MA 523: Homework 1. Yingwei Wang. Department of Mathematics, Purdue University, West Lafayette, IN, USA

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1 MA 523: Homework Yingwei Wang Department of Mathematics, Purdue University, West Lafayette, IN, USA One dimensional transport equation Question: Solve the transport equation using the method of characteristics. Solution: Let γ(s) = (x(s),t(s)), then u t +3u x = 0, (.) u(x,0) = x 2, (.2) { x (s) = 3, x(0) = x, t (s) =, t(0) = t, { x(s) = x+3s, t(s) = t+s, Let ψ(s) = u(γ(s)) = u(x+3s,t+s), then ψ (s) = 3u x +u t = 0, ψ(s) = constant, ψ(0) = ψ( t), u(x,t) = u(x 3t,0) = (x 3t) 2. address: wywshtj@gmail.com; Tel:

2 2 n dimensional transport equation Question: Let x R n,b R n,ϕ : R n R, f : R n [0,+ ) R. Solve the transport equation using formally Fourier transform. Solution: It is easy to know that u t + < b,du >= f(x,t), (2.) u(x,0) = ϕ(x). (2.2) F(< b,du >) = 2πi < b,ξ > û(ξ,t), ξ R n. After Fourier transform on Eqs.(2.)-(2.2), we can get û t (ξ,t)+2πi < b,ξ > û(ξ,t) = ˆf(ξ,t), (2.3) û(ξ,0) = ˆϕ(ξ). (2.4) For fixed ξ, we just need to solve the ODE (2.3)-(2.4). (û(ξ,t)e ) 2πi<b,ξ>t = e 2πi<b,ξ>t (û t +2πi < b,ξ > û(ξ,t)) = e 2πi<b,ξ>t ˆf(ξ,t) t û(ξ,t)e 2πi<b,ξ>t ˆϕ(ξ)e 2πi<b,ξ>t = û(ξ,t) = ˆϕ(ξ)e 2πi<b,ξ>t + u(x,t) = ϕ(x)(x tb)+ t 0 t 0 t 0 e 2πi<b,ξ>s ˆf(ξ,s)ds, e 2πi<b,ξ>(t s) ˆf(ξ,s)ds, f(x (t s)b,s)ds. 2

3 3 Wave equation Question: Suppose φ(x) C 2 (R),ψ(x) C (R). Prove that u(x,t) given by d Alembert formula Eq.(3.) is a function in C 2 (R [0,+ )). u(x,t) = 2 [φ(x+ct)+φ(x ct)]+ 2c Furthermore, it satisfies the Cauchy problem: Solution: From Eq.(3.), we can get x+ct x ct ψ(y)dy. (3.) c 2 u xx u tt = 0, (3.2) u(x,0) = φ(x), (3.3) u t (x,0) = ψ(x). (3.4) u x = 2 [φ (x+ct)+φ (x ct)]+ [ψ(x+ct) ψ(x ct)], (3.5) 2c u xx = 2 [φ (x+ct)+φ (x ct)]+ 2c [ψ (x+ct) ψ (x ct)], (3.6) u t = c 2 [φ (x+ct) φ (x ct)]+ [ψ(x+ct)+ψ(x ct)], (3.7) 2 u tt = c2 2 [φ (x+ct)+φ (x ct)]+ c 2 [ψ (x+ct) ψ (x ct)], (3.8) Since both of φ and ψ are continuous, Eq.(3.6) and Eq.(3.8) imply that u xx and u tt are continuous, which means u(x,t) C 2 (R [0,+ )). Furthermore, Eq.(3.6) and Eq.(3.8) also imply that c 2 u xx u tt = 0. Besides, Eq.(3.) gives that And Eq.(3.7) gives that u(x,0) = φ(x). u t (x,0) = ψ(x). Now we can conclude that u(x, t) given by d Alembert formula Eq.(3.) satisfies the Cauchy problem (3.2)- (3.4). 3

