In the elliptic case we shall take the slope of =constant as
(10.4.1) and that of ψ=constant as
(10.4.2)
Since the square root is purely imaginary, the two slopes are complex conjugates. It follows that and ψ are complex conjugates, i.e., and η=ξ*, the asterisk denoting a complex conjugate.
In terms of and ψ, (10.4.1) and (10.4.2) can be expressed as
(10.4.3) (10.4.4)
The characteristics have been chosen so that a1=0 and c1=0, but notice that neither a nor c can be zero because of the condition for ellipticity. For b1 substitute from (10.4.3) and (10.4.4) into (10.3.5) to obtain
Since and ψ are complex conjugates (10.4.5) may be rewritten as showing that b1 is real and has the same sign as a.
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Furthermore, (10.3.7) and (10.3.8) indicate that so that the partial differential equation has been transformed to
(10.4.6) In this form, sometimes called the characteristic form of an elliptic equation, the partial differential equation involves complex coordinates. To avoid these a further transformation is made. Let
Then λ is the real part of ξ and μ is the imaginary part of ξ so that both λ and μ are real. Moreover
so that (10.4.6) becomes By division by we derive
(10.4.7) where e2, f2 and g2 are all real; in particular,
(10.4.8) from (10.3.9) and (10.4.5).
Equation (10.4.7) is known as the normal form of an elliptic partial differential equation. The normal form appears in its simplest guise as
(10.4.9)
where k2 is a real constant. If k2=0 it reduces to Laplace’s equation and occurs in problems in potential theory. If k2 is positive it arises in the study of two-dimensional harmonic waves such as those produced by a vibrating membrane. The case when k2 is negative typically results from trying to solve the two-dimensional wave equation by certain methods.
Usually, solutions of elliptic partial differential equations are sought that satisfy prescribed boundary conditions. There are four of common occurrence.
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(1) Dirichlet problem. (a) In the interior problem u is given at every point of a closed curve C and has to be found inside C. (b) For the exterior problem u is given on a closed curve C and has to be determined outside C. A supplementary condition specifying the behaviour at infinity is normally imposed. For example, there might be a requirement that ∂u/∂r→0 as r=(x2+y2)1/2→∞ in potential theory.
(2) Neumann problem. (a) In the interior problem the derivative of u along the normal to a closed curve C, i.e., ∂u/∂n, is prescribed and u is required inside C. Often ∂u/∂n cannot be specified arbitrarily on C; for instance, if k2=0 in (10.4.9), ∫C(∂u/∂n)ds=0 by the divergence theorem so that any values ascribed to ∂u/∂n on C must be compatible with this relation. (b) ∂u/∂n is given on the closed curve C and u is to be found outside, usually subject to a condition at infinity as in (1b).
(3) Mixed problem. Here the closed curve C is split into two portions C1 and C2. On C1, u is given and on C2, ∂u/∂n is specified. Again, both the interior and exterior cases may arise.
(4) Impedance or Robin problem. In this case hu+∂u/∂n is prescribed on the closed curve C, h being a known function. It can be regarded as including the three preceding cases.
Before leaving the subject we give a theorem which guarantees that the solution to the boundary value problem is unique in suitable circumstances.
THEOREM 10.4.1
Let the partial differential equation (10.3.1) be elliptic (ac>b2) in a simply connected domain D with g≤0 and a, c positive. Then, if u=0 on a simple closed curve C in D, u=0 at all points inside C.
Remark that a and c must have the same sign in order to meet the condition for an elliptic equation.
PROOF Change the variables so that the equation goes over to its normal form (10.4.7). The transformation
maps D into a domain D1 and C into a simple closed curve C1. Note that the mapping is one-to-one because the Jacobian
If vanishes so does on account of (10.4.3), contrary to a characteristic having a definite slope. Therefore the Jacobian is non-zero and the mapping is one-to-one.
Also, from (10.4.8) and the hypotheses of the theorem, g2≤0.
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Suppose that u is continuous and bounded inside C. It will have the same properties inside C1. As a continuous function u attains both its upper bound M (which cannot be negative since u=0 on C1) and its lower bound m. Let M be achieved at λ=λ0, μ=μ0. It follows that
where (f)0 means the value of f at λ=λ0, μ=μ0. Putting λ=λ0, μ=μ0 in (10.4.7) we obtain (10.4.10) But, for a maximum
(10.4.11) (10.4.12)
If g=0 then g2=0 and we could satisfy (10.4.10) and (10.4.11) but not (10.4.12); the same is true if M=0. On the other hand, if M>0 and g<0, then g2M<0 and to satisfy (10.4.10) one at least of (∂2u/∂λ2)0, (∂2u/ ∂μ2)0 must be positive, contradicting (10.4.11). Thus the conditions for a maximum cannot be met inside C1 and so u must not be greater than zero inside C1.
A similar argument based on the lower bound m reveals that u cannot be less than zero inside C1. Hence u is zero inside C1 and therefore zero inside C. The theorem is proved.
COROLLARY 10.4.1 (Uniqueness Property)
Under the conditions of Theorem 10.4.1 there is only one u which solves (10.3.1) and takes given values on C.
Another way of describing this corollary is to say that the solution of the interior Dirichlet problem is urique for an elliptic partial differential equation in which a and g have opposite signs.
PROOF Suppose there were two possible solutions u1 and u2. Then u1−u2 is zero on C and also satisfies (10.3.1). Consequently, Theorem 10.4.1 implies that u1−u2 vanishes inside C and the corollary has been demonstrated.
It is important to observe that when a and g have the same signs Theorem 10.4.1 fails, in general. A counter- example is furnished by
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where m and n are integers. The solution
vanishes on the perimeter of the square 0≤x≤π, 0≤y≤π but is clearly not zero at all interior points.