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, which is desired. - No. To find a counterexample, we consider the initial condition u(x,y,0)\equiv0 and u_t(x,y,0)=h(x,y), where h satisfies h\equiv1 in B_1(0), 0\leq h\leq1 in B_2(0)-B_1(0) and h\equiv0 in \mathbb R^2-B_2(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)|\leq C/t for (x,y)\in\mathbb R^2. By (39), we have
\begin{aligned}u(x,y,t)&=\frac1{4\pi}\frac\partial{\partial t}\left(2t\int_{\xi_1^2+\xi_2^2<1}\!\frac{g(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)+\frac t{4\pi}\left(2\int_{\xi_1^2+\xi_2^2<1}\!\frac{h(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)\\&=\frac t{2\pi}\int_{\xi_1^2+\xi_2^2<1}\frac{h(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2=\frac1{2\pi c^2t}\int_{B_{ct}(x,y)}\!\frac{h(\eta,\mu)}{\sqrt{1-(ct)^{-2}((\eta-x)^2+(\mu-y)^2)}}\,\mathrm d\eta\,\mathrm d\mu.\end{aligned}
It is clear that B_1(0)\subseteq B_{ct+1}(ct,0) for t\geq0. Then we have\begin{aligned}u(ct,0,t+c^{-1})&=\frac1{2\pi c(ct+1)}\int_{B_{ct+1}(ct,0)}\!\frac{h(\eta,\mu)}{\sqrt{1-(ct+1)^{-2}((\eta-ct)^2+\mu^2)}}\,\mathrm d\eta\,\mathrm d\mu\\&\geq\frac1{2\pi c(ct+1)}\int_{B_1(0,0)}\!\frac{h(\eta,\mu)}{\sqrt{1-(ct+1)^{-2}((\eta-ct)^2+\mu^2)}}\,\mathrm d\eta\,\mathrm d\mu\\&=\frac1{2\pi c}\int_{B_1(0,0)}\!\frac1{\sqrt{(ct+1)^2-(\eta-ct)^2-\mu^2}}\,\mathrm d\eta\,\mathrm d\mu\\&=\frac1{2\pi c}\int_0^{2\pi}\!\int_0^1\!\frac r{\sqrt{1-r^2+2ct(1+r\cos\theta)}}\,\mathrm dr\,\mathrm d\theta\\&\geq\frac1{4\pi c}\int_0^{2\pi}\!\int_0^1\!\frac r{\sqrt{1+4ct}}\,\mathrm dr\,\mathrm d\theta=\frac1{4\sqrt{1+4ct}}.\end{aligned}
Thus, by |u(x,y,t)|\leq C/t for all (x,y)\in\mathbb R^2 and t>0, we get\displaystyle\frac1{4\sqrt{1+4ct}}\leq\frac Ct\quad\text{for all}~t>0,
which is equivalent to t^2-64c^3t-16c^2\leq0 for all t>0. This leads a contradiction. Thereofre, it is impossible to find a positive constant C such that |u(x,y,t)|\leq C/t for all (x,y)\in\mathbb R^2 and t>0. - Suppose n=2k+1 for a positive integer k and u satisfies (26).
- Show that V^x(r,t) given by (40) satisfies (32).
- Derive (42) from (35) and (41); note c_n=1\cdot3\cdots(n-2).
- Firstly, we prove the following identity by mathematical induction:
\displaystyle\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^k(r^{2k}\phi'(r)),
where k\in\mathbb N and \phi\in C^{k+1}(\mathbb R). Then we claim thatNow we can use (28), (40) and Claim to obtainProof of Claim
When k=1, it is easy to verify that\begin{aligned}\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))&=\frac{\mathrm d^2}{\mathrm dr^2}(r\phi(r))=\frac{\mathrm d}{\mathrm dr}(r\phi'(r)+\phi(r))=r\phi''(r)+2\phi'(r)\\&=\frac{r^2\phi''(r)+2r\phi'(r)}r=\frac1r\frac{\mathrm d}{\mathrm dr}(r^2\phi'(r)).\end{aligned}
This shows the identity holds true for k=1. Now we suppose the identity holds true for k=n, i.e.,\displaystyle\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n-1}(r^{2n-1}\phi(r))=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n}\phi'(r)).
