Diffusion equation in one spatial variable Cauchy problem. u(x, 0) = φ(x)

Transcription

1 Diffusion equation in one spatial variable Cauchy problem. u t (x, t) k u xx (x, t) = f(x, t), x R, t > u(x, ) = φ(x) 1

2 Some more mathematics { if x < Θ(x) = 1 if x > is the Heaviside step function. It holds Θ (x) = δ(x) where δ is the Dirac delta function. Some properties of delta function: Intuitively, δ(x) = { if x if x = For any continuous function f, it holds δ(y)f(y)dy = f() 2

3 The latter relation implies δ(x y)φ(y)dy = φ(x) for any continuous function φ. The integral g(x y)f(y)dy is called the convolution of functions g and f. The transformation of the step function Θ to an arbitary continuous function φ: Θ δ differentiation convolution φ 3

4 Cauchy problem for homogeneous equation with step function w t (x, t) k w xx (x, t) =, x R, t > w(x, ) = Θ(x) We transform the equation w t (x, t) k w xx (x, t) = to an ordinary differential equation by means of the change of variables w(x, t) = f(z), z = x 4

5 Compute: w t = f (z)z t = 1 x 2 f (z) 3 w x = f (z)z x = 1 f (z) w xx = x w x = 1 f (z) Substituting these formulas to the equation w t (x, t) k w xx (x, t) = we deduce = w t kw xx = 1 x 2 f (z) k 1 3 f (z) = 1 4t f (z) 1 x 2 f (z) = 3 = 1 [ f (z) + 4tx ] 4t 2 f (z) = 1 [ f (z) + 2x ] f (z) = 1 3 4t 4t [f (z) + 2zf (z)]. Thus, we obtain the ordinary ordinary differential equation f (z) + 2zf (z) =. ( ) Denote g = f. Then the equation is g (z) + 2zg(z) =. We solve it: dg dg = 2zg dz g = 2zdz dg g = 2zdz ln g = z 2 + c 1 g = c 1 e z2 Further, 5

6 f = c 1 e z2 z f(z) = c 1 e s2 ds + c 2 The general solution of the equation (*) is z f(z) = c 1 e s2 ds + c 2 where c 1, c 2 are arbitrary constants. Consequently, w(x, t) = c 1 x e s2 ds + c 2 We use the initial condition w(x, ) = Θ(x) to determine the constants c 1 and c 2. In case x < and t + = w(x, ) = c 1 e s2 ds + c 2 = c 1 e s2 ds + c 2 In case x > and t + 1 = w(x, ) = c 1 e s2 ds + c 2 6

7 It is known that e s2 ds = π 2. Therefore, { π 2 c 1 + c 2 = π 2 c 1 + c 2 = 1 Solving this linear system we get c 1 = 1 π, c 2 = 1 2. The solution of the posed Cauchy problem with step function is w(x, t) = π x e s2 ds Using the error function erf (z) = 2 π z e s2 ds the solution is written in the form w(x, t) = 1 2 [ ( )] x 1 + erf This not a classical solution (w is not continuous). 7

8 Cauchy problem for homogeneous equation with delta function G t (x, t) k G xx (x, t) =, x R, t > G(x, ) = δ(x) The previous problem was w t (x, t) k w xx (x, t) =, x R, t > w(x, ) = Θ(x) Since Θ = δ, it holds G = w x. From the formula w(x, t) = π we obtain the formula for G: x e s2 ds G(x, t) = w x (x, t) = 1 4πkt e x2 8

9 G is called the heat (diffusion) kernel or the fundamental solution of the diffusion equation. 9

10 Cauchy problem for homogeneous equation in general case u t (x, t) k u xx (x, t) =, x R, t > u(x, ) = φ(x) where φ is an arbitrary given function. Due to the properties of the delta function, G(x y, t)φ(y)dy = t= = δ(x y)φ(y)dy = φ(x) G(x y, )φ(y)dy = This suggests that u(x, t) = G(x y, t)φ(y)dy 1

11 Verification of the equation: We obtain ( ) ( ) t k 2 G(x y, t)φ(y)dy = x 2 t k 2 G(x y, t)φ(y)dy = x 2 ( ) because k 2 G(x, t) =. t x 2 Thus, indeed the solution of the Cauchy problem is u(x, t) = G(x y, t)φ(y)dy = 1 4πkt e (x y)2 φ(y)dy 11

12 Theorem. Let φ be a bounded continuous function on R. The Cauchy problem u t (x, t) k u xx (x, t) =, x R, t > u(x, ) = φ(x) has a classical solution given by the formula u(x, t) = G(x y, t)φ(y)dy = 1 4πkt e (x y)2 φ(y)dy 12

13 Cauchy problem for nonhomogeneous equation Theorem. Let f = f(x, t) and φ = φ(x) be bounded and continuous functions. The Cauchy problem for the nonhomogeneous diffusion equation u t (x, t) k u xx (x, t) = f(x, t), x R, t > u(x, ) = φ(x) has a classical solution given by the formula u(x, t) = t G(x y, t)φ(y)dy+ G(x y, t s)f(y, s)dyds (1) where G is the diffusion kernel. 13

14 Derivation of the formula (1) by means of the operator method. Firstly, we solve the following Cauchy problem for ordinary differential equation: d v(t) + Av(t) = f(t), t >, v() = φ (2) dt where A and φ are given numbers. The homogeneous equation is d dt v H(t) + Av H (t) = The characteristic equation λ + A = has the solution λ = A. The general solution of the homogeneous equation is v H (t) = Ce At We use the variation of constants to derive a particular solution of the inhomogeneous equation v P (t) = D(t)e At Putting it to the equation we get D e At DAe At + AD(t)e At = f Thus, D (t) = e At f(t) 14

15 Consequently, D(t) = t e As f(s)ds and t v P (t) = D(t)e At = e At e As f(s)ds = t e A(t s) f(s)ds The general solution of the ODE is v(t) = v H (t) + v P (t) = Ce At + t e A(t s) f(s)ds We find v() = C and by initial condition v() = φ we have C = φ. Thus, v(t) = φe At + t e A(t s) f(s)ds We have shown that the solution of (2) is v(t) = S(t)φ + where t S(t) = e ta S(t s)f(s)ds 15

16 In the particular case f, the solution of (2) is v(t) = S(t)φ We compare it with the formula of the solution of the homogeneous Cauchy problem for the diffusion equation: u(x, t) = S(t)φ = G(x y, t)φ(y)dy It holds t S(t s)f(x, s)ds = t G(x y, t s)f(y, s)dyds 16

17 Using the analogy with the solution of the Cauchy problem (2) for ODE, we put together the solution of the nonhomogeneous Cauchy problem for the diffusion equation: u(x, t) = S(t)φ + = t G(x y, t)φ(y)dy + S(t s)f(x, s)ds = t G(x y, t s)f(y, s)dyds The formula (1) is derived. The operator S(t) is called the semigroup generated by the operator A = k 2 x 2. Sometimes it is written S(t) = e ta 2 tk = e x 2 17