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B5b Applied Partial Differential Equations 3–
3 Second-order semilinear
equations
3.1 Introduction
Now we consider second-order scalar equations of the form
a(x, y) ∂^2 u ∂x^2
2b(x, y) ∂^2 u ∂x∂y
c(x, y) ∂^2 u ∂y^2
= f
x, y, u, ∂u ∂x
∂u ∂y
This is called a semilinear equation because the coefficients a, b and c are independent of u and its derivatives. Note that a second-order scalar equation like (3.1) may always be transformed into a first-order system by setting p = ∂u/∂x, q = ∂u/∂y. Where one of the independent variables is clearly supposed to represent time, we will also write (3.1) in the form
a(x, t) ∂^2 u ∂t^2
- 2b(x, t) ∂^2 u ∂x∂t
- c(x, t) ∂^2 u ∂x^2 = f
x, t, u, ∂u ∂x
∂u ∂t
We will concentrate on three canonical examples of particular importance, namely the wave equation ∂^2 u ∂t^2
∂^2 u ∂x^2
= 0, (3.3a)
the heat equation ∂u ∂t
∂^2 u ∂x^2 = 0, (3.3b)
and Laplace’s equation ∂^2 u ∂x^2
∂^2 u ∂y^2 = 0. (3.3c)
Before analysing PDEs of the general form (3.1), we begin by reviewing some standard meth- ods for solving these three simple special cases.
3.2 Summary of standard methods for second-order PDEs
Separation of variables
This is a method for solving linear PDEs on a fixed finite domain (in at least one variable). The idea is to write the solution in the form
u(x, y) =
∑^ ∞
n=
Xn(x)Yn(y), (3.4)
where the functions Xn and Yn satisfy ordinary differential equations.
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Example 3.1 The general solution of
∂^2 u ∂x^2 − ∂^2 u ∂y^2 = 0, (3.5)
subject to u(x, 0) = u(x, a) = 0 may be written in the form
u(x, y) =
∑^ ∞ n=
{ an cos
( (^) nπx a
)
( (^) nπx a
)} sin
( (^) nπy a
)
. (3.6)
The arbitrary constants an and bn may be found by Fourier analysis if u and ∂u/∂x are given on (say) x = 0.
Transforms
Fourier and Laplace transforms (and there are many others) are useful for solving linear PDEs on infinite or semi-infinite domains. They turn differential operators into algebraic operators and thus reduce PDEs to ODEs.
Example 3.2 The heat equation ∂u ∂t = ∂^2 u ∂x^2 , (3.7)
subject to u → 0 as x → ±∞ and u = u 0 (x) when t = 0 may be solved by taking a Fourier transform in x. The Fourier transform of u is defined by
ˆu(t, k) =
∫ (^) ∞
−∞
u(x, t)e−ikx^ dx, (3.8)
and u may be recovered from ˆu by using the inversion formula
u(x, t) = 1 2 π
∫ (^) ∞
−∞
ˆu(t, k)eikx^ dx. (3.9)
The heat equation is transformed to
∂ uˆ ∂t = −k^2 uˆ ⇒ uˆ = ˆu 0 e−k (^2) t , (3.10)
where uˆ 0 is the Fourier transform of u 0. Then the convolution theorem gives
u(x, t) = u 0 (x)? f (x, t) =
∫ (^) ∞
−∞
u 0 (ξ)f (x − ξ, t) dξ, (3.11)
where f^ ˆ (t, k) = e−k^2 t. (3.12)
It is straightforward to invert this transform and thus find
f (x, t) = 1 2
√ πt
exp
( − x^2 4 t
)
. (3.13)
(This is the Green’s function for the heat equation.)
