Lecture 29

Uniqueness Theorem

29.1 Uniqueness Theorem

The uniqueness of a solution to a linear system of equations is an important concept in

mathematics. Under certain conditions, ordinary differential equation partial differential

equation and matrix equations will have unique solutions. But uniqueness is not always

guaranteed as we shall see. This issue is discussed in many math books and linear algebra

books [69, 81]. The prove of uniqueness for Laplace and Poisson equations are given in

[29, 48] which is slightly different for electrodynamic problems. In this section, we will prove

uniqueness theorem for electrodynamic problems [31, 34, 47, 59, 75]. First, let us assume that

there exist two solutions in the presence of one set of common impressed sources Ji and Mi.

Namely, these two solutions are Ea, Ha, Eb, Hb. Both of them satisfy Maxwell’s equations

and the same boundary conditions. Are Ea = Eb, Ha = Hb?

To study the uniqueness theorem, we consider general linear anisotropic inhomogeneous

media, where the tensors μ and ε can be complex so that lossy media can be included, it

follows that

∇ × Ea = −jωμ · Ha − Mi (29.1.1)

∇ × Eb = −jωμ · Hb − Mi (29.1.2)

∇ × Ha = jωε · Ea + Ji (29.1.3)

∇ × Hb = jωε · Eb + Ji (29.1.4)

By taking the difference of these two solutions, we have

∇ × (Ea − Eb) = −jωμ · (Ha − Hb) (29.1.5)

∇ × (Ha − Hb) = jωε · (Ea − Eb) (29.1.6)

Or alternatively, defining δE = Ea − Eb and δH = Ha − Hb, we have

∇ × δE = −jωμ · δH (29.1.7)

∇ × δH = jωε · δE (29.1.8)

291

292 Electromagnetic Field Theory

The difference solutions satisfy the original source-free Maxwell’s equations.

By taking the left dot product of δH∗ with (29.1.7), and then the left dot product of δE∗

with the complex conjugation of (29.1.8), we obtain

δH∗ · ∇ × δE = −jωδH∗ · μ · δH

δE · ∇ × δH∗ = −jωδE · ε∗ · δE∗ (29.1.9)

Now, taking the difference of the above, we get

δH∗ · ∇ × δE − δE · ∇ × δH∗ = ∇ · (δE × δH∗)

= −jωδH∗ · μ · δH + jωδE · ε∗ · δE∗ (29.1.10)

Figure 29.1: Geometry for proving the uniqueness theorem. We like to know the boundary

conditions needed on S in order to guarantee the uniqueness of the solution in V .

Next, integrating the above equation over a volume V bounded by a surface S as shown

in Figure 29.1. Two scenarios are possible: one that the volume V contains the impressed

sources, and two, that the sources are outside the volume V . After making use of Gauss’

divergence theorem, we arrive at



V

∇ · (δE × δH∗)dV =



S

(δE × δH∗) · dS

=



V

[−jωδH∗ · μ · δH + jωδE · ε∗ · δE∗]dV (29.1.11)

And next, we would like to know the kind of boundary conditions that would make the

left-hand side equal to zero. It is seen that the surface integral on the left-hand side will be

Uniqueness Theorem 293

zero if:1

1. If ˆn × E is specified over S so that ˆn × Ea = ˆn × Eb, then ˆn × δE = 0 or the PEC boundary

condition for δE, and then2

S (δE × δH∗) · ˆndS = S (ˆn × δE) · δH∗dS = 0.

2. If ˆn × H is specified over S so that ˆn × Ha = ˆn × Hb, then ˆn × δH = 0 or the PMC

boundary condition for δH, and then

S (δE × δH∗) · ˆndS = − S (ˆn × δH∗) · δEdS = 0.

3. If ˆn × E is specified over S1, and ˆn × H is specified over S2 (where S1 ∪ S2 = S), then

ˆn × δE = 0 (PEC boundary condition) on S1, and ˆn × δH = 0 (PMC boundary condition)

on S2, then the left-hand side becomes

S (δE × δH∗) · ˆndS = 

S1 + 

S2 = 

S1 (ˆn × δE) · δH∗dS

− 

S2 (ˆn × δH∗) · δEdS = 0.

Thus, under the above three scenarios, the left-hand side of (29.1.11) is zero, and then

the right-hand side of (29.1.11) becomes



V

[−jωδH∗ · μ · δH + jωδE · ε∗ · δE∗]dV = 0 (29.1.12)

For lossless media, μ and ε are hermitian tensors (or matrices3), then it can be seen, using

the properties of hermitian matrices or tensors, that δH∗ · μ · δH and δE · ε∗ · δE∗ are purely

real. Taking the imaginary part of the above equation yields



V

[−δH∗ · μ · δH + δE · ε∗ · δE∗]dV = 0 (29.1.13)

The above two terms correspond to stored magnetic field energy and stored electric field

energy in the difference solutions δH and δE, respectively. The above being zero does not

imply that δH and δE are zero.

