week4 - Week 4 Lectures, Math 716, Tanveer 1 Diffusion in R...

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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 1-D 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 1-D, 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 half-line 2.1 Dirichlet Boundary condition We consider solution to heat equation in 1-D, 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...
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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.

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week4 - Week 4 Lectures, Math 716, Tanveer 1 Diffusion in R...

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