3–4 OCIAM Mathematical Institute University of Oxford
subject to u(x, 0) = u(∞, t) = 0, u(0, t) = tm^ for some constant m. Try rescaling the dependent and independent variables as follows:
u = a¯u, t = bt,¯ x = cx,¯ (3.24)
where a, b and c are constants. Then the problem becomes
a b
∂ ¯u ∂¯t = a c^2
∂^2 u¯ ∂ x¯^2 , (3.25)
subject to ¯u = 0 at ¯t = 0 and as ¯x → ∞, and
a¯u = bmt¯m^ on x¯ = 0. (3.26)
Hence the PDE and boundary conditions are invariant under this rescaling if a, b and c satisfy
c =
√ b, a = bm, (3.27)
which may be rewritten as
x ¯x =
√ t ¯t ,^
u ¯u =
( t ¯t
)m , (3.28)
or x ¯ √¯ t
= x √ t
, u¯ ¯tm^ =^
u tm^
. (3.29)
Thus, u, t and x may be rescaled and, provided the ratios u/tm^ and x/
√ t are preserved, the problem is invariant, which implies that the solution is invariant. Hence, the solution can only depend on these two groups and may, therefore, be written in the form
u tm^ = f
( √x t
) , or u = tmf (η), η = √x t
, (3.30)
for some function f. If this form of u is substituted into the heat equation, f is found to satisfy
d^2 f dη^2 +^
η 2
df dη −^ mf^ = 0,^ (3.31)
with the boundary conditions f (0) = 1, f → 0 as η → ∞. The simplest case is m = 0, when the solution is f (η) = √^2 π
∫ (^) ∞
η/ 2
e−s 2 ds = erfc(η/2), (3.32)
the complementary error function.
3.3 Cauchy data
Cauchy data for the general second-order quiasi-linear PDE (3.1) is to specify u and its normal derivative on some curve Γ in the (x, y) plane, i.e.
x = x 0 (s), y = y 0 (s), u = u 0 (s), ∂u ∂n
= v 0 (s), (3.33)
where s parametrises Γ. This is equivalent to specifying u and both its first derivatives on Γ, since du 0 ds
∂u ∂x
dx 0 ds
∂u ∂y
dy 0 ds
B5b Applied Partial Differential Equations 3–
and
v 0 (s) = ∂u ∂n
∂u ∂x
dy 0 ds
∂u ∂y
dx 0 ds
dx 0 ds
dy 0 ds
) 2 }−^1 /^2
may be solved simultaneously for ∂u/∂x and ∂u/∂y. We may therefore replace (3.33) with
x = x 0 (s), y = y 0 (s), u = u 0 (s), ∂u ∂x
= p 0 (s), ∂u ∂y
= q 0 (s). (3.36)
A necessary condition for the solution u(x, y) to be uniquely defined in a neighbourhood of Γ is for the second derivatives of u to be determined on Γ. Now, if we differentiate the initial data along Γ and use the chain rule, we obtain
dp 0 ds
dx 0 ds
∂^2 u ∂x^2
dy 0 ds
∂^2 u ∂x∂y
dq 0 ds
dx 0 ds
∂^2 u ∂x∂y
dy 0 ds
∂^2 u ∂y^2
Along with the PDE (3.1), this gives us a system of three equations for the three second partial derivatives of u, and the determinant of this system is
∣∣ ∣∣ ∣∣ ∣∣ ∣
a 2 b c dx 0 ds
dy 0 ds
dx 0 ds
dy 0 ds
= a
dy 0 ds
− 2 b
dx 0 ds
dy 0 ds
dx 0 ds
A necessary condition for the Cauchy data (3.36) to determine u locally is, therefore,
a
dy 0 ds
− 2 b dx 0 ds
dy 0 ds
dx 0 ds
3.4 Characteristics
Definition
As for first-order equations, we define characteristics to be curves in the (x, y) plane on which Cauchy data do not determine a unique solution. If such a curve is parametrised by x = x(τ ), y = y(τ ) then, from (3.39), we have
a
dy dτ
− 2 b dx dτ
dy dτ
dx dτ
so the slopes of the characteristics satisfy
dy dx
= λ where aλ^2 − 2 bλ + c = 0 (3.41)
Characteristics may also be defined as curves across which the first derivatives of u may be discontinuous. It is left as an exercise to show that this alternative definition leads to the same equation (3.41) for the characteristic slopes.
B5b Applied Partial Differential Equations 3–
and so forth. The PDE (3.1) is clearly transformed to an equation with an analogous form, that is,
α(ξ, η) ∂^2 u ∂ξ^2
- β(ξ, η) ∂^2 u ∂ξ∂η
- γ(ξ, η) ∂^2 u ∂η^2 = φ
ξ, η, u, ∂u ∂ξ
∂u ∂η
We know that the characteristics are given by ξ = const and η = const, so the roots of the quadratic form
α
dη dτ
− 2 β
dη dτ
dξ dτ
dξ dτ
must be dξ/dτ = 0 and dη/dτ = 0. It follows that α = γ = 0, so that (3.47) takes the form
∂^2 u ∂ξ∂η = φ
ξ, η, u, ∂u ∂ξ
∂u ∂η
This is the so-called canonical form for second-order hyperbolic PDEs.