For resonance solutions, the stored electric energy can balance the stored magnetic energy.

The above resonance solutions are those of the difference solutions satisfying PEC or PMC

boundary condition or mixture thereof. Therefore, δH and δE need not be zero, even though

(29.1.13) is zero. This happens when we encounter solutions that are the resonant modes of

the volume V bounded by surface S.

Uniqueness can only be guaranteed if the medium is lossy as shall be shown later. It is

also guaranteed if lossy impedance boundary conditions are imposed.4 First we begin with

the isotropic case.

1In the following, please be reminded that PEC stands for “perfect electric conductor”, while PMC stands

for “perfect magnetic conductor”. PMC is the dual of PEC. Also, a fourth case of impedance boundary

condition is possible, which is beyond the scope of this course. Interested readers may consult Chew, Theory

of Microwave and Optical Waveguides [75].

2Using the vector identity that a · (b × c) = c · (a × b) = b · (c × a).

3Tensors are a special kind of matrices.

4See Chew, Theory of Microwave and Optical Waveguides.

294 Electromagnetic Field Theory

29.1.1 Isotropic Case

It is easier to see this for lossy isotropic media. Then (29.1.12) simplifies to



V

[−jωμ|δH|2 + jωε∗|δE|2]dV = 0 (29.1.14)

For isotropic lossy media, μ = μ′ − jμ′′ and ε = ε′ − jε′′. Taking the real part of the above,

we have from (29.1.14) that



V

[−ωμ′′|δH|2 − ωε′′|δE|2]dV = 0 (29.1.15)

Since the integrand in the above is always negative definite, the integral can be zero only if

δE = 0, δH = 0 (29.1.16)

everywhere in V , implying that Ea = Eb, and that Ha = Hb. Hence, it is seen that uniqueness

is guaranteed only if the medium is lossy. The physical reason is that when the medium is

lossy, a pure time-harmonic solution cannot exist due to loss. The modes, which are the

source-free solutions of Maxwell’s equations, are decaying sinusoids.

Notice that the same conclusion can be drawn if we make μ′′ and ε′′ negative. This

corresponds to active media, and uniqueness can be guaranteed for a time-harmonic solution.

In this case, no time-harmonic solution exists, and the resonant solution is a growing sinusoid.

29.1.2 General Anisotropic Case

The proof for general anisotropic media is more complicated. For the lossless anisotropic

media, we see that (29.1.12) is purely imaginary. However, when the medium is lossy, this

same equation will have a real part. Hence, we need to find the real part of (29.1.12) for the

general lossy case.

About taking the Real and Imaginary Parts of a Complex Expression

To this end, we digress on taking the real and imaginary parts of a complex expression. Here,

we need to find the complex conjugate5 of (29.1.12), which is scalar, and add it to itself to

get its real part. The complex conjugate of the scalar

c = δH∗ · μ · δH

is6

c∗ = δH · μ∗ · δH∗ = δH∗ · μ† · δH

5Also called hermitian conjugate.

6To arrive at these expressions, one makes use of the matrix algebra rule that if D = A · B · C, then

Dt = Ct · Bt · At. This is true even for non-square matrices. But for our case here, A is a 1 × 3 row vector,

and C is a 3 × 1 column vector, and B is a 3 × 3 matrix. In vector algebra, the transpose of a vector is implied.

Also, in our case here, D is a scalar, and hence, its transpose is itself.

Uniqueness Theorem 295

Similarly, the complex conjugate of the scalar

d = δE · ε∗ · δE∗ = δE∗ · ε† · δE

is

d∗ = δE∗ · ε† · δE

Therefore,

=m (δH∗ · μ · δH) = 1

2j δH∗ · (μ − μ†) · δH

=m (δE · ε · δE∗) = 1

2j δE∗ · (ε − ε†) · δE

and similarly for the real part.

Finally, after taking the complex conjugate of the scalar quantity (29.1.12) and adding it

to itself, we have



V

[−jωδH∗ · (μ − μ†) · δH − jωδE∗ · (ε − ε†) · δE]dV = 0 (29.1.17)

For lossy media, −j(μ − μ†) and −j(ε − ε†) are hermitian negative matrices. Hence the

integrand is always negative definite, and the above equation cannot be satisfied unless δH =

δE = 0 everywhere in V . Thus, uniqueness is guaranteed in a lossy anisotropic medium.