4 MA 523: Homework 2 Yingwei Wang Department of Mathematics, Purdue University, West Lafayette, IN, USA Spherically symmetric function Definition.. Call an operator T : R n R n is orthogonal (T O(n)) if T is linear and T t T = I. Definition.2. f : R n R is called spherically symmetric if f(tx) = f(x), x R n, T O(n). Lemma.. Prove that f is spherically symmetric if and only if f : [0,+ ) R, such that f(x) = f ( x ), x R n. Proof. Suppose f is spherically symmetric. Since T O(n), T x = x. Besides, f(tx) = f(x) x R n. (.) Let x = r, and f(x) = c. Then Eq.(.) tells us that x S(0,r), f(x) = c. It implies that f : [0,+ ) R, such that f(x) = f ( x ), x R n. Conversely, suppose that f : [0,+ ) R, such that f(x) = f ( x ), x R n. Then we can find f : [0,+ ) R such that f( x 2 ) = f ( x ), x R n. Now we have f(x) = f( x 2 ), x R n Since x 2 = x t x, we know that T O(n), x R n, f(tx) = f( Tx 2 ) = f((tx) t (Tx)) = f(x t T t Tx) = f(x t x) = f( x 2 ) = f(x). It follows that f is spherically symmetric. address: wywshtj@gmail.com; Tel:

5 2 Spherical averaging function Definition 2.. Suppose f : R n R is continuous, then define the spherical averaging of f as follows: M f (x,r) = f(y)dσ(y). σ n r n S(x,r) Lemma 2.. Prove that lim M f(x,r) = f(x). (2.) r 0 Proof. Since the sphere S(x,r) is compact and f is continuous, then f has maximum and minimum values on S(x,r). Let By Definition 2., we know that M f (x,r) = σ n r n σ n r n M = max y S(x,r) f(y), m = min y S(x,r) f(y). S(x,r) f(y)dσ(y) mdσ(y) M f (x,r) Mdσ(y) S(x,r) σ n r n S(x,r) m M f (x,r) M. (2.2) It is easy to know that as From Eq.(2.2) and Eq.(2.3), we can get Eq.(2.). m = M = f(x), as r 0. (2.3) 2

6 MA 523: Homework 3 Yingwei Wang Department of Mathematics, Purdue University, West Lafayette, IN, USA Spherical averaging function Definition.. Suppose f : R n R is continuous, then define the spherical averaging of f as follows: M f (x,r) = f(y)dσ(y). σ n r n S(x,r) Lemma.. Prove that lim M f(x,r) = f(x). (.) r 0 Proof. Since the sphere S(x,r) is compact and f is continuous, then f has maximum and minimum values on S(x,r). Let M = max y S(x,r) f(y), m = min y S(x,r) f(y). By Definition., we know that M f (x,r) = f(y)dσ(y) σ n r n S(x,r) mdσ(y) M σ n r n f (x,r) Mdσ(y) S(x,r) σ n r n S(x,r) m M f (x,r) M. (.2) address: wywshtj@gmail.com; Tel:

7 It is easy to know that From Eq.(.2) and Eq.(.3), we can get Eq.(.). m = M = f(x), as r 0. (.3) 2 C k domain Definition 2.. Let Ω R n. We say Ω is of class C k if x 0 Ω, U open such that x 0 U, and φ(x) C (U), such that 2. C but not C 2 Ω U = Ω {x U φ(x) < 0}, (2.) Ω U = Ω {x U φ(x) = 0}, (2.2) Dφ(x) 0, x U. (2.3) Let Ω = {(x,y) R 2 y > x α }, < α < 2. Prove that Ω is of class C but not of class C 2. Proof. Let U = R 2, φ(x,y) = x α y. It is easy to check that (2.)-(2.3) are satisfied here. Besides, both φ = x α x α and φ = arecontinuous, so φ y C (R 2 ). However, 2 φ = α(α ) x α 2 is not continuous at x = 0, so φ C 2 (R 2 ). It indicates that Ω x 2 is of class C. Next to prove that Ω is not of class C 2. Suppose φ(x,y) C 2 satisfying (2.)-(2.3). Consider the curve S = Ω U. x,y S, since φ(x,y) C 2, we have φ(x,y) = 0, (2.4) dy dx = φ x is continuous, (2.5) φ y ( dy ) ( dx + dy ) 2 φyy dx is also continuous. (2.6) φ y d2 y dx 2 = φxx + φ xy 2

8 But we also know that x,y S, y = x α, < α < 2 (2.7) d2 y dx = 2 α(α ) x α 2 is not continuous at x = 0. (2.8) Then (2.6) contradicts with (2.8). So Ω is not of class C C k but not C k+ Let Ω = {(x,y) R 2 y > x α }, k < α < k +. Prove that Ω is of class C k but not of class C k+. Proof. Let U = R 2, φ(x,y) = x α y. It is easy to check that (2.)-(2.3) are satisfied here. Similarly as mentioned above, it is easy to know that k φ,(i+j = k), is continuous. So Ω is of class C k. x i y j Next to prove that Ω is not of class C k+. Suppose φ(x,y) C k+ satisfying (2.)-(2.3). Consider the curve S = Ω U. x,y S,from φ(x,y) C k+ weknowthat dn y,iscontinuousforeachn =,,k+ dx n. But we also know that x,y S, y = x α, k < α < k + dk+ y k = x α k (α n) is not continuous at x = 0. (2.9) dxk+ n=0 It follows that Ω is not of class C k+. 3 Minkowski formula Theorem 3. (Divergence theorem). Suppose Ω R n be bounded open set of class C, F C ( Ω,R n ), F = (F,,F n ), then div Fdx = < F,ν > dσ, (3.) Ω Ω 3