Example 3.6 For the PDE ∂^2 u ∂x^2 − ∂^2 u ∂y^2 = f (x, y), (3.50)
the characteristics are given by ( dy dx
) 2 − 1 = 0 ⇒ y ± x = const. (3.51)
So we set ξ = x − y, η = x + y and, by changing variables, obtain
∂^2 u ∂ξ∂η =^1 4 f
( ξ + η 2 , η^ −^ ξ 2
) = φ(ξ, η), say. (3.52)
This may now be integrated directly to give the so-called D’Alembert solution
u =
∫ ∫ φ(ξ, η) dξdη + h 1 (ξ) + h 2 (η). (3.53)
Elliptic equations
For elliptic equations, the roots of the characteristic equation (3.41) are complex conjugates, say dy dx = λR(x, y) ± iλI (x, y). (3.54)
Suppose that the integrals of these ODEs can be written in the form
ξ(x, y) ± iη(x, y) = const, (3.55)
for some functions ξ and η. Then we use ξ and η as new variables. Again, the transformed equation must be of the form (3.47) but, this time, the roots of
α
dη dτ
− 2 β
dη dτ
dξ dτ
dξ dτ
are dξ/dτ ± idη/dτ = 0, which implies that β = 0, γ = α. The canonical form for elliptic PDEs is, therefore, ∂^2 u ∂ξ^2
∂^2 u ∂η^2 = φ
ξ, η, u,
∂u ∂ξ
∂u ∂η
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Example 3.7 For the PDE
y^2 ∂
(^2) u ∂x^2
(^2) u ∂x∂y
(^2) u ∂y^2 = 0, (3.58)
the characteristics satisfy
dy dx = (1 ± i) x y ⇒ (1 ± i)x^2 − y^2 = const. (3.59)
If we choose ξ = x^2 , η = x^2 − y^2 , then the equation transforms to
∂^2 u ∂ξ^2
∂^2 u ∂η^2 = − 1 2 ξ
∂u ∂ξ
( ξ + η 2 ξ(ξ − η)
) ∂u ∂η
. (3.60)
Parabolic equations
For parabolic PDEs, there is one repeated characteristic slope
dy dx
b a
which we suppose has the solution η(x, y) = const. Then any convenient linearly independent function ξ(x, y) may be chosen, and the PDE (3.1) transforms to the canonical form
∂^2 u ∂ξ^2
= φ
ξ, η, u, ∂u ∂ξ
∂u ∂η
3.6 Hyperbolic equations
Non-Cauchy data
As stated in Section 3.3, Cauchy data for second-order equations specifies both u and its normal derivative on a plane curve Γ. In general, though, there may be boundaries on which it is appropriate to give just one boundary condition or even none.
Example 3.8 Consider a string of length L and density (i.e. mass per unit length) ρ, stretched to a tension T between the points x = 0 and x = L. It may be shown that small transverse displacements u(x, t) satisfy the wave equation ∂^2 u ∂t^2 = c^2 ∂^2 u ∂x^2 , (3.63)
where t is time and c^2 = T /ρ. Suppose we wish to determine u(x, t) in the region 0 ≤ x ≤ L, 0 ≤ t ≤ T. To do so, we need to specify the initial displacement u(x, 0) = u 0 (x) and the initial velocity ∂u/∂t(x, 0) = v 0 (x); these correspond to Cauchy data on the initial curve t = 0, 0 ≤ x ≤ L. On the boundary curves x = 0 and x = L, we give just one boundary condition, namely that the displacement is zero: u(0, t) = u(L, t) = 0. Finally, on the boundary t = T we give no boundary conditions at all. The number of boundary conditions applied on each boundary and the characterstics are illustrated in Figure 1.
Example 3.8 illustrates that, in general, the number of boundary conditions needed on any boundary is equal to the number of characteristic families travelling out of that boundary. Where two sets of characteristics travel out, a boundary is called time-like, and two conditions must be given. On a space-like boundary, with one characteristic family travelling in and one travelling out, just one condition must be given. Finally, no conditions may be given on a boundary where all the characteristics travel in.
