4.2.1: Case n=3
( \newcommand{\kernel}{\mathrm{null}\,}\)
The Euler-Poisson-Darboux equation in this case is
$$(rM)_{rr}=c^{-2}(rM)_{tt}.\]
Thus rM is the solution of the one-dimensional wave equation with initial data
(rM)(r,0)=rF(r) (rM)t(r,0)=rG(r).
From the d'Alembert formula we get formally
M(r,t)=(r+ct)F(r+ct)+(r−ct)F(r−ct)2r+12cr∫r+ctr−ct ξG(ξ) dξ.
The right hand side of the previous formula is well defined if the domain of dependence [x−ct,x+ct] is a subset of (0,∞). We can extend F and G to F0 and G0 which are defined on (−∞,∞) such that rF0 and rG0 are C2(R1)-functions as follows.
Set
$$
F_0(r)=\left\{F(r)r>0f(x)r=0F(−r)r<0\right.\
\]
The function G0(r) is given by the same definition where F and f are replaced by G and g, respectively.
Lemma. rF0(r), rG0(r)∈C2(R2).
Proof. From definition of F(r) and G(r), r>0, it follows from the mean value theorem
$$\lim_{r\to+0} F(r)=f(x),\ \ \ \lim_{r\to+0} G(r)=g(x).\]
Thus rF0(r) and rG0(r) are C(R1)-functions. These functions are also in C1(R1). This follows since F0 and G0 are in C1(R1). We have, for example,
F′(r)=1ωn∫∂B1(0) n∑j=1fyj(x+rξ)ξj dSξF′(+0)=1ωn∫∂B1(0) n∑j=1fyj(x)ξj dSξ=1ωnn∑j=1fyj(x)∫∂B1(0) nj dSξ=0.
Then, rF0(r) and rG0(r) are in C2(R1), provided F″ and G″ are bounded as r→+0. This property follows from
$$F''(r)=\dfrac{1}{\omega_n}\int_{\partial B_1(0)}\ \sum_{i,j=1}^n f_{y_iy_j}(x+r\xi)\xi_i\xi_j\ dS_\xi.\]
Thus
$$F''(+0)=\dfrac{1}{\omega_n}\sum_{i,j=1}^n f_{y_iy_j}(x)\int_{\partial B_1(0)}\ n_in_j\ dS_\xi.\]
We recall that f,g∈C2(R2) by assumption.
◻
The solution of the above initial value problem, where F and G are replaced by F0 and G0, respectively, is
M0(r,t)=(r+ct)F0(r+ct)+(r−ct)F0(r−ct)2r+12cr∫r+ctr−ct ξG0(ξ) dξ.
Since F0 and G0 are even functions, we have
$$\int_{r-ct}^{ct-r}\ \xi G_0(\xi)\ d\xi=0.\]
Thus
M0(r,t)=(r+ct)F0(r+ct)−(ct−r)F0(ct−r)2r+12cr∫ct+rct−r ξG0(ξ) dξ,
see Figure 4.2.1.1.
Figure 4.2.1.1: Changed domain of integration
For fixed t>0 and 0<r<ct it follows that M0(r,t) is the solution of the initial value problem with given initially data (4.2.1.1) since F0(s)=F(s), G0(s)=G(s) if s>0.
Since for fixed t>0
$$u(x,t)=\lim_{r\to 0} M_0(r,t),\]
it follows from d'Hospital's rule that
u(x,t)=ctF′(ct)+F(ct)+tG(ct)=ddt(tF(ct))+tG(ct).
Theorem 4.2. Assume f∈C3(R3) and g∈C2(R3) are given. Then there exists a unique solution u∈C2(R3×[0,∞)) of the initial value problem (4.2.2)-(4.2.3), where n=3, and the solution is given by the Poisson's formula
u(x,t)=14πc2∂∂t(1t∫∂Bct(x) f(y) dSy)+14πc2t∫∂Bct(x) g(y) dSy.
Proof. Above we have shown that a C2-solution is given by Poisson's formula. Under the additional assumption f∈C3 it follows from Poisson's formula that this formula defines a solution which is in C2, see F. John [10], p. 129.
◻
Corollary. From Poisson's formula we see that the domain of dependence for u(x,t0) is the intersection of the cone defined by |y−x|=c|t−t0| with the hyperplane defined by t=0, see Figure 4.2.1.2.
Figure 4.2.1.2: Domain of dependence, case n=3.
Contributors and Attributions
Integrated by Justin Marshall.