9 where div F = n i= F i x i. Proposition 3. (Minkowski formula). Let Ω R n be bounded piecewise C domain. Then < x,ν > dσ = nvol n (Ω). (3.2) Ω Proof. On one hand, by the definition of volume, Vol n (Ω) = dx. (3.3) On the other hand, choose F = x = (x,x 2,,x n ), then div F = n F i i= x i = n. By Theorem 3., we know that < x,ν > dσ = ndx = n dx. (3.4) Ω Ω Ω By Eq.(3.3) and Eq.(3.4), we can get Eq.(3.2). Ω 4

10 MA 523: Homework 4 Yingwei Wang Department of Mathematics, Purdue University, West Lafayette, IN, USA Wave equation in R 3 : Existence Consider the nonhomogeneous wave equation in R 3 : c 2 w w tt = f(x,t), (.) w(x,0) = ϕ(x), (.2) w t (x,0) = ψ(x), (.3) where x R 3,t > 0. In order to solve the nonhomogeneous equation, we consider the following two equations: and The solution to the equation (.4)-(.6) is given by c 2 u u tt = 0, (.4) u(x,0) = ϕ(x), (.5) u t (x,0) = ψ(x), (.6) c 2 v v tt = f(x,t), (.7) v(x,0) = 0, (.8) v t (x,0) = 0. (.9) u(x,t) = t (tm ϕ(x,ct))+tm ψ (x,ct), (.0) = M ϕ (x,ct)+ct M ϕ(x,ct) +tm ψ (x,ct), (.) t = [ϕ(y)+ < Dϕ(y),y x > +tψ(y)]dσ(y). (.2) 4πc 2 t 2 S(x,ct) address: wywshtj@gmail.com; Tel:

11 The solution to the equation (.7)-(.9) is given by t [ ] v(x, t) = f(y, s)dσ(y) ds, (.3) 0 4πc 2 (t s) S(x,c(t s)) = f(y,t y x ) c dy. (.4) 4πc 2 y x B(x,ct) Then, the solution to the equation (.)-(.3) is given by which is called Kirchhoff formula. w(x,t) = u(x,t)+v(x,t), (.5) Theorem.. The function given by (.5) satisfies the initial conditions (.2) and (.3). Proof. First, by Eq.(.), let us rewrite u(x, t) as u(x,t) = M ϕ (x,ct)+p(x,t)+tm ψ (x,ct), (.6) where p(x,t) = ct M ϕ(x,ct) t Then we know that = < Dϕ(y),y x > dσ(y). (.7) 4πc 2 t 2 S(x,ct) So limp(x,t) =< Dϕ(y),y x > y=x = 0. t 0 u(x,0) = lim t 0 (M ϕ (x,ct)+p(x,t)+tm ψ (x,ct)), = ϕ(x)+0+0ψ(x), = ϕ(x). (.8) Besides, by Eq.(.4), let us rewrite v(x,t) as v(x,t) = 4πc 2 = 4πc 2 4πc 2 y x <ct ct 0 ct 0 r f(y,t y x ) c y x y x =r dy, (.9) f(y,t r )dσ(y)dr, (.20) c r 4πr2 f(x,0)dr = f(x,0) 4πc 2 2 ct 0 r 2 dr 0, (.2)

