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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 πκt exp bracketleftbigg − x 2 4 κt 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 = κ Δ 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 ) = κ n summationdisplay j =1 ∂ 2 ∂x 2 j S ( x i , t ) productdisplay i negationslash = j S ( x i , t ) = κ Δ 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 κπt ) n/ 2 exp bracketleftbigg − x 2 1 + x 2 2 + ..x 2 n 4 κt bracketrightbigg = 1 (4 κπt ) n/ 2 exp bracketleftbigg − x 2 4 κt 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) = φ ( x ) for φ ∈ C ( R n ) (8) is given by u ( x , t ) = integraldisplay R n S ( x − y , t ) φ ( 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 − κv xx = 0 for x > , t > (10) v ( x, 0) = φ ( 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 φ ( x ) is only specified for x > 0. φ x φ( x) odd Figure 1: Odd Extension of φ ( x ) to x ∈ R We do an odd extension , i.e. define an extended function φ odd ( x ) in R (see Fig. 1) so that φ odd ( x ) = φ ( x ) for x > 0 ; φ odd ( x ) = − φ ( − x ) for x < (13) . Let u ( x, t ) be a solution to heat equation so as to satisfy u ( x, 0) = φ odd ( x ) for x ∈ R (14) Then, applying equation (15) of week 3 notes, u ( x, t ) = integraldisplay ∞ −∞ S ( x − y, t ) φ 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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 Spring '08
 Tanveer,S
 Scalar, Boundary value problem, Boundary conditions, Dirichlet boundary condition, Neumann boundary condition

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