Solutions To Exercises From Chapter 2 Of Lawrence C. Evans .

Solutions to exercises from Chapter 2 ofLawrence C. Evans’ book ‘Partial DifferentialEquations’Sümeyye YilmazBergische Universität WuppertalWuppertal, Germany, 42119February 21, 20161Write down an explicit formula for a function u solving the initialvalue problem)ut b · Du cu 0 in Rn (0, )u g on Rn {t 0}Solution. We use the method of characteristics; consider a solution to thePDE along the direction of the vector (b, 1): z(s) u(x bs, t s). We haveż(s) ut (x bs, t s) b·Du(x bs, t s) cu(x bs, t s) cz(s), thusthe PDE reduces to an ODE. The characteristic curves can be parametrizedby (x0 bs, t0 s); and all the characteristic curves are parallel to each other.Given any (x0 , t0 ) in Rn (0, ), we have u(x0 , t0 ) u(x0 bt0 , 0)e ct0 g(x0 bt0 )e ct0 ; and this is an explicit formula for the solutions to the PDE.1

2Prove that Laplace’s equation u 0 is rotation invariant; that is,if O is an orthogonal n n matrix and we definev(x) : u(Ox)(x Rn ),then v 0.P roof. By chain rule we haveDxi v(x) Σnk 1 Dxk u(Ox)oik ,thenDxi xj v(x) Σnl 1 Σnk 1 Dxk xl u(Ox)oik ojl .Since O is orthogonal, OOT I, that is,(1 if k loik oil 0 if k 6 l.Thus v Σni 1 Σnk 1 Σnl 1 Dxk xl u(Ox)oik oil u 0.3Modify the proof of the mean value formulas to show for n 3 thatˆ111( n 2 n 2 )f dx,u(0) gdS n(n 2)α(n) B(0,r) x r B(0,r)provided u f in B 0 (0, r)u g on B(0, r))P roof. Defineφ(s) : u(y)dS(y) B(x,s)u(x sz)dS(z) B(0,1)2

we shall have0Du(x sz) · zdS(z) φ (s) B(0,1) B(x,s)D(y) B(x,s)s udS(y) νn 2 u(y)dy B(x,s)sny zdS(y)s u(y)dyB(x,s)We have φ(r) φ( ) ˆ rˆ r s0φ (s)ds u(y)dy dsn B(0,s) ˆ r ˆ r ˆ s1 f (y)dy ds f(y)dydsn B(0,s)nα(n)sn 1 B(0,s) ˆ rˆ 1 ˆ r 11 f(y)dyf(y)dyds n(2 n)α(n) sn 2 B(0,s)sn 2 B(0,s) ˆˆ 111 f (y)dy n 2f (y)dy n 2n(n 2)α(n)r B(0,r)B(0, )ˆˆ r 1f (y)dy ds n 2s B(0,s) Now notice that1 n 2andˆ0r1sn 2ˆf (y)dy C 2 ;B(0, )ˆ rˆ f (y)f (y)dy ds dydsn 20 B(0,s) s B(0,s)ˆf (x) dxn 2B(0,r) x ˆAs 0 we haveˆˆ rˆˆ 11f (x)f (y)dy f (y)dy ds dxn 2n 2n 2 sB(0, ) B(0,s)B(0,r) x and φ( ) u(0). We have thus demonstrated3

1u(0) gdS n(n 2)α(n) B(0,r)ˆ(B(0,r)11 n 2 )f dx.n 2 x r4Give a direct proof that if u C 2 (U ) C ( Ū ) is harmonic within abounded open set U , then maxŪ u max U u.P roof. Define u : u x 2 . Suppose u is achieveing maximum in Ūat an interior point x0 , then D(u (x0 )) 0 and H Dij (u (x0 )) is negativedefinite. Yet, (u ) 2 0, a contradiction, as the Laplacian is the traceof the Hessian. Letting 0, we can conclude that u cannot attain aninterior maximum.5We say v C 2 (Ū ) is subharmonic if v 0 in U.(a) Prove for subharmonicvdy for all B(x, r) U.v(x) B(x,r)(b) Prove that therefore maxŪ v max U v.(c) Let φ : R 7 R be smooth and convex. Assume u is harmonicand v : φ(u). Prove v is subharmonic.(d) Prove v : Du 2 is subharmonic, whenever u is harmonic.Solution. (a) We setf (r) : u(y)dS(y) B(x,r)u(x rz)dS(z). B(0,1)4