12 as t 0. Now we can conclude that w(x,0) = u(x,0)+v(x,0) = ϕ(x). Second, let us consider the first derivative of u(x,t) and v(x,t). Recall the Euler-Poisson-Darboux formula 2 M f + n r 2 r where f C 2 (R n ). By Eq.(.7) and Eq.(.22), we can get u t (x,t) M f r = M f, (.22) = 2c M ϕ(x,ct) +c 2 t 2 M ϕ (x,ct) +ct M ψ(x,ct) +M r r 2 ψ (x,ct), r = 2c M ( ϕ r +c2 t M ϕ 2 ) M ϕ +ct M ψ ct r r +M ψ, = c 2 tm ϕ (x,ct)+ct M ψ(x,ct) +M ψ (x,ct), r ψ(x), as t 0. Besides, by Eq.(.20), we know that v t (x,t) = [ ct f(y,0)dσ(y)+ 4πc 2 ct y x =ct 0 r limv t (x,t) = [ ] limf(x,0)t+lim t 0 4πc 2 t 0 t 0 2 f t(x,0)t 2 = 0. Now we can conclude that y x =r w t (x,0) = u t (x,0)+v t (x,0) = ψ(x). ] f t (y,t r/c)dσ(y)dt 2 Laplacian operator and spherical averaging Lemma 2.. Suppose f(x) = 3 2 f, i= x 2 i M f (x,r) = 4πr 2 S(x,r) f(y)dσ(y), 3

13 where x = (x,x 2,x 3 ) R 3. Then M f (x,r) = M f. (2.) Proof. Considering the transformation of spherical coordinates, we can get M f (x,r) = f(y)dσ(y), 4πr 2 = 4πr 2 M f (x,r) = 4πr 2 = 4πr 2 = M f. y x =r π 2π 0 0 π 2π 0 0 y x =r f(x +rsinϕcosθ,x 2 +rsinϕsinθ,x 3 +rcosθ)dθdϕ, f(x +rsinϕcosθ,x 2 +rsinϕsinθ,x 3 +rcosθ)dθdϕ, f(y)dσ(y), 3 Wave equation in R 3 : uniqueness Lemma 3. (First speed propagation). Let (x 0,t 0 ) R n (0,+ ). Suppose u C 2 (c(x 0,t 0 )) C ( c(x 0,t 0 )) satisfy Then u 0 in the cone c(x 0,t 0 ), where c 2 u u tt = 0, in c(x 0,t 0 ) (3.) u(x,0) = u t (x,0) = 0, x x 0 ct 0. (3.2) c(x 0,t 0 ) = {(x,t) R n [0,+ ) 0 t t 0, x x 0 < c(t 0 t)}. (3.3) Using this lemma, we can prove this theorem. Theorem 3. (Uniqueness of Cauchy problem). Given ϕ C 3 (R 3 ),ψ C 2 (R 3 ),f C 2 (R 3 [0,+ )), there exists only one solution w C 2 (R 3 (0,+ )) C (R 3 [0,+ )) of the problem (.)-(.3). Such solution is given by Eq.(.5). Proof. Suppose there are two solutions of the problem (.)-(.3), namely w (x,t) and w 2 (x,t). Then it is easy to know that w 0 = w w 2 is the solution of (3.)-(3.2). By Lemma 3., we know that in each cone like (3.3), w 0 0. So w 0 0 in R 3 [0,+ ). It follows that w w 2 in R 3 [0,+ ). 4

14 MA 523: Homework 5 Yingwei Wang Department of Mathematics, Purdue University, West Lafayette, IN, USA Laplacian operator Lemma.. Let Ω R n be an open set, f C 2 (Ω) and g C 2 (R). Then g f : Ω R. Prove that (g f) = g (f) f 2 +g (f) f. (.) Proof. Let x = (x,x 2,,x n ) Ω, h = g f. Then h = g (f) f x i x i ( ) 2 2 h f = g (f) +g (f) 2 f x 2 i x i x 2 i n 2 h h = i= x 2 i = g (f) n ( ) 2 f +g (f) i= x i = g (f) f 2 +g (f) f. n i= 2 f x 2 i 2 Convex functions Theorem 2.. Let Ω R n be an open convex set, f C 2 (Ω). Then f is convex in Ω 2 f 0, where 2 f 0 means the Hessian matrix of f(x) is positive-semidefinite. address: wywshtj@gmail.com; Tel:

15 Proposition 2.. Prove that f(x) = x a, a 2, x R n is convex. Proof. By Theorem 2., in order to prove f(x) is convex, we just need to show that 2 f 0. Besides, f C 2 (R n ), so 2 f is symmetric. Let x = (x,x 2,,x n ) t R n then ξ R n, we have ( n a/2 f(x) = x a = xi) 2, i= ( n f = a a/2 x 2 x i 2 i) (2x i ) = ax i x a 2. i= 2 f = aδ ij x a 2 +a(a 2)x i x j x a 4. x j x i 2 f = a x a 4 ( x 2 I n +(a 2)xx t ). < 2 f(x)ξ,ξ > = ξ t 2 fξ = a x a 4 ξ t ( x 2 I n +(a 2)xx t )ξ = a x a 4 ( x 2 ξ t ξ +(a 2)ξ t xx t ξ) = a x a 4 ( x 2 ξ 2 +(a 2) < ξ,x > 2 ) 0, if a 2. Now we know that 2 f 0, which implies that f(x) is convex. 2

