HOMEWORK 4 1. P45. # 1.
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1 HOMEWORK 4 SHUANGLIN SHAO P45 # Proof By the maximum principle, u(x, t x kt attains the maximum at the bottom or on the two sides When t, x kt x attains the maximum at x, ie, x When x, x kt kt attains the maximum in the interval t T When x, x kt kt attains the maximum in the interval t T Thus the maximum of u is in the closed rectangle { x, t T } P45 # Proof (a Let M(T the maximum of u(x, t in the closed rectangle { x l, t T } M(T is a decreasing function of T Indeed, when T T, the rectangle R { x l, t T } is contained in the rectangle R { x l, t T }; the bottom and the two lateral sides of R are contained in those of R On the other hand, the uniqueness of solutions to the diffusion equation shows that the solution u on R is an extension of u on R So by the maximum principle, M(T M(T (b Similarly as in proving (a, m(t is an increasing function of T 3 P45 #3 Proof (a By the strong maximum principle, u(x, t > in the interior points < x <, < t < because the minimum value of u,, is attained at the boundary point, and at the two lateral sides
2 (b We follow the hint Let µ(t the maximum of u(x, t over x Let X(t [, ] such that µ(t u(x(t, t On the two lateral sides, the value of u is By part (a, µ(t > So X(t (, On the point (X(t, t, following the proof of the maximum principle in the book, So we differentiate µ in t, u x (X(t, t, u t (X(t, t µ (t u x X (t + u t u t We can establish it by a different method For each t >, let µ(t the maximum of u(x, t over x By part (a, u(x, t for all x, t Then µ(t for t > Suppose that < t < t Define R(t, t { x, t t t } Since the value of u is on the two lateral sides, by the maximum principle, the maximum of u on R(t, t is µ(t This implies that µ(t µ(t So µ is decreasing in t > 4 P46 # 4 Proof (a On the two lateral sides and on the bottom, the minimum and the maximum of u is and 4, respectively So by the strong maximum principle, < u(x, t < (b Both u(x, t and u( x, t satisfy the equation u t ku xx and the two lateral side conditions, and the initial condition By the uniqueness theorem for the diffusion equation, u(x, t u( x, t (c Let E(t (u(x, t dx
3 We differentiate it in t, de(t dt k uu t dx uu xx dx uu x x x k (u x dx k (u x dx So E(t is decreasing in t 5 P46 # 6 Proof We prove it by considering the difference w v u The function w(x, t satisfies the equation w t kw xx, with w being nonnegative when either t, orx or x l By the maximum principle, the minimum value of w is attained when either t, or x or x l So Thus for t < and x l w u v 6 P5 # Proof The function φ satisfies that So the solution u is u(x, t φ(x, x < l; φ(x, x > l l l l e (x y φ(ydy e (x y dy l e (x y dy 3
4 For the first integral, l l x e ( y x dy l x π l x Erf( Similarly for the second integral, e z dz e z dz Hence l l+x e ( y x dy l+x π l+x Erf( u(x, t e z dz e z dz Erf( l x l+x Erf( 7 P5 # Proof The function φ satisfies that So the solution u is u(x, t φ(x, for x > ; φ(x 3, for x < e (x y φ(ydy e (x y dy + 3 e (x y dy 4
5 For the first integral, π Erf( x x e ( x y dy e z ( dz x e z dz Similarly for the second integral, π e ( y x dy x x 3Erf( x e z dz e z dz Hence u(x, t Erf( x + 3Erf( x 5
6 8 P5 # 3 Proof u(x, t e3x 9kt e3x 9kt π e 3x 9kt e (x y φ(ydy e (x y +3y dy e x xy+y +kyt dy e (y x+6kt +kxt 36k t dy e (y x+6kt +3x 9kt dy ( e y x+6kt dy e y dy 9 P5 # 6 Proof Let A e x dx e x dx Thus A e x y dxdy 4 e (x +y dxdy 4 R 4 π 4 π 4 π ( 4 e r πrdr e r d(r e x dx π 4 6
7 So A π P5 #7 Proof From Exercise # 6, If setting p x, we have e p dp e x dx π which implies that π e x dx π S(x, tdx, S(x, tdx P53 # 8 Proof Let Since S is even function in x, S(x, t πkt e x S(x, t S( x, t Thus S(x, t S( x, t max S(x, t max S(x, t δ x < δ x< For each t >, the function S(x, t is decreasing on the interval [δ, Hence max S(x, t δ x< πkt e δ 7
8 We will apply the L Hospital rule to show that e δ / goes to zero as t goes to zero πk lim t / t e δ πk lim t δ t/ e δ k which goes to zero as both t / and e δ This proves the claim /t 3/ e δ δ go to zero when t goes to zero P53 # 5 Proof Let u, v be two solutions to the diffusion equation with Neumann boundary conditions Let w u v Then w satisfies the following system of equations: w t kw xx w(x,, w x (, t, w x (l, t We multiply the equation by w to obtain w(w t kw xx Thus d(w k(w x w x + kwx dt Integrating both sides and using the boundary conditions, we see that d w dx + k w dt xdx Since k, we see that the quantity w dx is decreasing for t > Thus (u(x, t v(x, t dx w (x, dx This implies that u(x, t v(x, t for all x, t This proves the uniqueness 8
9 3 P 54 # 6 Proof We set u(x, t e bt v(x, t We substitute u into the equation u t ku xx + bu e bt ( bv + v t kv xx + bv Thus v t kv xx The initial condition changes to v(x, φ(x formula for the homogeneous diffusion equation, where v(x, t S(x, t S(x y, tφ(ydy, e x Hence by the solution Therefore u(x, t e bt S(x y, tφ(ydy This is the solution Department of Mathematics, KU, Lawrence, KS address: slshao@mathkuedu 9
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