 2 ( xx)n+ ] = 0 j x u : (2.2.2a) Neglecting the second-order derivative terms in the local discretization errors, we obtain the nite di erence equation n Uj +1 ; U n j t Solving for +( ) n a Uj +1 ; Uj n Uj n x =0 : +1 , we obtain n Uj +1 n Uj n = (1 + j ) n Uj ; nn j Uj +1 (2.2.2b) 2.2. A Kinematic Wave Equation 9 n+1 n j j+1 Figure 2.2.1: Computational stencil of the forward time-forward space nite di erence scheme (2.2.2b). where n j ( n) a Uj t x (2.2.3) : n The parameter j is called the Courant number. The nite di erence equation (2.2.2b) involves three points as indicated in the stencil of Figure 2.2.1. It is easy to solve using the prescribed initial data (2.2.1b). Knowing 0 ) for all , we calculate j1 for all of interest. Then, knowing j1 for all , j= ( we repeat the process to obtain j2 , etc. Forward di erences are appropriate for approximating time derivatives in this type of marching procedure, but it seems reasonable to also consider backward di erences or centered di...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online