This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MA 2930, April 13, 2011 Worksheet 11 1. Solve the heat equation u xx = 4 u t for the following boundary conditions: u ( x, 0) = x, ≤ x ≤ 5 and (1) u (0 ,t ) = 0 = u (5 ,t ) (2) u x (0 ,t ) = 0 = u x (5 ,t ) (3) u (0 ,t ) = 0 = u x (5 ,t ) (4) u x (0 ,t ) = 0 = u (5 ,t ) (5) u (0 ,t ) = 10 , u (5 ,t ) = 30 (6) u (0 ,t ) = 10 , u x (5 ,t ) = 30 (7) u x (0 ,t ) = 10 , u (5 ,t ) = 30 Why is an obvious one missing from this list? First note when comparing with the standard equation α 2 u xx = u t that α 2 = 1 / 4. To find general solution we’ll use that fact that x part of the solution has the form X ( x ) = c 1 cos √ λx + c 2 sin √ λx, X (0) = u (0 ,t ) ,X (5) = u (5 ,t ) Thus, for the homogeneous boundary conditions the solutions consists of all those sines or cosines that satisfy the boundary conditions. To find these it’s best to draw the graphs of sine and cosine and see which values of the period fit the boundary condition. The corresponding tpart of the solution is then completely determined by the coefficient (angular frequency) of the xpart and α 2 . Thus, (1) u (0 ,t ) = 0 means sines. u (5 ,t ) = 0 means these sines must fit an integral multiple of their halfperiods in the distance 5. So possible half periods are 5 /n...
View
Full
Document
This note was uploaded on 06/10/2011 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 TERRELL,R
 Differential Equations, Equations

Click to edit the document details