partial differential equations (by Lawrance C. Evans)答案

【实例简介】
Graduate studies in Mathematics part one: representation for solutons part two: theory for partial differential equations part three: theory for nonlinear partial differential equations
Let∈>Obeg (1) P(e)=o(r)-/(t)dt gds B(0,r) Using integration by parts, we compute or(t)dt ∈nQ(n)n-T B(0,n) fdydt b(0,1) I I fds na(n)12-n √B(0.t) 2-nt lB(0,1) fas dt fd fd n(n-2)q(n) aB(0. 4) B(0,r) B(0,e fdy+d) b(0,r) Observe that fly≤C·∈2, for some constant C>0 B(0. an f(rdx dt B(0,∈) 0 o0,-2/aS As∈→0,+J→o6Wf(x)x.Thus lim or(t)dt n(n-2)a(n)JB(O, r)/ f(xdr.I fdy) B(0,r) (n-2c( B(0,n) Thcrcforc, letting∈→0, we have from(1) na(n)rn-1 gds )fdx aB(O, r) n(n-2)a(n)JRon 1x/m-2 r Problem 4. We say vE C(U) is subharmonic if △υ<0inU (a) prove for subharmonic v that v(x)≤ v dy for all B(x, r)cU. B(x, r) (b) prove that therefore maxo v= manau v. (c)Let :Rr-R be smooth and convex. Assume u is harmonic and v: (u). Prove v is subharmonic ( d)Prove v:=Dul 2 is subharmonic, whenever u is harmonic Solution (a)As in the proof of Theorem 2, set o(r): =aBcx. v dS()and obtain △w(y)dy≥0 For0<∈<r. (S)d=d()-(e)≥0. Hencc,db(r)≥limφ(∈)=v(x). Thercfore, v(z ds (z ds B(x, r) c(n BCr, r) c(nrh aB(xs na(n)s s)dS≥ ns"-v(r)ds=v(r) (b) We assume that U CR is open and bounded. For a moment, we assumc also that U is connected. Suppose that xo e U is such a point that v(xo)=M: maxi v. Then for 0<r< dist(xo, aU) M=1(x0)≤ v dv< m B(xo, r) Due to continuity of v, an equality holds only if v=M within B(xo, r). Therefore, the set u(M))0U=xE Ulu(x)=M is both open and relatively closed in U. by the connect edness of u. v is constant within the set u. hence it is constant within u and we conclude that maxo v= manau v Now let Uili e I be the connected components of U. Pick any x E U and findjE I such thatxEU. We obtain vx)≤maxv=maxw≤maxv and conclude that maxi v= maxau v )Forx=(x1,…,xn)∈ C and 1≤i,j≤n, LL xax -b(u(x)=d"(u(x) xdx ax (x)+φ(l(x) Ox: ax Since is convex, then "(x)20 for any x E R. Recall that u is harmonic and obtain △v=d"() ∑ O +△u=db"(u) dx ∑(m ( d)We set 1:=|D2=∑(m)Fox=(x,…,x)∈Uand1ij≤n 02y dxd 2∑ au ox;dx xdx rox Therefore, ∑ l ol L dxd k=1 k △v=2∑ L +∑ 0u0 u Xox (△)=2∑ l≤ik≤n k dx: ox 1≤ik≤n Problem 5: Prove that there exists a constant C, depending only on n, such that max ul s Cl max lgl +max lfl B(0,1) 0B(0,1 B(0,1) whenever u is a smooth solution of △L=J L OnB(0,1) Proof: Let M: = maxB(O 1) Ifl, then we define v(x)=u(x)+2nlxl and w(x)=-u(x)+2lx12.We first consider v(x). Note that △v=-△l-M=f-M≤0 So, v(x)is a subharmonic funcion From Problem 4(b), we have maxv(x)= max v(r)< max lg B(0,1) bB(0,1) OB(0,1) 2n That is maxl(x)≤max(x)≤max|8l+ max I B(0,1) B(0,1) B(0,1) 2nB(.1 Then, for w(x), we have -△w=△l-M=-f-M≤0 Again, we can get M max w(x)= max w(x)< max gl B0,1) 2n max-l(x)≤maxw(x)≤max|!|+xmax B(0,1) 2n B(O, 1) If 0B(0,1) Combining these two together, we finally proved the problem Problem 6. Use Poisson's formula for the ball to prove 厂十 l(0)≤l(x)≤ (0) (r+x) whenever u is positive and harmonic in B(0, r). This is an explicit form of Harnack's inequality Solution Since y∈aB(0,r), then x-y≤lx+r. Therefore, g() na(n)r JaB(o, rIx-yn 3(y) 8y) F dso na(n)r JoB(.r(r+Ixl) (r+Ixdi-I na(n )r-l g) y 0B(0,r) (r+|x|)n-1 g(S(y)=22r-1x B(0,r) r+|xD)-1 The inequality u(x)≤ (r-yi-ru(o) can be proven in a similar way Problem 7. Prove Poisson's formula for a ball: Assume g E C(aB(O, r) and let L(x)三 na(n) JaB(o.rlx-yl S()forx∈B(O,n Show that Proof Problem g Let u be the solution of u=o In u=8 n oRn given by Poisson's formula for the half-space. Assume g is bounded and g(x)=lxl for x E ORn Lx| le1. Show Du is not bounded near x=0.