3–10 OCIAM Mathematical Institute University of Oxford
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....................... .................
......................
............................... ............................... ............................. ............................ ............................ ......................... ........................ ....................... ..................... ..................... .................... ..................... .................... ..................... ..................... ................... ....................... .......................... ............................... ........................................ ................................................... ........................................................... ..............................................................
B
A
η P (ξ, η)
x
y
D
Figure 2: The integration region D for Riemann’s method.
Now we integrate over the region illustrated in Figure 2, namely the region bounded by Γ and the lines x = ξ and y = η, where ξ and η are constants. The intersection between y = η and Γ is labelled A, the point where x = ξ intersects Γ is labelled B, and the point (ξ, η) is labelled P. By using (3.66) and Green’s theorem, we obtain ∫ ∫
D
{RL[u] − uL?[R]} dxdy =
∂D
R
∂u ∂y
dy + u
∂R
∂x
− bR
dx
where R is called the Riemann function. The idea is to choose the properties of R to simplify (3.67) as much as possible. On the left-hand side of (3.67), we use the fact that L[u] = f and choose R to satisfy L?[R] = 0. On the right-hand side, we note that dy = 0 on AP and that dx = 0 on P B. The latter integral along P B we integrate by parts:
∫ ∫
D
Rf dxdy =
∫ A
P
u
∂R
∂x
− bR
dx +
∫ P
B
u
aR −
∂R
∂y
dy + [Ru]PB
∫ B
A
R
∂u ∂y
dy + u
∂R
∂x − bR
dx
Now we suppose that ∂R/∂x = bR on AP and that ∂R/∂y = aR on BP , to eliminate the first two integrals. If we also choose R = 1 at the point P , we end up with
u(ξ, η) =
D
Rf dxdy + R(B)u(B) −
∫ B
A
R
∂u ∂y
dy + u
∂R
∂x
− bR
dx
This gives us the solution at any arbitrary point (ξ, η) in terms of f , u and its derivatives on Γ (i.e. the Cauchy data), and the Riemann function R, so if we can find R then the problem is solved in principle.
B5b Applied Partial Differential Equations 3–
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.................................. .............................. ............................. ............................. ............................. ............................ ............................ ............................. ............................ ........................... ........................... ........................... .......................... ......................... ......................... ...................... ........................ ...................... ...................... ...................... ..................... .................... ..................... ..................... .................... ................... .................... ..................... .................... ..................... ..................... ............
x
y
D
R = 0
R 6 = 0
Figure 3: Schematic showing the behaviour of the Riemann function.
To summarise, the properties we require of R are
L?[R] =
∂^2 R
∂x∂y
∂x
(aR) −
∂y
(bR) + cR = 0, x < ξ, y < η
∂R ∂x = bR y = η,
∂R ∂y = aR x = ξ,
R = 1 (x, y) = (ξ, η).
It may be shown that the problem (3.69) determines R(x, y; ξ, η) uniquely. Notice that R is independent of the function f and the boundary data applied to u: it depends only on the orginal differential operator L. Thus, it may be found “once and for all” and then applied to any boundary data and any right-hand side f. However, finding R explicitly is very difficult except in simple special cases. It is generally most useful for theoretical purposes, for example proving that u depends continuously on its initial data. Finally, we note an alternative definition of R in terms of generalised functions. Suppose we form a domain S, in which the solution is to be found, between Γ and a second curve γ. Now we solve the problem
L?[R] = δ(x − ξ)δ(y − η) in S,
R =
∂R
∂n = 0 on γ,
for R backwards, starting from γ, where δ is the Dirac delta function. As illustrated in Figure 3, since R and its first derivatives are zero on γ, R is identically zero outside the
B5b Applied Partial Differential Equations 3–
and F satisfies
d^2 F ds^2
1 s
dF ds
with F (0) = 1. The solution is the Bessel function F = J 0 (s),^1 so the Riemann function for the telegraph equation is
R =
{ J 0
( 2
√ (ξ − x)(η − y)
) x < ξ, y < η, 0 otherwise.