Then we can observe that\begin{aligned}\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n+1}\phi(r))&=\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n-1}\left[\frac{(2n+1)r^{2n}\phi(r)+r^{2n+1}\phi'(r)}r\right]\\&=(2n+1)\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n-1}(r^{2n-1}\phi(r))+\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n-1}(r^{2n-1}(r\phi'(r)))\\&=(2n+1)\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n}\phi'(r))+\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n}(r\phi'(r))')\\&=\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n+1}r^{2n}\phi'(r)+r^{2n}(r\phi''(r)+\phi'(r)))\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n\frac1r\frac{\mathrm d}{\mathrm dr}(r^{2n+2}\phi'(r))\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n+1}(r^{2n+2}\phi'(r)).\end{aligned}
Hence by the mathematical induction, the identity holds true for all n\in\mathbb N.\begin{aligned}\frac{\partial^2V^x}{\partial t^2}(r,t)&=\frac{\partial^2}{\partial t^2}\left(\frac1r\frac{\partial}{\partial r}\right)^{k-1}(r^{2k-1}M_u(x,r,t))=\left(\frac1r\frac{\partial}{\partial r}\right)^{k-1}(r^{2k-1}\frac{\partial^2M_u}{\partial t^2}(x,r,t))\quad\text{(by (28))}\\&=\left(\frac1r\frac\partial{\partial r}\right)^{k-1}\left[r^{2k-1}\cdot c^2\left(\frac{\partial^2}{\partial r^2}+\frac{n-1}r\frac\partial{\partial r}\right)M_u(x,r,t)\right]\quad\text{(by (40))}\\&=c^2\left(\frac1r\frac\partial{\partial r}\right)^{k-1}\cdot\frac1r\left(r^{n-1}\frac{\partial^2}{\partial r^2}+(n-1)r^{n-2}\frac\partial{\partial r}\right)M_u(x,r,t)\\&=c^2\left(\frac1r\frac\partial{\partial r}\right)^{k-1}\cdot\frac1r\frac\partial{\partial r}\left(r^{n-1}\frac\partial{\partial r}\right)M_u\\&=c^2\left(\frac1r\frac\partial{\partial r}\right)^k\left(r^{2k}\frac{\partial M_u}{\partial r}(x,r,t)\right)\\&=c^2\frac{\partial^2}{\partial r^2}\left(\frac1r\frac\partial{\partial r}\right)^{k-1}(r^{2k-1}M_u(x,r,t))\quad\text{(by Claim)}\\&=c^2\frac{\partial^2V^x}{\partial r^2}(x,r,t)\quad\text{(by (28))}.\end{aligned}
The proof is complete. - For any k\in\mathbb N and \phi\in C^{k-1}(\mathbb R), we prove the following identity by the mathematical induction:
\displaystyle\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))=\sum_{j=0}^{k-1}\beta_j^kr^{j+1}\phi^{(j)}(r),
where the constants \beta_j^k (j=0,\dots,k-1) are independent of \phi.Then we claim that \beta_0^k=c_{2k+1}=c_n for k\in\mathbb N.Proof of Claim 1
When k=1, it is easy that\displaystyle\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))=r\phi(r)=\beta_0^1\cdot r^{0+1}\phi^{(0)}(r),
where \beta_0^1=1=c_3. This shows the claim holds true for k=1. Now we suppose the claim holds true for k=m, i.e., there exists \beta_j^m (j=0,1,\dots,m-1) such that\displaystyle\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{m-1}(r^{2m-1}\phi(r))=\sum_{j=0}^{m-1}\beta_j^mr^{j+1}\phi^{(j)}(r),
where \beta_0^m=c_{2m+1}. Then we can observe that\begin{aligned}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^m(r^{2m+1}\phi(r))&=\frac1r\frac{\mathrm d}{\mathrm dr}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{m-1}(r^{2m-1}(r^2\phi(r)))\\&=\frac1r\frac{\mathrm d}{\mathrm dr}\sum_{j=0}^{m-1}\beta_j^mr^{j+1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))\\&=\frac1r\sum_{j=0}^{m-1}\beta_j^m\left[(j+1)r^j\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))+r^{j+1}\frac{\mathrm d^{j+1}}{\mathrm dr^{j+1}}(r^2\phi(r))\right]\\&=\sum_{j=0}^{m-1}(j+1)\beta_j^mr^{j-1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))+\sum_{j=0}^{m-1}\beta_j^mr^j\frac{\mathrm d^{j+1}}{\mathrm dr^{j+1}}(r^2\phi(r))\\&=\sum_{j=0}^{m-1}(j+1)\beta_j^mr^{j-1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))+\sum_{j=1}^m\beta_{j-1}^mr^{j-1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))\\&=\beta_0^mr\phi(r)+\sum_{j=1}^{m-1}[(j+1)\beta_j^m+\beta_{j-1}^m]r^{j-1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))+\beta_{m-1}^mr^{m-1}\frac{\mathrm d^m}{\mathrm dr^m}(r^2\phi(r)).\end{aligned}
- For j=1, \displaystyle\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))=2r\phi(r)+r^2\phi'(r);
- for j\geq2, \displaystyle\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))=r^2\phi^{(j)}(r)+2jr\phi^{(j-1)}(r)+j(j-1)\phi^{(j-2)}(r).
