Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 3.2
- If F(s) is a C2-function of the one-variable s, find a condition on the vector α=(α1,α2,α3) so that
u(x1,x2,x3,t)=F(α1x1+α2x2+α3x3−t)
is a solution of (26). (Such solutions are called plane waves and are constant on the planes α⋅x−t=constants.) - Find the relationship which must hold between the initial data g(x) and h(x) for a plane wave solution.
- Find all plane wave solutions of (26) with the initial condition u(x1,x2,x3,0)=x1−x2+1.
- If F(s) is a C2-function of the one-variable s, find a condition on the vector α=(α1,α2,α3) so that
- Plugging the form of solution into (26), we have
F″(α⋅x−t)=utt(x,t)=c2Δu(x,t)=c2(α21+α22+α23)F″(α⋅x−t).
Thus, the vector α must satisfy |α|=√α21+α22+α23=1/c. - The initial data g and h must satify
g(x)=F(α⋅x),h(x)=−F′(α⋅x).
By differentiating g with respect to x, we find that g and h satisfies the following relationship:∇g(x)=−∇xF(α⋅x)=−α⋅F′(α⋅x)=−α⋅h(x)for x∈R3,
which also means h(x)=−c2α⋅∇g(x) for x∈R3. - Let g(x)=g(x1,x2,x3)=x1−x2+1 for x=(x1,x2,x3)∈R3. Then by part (b), we get
h(x)=h(x1,x2,x3)=−c2α⋅∇g(x1,x2,x3)=−c2(α1,α2,α3)⋅(1,−1,0)=−c2(α1−α2)for (x1,x2,x3)∈R3,
which is a constant function. Now we employ the Kirchhoff's formula (37), we haveu(x,t)=14π∂∂t(t∫|ξ|=1g(x+ctξ)dSξ)+t4π∫|ξ|=1h(x+ctξ)dSξ=14π∫|ξ|=1g(x+ctξ)dSξ+t4π∫|ξ|=1cξ⋅∇g(x+ctξ)dSξ+t4π∫|ξ|=1h(x+ctξ)dSξ=14π∫|ξ|=1g(x+ctξ)dSξ+t4π∫|ξ|=1(1−cα⋅ξ)h(x+ctξ)dSξ=14π∫|ξ|=1[x1−x2+ct(ξ1−ξ2)+1]dSξ−c2(α1−α2)t4π∫|ξ|=1(1−cα1ξ1−cα2ξ2−cα3ξ3)dSξ==14π∫|ξ|=1(x1−x2+1)dSξ−c2(α1−α2)t4π∫|ξ|=11dSξ=x1−x2+1−c2(α1−α2)tfor x=(x1,x2,x3)∈R3,t≥0.
Here we have used the symmetry of integral, i.e., ∫|ξ|=1ξkdSξ=0 for k=1,2,3 and the surface area of unit ball in R3 is 4π. - Find the solution of the initial value problem
{utt=uxx+uyy+uzz,u(x,y,z,0)=x2+y2,ut(x,y,z,0)=0.
- by using (37);
- by using (39).
- Let g(x,y,z)=x2+y2 and h(x,y,z)=0 for (x,y,z)∈R3 and c=1. Using the Kirchhoff's formula (37), we have
u(x,y,z,t)=14π∂∂t(t∫|ξ|=1g(x+ctξ1,y+ctξ2,z+ctξ3)dSξ)+t4π∫|ξ|=1h(x+ctξ1,y+ctξ2,z+ctξ3)dSξ=14π∫|ξ|=1g(x+tξ1,y+tξ2,z+tξ3)dSξ+t4π∫|ξ|=1ξ⋅∇g(x+tξ1,y+tξ2,z+tξ3)dSξ=14π∫|ξ|=1[(x+tξ1)2+(y+tξ2)2]dSξ+t4π∫|ξ|=1(ξ1,ξ2,ξ3)⋅(2x+2tξ1,2y+2tξ2,0)dSξ=14π∫|ξ|=1(x2+y2)dSξ+2t4π(xξ1+yξ2)dSξ+t24π∫|ξ|=1(ξ21+ξ22)dSξ+2t4π∫|ξ|=1(xξ1+yξ2)dSξ+2t24π∫|ξ|=1(ξ21+ξ22)dSξ=x2+y2+3t24π∫|ξ|=1(ξ21+ξ22)dSξ=x2+y2+3t24π⋅23∫|ξ|=1(ξ21+ξ22+ξ23)dSξ=x2+y2+2t2for (x,y,z)∈R3,t≥0.