(3.80)
3.7 Elliptic equations
Well-posed boundary data
The canonical form for second-order elliptic PDEs is
∂^2 u ∂x^2
∂^2 u ∂y^2 = f
x, y, u, ∂u ∂x
∂u ∂y
The operator on the left-hand side is referred to as the Laplacian, for which the symbols ∇^2 u or 4 u are often used as shorthand. Recall the Cauchy–Kowalevski theorem, which states that a unique solution to (3.81) exists in a neighbourhood of an initial curve Γ if u, ∂u/∂x and ∂u/∂y are specified analytic functions on Γ and satisfy the condition (3.39). However, the Cauchy problem is ill posed for elliptic equations. This manifests itself in several ways, including the following.
- Small changes in the initial data may lead to arbitrarily large changes in the solution. This makes it impossible to compute meaningful solutions numerically.
- Although a unique solution exists in a neighbourhood of Γ, that neighbourhood may be arbitrarily small, in that a singularity may form arbitrarily close to Γ.
It transpires that appropriate boundary data for (3.81) is to give just one boundary condition on u everywhere on a closed curve. We show below that this does indeed give a unique solution for Poisson’s equation, a special case of (3.81). This should be constrasted with the situation for hyperbolic equations where, as illustrated in Figure 1, either 0, 1 or 2 boundary conditions are given at each point on a closed curve, depending on the number of inward-travelling characteristics.
Uniqueness theorems for Poisson’s equation
Poisson’s equation is a special case of (3.81), in which the right-hand side f depends only on x and y. Consider the so-called Dirichlet problem, namely Poisson’s equation for u(x, y) in some domain D, with u given on the boundary of D:
∇^2 u = ∂^2 u ∂x^2
∂^2 u ∂y^2
= f (x, y) (x, y) ∈ D,
u = g(x, y) (x, y) ∈ ∂D.
(^1) the other linearly independent solution to this second-order ODE is Y 0 (s), which is singular as s → 0
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We will now show that, if a solution of (3.82) exists, then it is unique. Suppose there exist that two solutions u 1 and u 2 of (3.82). If φ = u 1 − u 2 , then φ satisfies the homogeneous Dirichlet problem
∇^2 φ = 0 (x, y) ∈ D, φ = 0 (x, y) ∈ ∂D.
Now consider the Dirichlet integral ∫ ∫
D
∇ · (φ∇φ) dxdy =
D
∂x
φ
∂φ ∂x
∂y
φ
∂φ ∂y
dxdy. (3.84)
This may be written in two ways (i) by expanding out the derivatives on the left-hand side, (ii) by using Green’s theorem: ∫ ∫
D
φ∇^2 φ + |∇φ|^2
dxdy =
∂D
φ ∂φ ∂n
ds, (3.85)
where n and s refer to the outward-pointing normal and arclength respectively on ∂D. Now we simplify by using the properties (3.83) of φ: ∫ ∫
D
|∇φ|^2 dxdy = 0. (3.86)
Since the integrand is non-negative (and assumed to be continuous), it must be identically zero in D. It follows that φ is constant in D where, since φ = 0 on ∂D, that constant must be zero. Hence u 1 ≡ u 2 , so the solution to (3.82), if it exists, is unique. Next consider the Neumann problem, in which the normal derivative of u, rather than u itself, is specified on ∂D: ∇^2 u = f (x, y) (x, y) ∈ D, ∂u ∂n
= g(x, y) (x, y) ∈ ∂D.
By Green’s theorem, ∫ ∫
D
∇^2 u dxdy =
∂D
∂u ∂n
ds ⇒
D
f dxdy =
g ds. (3.88)
This is the solvability condition for (3.87); if (3.88) is not satisfied, then (3.87) has no solution. Now suppose that (3.87) is satisfied and let u 1 and u 2 be two solutions of (3.87). Then φ = u 1 − u 2 satisfies the homogeneous Neumann problem
∇^2 φ = 0 (x, y) ∈ D, ∂φ ∂n
= 0 (x, y) ∈ ∂D,
and it is straightforward to show (using the same approach as for the Dirichlet problem) that φ must therefore be constant. Thus, if the solvability condition (3.88) is satisfied, then the solution to (3.87), if it exists, is unique up to the addition of an arbitrary constant. This behaviour of the solutions of the Neumann problem is an instance of the Fredholm alternative. The homogeneous problem (3.89) has the nontrivial solution φ = const. Thus there is no solution to the inhomogeneous version (3.87) unless an orthogonality condition (3.88) is satisfied. If so, then the solution to (3.87) is nonunique since an arbitrary multiple of the homogeneous solution may be added.
