Global Journal of Mathematical Analysis, 1 (2) (2013) 48-52
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°Science
Publishing Corporation
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The Cauchy problem and Hadamard’s example
E. Mogileva* and O. Yaremko
Penza State University, 440026, Penza, Russia
*Corresponding author E-mail: ElenaSergIvan@yandex.ru
Abstract
Integral representation for harmonic function in the ring. We prove the existence and uniqueness of solutions of
the Cauchy problem for the Laplace equation in the circle. Integral representation for the solution of the Cauchy
problem was found.
Keywords: Cauchy problem, examine Hadamard’s.
1
Introduction
We now examine Hadamard’s example of an ill-posed problem [1, 2]. Consider Cauchy problem for the Laplace’s
equation
µ
¶
1 ∂
∂u
1 ∂2u
∆u ≡
r
+ 2
=0
r ∂r
∂r
r ∂ϕ2
¯
√ cosnϕ
∂u ¯¯
u|r=1 = 0,
,
= e− n
¯
∂n r=1
n
n is an add integer. Then
u(r, ϕ) = e−
√ rn
n
− r−n
cos(nϕ)
2n2
is the unique solution Cauchy problem. But now consider large n. We have
√
||e−
n cosnϕ
n
|| → 0
for any Sobolev norm [3] because the beats all polynomials in n (from any finite order of derivatives). Hence we
can choose n so large that the boundary condition is as close to 0 as we please. But the solution u(r, ϕ) grows large
at any fixed line r = constant as n → ∞ ;
||u(r, ϕ) − u0 || → ∞
hence u0 ≡ 0 solves
µ
¶
∂u
1 ∂2u
1 ∂
r
+ 2
=0
∆u ≡
r ∂r
∂r
r ∂ϕ2
¯
∂u ¯¯
u|r=1 = 0,
= 0,
∂n ¯r=1
uniquely. So this problem does not depend continuously on the data, and hence is not well-posed. Therefore it is
not something we can safely ignore, well- posedness.
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Global Journal of Mathematical Analysis
In this paper the analytical solution of the Hadamard problem [4] in a ring is resulted. The received results can
be applied for solving inverse boundary problems, type of the Hadamard problem. Let us arrive at the function
u (r, ϕ) such that u is the solution of the Laplace’s equation
µ
¶
1 ∂
∂u
1 ∂2u
∆u ≡
r
+ 2
=0
(1)
r ∂r
∂r
r ∂ϕ2
in the ring K: r < |z| < 1; z = reiϕ , u(r, ϕ is continuous in the closed ring K̄. Moreover, for the function u the
conditions hold
¯
∂u ¯¯
u|r=1 = g(ϕ),
= h (ϕ) ,
(2)
∂n ¯r=1
h (ϕ) is continuous, 2π- periodic function. Let us show that the Hadamard problem solution is
1
u (r, ϕ) =
2π
1
+
2π
1
+
2π
2
Z
Z
∞
e
0
Z2π
∞
−ε
e
0
Z∞
0
Z2π
−ε
0
£
¡
¢¤
Re exp εze−it g (t) dt+
0
·
µ it ¶¸
Z2π
e
1
1
Re
exp ε
g (t) dt −
h (ψ) dψ · ln r+
z
z
2π
0
´
1 ³ −λ
e r − e−λr
λ
Z2π
eλ cos(ϕ−ψ) cos (λ sin (ϕ − ψ)) h (ψ) dψ dλ.
(3)
0
The main lemma
Lemma 2.1 If the function w = f (z) is analytic in the ring r < |z| < 1, and continuous in the closed ring
r ≤ |z| ≤ 1, then
Z ∞
Z
z
1
dς
−ε
f (z) =
e
eε ς f (ς) +
2πi 0
ς
Cr1
1
+
2πi
Z
Z
∞
e
−ε
0
1 ες
e z f (ς) dς, r < |z| < 1.
z
(4)
Cr2
Proof: Consider the sequence of function
Z N
Z
z
1
dς
fN (z) =
e−ε
eε ς f (ς) +
2πi 0
ς
Cr1
1
+
2πi
Z
Z
N
e
0
−ε
1 ες
e z f (ς) dς, r < |z| < 1.
z
Cr2
From the Cauchy formula [6] it follows that the contour of integration can be transferred to a circles C1 , Cr . We
get
Z
Z N
z
1
dς
fN (z) =
e−ε eε ς f (ς) +
2πi 0
ς
C1
+
1
2πi
Z
0
N
Z
e−ε
C2
1 ες
e z f (ς) dς, r < |z| < 1.
z
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Global Journal of Mathematical Analysis
Interchanging the order of integration in each summand and integrating the inner integrals with respect to ε, we
obtain
#
ς
Z "
1
1
e−N (1− z )
−
−
f (ς) dς, r < |z| < 1.