Here we have used the symmetry of integral, i.e., ∫|ξ|=1ξkdSξ=0 for k=1,2,3 and∫|ξ|=1ξ21dSξ=∫|ξ|=1ξ22dSξ=∫|ξ|=1ξ31dSξ=4π3.
In addition, the integral ∫|ξ|=1ξ⋅∇g(x+tξ)dSξ also can be computed by the Gauss' divergence theorem as follows.∫|ξ|=1ξ⋅∇g(x+tξ1,y+tξ2,z+tξ3)dSξ=∫|ξ|≤1Δξ[g(x+tξ1,y+tξ2,z+tξ3)]dSξ=t∫|ξ|≤1(Δg)(x+tξ1,y+tξ2,z+tξ3)dSξ=t∫|ξ|≤14dSξ=t⋅4⋅4π3=16πt3.
- Let g(x,y)=x2+y2 and h(x,y)=0 for (x,y)∈R2 and c=1. Then using (39), we have
u(x,y,t)=14π∂∂t(2t∫ξ21+ξ22<1g(x+ctξ1,y+ctξ2)√1−ξ21−ξ22dξ1dξ2)+t4π(2∫ξ21+ξ22<1h(x+ctξ1,y+ctξ2)√1−ξ21−ξ22dξ1dξ2)=24π∫ξ21+ξ22<1(x+tξ1)2+(y+tξ2)2√1−ξ21−ξ22dξ1dξ2+24π∫ξ21+ξ22<12tξ1(x+tξ1)+2tξ2(y+tξ2)√1−ξ21−ξ22dξ1dξ2=24π∫ξ21+ξ22(x+tξ1)2+(y+tξ2)2+2t[ξ1(x+tξ1)+ξ2(y+tξ2)]√1−ξ21−ξ22dξ1dξ2.
Using the polar coordinate, we set ξ1=rcosθ and ξ2=rsinθ for 0≤r<1 and 0≤θ<2π. Then the double integral can be transformed and we haveu(x,y,t)=24π∫2π0∫10x2+y2+4rt(xcosθ+ysinθ)+3t2r2√1−r2⋅rdrdθ=∫10r(x2+y2)+3t2r3√1−r2dr=(x2+y2)∫10r√1−r2dr+3t2∫10r3√1−r2dr=(x2+y2)⋅(−√1−r2)|r=1r=0+3t2⋅(−(2+r2)√1−r23)|r=1r=0=x2+y2+3t2⋅23=x2+y2+2t2for (x,y)∈R2,t≥0.
- Use Duhamel's principle to find the solution of the nonhomogeneous wave equation for three space dimensions utt−c2Δu=f(x,t) with initial conditions u(x,0)=0=ut(x,0). What regularity in f(x,t) is required for the solution u to be C2?
- Let Ω={(x,y)∈R2:0<x<a and 0<y<b}, and use separation of variables to solve the initial/boundary value problem
{utt=uxx+uyyfor (x,y)∈Ω and t>0,u(x,y,t)=0for (x,y)∈∂Ω and t>0,u(x,y,0)=sinπxasin2πyb,and ut(x,y,0)=0for (x,y)∈Ω.
- Find a formula for the solution v(x,t)=v(x1,x2,t) of the Cauchy problem for the two-dimensional Klein-Gordon equation:
{vtt=c2Δv−m2vfor x∈R2 and t>0,v(x,0)=g(x),vt(x,0)=h(x)for x∈R2.
- For n=3, suppose that g,h∈C∞0(R3) and consider the solution of (26), (27) given by (37). Show that there is a constant C so that |u(x,t)|≤C/t for all x∈R3 and t>0.
- Is a similar result true for n=2?