\begin{aligned}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^m(r^{2m+1}\phi(r))&=\beta_0^mr\phi(r)+(2\beta_1^m+\beta_0^m)(2r\phi(r)+r^2\phi'(r))\\&\quad+\sum_{j=2}^{m-1}[(j+1)\beta_j^m+\beta_{j-1}^m]r^{j-1}[r^2\phi^{(j)}+2jr\phi^{(j-1)}(r)+j(j-1)\phi^{(j-2)}(r)]\\&\quad+\beta_{m-1}^mr^{m-1}(r^2\phi^{(m)}+2mr\phi^{(m-1)}(r)+m(m-1)\phi^{(m-2)}(r))\\&=(3\beta_0^m+4\beta_1^m)r\phi(r)+(2\beta_1^m+\beta_0^m)r^2\phi'(r)+\sum_{j=2}^{m-1}[(j+1)\beta_j^m+\beta_{j-1}^m]r^{j+1}\phi^{(j)}\\&\quad+\sum_{j=1}^{m-2}2[(j+2)(j+1)\beta_{j+1}^m+(j+1)\beta_j^m]r^{j+1}\phi^{(j)}(r)\\&\quad+\sum_{j=0}^{m-3}[(j+3)(j+2)(j+1)\beta_{j+2}^m+(j+2)(j+1)\beta_{j+1}^m]r^{j+1}\phi^{(j)}(r)\\&\quad+m(m-1)\beta_{m-1}^mr^{m-1}\phi^{(m-2)}(r)+2m\beta_{m-1}^mr^m\phi^{(m-1)}(r)+\beta_{m-1}^mr^{m+1}\phi^{(m)}\\&=\sum_{j=0}^m\beta_j^{m+1}r^{j+1}\phi^{(j)}(r),\end{aligned}
where \beta_j^{m+1} are given by\begin{aligned} &\beta_0^{m+1}=3\beta_0^m+6\beta_1^m+6\beta_2^m,\\&\beta_1^{m+1}=\beta_0^m+4\beta_1^m+18\beta_2^m+24\beta_3^m,\\&\beta_j^{m+1}=\beta_{j-1}^m+3(j+1)\beta_j^m+3(j+2)(j+1)\beta_{j+1}^m+(j+3)(j+2)(j+1)\beta_{j+2}^m\quad\text{for}~2\leq j\leq m-3,\\&\beta_{m-2}^{m+1}=\beta_{m-3}^m+2(m-1)\beta_{m-2}^m+3m(m-1)\beta_{m-1}^m,\\&\beta_{m-1}^{m+1}=3m\beta_{m-1}^m+m\beta_{m-1}^m,\\&\beta_m^{m+1}=\beta_{m-1}^m.\end{aligned}
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
\begin{aligned}\sum_{j=0}^{k-1}\beta_j^kr^{j+1}\phi^{(j)}(r)=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-2}[(2k-1)r^{2k-3}\phi(r)+r^{2k-2}\phi'(r)]\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-3}[(2k-1)(2k-3)r^{2k-5}\phi(r)+\cdots]\\&=\cdots=(2k-1)(2k-3)(2k-5)\cdots3\cdot r^1\phi(r)+\cdots.\end{aligned}
It is clear that \beta_0^k=(2k-1)(2k-3)(2k-5)\cdots3\cdot1=c_{2k+1}=c_n. The proof is complete.\displaystyle V^x(r,t)=\left(\frac1r\frac\partial{\partial r}\right)^{k-1}(r^{2k-1}M_u(x,r,t))=\sum_{j=0}^{k-1}\beta_j^kr^{j+1}\frac{\partial^jM_u}{\partial r^j}(x,r,t).