16 MA 523: Homework 6 Yingwei Wang Department of Mathematics, Purdue University, West Lafayette, IN, USA Poisson equation Lemma.. Let Ω = B(0,R) R n be an open ball, ϕ C(Ω) be spherically symmetric. Consider the Dirichlet problem f = ϕ, x Ω, f S(0,R) = C, where C is a constant. Let ϕ(x) = Φ( x ) = Φ(t). Then the solution to this problem is f(x) = F( x ) = C Question: Solve the problem R x t t n s n Φ(s)dsdt. (.) f = x a, x B(0,R), f S(0,R) = C, where a > 0. Further, check that if k < a < k +, then f C k+2 (R n ). address: wywshtj@gmail.com; Tel:

17 Solution: By the formula (.), we know that the solution to this problem is f(x) = C = C R x R x = C n+a t t n s n s a dsdt, 0 t n n+a tn+a dt, R x t +a dt, = C Ra+2 x a+2 (n+a)(a+2). Since x a+2 C k+2 (R n ) if k < a < k +, we know that f C k+2 (R n ). 2 Laplacian equation Question: If f solves the problem { f = 0, x Q, f Q = ϕ, (2.) where Q is a quadrilateral, ϕ = on one of Q and ϕ = 0 on other sides of Q. Find f(0). Solution: Without of generality, we can assume that Q = (,) (,) and ϕ = on the top of Q and ϕ = 0 on other sides of Q. Then the problem (2.) becomes f(x,y) = 0, f(x,) =, f(x, ) = f(,y) = f(,y) = 0. (x,y) (,) (,), If we do an orthogonal rotation on f, we can can f = f T, where ( ) sin(π/4) cos(π/4) T =. cos(π/4) sin(π/4) (2.2) Then f solves this problem f (x,y) = 0, (x,y) (,) (,), f (,y) =, f (x, ) = f (x,) = f (,y) = 0. (2.3) 2

18 Besides, f (0,0) = f(0,0). Similarly, f 2 = f T solves this problem f 2 (x,y) = 0, f 2 (x, ) =, f 2 (x,) = f 2 (,y) = f 2 (,y) = 0. Besides, f 2 (0,0) = f (0,0) = f(0,0). Also, f 3 = f 2 T solves this problem (x,y) (,) (,), f 3 (x,y) = 0, f 3 (,y) =, f 3 (x, ) = f 3 (x,) = f 3 (,y) = 0. (x,y) (,) (,), Besides, f 3 (0,0) = f 2 (0,0) = f (0,0) = f(0,0). Consider the function f = f +f +f 2 +f 3, which solves this problem { f(x,y) = 0, (x,y) (,) (,), f(x, ) = f(x,) = f(,y) = f(,y) =. (2.4) (2.5) (2.6) On one hand, it is easy to know that the solution to the problem (2.6)is that It implies that f(0,0) =. On the other hand, f(x,y) =, (x,y) (,) (,). (2.7) f(0,0) = f(0,0)+f (0,0)+f 2 (0,0)+f 3 (0,0), f(0,0) = 4 f(0,0) = 4. Finally, the numerical solution to problem (2.2) is shown in Figs

19 Numerical Solution to Poisson Equation, h= u(x,y) y x Figure : Numerical solution to problem (2.2) (overall) Numerical Solution to Poisson Equation at x= u(0,y) y Figure 2: Numerical solution to problem (2.2) (at a surface) 4

20 MA 523: Homework 7 Yingwei Wang Department of Mathematics, Purdue University, West Lafayette, IN, USA Bump function Question: Let Prove that F(t) C (R). F(t) = { exp{ t 2 }, t 0, t >. (.) Solution: The only possible discontinuities of F(x) is at t = ±. Let us focus on t =. First, let g(x) = e x and h(x) =, then F = g h. x 2 Second, we know that as x, the rate of e x is faster than any polynomial. Now we can get Similarly, for any n N, F (t) = g (h(t))h (t), lim (t) = e /( t2 ) 2x t F x = 0. 2 lim (t) = 0. t F(n) It follows that F (n) is continuous at t =. Similarly, F (n) is also continuous at t =. Then F(t) C (R). address: wywshtj@gmail.com; Tel:

21 MA 523: Homework 8 Yingwei Wang Department of Mathematics, Purdue University, West Lafayette, IN, USA Fundamental solution Definition. (Fundamental solution I). Suppose E(x) C (R n \{0}). Say E(x) is a fundamental solution of if. compact set K R n, E(x) dx < ; K 2. ϕ C 0 (Rn ), R n E(x) ϕ(x)dx = ϕ(0). Lemma.. Let { (n 2)σ n, n 3, x E(x) = n 2 log x, n = 2. (.) 2π Then E(x) is fundamental solution of. Proof. The case of n = 3 is done in class. Just need to prove the case of n = 2.. Claim that compact set K R n, E(x) dx <. K If K {0} =, since max x K E(x) <, it is obvious that E(x) dx <. K If K = B(0,R), then B(0,R) E(x) dx = R 0 rlogrdr <. 2. Claim that ϕ C 0 (Rn ), R n E(x) ϕ(x)dx = ϕ(0). Let K B(0,R) and B(0,ε) K, then just need to show E(x) ϕ(x)dx = lim E(x) ϕ(x)dx = ϕ(0). (.2) B(0,R) ε 0 ε< x <R address: wywshtj@gmail.com; Tel:

22 Let Ω ε = B(0,R)\ B(0,ε). Recall that since E = 0. Now we can get Ω ε E(x) ϕ(x)dx = div(e ϕ ϕ E) = E ϕ ϕ E = E ϕ, (.3) = = div(e ϕ ϕ E)dx Ω ε [E < ϕ,ν > ϕ < E,ν >]dσ Ω ( ε ) [E < ϕ,ν > ϕ < E,ν >]dσ. B(0,R) B(0,ε) Since ϕ 0 on B(0,R), then [E < ϕ,ν > ϕ < E,ν >]dσ = 0. B(0,R) Now we just need to compute [E < ϕ,ν > ϕ < E,ν >]dσ. B(0,ε) Choose E(x) = log x, then 2π E(x) = 2π x x B(0,ε)= x 2 2πε 2, < E,ν > B(0,ε) = 2π x/ε2,x/ε = 2πε, ϕ < E,ν > dσ = ϕ(x)dσ ϕ(0), as ε 0. (.4) 2πε S(0,ε) S(0,ε) Besides, since E(x) B(0,ε) = logε, we can get 2π [E < ϕ,ν > dσ, S(0,ε) = 2π logε < ϕ,ν > dσ, S(0,ε) = 2π logε ϕ(x)dx, B(0,ε) ( ) max 2 ϕ(x) ε 2 logε 0, as ε 0. (.5) B(0,) By Eqs.(.4)-(.5), we can get Eq.(.2). 2

23 2 Fundamental solution Definition 2. (Fundamental solution II). Say E x is a fundamental solution of with singularity at x R n if. E x C (R n \{x}); 2. compact set K R n, K E x dy < ; 3. ϕ C 0 (Rn ), R n E x ϕdy = ϕ(x). Lemma 2.. Let { (n 2)σ E x (y) = E(y x) n, n 3, y x n 2 log y x, n = 2. (2.) 2π Then E x (y) is a fundamental solution of with singularity at x R n. Proof. It is easy to know that E x C (R n \{x}) and compact set K R n, E K x dy <. Let z = y x, then E x (y) = Ẽ(z) is a fundamental solution of with singularity at z = 0. Then we have E(y x) ϕ(y)dy, R n = Ẽ(z) ϕ(z +x)dz, R n = ϕ(z +x) z=0 = ϕ(x). Now we are done. 3 Represent functions using fundamental solutions Lemma 3.. Let Ω R n be a bounded, open, piecewise C domain. Suppose f C 2 ( Ω), then x Ω, we have E x (y) f(y)dy = f(x)+ [f E x,ν E x f,ν ]dσ. (3.) Ω Theorem 3.. Prove that C(n) > 0 such that f C0 2(Rn ), f(y) f(x) C(n) y x dy, x n Rn, n 3. (3.2) R n 3 Ω

24 Proof. Let Ω be the support of f(x). Since f C0 (Rn ), then R > 0 such that Ω B(0,R). By Lemma 3., we know that for f C0(R 2 n ), we know that f(x) = E x (y) f(y)dy, (3.3) B(0,R) since f 0, f 0 on B(0,R). First, claim that inequality (3.2) is true if x = 0. By Eq.(3.3), we can get f(0) = E(y) f(y)dy, B(0,R) = (n 2)σ n y 2 n f(y)dy, B(0,R) = (n 2)σ n f(y), ( y 2 n ) dy, B(0,R) 2 f(y) (n 2)σ n B(0,R) y n dy, f(y) = C(n) dy. R y n n Second, for x R n, we define f(t) = f(x t), then Now we are done. f(x) = f(0), f(y) C(n) dy, R y n n f(x y) = C(n) dy, R y n n f(z) = C(n) dz. z x n R n 4