- Since g and h are in C∞0(R3), there exist two positive constants Rg and Rh such that supp(g)⊆ˉBRg(0) and supp(h)⊆ˉBRh(0). By the extreme value theorem, the continuous function must attain its maximum value in any given compact set, which means there exist two positive constants Mg and Mh such that |g(x)|≤Mg and |h(x)|≤Mh for x∈R3. Moreover, the support of |∇g| is a subset of one of g, and |∇g| is also a continuous function, which means there exists a positive constant M∇g such that |∇g(x)|≤M∇g for x∈R3. Then we use (37) and the change of variable y=x+ctξ (with dSξ=(ct)2dSy) to get
u(x,t)=14π∫|ξ|=1(g(x+ctξ)+th(x+ctξ))dSξ+t4π∫|ξ|=1cξ⋅∇g(x+ctξ)dSξ=14πc2t2∫∂Bct(x)(g(y)+th(y))dSy+t4πc2t2∫∂Bct(x)y−xt⋅∇g(y)dSy=14πc2t2∫∂Bct(x)(g(y)+th(y))dSy+14πc2t2∫∂Bct(x)(y−x)⋅∇g(y)dSy.
Now we can estimate |u(x,t)| as follows.|u(x,t)|≤14πc2t2∫∂Bct(x)(|g(y)|+t|h(y)|)dSy+14πc2t2∫∂Bct(x)|y−x||∇g(y)|dSy=14πc2t2∫∂Bct(x)∩BRg(0)|g(y)|dSy+14πc2t∫∂Bct(x)∩BRh(0)|h(y)|dSy+14πct∫∂Bct(x)∩BRg(0)|∇g(y)|dSy≤Mg4πc2t2H2(∂Bct(x)∩BRg(0))+Mh4πc2tH2(∂Bct(x)∩BRh(0))+M∇g4πctH2(∂Bct(x)∩BRg(0)),
where H2 denotes the two-dimensional Hausdorff measure in R3. Here we have used the fact that |y−x|=ct for y∈∂Bct(x).
Now we claim that H2(∂Bct(x)∩BR(0))≤4πR2 for R>0. When ct≤R, it is easy to know ∂Bct(x)∩BR(0)⊆∂Bct(x) andH2(∂Bct(x)∩BR(0))≤H2(∂Bct(x))=4π(ct)2≤4πR2.
When ct>0, the value of H2(∂Bct(x)∩BR(0)) can be attained as ∂Bct(x) passes the equatorial of BR(0). As the radius of ∂Bct(x) increases, the intersection ∂Bct(x)∩BR(0) would become more flat, which means H2(∂Bct(x)∩BR(0))≥πR2, the area of the circle formed by the equatorial of BR(0). On the other hand, as the radius of ∂Bct(x) decreases to R and ∂Bct(x) passes the equatorial of BR(0), the intersection ∂Bct(x)∩BR(0) may become ∂BR(0) and H2(∂Bct(x)∩BR(0))≤H2(∂BR(0))=4πR2. This completes the proof of Claim.
If t≥1, we can yse Claim to get|u(x,t)|≤MgR2gc2t2+MhR2hc2t+M∇gR2gct=1t(M∇gR2gc+MhR2hc2+MgR2gc2t)≤1t(M∇gR2gc+MhR2hc2+MgR2gc2):=C1t,
where C1=M∇gR2gc+MhR2hc2+MgR2gc2. On the other hand, if 0<t<1, we have t2<t<1 and|u(x,t)|≤Mg4πc2t2H2(∂Bct(x))+Mh4πc2tH2(∂Bct(x))+M∇g4πctH2(∂Bct(x))=Mg4πc2t2(4πc2t2)+Mh4πc2t(4πc2t2)+M∇g4πct(4πc2t2)<Mg4πc2t2(4πc2t)+Mh4πc2t(4πc2)+M∇g4πct(4πc2)=Mg+Mh+cM∇gt:=C2t,
where C2=Mg+Mh+cM∇g. Hence we choose C=max{C1,C2}, which is desired. - No. To find a counterexample, we consider the initial condition u(x,y,0)≡0 and ut(x,y,0)=h(x,y), where h satisfies h≡1 in B1(0), 0≤h≤1 in B2(0)−B1(0) and h≡0 in R2−B2(0). The existence of function h follows from Urysohn's lemma. Suppose by contradiction that there exists a positive constant C such that |u(x,y,t)|≤C/t for (x,y)∈R2. By (39), we have
u(x,y,t)=14π∂∂t(2t∫ξ21+ξ22<1g(x+ctξ1,y+ctξ2)√1−ξ21−ξ22dξ1dξ2)+t4π(2∫ξ21+ξ22<1h(x+ctξ1,y+ctξ2)√1−ξ21−ξ22dξ1dξ2)=t2π∫ξ21+ξ22<1h(x+ctξ1,y+ctξ2)√1−ξ21−ξ22dξ1dξ2=12πc2t∫Bct(x,y)h(η,μ)√1−(ct)−2((η−x)2+(μ−y)2)dηdμ.
