This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Week 4 Lectures, Math 716, Tanveer 1 Diffusion in R n Recall that for scalar x , S ( x, t ) = 1 4 t exp bracketleftbigg x 2 4 t bracketrightbigg (1) is a special solution to 1D heat equation with properties integraldisplay R S ( x, t ) dx = 1 for t > , and yet lim t + S ( x, t ) = 0 for fixed x negationslash = 0 (2) This was called a source solution of heat equation with source at the origin. We now claim that the product S ( x , t ) S ( x 1 , t ) S ( x 2 , t ) S ( x 3 , t ) ...S ( x n , t ) is a solution to the heat equation in R n : u t = for x R n (3) and satisfies property integraldisplay R n S ( x , t ) d x = 1 for t > , and yet lim t + S ( x , t ) = 0 for fixed x negationslash = (4) We note that by using product rule S t = n summationdisplay j =1 t S ( x j , t ) productdisplay i negationslash = j S ( x i , t ) = n summationdisplay j =1 2 x 2 j S ( x i , t ) productdisplay i negationslash = j S ( x i , t ) = S (5) Further, integraldisplay R n S ( x , t ) d x = integraldisplay R integraldisplay R ... integraldisplay R S ( x 1 , t ) S ( x 2 , t ) ...S ( x n , t ) dx 1 dx 2 ...dx n = 1 (6) Further if x negationslash = 0, then direct examination of S ( x , t ) = 1 (4 t ) n/ 2 exp bracketleftbigg x 2 1 + x 2 2 + ..x 2 n 4 t bracketrightbigg = 1 (4 t ) n/ 2 exp bracketleftbigg x 2 4 t bracketrightbigg (7) shows that lim t + S ( x , t ) = 0 for fixed x negationslash = 0. The solution S is the source solution in R n . Analogous to 1D, we have the following theorem: Theorem 1 The solution to the heat equation in R n that satisfies initial condition u ( x , 0) = ( x ) for C ( R n ) (8) is given by u ( x , t ) = integraldisplay R n S ( x y , t ) ( y ) d y (9) 1 2 Diffusion in the halfline 2.1 Dirichlet Boundary condition We consider solution to heat equation in 1D, with x R + and take the Dirichlet boundary condition at x = 0. So the problem is v t v xx = 0 for x > , t > (10) v ( x, 0) = ( x ) (11) v (0 , t ) = 0 (12) We seek to find a solution to this problem explicitly. If it exists, the classical solution for which v ( x, t ) 0 as x is unique by applying maximum principle or energy method. Now the initial data ( x ) is only specified for x > 0. x ( x) odd Figure 1: Odd Extension of ( x ) to x R We do an odd extension , i.e. define an extended function odd ( x ) in R (see Fig. 1) so that odd ( x ) = ( x ) for x > 0 ; odd ( x ) = ( x ) for x < (13) . Let u ( x, t ) be a solution to heat equation so as to satisfy u ( x, 0) = odd ( x ) for x R (14) Then, applying equation (15) of week 3 notes, u ( x, t ) = integraldisplay S ( x y, t ) odd ( y ) dy (15) 2 Breaking up the integral into two parts integraltext and integraltext and changing variables y y in the first and using (14), we note that (15) implies that u ( x, t ) = integraldisplay...
View
Full
Document
This note was uploaded on 07/26/2011 for the course MATH 716 taught by Professor Tanveer,s during the Spring '08 term at Ohio State.
 Spring '08
 Tanveer,S
 Scalar

Click to edit the document details