Diffusion on the half-line. The Dirichlet problem

Transcription

1 Diffusion on the half-line The Dirichlet problem Consider the initial boundary value problem (IBVP) on the half line (, ): v t kv xx = v(x, ) = φ(x) v(, t) =. The solution will be obtained by the reflection method with odd extensions. For the extended problem on u t ku xx = u(x, ) = φ o (x), the solution is already known to be u(x, t) = S(x y, t)φ o (y)dy. The restriction of u to the half-line v(x, t) = u(x, t) for x will be the solution of the half-line problem. An explicit formula for v can be deduced as following: u(x, t) = S(x y, t)φ(y)dy+ S(x y, t)[ φ( y)]dy 1

2 = S(x y, t)φ(y)dy S(x + y, t)φ(y )dy. So that for x 1 v(x, t) = [e (x y)2 /4kt e (x+y)2 /4kt ]φ(y)dy. 4πkt It is easy to see that the boundary condition is satisfied. The Neumann problem Next, consider the initial boundary value problem w t kw xx = for x > w(x, ) = φ(x) w x (, t) =. This time, use the even extension. Following a similar argument as before, one gets the solution w(x, t) = 1 [e (x y)2 /4kt + e (x+y)2 /4kt ]φ(y)dy. 4πkt The only difference from that of v is the plus sign of the second term in the integral. Source on a half-line With x = being a boundary, it is necessary to consider a boundary condition. 2

3 If the boundary condition is to specify the value of the solution at x =, the BIVP is in the form v t kv xx = f(x, t) for < x <, < t < v(, t) = h(t) v(x, ) = φ(x). The problem can be reduced to a simpler one. The boundary condition can be turned into the Dirichlet type by a subtraction technique. Define V (x, t) = v(x, t) h(t). Then V (x, t) will satisfy V t kv xx = f(x, t) h (t) V (, t) = V (x, ) = φ(x) h(). If the boundary condition is to specify spatial partial derivative of the solution at x =, the BIVP is in the form w t kw xx = f(x, t) for < x <, < t < w x (, t) = h(t) w(x, ) = φ(x). This boundary condition can be reduced to the Neumann type, again by a subtraction technique. The new function to use is W(x, t) = w(x, t) xh(t). 3

4 Diffusion in a finite interval The Dirichlet condition pde: u t ku xx = for < x < L (homogeneous) BC: u(, t) = u(l, t) = (Dirichlet) IC: u(x, ) = φ(x). Let u = T(t)X(x). The equation becomes T kt = X = λ = constant. X For X, the problem is X = λx in < x < l with X() = X(L) =. This is exactly the same problem for X(x) as before, and the conclusions are the same. Namely, λ = (nπ/l) 2 > and X(x) = sin( λx). The equation for T yields T(t) = Ae λkt. Its magnitude is monotonically decreasing. The infinite series u(x, t) = n=1 is the solution provided that φ(x) = A n sin nπx L. n=1 A n e (nπ/l)2 kt sin nπx L 4

5 The Neumann condition Applying the separation of variables method to both the wave equation and the diffusion equation produces the same twopoint boundary value problem for X(x) X = λx, X () = X (L) =. All the eigenvalues can be written collectively as ( nπ ) 2 λ n = βn 2 = n =, 1, 2, 3,... L with n now starting from. X n (x) = C n cos nπx L eigenfunctions. are the The equation for T(t) is T = βn 2 kt. For n >, the solution is T(t) = e β2 n kt. For n =, the solution is T(t) = a constant. Therefore, the diffusion equation with the Neumann BCs has the solution u(x, t) = 1 2 A + A n e β2 nkt cos β n x. The IC requires that φ(x) = 1 2 A + n=1 A n cos β n x. n=1 Note that all the β n modes are being damped. Only the β = mode (that is the mean of u) remains unaffected, i.e. lim u(x, t) = 1 t 2 A. Problem Set 8 5 Compare with the Dirichlet case.

6 The Maximum Principle For convenience, let us define some handy symbols. Let S be a set in n. Then S denotes the boundary of S, S denotes S S. For the open rectangle = {(x, t) < x < L, < t < T }, introduce two more symbols. U = {(x, t) < x < L, t = T } (top boundary of the rectangle excluding the two end points) and L = U (the bottom and side boundaries). Theorem: (The Weak Maximum Principle) Let = {(x, t) < x < L, < t < T } be an open rectangle in 2. Let u(x, t) be continuous on, u t, u xx exist and be continuous on, and satisfies u t ku xx =. Then max u = max L u. In other words, the maximum value of u(x, t) is assumed either initially (t =, < x < L) or on the lateral sides (x = or x = L, t T). Proof: Note that maximum values exist on,, and L as the sets are closed and the function is continuous. As L, max u max u is always true. The L theorem wants the equality to hold. We are going to proceed with prove by contradiction : 6