Then we use (30) and (35) to get\begin{aligned}u(x,t)&=\lim_{r\to0^+}M_u(x,r,t)\quad\text{(by (30))}\\&=\frac1{c_n}\lim_{r\to0^+}(c_nM_u(x,r,t))\\&=\frac1{c_n}\lim_{r\to0^+}\frac1r\sum_{j=0}^{k-1}\beta_j^kr^{j+1}\frac{\partial^jM_u}{\partial r^j}(x,r,t)\\&=\frac1{c_n}\lim_{r\to0^+}\frac{V^x(r,t)}r\\&=\frac1{c_n}\lim_{r\to0^+}\left[\frac{G^x(r+ct)+G^x(r-ct)}{2r}+\frac1{2cr}\int_{r-ct}^{r+ct}\!H^x(\rho)\,\mathrm d\rho\right]\quad\text{(by (41))}\\&=\frac1{c_n}\lim_{r\to0^+}\left[\frac{(G^x)'(r+ct)+(G^x)'(r-ct)}2+\frac{H^x(r+ct)-H^x(r-ct)}{2c}\right]\quad\text{(by L'Hospital's Rule)}\\&=\frac{(G^x)'(ct)+(G^x)'(-ct)}{2c_n}+\frac{H^x(ct)+H^x(-ct)}{2c_nc}.\end{aligned}
From (21) and (41) with replacing the variable r by t, we note that\begin{aligned}G^x(t)&=\left(\frac1t\frac\partial{\partial t}\right)^{k-1}(t^{2k-1}M_g(x,t))=\left(\frac1t\frac\partial{\partial t}\right)^{k-1}\left(\frac{t^{2k-1}}{\omega_n}\int_{|\xi|=1}\!g(x+t\xi)\,\mathrm dS_\xi\right)\\&=\frac1{\omega_n}=\left(\frac1t\frac\partial{\partial t}\right)^{\frac{n-3}2}t^{n-2}\int_{|\xi|=1}\!g(x+t\xi)\,\mathrm dS_\xi,\\H^x(t)&=\left(\frac1t\frac\partial{\partial t}\right)^{k-1}(t^{2k-1}M_h(x,t))=\left(\frac1t\frac\partial{\partial t}\right)^{k-1}\left(\frac{t^{2k-1}}{\omega_n}\int_{|\xi|=1}h(x+t\xi)\,\mathrm dS_\xi\right)\\&=\frac1{\omega_n}\left(\frac1t\frac\partial{\partial t}\right)^{\frac{n-3}2}t^{n-2}\int_{|\xi|=1}\!h(x+t\xi)\,\mathrm dS_\xi.\end{aligned}
This implies\begin{aligned}(G^x)'(ct)&=\frac1{\omega_n}\left(\frac\partial{\partial(ct)}\left(\frac1{ct}\frac\partial{\partial(ct)}\right)^{\frac{n-3}2}(ct)^{n-2}\int_{|\xi|=1}\!g(x+ct\xi)\,\mathrm dS_\xi\right)\\&=\frac1{\omega_n}\frac\partial{\partial t}\left(\frac1t\frac\partial{\partial t}\right)^{\frac{n-3}2}t^{n-2}\int_{|\xi|=1}\!g(x+ct\xi)\,\mathrm dS_\xi,\\H^x(ct)&=\frac1{\omega_n}\left(\frac1{ct}\frac\partial{\partial(ct)}\right)^{\frac{n-3}2}(ct)^{n-2}\int_{|\xi|=1}\!h(x+ct\xi)\,\mathrm dS_\xi\\&=\frac c{\omega_n}\left(\frac1t\frac\partial{\partial t}\right)^{\frac{n-3}2}t^{n-2}\int_{|\xi|=1}\!h(x+ct\xi)\,\mathrm dS_\xi.\end{aligned}
Also note that (G^x)'(ct)=(G^x)'(-ct) and H^x(ct)=-H^x(-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.
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