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Notice that the Laplacian is a self-adjoint operator, so we consider the integral ∫ ∫
S
u∇^2 G − G∇^2 u
dxdy =
S
∇ · {u∇G − G∇u} dxdy
∂S
u
∂G
∂n
− G
∂u ∂n
ds, (3.96)
where n and s refer to the outward-pointing normal and arclength on ∂D, and G is the Green’s function. If we choose G to satisfy
∇^2 G = δ(x − ξ)δ(y − η) in D, G = 0 on ∂D,
where δ is the Dirac delta-function, then we have
u(ξ, η) =
D
Gf dxdy +
∂D
g
∂G
∂n ds. (3.98)
If the Green’s function G can be found from (3.97), then (3.98) gives the solution at any point (ξ, η) ∈ D in terms of G and the given functions f and g. Now we have to find a function G whose Laplacian is equal to a delta function. Thus G should satisfy ∇^2 G = 0 when (x, y) 6 = (ξ, η) and should have some sort of singularity as (x, y) → (ξ, η), and it turns out that the correct singularity is
G ∼
2 π log
(x, y) − (ξ, η)
- O(1) as (x, y) → (ξ, η). (3.99)
It is readily verified that log
(x, y) − (ξ, η)
satisfies Laplace’s equation for all (x, y) 6 = (ξ, η). Hence (^) ∫ ∫
D
φ∇^2 G dxdy =
S
φ∇^2 G dxdy, (3.100)
where S is a disc of radius centred at (ξ, η) and φ(x, y) is any suitably differentiable test function. Recalling that ∇^2 G is a distribution, we evaluate the right-hand side as
∫ ∫
S
φ∇^2 G dxdy =
S
G∇^2 φ dxdy +
∂S
φ
∂G
∂n
− G
∂φ ∂n
ds. (3.101)
Now introducing local polar coordinates x = ξ + r cos θ, y = η + r sin θ, letting → 0 and using the asymptotic behaviour (3.99), we find that
∫ ∫
S
φ∇^2 G dxdy ∼
∫ (^2) π
0
0
2 π
log(r)∇^2 φ r drdθ
∫ (^2) π
0
2 πr
φ(ξ, η)r drdθ −
∫ (^2) π
0
2 π
log() ∂φ ∂r
dθ. (3.102)
The first and third integrals on the right-hand side vanish as → 0, so we are left with ∫ ∫
S
φ∇^2 G dxdy → φ(ξ, η) as → 0 , (3.103)
B5b Applied Partial Differential Equations 3–
and hence (3.100) reduces to ∫ ∫
D
φ∇^2 G dxdy ≡ φ(ξ, η). (3.104)
Thus ∇^2 G is indeed equal to a delta function. An alternative statement of (3.97) is, therefore,
∇^2 G = 0 in D(ξ, η), G = 0 on ∂D,
G ∼
2 π
log
∣(x, y) − (ξ, η)
∣ (^) as (x, y) → (ξ, η).
The trick to obtaining G is thus to find a function that satisfies Laplace’s equation inside D and is equal to − 1 /(2π) log
(x, y) − (ξ, η)
on ∂D.
Example 3.11 Green’s function for a half-plane For the problem ∇^2 u = f (x, y) y > 0 , u = g(x) y = 0, u → 0 y → ∞ ,
(3.106)
the Green’s function satisfies
∇^2 G = 0 y > 0 , (x, y) 6 = (ξ, η), G = 0 y = 0, G ∼ 1 2 π log
∣∣ (x, y) − (ξ, η)
∣∣ , (x, y) → (ξ, η).