Taking derivative we have0Du(x rz) · zdS(z)f (r) B(0,1)and consequently using Green’s formula we havey x0Du(y) ·dS(z)f (r) r B(0,1) u dS(y) B(x,r) νr u(y)dy 0 n B(x,r)This means that f (r) is non-decreasing, therefore given r 0u(x) limf (t) limt 0t 0 u(y)dS(y) B(x,t)u(y)dS(y) B(x,r)And this impliesˆrˆn u(y)dS(y) dsα(n)r u(x) 0 B(x,s)u(x) u(y)dyB(x,r)0(b) Assume the subharmonic function v attainsffl maximum at x0 U , forany ball B(x0 , r) U0 we have by (a) v(x0 ) B(x0 ,r) v(y)dy. Yet, v(x0 ) fflv(y) for any y B(x0 , r), thus v(x0 ) B(x0 ,r) v(y)dy, and v(x0 ) v(y) forany y B(x0 , r). We can choose r so that Ū B(x0 , r) {x1 }. We havethus shown that maxv maxv.x Ūx U00(c) As φ is convex and smooth, φ 0; and u 0. HencenX000 φ(u) (φ · (uxi )2 φ · uxi xi ) 0,i 1we can conclude that φ(u) is subharmonic.(d) If u is harmonic then u 0, thus uxi ( u)xi 0, hence Du isharmonic. Since Du 2 is convex with respect to Du, the result follows from(c).5

6Let U be a bounded open subset of Rn . Prove that there exists aconstant C, depending only on U, such thatmax u C(max g max f ) UŪŪwhere u is a smooth solution of( u f in Uu g on U.(Hint: (u x 2λ)2n 0, for λ : maxŪ f .)Solution. We consider the function v(x) : u(x) maxŪ f . Indeed v is a subharmonic function since v u x 2λ,2nwhere λ x 2 λ (f max f ) 0.2nŪThe weak maximum principle holds for subharmonic functionsmax u max v max vŪŪ2 max u U U2 x x λ max u maxλ max g C max f U U 2n U2nŪwhere C is some constant depending on U.Replacing u by u, we can conclude that there exists some constant Cdepending on U, such that max u C max g max f .Ū UŪ7Use Poisson’s formula for the ball to provern 2r x r x u(0) u(x) rn 2u(0)n 1(r x )(r x )n 16

whenever u is positive and harmonic in B 0 (0, r). This is an explicitform of Harnack’s inequality.Solution. We recall Poisson’s formula for a ballˆg(y)r2 x 2dS(y)u(x) nω(n)r B(0,r) x y ng(y) rn 2 (r2 x 2 )dS(y),n B(0,r) x y therefore,u(0)u(0)n 2 22 u(x) r(r x ),(r x )n(r x )nr x r x n 2rn 2u(0) u(x) ru(0).(r x )n 1(r x )n 1rn 2 (r2 x 2 )8Prove Theorem 15 in section 2.2.4. (Hint: Sinceu 1 solves (44) for g 1, the theory automatically implies B(0,1) K(x, y)dS(y) 1for each x B 0 (0, 1).)Let’s recall theorem 15 in section 2.2.4. (Poisson’s formula for a ball)Assume g C( B(0, r)) and define u byˆg(y)r2 x 2dS.u(x) nα(n)r B(0,r) x y nThen(i) u C (B 0 (0, r))(ii) u 0 in B(0, r), and(iii) limx x0 u(x) g(x0 ) for each point x0 B(0, r).x B 0 (0,r)P roof. (1) For each fixed x, the mapping y 7 G(x, y) is harmonic, exceptfor x y. As G(x, y) G(y, x), x 7 G(x, y) is harmonic, except for x y. G(x, y) K(x, y) for x, y B 0 (0, r).Thus x 7 yn7

