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Mathematics LibreTexts

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)+(rct)F(rct)2r+12crr+ctrct ξG(ξ) dξ.

The right hand side of the previous formula is well defined if the domain of dependence [xct,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)=1ωnB1(0) nj=1fyj(x+rξ)ξj dSξF(+0)=1ωnB1(0) nj=1fyj(x)ξj dSξ=1ωnnj=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,gC2(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)+(rct)F0(rct)2r+12crr+ctrct 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)(ctr)F0(ctr)2r+12crct+rctr ξG0(ξ) dξ,

see Figure 4.2.1.1.

Changed domain of integration


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 fC3(R3) and gC2(R3) are given. Then there exists a unique solution uC2(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)=14πc2t(1tBct(x) f(y) dSy)+14πc2tBct(x) g(y) dSy.

Proof. Above we have shown that a C2-solution is given by Poisson's formula. Under the additional assumption fC3 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 |yx|=c|tt0| with the hyperplane defined by t=0, see Figure 4.2.1.2.

Domain of dependence, case \(n=3\).

Figure 4.2.1.2: Domain of dependence, case n=3.

Contributors and Attributions


This page titled 4.2.1: Case n=3 is shared under a not declared license and was authored, remixed, and/or curated by Erich Miersemann.

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