It is clear that B1(0)⊆Bct+1(ct,0) for t≥0. Then we haveu(ct,0,t+c−1)=12πc(ct+1)∫Bct+1(ct,0)h(η,μ)√1−(ct+1)−2((η−ct)2+μ2)dηdμ≥12πc(ct+1)∫B1(0,0)h(η,μ)√1−(ct+1)−2((η−ct)2+μ2)dηdμ=12πc∫B1(0,0)1√(ct+1)2−(η−ct)2−μ2dηdμ=12πc∫2π0∫10r√1−r2+2ct(1+rcosθ)drdθ≥14πc∫2π0∫10r√1+4ctdrdθ=14√1+4ct.
Thus, by |u(x,y,t)|≤C/t for all (x,y)∈R2 and t>0, we get14√1+4ct≤Ctfor all t>0,
which is equivalent to t2−64c3t−16c2≤0 for all t>0. This leads a contradiction. Thereofre, it is impossible to find a positive constant C such that |u(x,y,t)|≤C/t for all (x,y)∈R2 and t>0. - Suppose n=2k+1 for a positive integer k and u satisfies (26).
- Show that Vx(r,t) given by (40) satisfies (32).
- Derive (42) from (35) and (41); note cn=1⋅3⋯(n−2).
- Firstly, we prove the following identity by mathematical induction:
d2dr2(1rddr)k−1(r2k−1ϕ(r))=(1rddr)k(r2kϕ′(r)),
where k∈N and ϕ∈Ck+1(R). Then we claim thatNow we can use (28), (40) and Claim to obtainProof of Claim
When k=1, it is easy to verify thatd2dr2(1rddr)k−1(r2k−1ϕ(r))=d2dr2(rϕ(r))=ddr(rϕ′(r)+ϕ(r))=rϕ″(r)+2ϕ′(r)=r2ϕ″(r)+2rϕ′(r)r=1rddr(r2ϕ′(r)).
This shows the identity holds true for k=1. Now we suppose the identity holds true for k=n, i.e.,d2dr2(1rddr)n−1(r2n−1ϕ(r))=(1rddr)n(r2nϕ′(r)).
Then we can observe thatd2dr2(1rddr)n(r2n+1ϕ(r))=d2dr2(1rddr)n−1[(2n+1)r2nϕ(r)+r2n+1ϕ′(r)r]=(2n+1)d2dr2(1rddr)n−1(r2n−1ϕ(r))+d2dr2(1rddr)n−1(r2n−1(rϕ′(r)))=(2n+1)(1rddr)n(r2nϕ′(r))+(1rddr)n(r2n(rϕ′(r))′)==(1rddr)n(r2n+1r2nϕ′(r)+r2n(rϕ″(r)+ϕ′(r)))=(1rddr)n1rddr(r2n+2ϕ′(r))=(1rddr)n+1(r2n+2ϕ′(r)).