^ (3.107)
This problem is readily solved using the method of images. Consider an image singularity at the point (ξ, −η), that is the function 1 /(2π) log
∣∣ (x, y) − (ξ, −η)
∣∣
. This function clearly satisfies Laplace’s equation away from the singularity (which is outside the half-plane in which G is to be defined). It is also equal to 1 /(2π) log
∣∣ (x, y) − (ξ, η)
∣∣ on the line y = 0. Hence the Green’s function for this problem is
G(x, y; ξ, η) = 1 2 π log
∣∣ (x, y) − (ξ, η)
∣∣ − 1 2 π log
∣∣ (x, y) − (ξ, −η)
∣∣
= 1 4 π log
( (x − ξ)^2 + (y − η)^2 (x − ξ)^2 + (y + η)^2
)
. (3.108)
Example 3.12 Green’s function for a circle For a Dirichlet problem on a circular disc of radius a, the Green’s function satisfies
∇^2 G = 0 x^2 + y^2 < a^2 , (x, y) 6 = (ξ, η), G = 0 x^2 + y^2 = a^2 , G ∼ 1 2 π log
∣∣ (x, y) − (ξ, η)
∣∣ , (x, y) → (ξ, η).
(3.109)
The point (ξ, η) inside the disc has a corresponding image point (ξ′, η′) outside the disc, defined by ( ξ′ η′
)
a^2 ξ^2 + η^2
( ξ η
)
. (3.110)
B5b Applied Partial Differential Equations 3–
for some arbitrary functions h 1 and h 2. If we require u to be a real-valued function of x and y, then we must have h 2 = h 1 , so the general real-valued solution of Laplace’s equation is
u = <
[
f (z)
]
(where f = 2h 1 ). Solutions of Laplace’s equation may sometimes by found by spotting a function f (z) whose real part is equal to a given function on a given curve.
Example 3.13 Find a solution u(x, y) of Laplace’s equation that is equal to |x| on y = 0. The complex-valued function f (z) = z + 2i π z log z (3.119)
may be split into its real and imaginary parts as follows:
f (z) =
{( 1 − 2 θ π
) x − 2 y π log^ r
}
{( 1 − 2 θ π
) y + 2 x π log^ r
} , (3.120)
where (r, θ) are plane polar coordinates, i.e. x = r cos θ, y = r sin θ. Notice that, on y = 0, <(f ) is equal to x when θ = 0 and −x when θ = π; in other words, <(f ) is equal to |x| when y = 0. A suitable solution is, therefore,
u(x, y) = <
[ f (z)
]
( 1 − 2 π tan−^1 (y/x)
) x − y π log
( x^2 + y^2
)
. (3.121)
Another useful technique from complex variable theory is conformal mapping. This is a mapping from one complex variable z = x + iy to another ζ = ξ + iη, given by a functional relation ζ = g(z), where g is analytic and g′(z) 6 = 0. Conformal mapping may be used to transform a complicated solution domain to a simple one (such as a half-plane or a disc). This technique works because Laplace’s equation is invariant under conformal mapping; see a textbook on complex analysis e.g. Priestley^3 for more details.
Example 3.14 Consider the solution of Laplace’s equation for u(x, y) with u → 0 as x^2 + y^2 → ∞ and u = x on the line segment y = 0, − 1 ≤ x ≤ 1. This line segment is the image of the unit circle |ζ| = 1 under the conformal map
x + iy = z = 1 2
( ζ + 1 ζ
) ⇒ ζ = z +
√ z^2 − 1. (3.122)
The boundary conditions are mapped to
u = <(ζ) on |ζ| = 1, u → 0 as ζ → ∞, (3.123)
and the solution may then easily be spotted. On the unit circle, we have ζ¯ = ζ−^1 and, hence, <(ζ−^1 ) = <(ζ). A suitable solution for u is thus
u = <
( ζ−^1
) = <
( z −
√ z^2 − 1
)
. (3.124)
Contour plots of u in the ζ- and z-planes are shown in Figure 4.
(^3) H. A. Priestley, 1990 Introduction to Complex Analysis, revised edition. Oxford University Press.
3–20 OCIAM Mathematical Institute University of Oxford
-3 -2 -1 1 2 3
1
2
3
-3 -2 -1 1 2 3
1
2
3
x
y
ï0.
ï0.
ï0.
ï0.1 0 0.
d
j
Figure 4: Contour plots of the solution u given by (3.124) versus ξ = <(ζ) and η = =(ζ) (left plot); x and y (right plot). The contour values are u = − 0. 9 , − 0. 8 , · · · , 0. 8 , 0 .9.
|ω| = 1
ω = g(z)
<(z) <(ω) ζ
=(z) =(ω)
D
Figure 5: Schematic of the conformal transformation ω = g(z) mapping D to the unit disc and ζ to the origin.