7 Suppose that max u max u = 2ɛ >. Then, since u L is continuous, there exist a smaller rectangle S = {(x, t) < x < L, < t < T δ} with δ > so that max S u max L S u > ɛ. (1) (If max u occurs at an interior point of, include this point in S. If max u occurs at a point (x m, t m ) on U, δ m > so that max u u(x, t) < ɛ (x, t) with (x, t) (x m, t m ) < δ m. Set δ = δ m /2.) Therefore, the statement of the theorem is also violated in S. Define w(x, t) = u(x, t) + ɛ 2L 2x2. (2) Then max S w max S u and max u > (Eq.1) max u + ɛ/2 (Eq.2) max w S L S L S max S w > max L S w. (3) 7

8 Thus, the maximum value of w occurs at a point (x 1, t 1 ) in the interior of S or on U S. In either case, w xx (x 1,t 1 ) and w t (x 1,t 1 ) so that w t (x 1, t 1 ) kw xx (x 1, t 1 ). On the other hand, w t (x 1, t 1 ) kw xx (x 1, t 1 ) = u t (x 1,t 1 ) ku xx (x 1, t 1 ) ɛ L 2 = ɛ L 2 <. This is a contradiction. Therefore, max u = max L u. Well-posedness of the diffusion equation A. Existence Solutions of the diffusion equation on or half- are given by convolutions of the diffusion kernel. For a finite-interval problem, the solution of the diffusion equation is provided by an appropriate kind of Fourier series. B. Uniqueness The maximum principle can be used to give a proof of uniqueness for the non-homogeneous Dirichlet problem of the diffusion equation: u t ku xx = f(x, t) for < x < L and t > 8

9 u(x, ) = φ(x) u(, t) = g(t) u(l, t) = h(t) with four given functions f, φ, g, and h. Proof: Let u 1 and u 2 be two solutions. Define w = u 1 u 2. Then w t w xx = and w(x, ) =, w(, t) =, and w(l, t) =. Inside any rectangle [, L] [, T] (T > ), the maximum principle says that w everywhere. The same argument can be applied to u 2 u 1 = w. Therefore, w =. A second proof of uniqueness can be obtained by the energy method: Let w be the solution of the Dirichlet (or Neumann) problem: w t = kw xx w(x, ) = φ(x), x L w(, t) = w(l, t) =. Then d L w 2 (x, t) dx. dt Proof: Multiplying the equation by w, one gets w(w t kw xx ) = ( 1 2 w2 ) t + ( kw x w) x + kw 2 x =. 9

10 Integrating over the interval (, L), one gets 1 2 d dt L w 2 dx = k L Therefore, w 2 dx is decreasing. w 2 xdx. If w is the difference of two solutions of the inhomogeneous diffusion equation, w(x, ) = and the boundary values of w (or w x ) are. Then L [w(x, t)] 2 dx L [w(x, )] 2 dx =. For continuous w, this implies that w(x, t) =. For a problem over, the energy method still works for the boundary conditions lim u if u 2 dx <. x ± C. Stability Consider the dependence of the solution on the initial condition. Let u 1, u 2 be the solutions with the initial conditions φ 1 and φ 2, respectively. Again, both the maximum principle and energy method can be applied to show stability, but they use different ways to measure distance between functions. Energy method: For w = u 1 u 2, the energy inequality can be written as L [u 1 (x, t) u 2 (x, t))] 2 dx 1 L [φ 1 (x) φ 2 (x)] 2 dx.

11 The left side is a quantity that measures the nearness of the solution, and the right side measures the nearness of the initial data. Thus, if the right side is small, the left side is also small. This is stability in the square integral sense. Maximum principle method: Consider two solutions u 1, u 2 on a rectangle. w = u 1 u 2 = on the lateral sides of the rectangle and w = φ 1 φ 2 on the bottom. The maximum principle says that for all (x, t), u 1 (x, t) u 2 (x, t) max(φ 1 φ 2 ) max φ 1 φ 2. Similarly, u 2 (x, t) u 1 (x, t) max φ 1 φ 2. Therefore, u 1 (x, t) u 2 (x, t) max φ 1 φ 2 (x, t) which implies that max u 1(x, t) u 2 (x, t) max φ 1 φ 2 t >. x L x L This is stability in the uniform sense. This stability analysis can obviously be extended to include the boundary conditions. An ill-posed problem The diffusion equation is not well-posed for t < (going backward in time). Consider the sequence of functions u n = (1/n) sin(nxπ/l)e n2 π 2 kt/l 2 < t <. 11

12 Each of them satisfies the diffusion equation on (, L) with zero boundary conditions u(, t) = u(l, t) =. Also, the functions specifying the initial condition u n (x, ) = (1/n) sin(nxπ/l) as n in both the uniform sense and the square integral sense. However, at a negative t, say 1, the sequence u n (x, 1) = (1/n) sin(nxπ/l)e n2 π 2 kt/l 2 diverges in both the uniform sense and the square integral sense. Thus, while the measures of u n (x, ) approach zero for n, the measures of u n (x, 1) become unbounded. This violates the stability cirterion for well-posedness. 12