2πi
ς −z
ς −z
Cr
We pass to the limit as N → ∞ and get the following estimates:
¯
¯
¯
¯
Z " −N (1− zς ) #
Z −N (1−|z|)
¯ 1
¯
e
e
e−N (1−|z|)
¯
¯≤ 1
f
(ς)
dς
|f (ς)| |dς| ≤
M,
¯ 2πi
¯
ς −z
1 − |z|
1 − |z|
¯
¯ 2π
C1
C1
Cr
Cr
¯
¯
¡ r¢
¡ r¢
¯
¯
Z " −N (1− zς ) #
Z −N 1− |z|
−N 1− |z|
¯ 1
¯
e
1
e
e
¯
|f (ς)| |dς| ≤
M,
f (ς) dς ¯¯ ≤
¯ 2πi
ς −z
|z| − r
|z| − r
¯
¯ 2π
M = max |f (z)| .
r≤|z|≤1
From the estimates it follows that the summands
Z " −N (1− zς ) #
Z −N (1− zς )
1
e
1
e
f (ς) dς,
f (ς) dς, r < |z| < 1,
2πi
ς −z
2πi
ς −z
C1
Cr
tend to 0. This is uniform convergence in z in any of the closed ring r1 ≤ |z| ≤ r2 . Thus we have lim fN (z) = f (z) .
N →∞
3
Main results
Theorem 3.1 Let us arrive at the function u (r, ϕ) such that u is the solution of the Laplace’s equation
µ
¶
1 ∂
∂u
1 ∂2u
∆u ≡
r
+ 2
=0
r ∂r
∂r
r ∂ϕ2
(5)
in the ring K:r < |z| < 1, u is continuous in the closed ring K̄. Moreover, for the function u the conditions hold
¯
∂u ¯¯
u|r=1 = g (ϕ) ,
= 0,
(6)
∂n ¯r=1
g (ϕ) is continuous, 2π- periodic function, the solution of the problem (7)-(8) is expressed in the form :
1
u (r, ϕ) =
2π
1
+
2π
Z
Z
∞
e
0
−ε
e
0
£
¡
¢¤
Re exp εze−it g (t) dt+
0
Z2π
∞
Z2π
−ε
0
µ it ¶¸
e
1
Re
exp ε
g (t) dt, r < |z| < 1,
z
z
·
z = reiϕ .
/
/
Proof: By the Cauchy – Riemann equations [5], so that −ivϕ = rur .We get v (1, ϕ) = 0. It follows that we get
f |r=1 = g (ϕ) for the function f = u + iv.
Substituting r1 = r2 = 1 in (4), we obtain
1
f (z) =
2π
+
1
2π
Z
0
∞
Z
Z2π
∞
e
0
−ε
¡
¢
exp εze−it g (t) dt+
0
Z2π
e−ε
0
µ it ¶
1
e
exp ε
g (t) dt, r < |z| < 1.
z
z
We take a real part of f (z). Theorem is proved.
51
Global Journal of Mathematical Analysis
Theorem 3.2 Let us arrive at the function u (r, ϕ) such that u is the solution of the Laplace’s equation
µ
¶
∂u
1 ∂2u
r
+ 2
=0
∂r
r ∂ϕ2
1 ∂
∆u ≡
r ∂r
(7)
in the ring K:r < |z| < 1,u is continuous in the closed ring K̄. Moreover, for the function u the conditions hold
¯
∂u ¯¯
= h(ϕ),
∂n ¯r=1
u|r=1 = 0,
g (ϕ) is continuous, 2π- periodic function, The solution of the problem (7)-(8) is expressed in the form:
Z2π
1
u (r, ϕ) = −
2π
1
+
2π
Z∞
0
h (ψ) dψ · ln r+
0
´
1 ³ −λ
e r − e−λr
λ
Z2π
eλ cos(ϕ−ψ) cos (λ sin (ϕ − ψ)) h (ψ) dψ dλ.
0
z = reiϕ .
Proof: Consider the Cauchy auxiliary problem:
µ
¶
∂ ũ
1 ∂ 2 ũ
r
+ 2
= 0,
∂r
r ∂ϕ2
1 ∂
∆ũ ≡
r ∂r
ũ|r=1
¯
∂ ũ ¯¯
= g(ϕ),
= 0,
∂n ¯r=1
where
Z2π
1
g(ϕ) =
2π
ψh (ϕ + ψ) dψ.
0
The solution of the problem (7)-(8) is by theorem 2:
Z
1
u (r, ϕ) =
2π
1
+
2π
Z
∞
∞
e
0
e
0
£
¡
¢¤
Im exp εze−it g(t)dt+
0
Z2π
−ε
0
Z2π
−ε
·
µ it ¶¸
1
e
Im
exp ε
g(t)dt, r < |z| < 1.
z
z
Integrating by parts, and using
g 0 (ϕ) = h (ϕ) ,
1
u
e(r, ϕ) =
2π
we get (3).
Z2π
ψu (r, ϕ + ψ) dψ,
0
(8)
52
Global Journal of Mathematical Analysis
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