25 MA 523: Homework 9 Yingwei Wang Department of Mathematics, Purdue University, West Lafayette, IN, USA Poisson kernel Let P(ξ,y) be Poisson kernel P(ξ,y) = ξ 2 σ n ξ y n, (.) where ξ B(0,),y B(0,). Let ϕ C( B(0,)), and define f(ξ) be f(ξ) = P(ξ, y)ϕ(y)dσ(y). (.2) Prove that f C(B(0,)). B(0,) Proof. First, it is easy to know that for fixed ξ B(0,), the function P(ξ,y) defined by Eq.(.) is continuous and bounded on the compact set y B(0,). Second, for fixed y B(0,), the function P(ξ,y) defined by Eq.(.) is continuous on ξ B(0,). Third, for any fixed ξ 0 B(0,), we can choose a sequence {ξ n } n= B(0,) such address: wywshtj@gmail.com; Tel:

26 that lim n ξ n = x 0. By Lebesgue s dominated convergence theorem, we can get = lim = = lim f(ξ n) n n B(0,) B(0,) B(0,) = f(ξ). P(ξ n,y)ϕ(y)dσ(y), lim P(ξ n,y)ϕ(y)dσ(y), n P(ξ, y)ϕ(y)dσ(y), It follows that the function defined by Eq.(.2) is C(B(0,)). 2

27 MA 523: Homework 0 Yingwei Wang Department of Mathematics, Purdue University, West Lafayette, IN, USA Heat kernel Definition. (Heat kernel). Define the function G(x,t) = (4πt) n/2 exp Lemma.. The heat kernel defined by (.) satisfies ) ( x 2. (.) 4t x G G t = 0, (.2) G(x,t)dx =, t (0,+ ). R n (.3) Lemma.2. The Fourier transform of Heat kernel defined by (.) is Ĝ(ξ,t) = e 4π2 t ξ 2. (.4) Definition.2 (Temperaturetransform). For f C 0 (R n ), define the temperature transform as P t f(x) = G(x y,t)f(y)dy. (.5) R n Theorem.. Prove that P t (P s f) = P t+s f. (.6) Proof. First, we claim that G(x y,t+s) = G(x z,t)g(y z,s)dz, x,y R n, t,s > 0. (.7) R n address: wywshtj@gmail.com; Tel:

28 Thanks to Lemma.2, if we do the Fourier transform on both sides of (.7), we can get F x ξ (G(x y,t+s)) = e 2πi ξ,y e 4π2 (t+s) ξ 2, (.8) and = = ) F x ξ G(x z,t)g(y z,s)dz, (R n F x ξ (G(x z,t))g(y z,s)dz, R n e 2πi ξ,z G(y z,s)dz, 4π2t ξ 2 e R n = e 4π2 t ξ 2 F z ξ (G(z y,s)), = e 4π2 t ξ 2 e 4π2 s ξ 2 e 2πi ξ,y, = e 2πi ξ,y e 4π2 (t+s) ξ 2. (.9) By Eqs.(.8)-(.9), we know that ) F x ξ (G(x y,t+s)) = F x ξ G(x z,t)g(y z,s)dz, (R n which implies (.7). Second, from (.7) and the definition of P t, we can get the conclusion. On one hand, ( ) P t (P s f) = G(x y,t) G(y z,s)f(z)dz dy, R n R n = G(x y,t)g(y z,s)f(z)dzdy, R n R ( n ) = G(x y,t)g(y z,s)dy f(z)dz, R n R ( n ) = G(x z,t)g(y z,s)dz f(y)dy. (.0) R n R n On the other hand, P t+s f = G(x y,t+s)f(y)dy. (.) R n Since we have already known that Eq.(.7) is true, now we can get the conclusion (.6) according to Eqs.(.0)-(.). 2