Hence by the mathematical induction, the identity holds true for all n∈N.∂2Vx∂t2(r,t)=∂2∂t2(1r∂∂r)k−1(r2k−1Mu(x,r,t))=(1r∂∂r)k−1(r2k−1∂2Mu∂t2(x,r,t))(by (28))=(1r∂∂r)k−1[r2k−1⋅c2(∂2∂r2+n−1r∂∂r)Mu(x,r,t)](by (40))=c2(1r∂∂r)k−1⋅1r(rn−1∂2∂r2+(n−1)rn−2∂∂r)Mu(x,r,t)=c2(1r∂∂r)k−1⋅1r∂∂r(rn−1∂∂r)Mu=c2(1r∂∂r)k(r2k∂Mu∂r(x,r,t))=c2∂2∂r2(1r∂∂r)k−1(r2k−1Mu(x,r,t))(by Claim)=c2∂2Vx∂r2(x,r,t)(by (28)).
The proof is complete. - For any k∈N and ϕ∈Ck−1(R), we prove the following identity by the mathematical induction:
(1rddr)k−1(r2k−1ϕ(r))=k−1∑j=0βkjrj+1ϕ(j)(r),
where the constants βkj (j=0,…,k−1) are independent of ϕ.Then we claim that βk0=c2k+1=cn for k∈N.Proof of Claim 1
When k=1, it is easy that(1rddr)k−1(r2k−1ϕ(r))=rϕ(r)=β10⋅r0+1ϕ(0)(r),
where β10=1=c3. This shows the claim holds true for k=1. Now we suppose the claim holds true for k=m, i.e., there exists βmj (j=0,1,…,m−1) such that(1rddr)m−1(r2m−1ϕ(r))=m−1∑j=0βmjrj+1ϕ(j)(r),
where βm0=c2m+1. Then we can observe that(1rddr)m(r2m+1ϕ(r))=1rddr(1rddr)m−1(r2m−1(r2ϕ(r)))=1rddrm−1∑j=0βmjrj+1djdrj(r2ϕ(r))=1rm−1∑j=0βmj[(j+1)rjdjdrj(r2ϕ(r))+rj+1dj+1drj+1(r2ϕ(r))]=m−1∑j=0(j+1)βmjrj−1djdrj(r2ϕ(r))+m−1∑j=0βmjrjdj+1drj+1(r2ϕ(r))=m−1∑j=0(j+1)βmjrj−1djdrj(r2ϕ(r))+m∑j=1βmj−1rj−1djdrj(r2ϕ(r))=βm0rϕ(r)+m−1∑j=1[(j+1)βmj+βmj−1]rj−1djdrj(r2ϕ(r))+βmm−1rm−1dmdrm(r2ϕ(r)).
- For j=1, djdrj(r2ϕ(r))=2rϕ(r)+r2ϕ′(r);
- for j≥2, djdrj(r2ϕ(r))=r2ϕ(j)(r)+2jrϕ(j−1)(r)+j(j−1)ϕ(j−2)(r).
(1rddr)m(r2m+1ϕ(r))=βm0rϕ(r)+(2βm1+βm0)(2rϕ(r)+r2ϕ′(r))+m−1∑j=2[(j+1)βmj+βmj−1]rj−1[r2ϕ(j)+2jrϕ(j−1)(r)+j(j−1)ϕ(j−2)(r)]+βmm−1rm−1(r2ϕ(m)+2mrϕ(m−1)(r)+m(m−1)ϕ(m−2)(r))=(3βm0+4βm1)rϕ(r)+(2βm1+βm0)r2ϕ′(r)+m−1∑j=2[(j+1)βmj+βmj−1]rj+1ϕ(j)+m−2∑j=12[(j+2)(j+1)βmj+1+(j+1)βmj]rj+1ϕ(j)(r)+m−3∑j=0[(j+3)(j+2)(j+1)βmj+2+(j+2)(j+1)βmj+1]rj+1ϕ(j)(r)+m(m−1)βmm−1rm−1ϕ(m−2)(r)+2mβmm−1rmϕ(m−1)(r)+βmm−1rm+1ϕ(m)=m∑j=0βm+1jrj+1ϕ(j)(r),
where βm+1j are given byβm+10=3βm0+6βm1+6βm2,βm+11=βm0+4βm1+18βm2+24βm3,βm+1j=βmj−1+3(j+1)βmj+3(j+2)(j+1)βmj+1+(j+3)(j+2)(j+1)βmj+2for 2≤j≤m−3,βm+1m−2=βmm−3+2(m−1)βmm−2+3m(m−1)βmm−1,βm+1m−1=3mβmm−1+mβmm−1,βm+1m=βmm−1.