(2) Note that we haver2 x 21 nα(n)rˆ B(0,r)1dS x y nfor x B 0 (0, r). And as g is bounded, u is likewise bounded. Since x 7 K(x, y) is smooth, for x 6 y, we can verify that u C (B 0 (0, r)) withˆ u(x) x K(x, y)g(y)dS 0 B(0,r).(3) Now fix x0 B(0, r), 0. Choose δ 0 such that g(y) g(x0 ) if y x0 δ, y B(0, r). Then if x x0 2δ , x B 0 (0, r),ˆ0 u(x) g(x ) K(x, y)[g(x) g(x0 )]dy B(0,r)ˆ K(x, y)[g(x) g(x0 )]dy B(0,r) B(x0 ,δ)ˆK(x, y)[g(x) g(x0 )]dy : I J B(0,r)/B(x0 ,δ) We have I B(0,r) K(x, y)dy . Furthermore if x x0 2δ and y x0 δ, we have y x0 y x 2δ y x 12 y x0 . Hence1 y x y x0 .2ThusˆJ 2kgkL K(x, y)dy B(0,r)/B(x0 ,δ)2n 1 (r2 x 2 )kgkL nα(n)rˆ y x0 n dy 0 B(0,r)/B(x0 ,δ) as x x0 .Combining this two estimates we can conclude that limx x0 u(x) g(x0 )x B 0 (0,r)0for each point x B(0, r).8

9( u 0 in Rn given by Poisson’s formulau g on Rn for the half-space. Assume g is bounded and g(x) x for x Rn , x 1. Show Du is not bounded near x 0. (Hint: Estimateu(λen ) u(0).)λLet u be the solution ofSolution. We recall Poisson’s formula for the half-spaceˆ2xng(y)dS(y).u(x) nα(n) Rn x y nNotice that u(0) g(0) 0; following the hint we estimate u(λenλ) u(0) :ˆ 1 2en y pu(λen ) u(0) n dS(y).λnα(n) Rn (λ2 y 2 )We consider the integral2ennα(n)ˆ Rn { y 1} y p n dS(y)(λ2 y 2 )ˆ1 (n 1)α(n 1)0ˆ1 (n 1)α(n 1)0r n rn 2 drλ2 r 21r (λ/r)2 1 n/2 drthus the integral diverges. Hence lim u(λenλ) u(0) cannot be bounded, whichλ 0implies Du · en is unbounded at 0. Therefore Du is unbounded at 0.10(Reflection principle)(a) Let U denote the open half-ball {x Rn x 1, xn 0}.Assume u C 2 (U ) is harmonic in U , with u 0 on U xn 0.Set(u(x) if xn 0v(x) : u(x1 , ., xn 1 , xn ) if xn 09

for x U B 0 (0, 1). Prove v C 2 (U ) and thus v is harmonic withinU.(b) Now assume only that u C 2 (U ) C(U ). Show that v isharmonic within U. (Hint: Use Poisson’s formula for the ball.)Solution. (a) Apparently v(x) is of class C 2 in each of U and U : B(0, 1)/U . We can check that for i 6 nlim xi xn v(x1 , x2 , ., xn ) xi xn v(x1 , x2 , ., 0)xn 0 xi xn u(x1 , x2 , ., 0) lim xi xn v(x1 , x2 , ., xn );xn 0while xn xn v(x1 , x2 , ., 0) 0, and all other derivatives which do not involvexn vanishes. So we can assure that D2 v exists and is continuous in U .Since v(x) u(x) in U and v(x) u(x1 , x2 , ., xn ) if xn 0. Andwe have already known that u is harmonic in U . Given any x with xn 0,we have v(x) satisfies a local version of mean value theoremv(x) v(y)dy B(x,δ)for small enough δ 0. The local mean value property is equivalent toharmonicity. Finally, if xn 0, we also havev(y)dy 0v(y)dy v(x) B (x,δ) B (x,δ)since v(x) v(x1 , x2 , ., xn ). We therefore can conclude that v(x) is aharmonic function in U .(b) Now assume that u is merely continuous on the boundary, we wouldlike to find a harmonic function w such that w v on B(0, 1) by thePoisson’s formula for the ballˆr2 x 2v(y)w(x) dS(y),nα(n)r B(0,1) x y nwe shall see that w(x) 0 on xn 0. Therefore, w u on U and by themaximum principle we have u w 0 throughout U . So w v in U andwe can similarly show w v in U . Therefore, v is harmonic within U.10