(Hint: Estimate u(en-i(o).) Proof: From formula(33 )on page 37, we have l(x)= gby na(n -y and u(0)=g(0)=o. Thus, using hint, we get (en)-l(0 nc(n C g(y) dy+ g0 na(n n)Jlvls1 O ORn Men-yln na(n)Jv>10oR! Ren-yrx Taking absolute value on both sides we have u(en)-u(o g() (y n(7)Jws1∩a|Aen=y n(m)J1een-乙 Since g is bounded, so it is obvious that I2 is bounded and independent of n. For I1, in this case, g(y)=lyl, So nQ(n)Js∩ aRn en-y nc(n lv|s∩OR4 (是+|y Note that for fixed y, Q+li Is increasing when d is decreasing to O, so by Monotone Convergence theorem we have 2 lim o na(n l1∩O4 (+|y ds(y=c 0 0Bn-1(0,r) 0 So Du is unbounded nearx=0 Problem 1o Suppose u is smooth nd solves u, -Au=o"X(O, co) (i) Show ua(r, t): =u(x, 't)also solves the heat equation for each E R (1 Use(i)to show v(r, 1): =r. Du(x, 0)+ 2lu (x, t) solves the heat equation as well (i)ua(x,t)=22u,ax, 2t)and uix (x, t)=u(x, 1D)for each i. Then uxxx(x, t)=22uxax,2t) Consequently,△n=2△ u and u-△ax=2(a-△), so ua solves the heat equation for all∈R. (ii)We differentiate u(x, t)=u(xI,., xn, t)with respect to a we get xM42(x1,…,x,21)+2.(x1,,xn,21)=x:D(lx,21)+2tu1(x,) Taking =1, we then have that v(x, t)=x. Du(x, t)+2tu, (r, t). u is smooth, so the second derivatives of u(x, nt)are continuous, meaning the mixed partials are equal. Therefore v-△v atad x,2)-△a14(x,21)=m1(x,2)-a△(x,2)=a(x-△m)=0, since ua satisfies the heat equation for all a. Thus v does as well Problem 11: Assume n= l and u(,t)=v() a) Show XX if and only if 4xy”(x)+(2+x)v(x)=0(x>0) b) Show that the general solution of (1)is v(O=C dstd 0 c)Differentiate v()with respect to x and select the constant c properly, so as to obtain the funda- mental solution g for n= 1 a)assume that u, = urr. Then 21 +4 So u,= uxx implies that 2V +4x 4 0 If we let z=t, we get (z)+=+vy(x)=0 Multiplying this equation by t gives the desired equality For the other direction, reverse the steps, and hence our proof is done 4xy"+(2+ (by integrating log(v)=-log vz Cz dstd as is desired /4。-1/2 S+十 0 4。-1/2 dstd 0 X r2 C—e C Now we want to integrate over R and set the integral equal to 1. Thus we get 2c Vt etting y v4r, we get dy=(4t-1i2dx and substituting, we get Employing the identity dy =vr and solving for c, we 4VIT Thus d(x,): 2 vrt 1s easily shown to solve the equation Problem 12. Write down an explicit formula for a solution of l1-△a+cut=finR"x(O,∞) u=g on R'x[t=01, where c∈R. 10 Solution: Set v(x, t)=u(x, t)eCt. Then, v,=ueCt + CeCiu and Vx.x: =uxxeCr △ uei t ce △ (u1-△u+Cu) f So. v is a solution of v-△v=ef on rxt=o) By(17)(p51) v(, t)= (x-y, tg(y) o(x-y, t-s)ef(, s)dyds where g is the fundamental solution of the hear equation. Since v(r, t)=u(r, tec, we have x Φ(x-y,t)g(y)dy+ (x-y, t-s)ecf(y, s)dyds Problem 13: Given8: [0, 0]R, with g(0)=0, derive the formula A- u(x (sds,x>o V47J(t-5)32 for a solution of the initial/ boundary-value problem =0intl+×(0,∞) l=0On+×{t=O}, On{x=0}×[0,∞). Proof. we define ∫a(x-g( D=1-(-x.)+g(0 So. we have l(x,1)-8(1) >0 v (,t)= 1-(-x,)+8()x≤0 uxx(x, t) 1)= X 0. 【实例截图】
【核心代码】