29 MA 523: Homework Yingwei Wang Department of Mathematics, Purdue University, West Lafayette, IN, USA Mean value formula for temperature Definition. (Intersecting surface of heat ball). Ir s (z 0) = {x R n x x 0 2 < R r (t 0 s)}, (.) ( ) where R r (t 0 s) = 2n(t 0 s)log. r t 0 s Definition.2 (Heat kernel). Define the function G(x,t) = (4πt) n/2 exp ) ( x 2. (.2) 4t Definition.3 (Generalized Heat kernel). { ( ) G(z 0, ) = G(x 0 y,t s) = (4π(t 0 s)) n/2 exp x 0 y 2 4(t 0 s), t 0 > s, 0, t 0 s. (.3) Lemma.. The generalized heat kernel defined by (.3) satisfies Lemma.2. The heat kernel defined by (.2) satisfies H (G(z 0, )) = y G+G s = 0, (.4) G(z 0,s)dy =, s (0,t 0 ). R n (.5) lim (x,t) (x 0,0) where ϕ C(R n ) and bounded. address: wywshtj@gmail.com; Tel: R n G(x y,t)ϕ(y)dy = ϕ(x 0 ). (.6)

30 Theorem.. Suppose u(z) = u(x,t) C 2, (R n+ ), prove that u(z)g(z 0, )dσ = u(z 0 ). (.7) lim s t 0 I s r (z 0) Proof. We know that u(z)g(z 0, )dσ Ir s(z 0) = u(x,s)g(x 0 x,t 0 s)dx, x x 0 2 <R r(t 0 s) = [(u(x,s) u(x 0,t 0 ))+u(x 0,t 0 )]G(x 0 x,t 0 s)dx, = P +Q, x x 0 2 <R r(t 0 s) where P = Q = x x 0 2 <R r(t 0 s) x x 0 2 <R r(t 0 s) (u(x,s) u(x 0,t 0 ))G(x 0 x,t 0 s)dx, u(x 0,t 0 )G(x 0 x,t 0 s)dx, Let us consider P and Q respectively. On one hand, since u(z) = u(x,t) C 2, (R n+ ), we know that ε > 0, δ > 0 such that s (t 0 δ,t 0 ), u(x,s) u(x 0,t 0 ) < ε. By Lemma., we can get P lim s t 0 x x 0 2 <R r(t 0 s) P = 0. u(x,s) u(x 0,t 0 ) G(x 0 x,t 0 s)dx ε G < ε, R n On the other hand, let y = x x 0 2, then dx = t 0 s 2n (t 0 s) n/2 dy. We know that Q = u(x 0,t 0 )G(x 0 x,t 0 s)dx, x x 0 2 <R r(t 0 s) = u(x 0,t 0 )(π) n/2 ( y 2 < n 2 log r )e y 2 dy, t 0 s u(x 0,t 0 )(π) n/2 dy, as s t 0. It follows that lim s t 0 R n e y 2 By Eqs.(.8) - (.9), we can get the conclusion (.7). (.8) Q = u(x 0,t 0 ). (.9) 2

31 MA 523: Homework 2 Yingwei Wang Department of Mathematics, Purdue University, West Lafayette, IN, USA Euler s theorem for homogeneous functions Definition. (Homogeneous function). Let f : R n R. We say that f is homogeneous of degree k if for x R n, λ > 0, f(λx) = λ k f(x). Theorem. (Euler). Let f C (R n ). Then f is homogeneous of degree k if and only if x, f(x) = kf(x). (.) Proof. ): Suppose f is homogeneous of degree k. Fix x R n and define a function g : [0,+ ) R by g(λ) = f(λx) λ k f(x). On one hand, since f is homogeneous of degree k, we have On the other hand, by chain rule, we also have By Eqs.(.2)-(.3), we can get Let λ = in (.4), we can get (.). g(λ) = 0, λ > 0, g (λ) = 0, λ > 0. (.2) g (λ) = x, f(λx) kλ k f(x), λ > 0. (.3) x, f(λx) = kλ k f(x), λ > 0. (.4) address: wywshtj@gmail.com; Tel:

32 ) Suppose Eq.(.) is true. Let us just modify Eq.(.) to get λx, f(λx) = kf(λx). (.5) Fix x R n and again define a function g : [0,+ ) R by g(λ) = f(λx) λ k f(x). Take the derivative to λ, we can get g (λ) = x, f(λx) kλ k f(x), = λ λx, f(λx) kλ k f(x), = λ kf(λx) kλ k f(x), λg (λ) = kf(λx) kλ k f(x), = k(f(λx) λ k f(x)) = kg(λ), g (λ) k g(λ) = 0. λ Besides, it is obvious that g() = 0. Therefore, g(λ) satisfies this initial value problem { g (λ) k g(λ) = 0, λ (.6) g() = 0. It is easy to know that the solution to the Cauchy problem (.6) is g(λ) 0. It follows that f(x) is homogeneous of degree k. 2