Similar statement can be made as the isotropic case if the medium is active. Then the

integrand is positive definite, and the above equation cannot be satisfied unless δH = δE = 0

everywhere in V and hence, uniqueness is satisfied.

29.1.3 Hind Sight

The proof of uniqueness for Maxwell’s equations is very similar to the proof of uniqueness for

a matrix equation [69]

A · x = b (29.1.18)

If a solution to a matrix equation exists without excitation, namely, when b = 0, then the

solution is the null space solution [69], namely, x = xN . In other words,

A · xN = 0 (29.1.19)

These null space solutions exist without a “driving term” b on the right-hand side. For

Maxwell’s Equations, b corresponds to the source terms. They are like the homogeneous

solution of an ordinary differential equation or a partial differential equation [81]. In an

enclosed region of volume V bounded by a surface S, homogeneous solutions are the resonant

solutions of this Maxwellian system. When these solutions exist, they give rise to non-

uniqueness.

Also, notice that (29.1.7) and (29.1.8) are Maxwell’s equations without the source terms.

In a closed region V bounded by a surface S, only resonance solutions for δE and δH with

the relevant boundary conditions can exist when there are no source terms.

296 Electromagnetic Field Theory

As previoulsy mentioned, one way to ensure that these resonant solutions are eliminated

is to put in loss or gain. When loss or gain is present, then the resonant solutions are decaying

sinusoids or growing sinusoids. Since we are looking for solutions in the frequency domain,

or time harmonic solutions, we are only looking for the solution on the real ω axis on the

complex ω plane. These non-sinusoidal solutions are outside the solution space: They are

not part of the time-harmonic solutions we are looking for. Therefore, there are no resonant

null-space solutions.

29.1.4 Connection to Poles of a Linear System

The output to input of a linear system can be represented by a transfer function H(ω) [45,153].

If H(ω) has poles, and if the system is lossless, the poles are on the real axis. Therefore, when

ω = ωpole, the function H(ω) becomes undefined. This also gives rise to non-uniqueness of the

output with respect to the input. Poles usually correspond to resonant solutions, and hence,

the non-uniqueness of the solution is intimately related to the non-uniqueness of Maxwell’s

equations at the resonant frequencies of a structure. This is illustrated in the upper part of

Figure 29.2.

Figure 29.2: The non-uniqueness problem is intimately related to the locations of the poles

of a transfer function being on the real axis.

Uniqueness Theorem 297

If the input function is f (t), with Fourier transform F (ω), then the output y(t) is given

by the following Fourier integral, viz.,

y(t) = 1

2π

 ∞

−∞

dωejωtH(ω)F (ω) (29.1.20)

where the Fourier inversion integral path is on the real axis on the complex ω plane. The

Fourier inversion integral is undefined or non-unique.

However, if loss is introduced, these poles will move away from the real axis as shown

in the lower part of Figure 29.2. Then the transfer function is uniquely determined for all

frequencies on the real axis. In this way, the Fourier inversion integral in (29.1.20) is well

defined, and uniqueness of the solution is guaranteed.

29.1.5 Radiation from Antenna Sources

The above uniqueness theorem guarantees that if we have some antennas with prescribed

current sources on them, the radiated field from these antennas are unique. To see how this

can come about, we first study the radiation of sources into a region V bounded by a large

surface Sinf as shown in Figure 29.3 [34].

Even when ˆn × E or ˆn × H are specified on the surface at Sinf, the solution is nonunique

because the volume V bounded by Sinf, can have many resonant solutions. In fact, the

region will be replete with resonant solutions as one makes Sinf become very large. The way

to remove these resonant solutions is to introduce an infinitesimal amount of loss in region

V . Then these resonant solutions will disappear. Now we can take Sinf to infinity, and the

solution will always be unique.

Notice that if Sinf → ∞, the waves that leave the sources will never be reflected back

because of the small amount of loss. The radiated field will just disappear into infinity. This

is just what radiation loss is: power that propagate to infinity, but never to return. In fact,

one way of guaranteeing the uniqueness of the solution in region V when Sinf is infinitely large,

or that V is infinitely large is to impose the radiation condition: the waves that radiate to

infinity are outgoing waves only, and never do they return. This is also called the Sommerfeld

radiation condition [154]. Uniqueness of the field outside the sources is always guaranteed if

we assume that the field radiates to infinity and never to return. This is equivalent to solving

the cavity solutions with an infinitesimal loss, and then letting the size of the cavity become

infinitely large.

298 Electromagnetic Field Theory

Figure 29.3: The solution for antenna radiation is unique because we impose the Sommerfeld

radiation condition when seeking the solution. This is equivalent to assuming an infinitesimal

loss when seeking the solution in V .

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