Therefore, we complete the proof of Claim 1.By (40) and Claims 1 and 2, we haveProof of Claim 2
Clearly, the claim holds true for k=1. From Claim 1, a direct computation gives
k−1∑j=0βkjrj+1ϕ(j)(r)=(1rddr)k−1(r2k−1ϕ(r))=(1rddr)k−2[(2k−1)r2k−3ϕ(r)+r2k−2ϕ′(r)]=(1rddr)k−3[(2k−1)(2k−3)r2k−5ϕ(r)+⋯]=⋯=(2k−1)(2k−3)(2k−5)⋯3⋅r1ϕ(r)+⋯.
It is clear that βk0=(2k−1)(2k−3)(2k−5)⋯3⋅1=c2k+1=cn. The proof is complete.Vx(r,t)=(1r∂∂r)k−1(r2k−1Mu(x,r,t))=k−1∑j=0βkjrj+1∂jMu∂rj(x,r,t).
Then we use (30) and (35) to getu(x,t)=limr→0+Mu(x,r,t)(by (30))=1cnlimr→0+(cnMu(x,r,t))=1cnlimr→0+1rk−1∑j=0βkjrj+1∂jMu∂rj(x,r,t)=1cnlimr→0+Vx(r,t)r=1cnlimr→0+[Gx(r+ct)+Gx(r−ct)2r+12cr∫r+ctr−ctHx(ρ)dρ](by (41))=1cnlimr→0+[(Gx)′(r+ct)+(Gx)′(r−ct)2+Hx(r+ct)−Hx(r−ct)2c](by L'Hospital's Rule)=(Gx)′(ct)+(Gx)′(−ct)2cn+Hx(ct)+Hx(−ct)2cnc.
From (21) and (41) with replacing the variable r by t, we note thatGx(t)=(1t∂∂t)k−1(t2k−1Mg(x,t))=(1t∂∂t)k−1(t2k−1ωn∫|ξ|=1g(x+tξ)dSξ)=1ωn=(1t∂∂t)n−32tn−2∫|ξ|=1g(x+tξ)dSξ,Hx(t)=(1t∂∂t)k−1(t2k−1Mh(x,t))=(1t∂∂t)k−1(t2k−1ωn∫|ξ|=1h(x+tξ)dSξ)=1ωn(1t∂∂t)n−32tn−2∫|ξ|=1h(x+tξ)dSξ.
This implies(Gx)′(ct)=1ωn(∂∂(ct)(1ct∂∂(ct))n−32(ct)n−2∫|ξ|=1g(x+ctξ)dSξ)=1ωn∂∂t(1t∂∂t)n−32tn−2∫|ξ|=1g(x+ctξ)dSξ,Hx(ct)=1ωn(1ct∂∂(ct))n−32(ct)n−2∫|ξ|=1h(x+ctξ)dSξ=cωn(1t∂∂t)n−32tn−2∫|ξ|=1h(x+ctξ)dSξ.
Also note that (Gx)′(ct)=(Gx)′(−ct) and Hx(ct)=−Hx(−ct). Therefore, we obtain (42) by combining above equations.
Solution
Solution
Solution
Consider the following wave equation{Utt(x,t,s)−c2ΔU(x,t,s)=0for x∈R3,t>0,s>0U(x,0,s)=0for x∈R3,s>0Ut(x,0,s)=f(x,s)for x∈R3,s>0.
By Kirchhoff's formula, we haveU(x,t,s)=t4π∫|ξ|=1f(x+ctξ,s)dSξfor x∈R3,t≥0,s>0.
Then by Duhamel's principle, the solution isu(x,t)=∫t0U(x,t−s,s)ds=14π∫t0(t−s)∫|ξ|=1f(x+c(t−s)ξ,s)dSξds
According to this formula for solution, when f is C2 in x and C0 in t, u will be a C2-function.Solution
Using the separation of variables, we suppose that the solution u(x,t) is in the formu(x,y,t)=∞∑m,n=1um,n(x,y,t)=∞∑m,n=1Xm(x)Yn(y)Tm,n(t)for (x,y)∈Ω,t>0.
