Subtracting again, we have Eo·(V×H0)-Ho·(V×Eo)=H v·(Eo×Ho)=Eo + Ho by(B. 44). Integrating both sides throughout V and using the divergence theorem on the left-hand side, we get +H0 dv Breaking S into two arbitrary portions and using(B6), we obtain E0·(×Ho)dS-/Ho(xEo)dS dv Now if nx Eo=0 or n x Ho=0 over all of S, or some combination of these conditions holds over all of s. then ar+Ho dBo dv=o (213) This expression implies a relationship between Eo, Do, Bo, and Ho. Since V is arbitrary, we see that one possibility is simply to have Do and Bo constant with time. However since the fields are identically zero for t <0, if they are constant for all time then those constant values must be zero. Another possibility is to have one of each pair (Eo, Do) and(Ho, Bo) equal to zero. Then, by(2.9) and(2.10), Eo =0 implies Bo=0, and Do =0 implies Ho =0. Thus El= E, BI= B2, and so on, and the solution is unique throughout V. However, we cannot in general rule out more complicated relationships The number of possibilities depends on the additional constraints on the relationship between Eo, Do, Bo, and Ho that we must supply to describe the material supporting the field -i. e, the constitutive relationships. For a simple medium described by the time-constant permittivity e and permeability u,(13)becomes aHo 0, 1 a 2 d (∈Eo·Eo+puHo·Ho)dV=0. the integrand is always positive or zero(and not constant with time, as mentioned ) the only possible conclusion is that Eo and Ho must both be zero, and thus the fields are unique. When establishing more complicated constitutive relations, we must be careful to en- sure that they lead to a unique solution, and that the condition for uniqueness is un- derstood. In the case above, the assumption n x Eos=0 implies that the tangential components of E and Eg are identical over S- that is, we must give specific values of these quantities on S to ensure uniqueness. A similar statement holds for the condition f x Hols =0. Requiring that constitutive relations lead to a unique solution is known @2001 by CRC Press LLCSubtracting again, we have E0 · (∇ × H0) − H0 · (∇ × E0) = H0 · ∂B0 ∂t + E0 · ∂D0 ∂t , hence −∇ · (E0 × H0) = E0 · ∂D0 ∂t + H0 · ∂B0 ∂t by (B.44). Integrating both sides throughout V and using the divergence theorem on the left-hand side, we get −  S (E0 × H0) · dS =  V E0 · ∂D0 ∂t + H0 · ∂B0 ∂t dV. Breaking S into two arbitrary portions and using (B.6), we obtain  S1 E0 · (nˆ × H0) d S −  S2 H0 · (nˆ × E0) d S =  V E0 · ∂D0 ∂t + H0 · ∂B0 ∂t dV. Now if nˆ × E0 = 0 or nˆ × H0 = 0 over all of S, or some combination of these conditions holds over all of S, then  V E0 · ∂D0 ∂t + H0 · ∂B0 ∂t dV = 0. (2.13) This expression implies a relationship between E0, D0, B0, and H0. Since V is arbitrary, we see that one possibility is simply to have D0 and B0 constant with time. However, since the fields are identically zero for t < 0, if they are constant for all time then those constant values must be zero. Another possibility is to have one of each pair (E0, D0) and (H0,B0) equal to zero. Then, by (2.9) and (2.10), E0 = 0 implies B0 = 0, and D0 = 0 implies H0 = 0. Thus E1 = E2, B1 = B2, and so on, and the solution is unique throughout V. However, we cannot in general rule out more complicated relationships. The number of possibilities depends on the additional constraints on the relationship between E0, D0, B0, and H0 that we must supply to describe the material supporting the field — i.e., the constitutive relationships. For a simple medium described by the time-constant permittivity and permeability µ, (13) becomes  V E0 · ∂E0 ∂t + H0 · µ ∂H0 ∂t dV = 0, or 1 2 ∂ ∂t  V ( E0 · E0 + µH0 · H0) dV = 0. Since the integrand is always positive or zero (and not constant with time, as mentioned above), the only possible conclusion is that E0 and H0 must both be zero, and thus the fields are unique. When establishing more complicated constitutive relations, we must be careful to en￾sure that they lead to a unique solution, and that the condition for uniqueness is un￾derstood. In the case above, the assumption nˆ × E0   S = 0 implies that the tangential components of E1 and E2 are identical over S — that is, we must give specific values of these quantities on S to ensure uniqueness. A similar statement holds for the condition nˆ × H0   S = 0. Requiring that constitutive relations lead to a unique solution is known