11(Kelvin transform for Laplace’s equation) The Kelvin transf ormKu ū of a function u : Rn R isu(x) : u(x̄) x̄ n 2 u(x/ x 2 ) x 2 n , (x 6 0),where x̄ x/ x 2 . Show that if u is harmonic, then so is ū. (Hint:First show that Dx x̄(Dx x̄)T x 4 I. The mapping x x̄ is conf ormal,meaning angle preserving.)P roof. We denote ψ(x) x̄, and computeδij2xi xj jψ (x) 2 , xi x x 4where δij is Kronecker’s delta symbol. We write Dx x̄ in coordinate freeexpression:xxTDx x̄ x 2 (I 2 2 ), x thus Dx x̄(Dx x̄)T x 4 (I 4 x 2 xxT 4 x 4 xxT xxT ) x 4 I.We now calculate ψ; since careful computations shall show thatψxj i xi 2 x 4 (xj 2δij xi ) 8 x 6 x2i xj .Thus, ψ n(2 n) x x4 .Hence T u(x/ x 2 ) x 2 n x 2 n u(x̄) 2DuDx̄ D x 2 n x 2 n Du · x̄ x 2 n Tr (Dx̄)T D2 uDx̄ .Note that the firstis 0 for x 6 0; and the last term is also 0 as termT 2 4Tr (Dx̄) D uDx̄ x u 0. While we havexxT ) (2 n) x n x x 2 2(n 2) x 2 n Du · x x 2 n Du · x̄.2DuDx̄ D x 2 n T 2Du x 2 (I 2Therefore the Kelvin transform preserves harmonic functions.11

12Suppose u is smooth and solves ut u 0 in Rn (0, ).(a) Show uλ (x, t) : u(λx, λ2 t) also solves the heat equation foreach λ R.(b) Use (a) to show v(x, t) : x · Du(x, t) 2tut (x, t) solves the heatequation as well.P roof. (a) We compute to show that the dilation of u also satisfies theheat equation, since uλ (x, t) u(λx, λ2 t) λ2 ut t tand 2uλ (x, t) λ2 uxi xi . x2iWe see that uλ solves the heat equation.(b) Differentiating uλ with respect to λ we shall have uλ x · Du(λx, λ2 t) 2tλut (λx, λ2 t). λNow let λ 1 we then see that uλ x · Du(x, t) 2tut (x, t) v(x, t). λSince u is smooth, the mixed partials are equal under exchanges of orders ofdifferentiation. And uλ solves the heat equation, that is(uλ )t (uλ ) 0.We differentiate both sides with respect to λ and conclude that v(x, t) is alsoa solution to the heat equation.13Assume n 1 and u(x, t) v( xt ).000(a) Show ut uxx if and only if v z2 v 0. Show that the generalsolution is12

ˆze sv(z) c2 /4ds d.0 v( xt )(b) Differentiate u(x, t) with respect to x and select theconstant c properly to obtain the fundamental solution Φ for n 1.Explain why this procedure produces the fundamental solution.(Hint: What is the initial condition for u?)3000Solution. (a) Differentiate v we have ut 12 xt 2 v and uxx 1t v . By the000xv 0. Vice versa, if v satisfies the ODE thenheat equation we get v 2 tz20 0u solves the heat equation. We solve the ODE, notice that (e 4 v(z) ) 0,then the solution shall beˆ z2e s /4 ds d.v(z) c0(b) The initial condition is the δ distribution. Differentiate u(x, t) v(z)x2with respect to x we have ux (x, t) ct e 4t , which should also be a solutionto the heat equation. Since the integral of δ function is 1,ˆ s2e 4 ds 2c π,1 c we must have c for n 1.1 ,2 π2thus Φ(x, t) x 1 e 4t4πtis the fundamental solution14Write down an explicit formula for a solution of(ut u cu f in Rn (0, )u g on Rn {t 0}where c R.Solution. First we consider the homogeneous equation with initial boundary data(vt v cv 0 in Rn (0, )v h on Rn {t 0}13