Due to the superposition principle, we firstly consider u(x,y,t)=X(x)Y(y)T(t) to fulfill the wave equation with the boundary condition, which givesX(x)Y(y)T″(t)=X″(x)Y(y)T(t)+X(x)Y″(y)T(t),X(0)Y(y)T(t)=X(a)Y(y)T(t)=X(x)Y(0)T(t)=X(x)Y(b)T(t)=0.
Surely, we want to find a general solution so X, Y and T are not identically zero and X(0)=X(a)=Y(0)=Y(b)=0. From the differential equation, we haveT″(t)T(t)=X″(x)X(x)+Y″(y)Y(y)for (x,y)∈Ω,t>0.
Clearly, X″(x)/X(x), Y″(y)/Y(y) and T″(t)/T(t) are all constants. By solving the eigenvalue problems for X″(x)=λX(x) in (0,a) with X(0)=X(a)=0, Y″(y)=μY(y) in (0,b) with Y(0)=Y(b)=0, we obtainXm(x)=sinmπxafor x∈(0,a),Yn(x)=sinnπybfor y∈(0,b),
where λm=−(mπ/a)2 and μn=−(nπ/b)2 for m,n∈N. Then T satisfiesT″(t)=(λm+μn)T(t)=−(m2π2a2+n2π2b2)T(t),
which givesTm,n(t)=Am,ncos(tωm,n)+Bm,nsin(tωm,n)andωm,n=π√m2a2+n2b2.
Hence the solution has the following form:u(x,t)=∞∑m,n=1[Am,ncos(tωm,n)+Bm,nsin(tωm,n)]sinmπxasinnπybfor (x,y)∈Ω,t>0.
From the intial condition, it is easy to know A1,2=1, Am,n=0 for (m,n)≠(1,2), and Bm,n=0 for all (m,n)∈N2. Therefore, the solution isu(x,y,t)=cos(πt√1a2+4b2)sinπxasin2πybfor (x,y)∈ˉΩ,t≥0.
Solution
Define u:R3×ˉR+ byu(x,y,z,t)=cos(mz/c)v(x,y,t)for (x,y,z,t)∈R3×ˉR+.
Then it can be checked thatutt(x,y,z,t)=cos(mz/c)vtt(x,y,t)=cos(mz/c)[c2vxx(x,y,t)+c2vyy(x,y,t)−m2v(x,y,t)]=c2uxx(x,y,z,t)+c2uyy(x,y,z)+c2uzz(x,y,z)=c2Δu(x,y,z).
This means u satisfies the three-dimensional wave equation with the initial conditionu(x,y,z,0)=cos(mz/c)g(x,y),ut(x,y,z,0)=cos(mz/c)h(x,y)for (x,y,z)∈R3.
Then by Kirchhoff's formula (37), we haveu(x,y,z,t)=14π∂∂t(t∫|ξ|=1cos(m(z+ctξ3)/c)g(x+ctξ1,y+ctξ2)dSξ)+t4π∫|ξ|=1cos(m(z+ctξ3)/c)h(x+ctξ1,y+ctξ2)dSξ.
The soluion v isv(x,y,t)=u(x,y,0,t)=14π∂∂t(t∫|ξ|=1cos(mtξ3)g(x+ctξ1,y+ctξ2)dSξ)+t4π∫|ξ|=1cos(mtξ3)h(x+ctξ1,y+ctξ2)dSξ=12π∂∂t(t∫ξ21+ξ22<1cos(mt√1−ξ21−ξ22)g(x+ctξ1,y+ctξ2)√1−ξ21−ξ22dξ1dξ2)+t2π∫ξ21+ξ22<1cos(mt√1−ξ21−ξ22)h(x+ctξ1,y+ctξ2)√1−ξ21−ξ22dξ1dξ2=12π∂∂t(t∫2π0∫10rcos(mt√1−r2)g(x+ctrcosθ,y+ctrsinθ)√1−r2drdθ)+t2π∫2π0∫10rcos(mt√1−r2)h(x+ctrcosθ,y+ctrsinθ)√1−r2drdθ.
沒有留言:
張貼留言