The decomposition of a solution of the quasilinear stochastic parabolic equation with weak source
We obtain conditions which guarantee the existence of a decomposition of a solution of the quasilinear stochastic parabolic equation with a weak source.
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| Cite this: | The decomposition of a solution of the quasilinear stochastic parabolic equation with weak source / S. Melnik // Theory of Stochastic Processes. — 2007. — Т. 13 (29), № 3. — С. 55–64. — Бібліогр.: 8 назв.— англ. |
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| citation_txt | The decomposition of a solution of the quasilinear stochastic parabolic equation with weak source / S. Melnik // Theory of Stochastic Processes. — 2007. — Т. 13 (29), № 3. — С. 55–64. — Бібліогр.: 8 назв.— англ. |
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Theory of Stochastic Processes
Vol. 13 (29), no. 3, 2007, pp. 55–64
UDC 519.21
SERGEY MELNIK
THE DECOMPOSITION OF A SOLUTION OF THE QUASILINEAR
STOCHASTIC PARABOLIC EQUATION WITH WEAK SOURCE
We obtain conditions which guarantee the existence of a decomposition of a solution
of the quasilinear stochastic parabolic equation with a weak source.
1. Definitions, notations, formulation of the problem.
On a complete probability space (Ω,F, (Ft)t≥0,P), we consider the Cauchy problem
(1) du(t, x) = auxx(t, x)dt + b|u(t, x)|γ−1u(t, x)dw(t),
t ∈ [0; τ(ω)), x ∈ R1, u(0, x) = u0(x) ≥ 0.
Here: w(t) is a standard (Ft)t≥0 - adapted Wiener process in R1; a, b are positive con-
stants, γ ∈ (0; 1); τ(ω) is a stopping time; u0(x) is a nonrandom nonnegative function,
u0 ∈ W = W 1
2 (R1) ∩ Lγ+1(R1).
The alphabetic subindex by a function symbol means a partial derivative in the sense
of distributions with respect to appropriate variable.
We will use the next notations.
W 1
2 (R1) is the usual Sobolev space consisting of functions having the first derivative
(in the sense of distributions) square integrable on R1, and the space Lα(R1) consists of
functions integrable on R1 with degree α,
p = ‖v‖2 =
∫
|v(y)|2dy, z = ‖vy‖2 =
∫
|vy(y)|2dy, q = |‖v‖|γ+1 =
∫
|v(y)|γ+1dy.
f(t; u) =
1
2
‖u(t, •)‖2 − 1
2
‖u(0, •)‖2 +
a
2
t∫
0
‖ux(s, •)‖2ds − b
γ + 1
t∫
0
|‖u(s, •)‖|γ+1dw(s).
M =
(
(γ + 1)(3 − γ)B
2b
) 1
γ−1
, N = 2A − (γ − 2)B2, μ =
2aM2(γ−1)
(γ − 3)N
.
Definition 1.
A stochastic process u(t, x) is the solution of problem (1) if there exists a stopping
time τ(ω) such that ∀T ≥ 0
u ∈ L1(Ω; C([0; T ∧ τ(ω)); L2(R1))) ∩ L2([0; T ∧ τ(ω)) × Ω; W 1
2 (R1) ∩ Lγ+1(R1))
2000 AMS Mathematics Subject Classification. Primary 60H15, 35R60.
Key words and phrases. Decomposition, stochastic parabolic equation.
55
56 SERGEY MELNIK
and ∀g ∈ W, ∀t ≥ 0 the following equality holds with probability 1:
∫
u(t ∧ τ, x)g(x)dx −
∫
u0(x)g(x)dx =
= −a
t∧τ∫
0
∫
ux(s, x)gx(x)dxds + b
t∧τ∫
0
∫
|u(s, x)|γ−1u(s, x)g(x)dxdw(s).
Definition 2. A stochastic process u(t, x) admits a decomposition, if it can be written
down in such a form: u(t, x) = r(t)v (xrm(t)). Here, r(t) is a nonnegative stochastic
process, m ∈ R1, v ∈ W 1
2 (R1) ∩ Lγ+1(R1).
The process r(t) is called an amplitude, the function v(y) is called the space form of
the process u(t, x). The process xf (t) = min{x > 0 : v(xrm(t)) = 0} is called a front
point.
We introduce: τ0(ω) = inf{t ≥ 0 : r(t) = 0}, τ∞(ω) = inf{t ≥ 0 : r(t) /∈ R1},
τ(ω) = min{τ0(ω), τ∞(ω)}.
Remark 1. In this work, we are based on the ideas of work [1].
Remark 2. In the deterministic theory of partial differential equations with power non-
linearities, the main space form of the solutions is a nonnegative even function which
decreases on y ≥ 0, and its derivative is equal to zero if y = 0 ( [2, p.173-174]). In this
work, we consider the same space forms.
Formulation of the problem. Suppose that the parameters of problem (1) satisfy
the previous conditions. We will obtain conditions which guarantee the existence of a
decomposition of the solution of problem (1).
2. Main results.
Theorem 1. Suppose that the following conditions are satisfied.
1) γ ∈ (0; 1), N �= 0, B > 0.
2) r(t) is the solution of the problem:
(2) dr(t) = Ar2γ−1(t)dt + Brγ(t)dw(t), t ∈ [0; τ(ω)), r(0) > 0.
3) v(y) is the solution of the problem:
(3) μ
(
p
q
)2
vyy(y) + (γ + 1)
p
q
vγ(y) − v(y) = 0,
y ∈ R1, v(y) ≥ 0, v(−y) = v(y), vy(0) = 0, v(+∞) = 0.
4)
u0(x) = r(0)M
(
p
q
) 1
γ−1
v
(
xrγ−1(0)Mγ−1 p
q
)
.
Then the solution of problem (1) admits such a form of decomposition:
u(t, x) = r(t)M
(
p
q
) 1
γ−1
v
(
xrγ−1(t)Mγ−1 p
q
)
.
THE DECOMPOSITION OF A SOLUTION 57
3. Subsidiary results.
We suppose here that the conditions of Theorem 1 are satisfied.
Remark 3. According to Theorem 6 [3, p.246] for all A and B there exists a stopping
time τ(ω) such that the solution of problem (2) exists and is unique for all t ∈ [0; τ(ω)).
Moreover, according to [4, p.73], the solution of problem (2) is a nonnegative stochastic
process.
Lemma 1. Let δ > 0. There exists a unique nonnegative generalized solution of the
problem:
(4) VY Y (Y ) + (γ + 1)δ|V (Y )|γ−1V (Y ) − V (Y ) = 0,
Y ≥ 0, VY (0) = 0, V (0) = V0 ∈ (0; ((γ + 1)δ)
1
1−γ ).
This solution has the following properties.
1. There exists a front point Y0 = inf {Y ≥ 0 : V (Y ) ≤ 0} < +∞ such that V (Y ) > 0,
if Y ∈ [0; Y0) and V (Y ) = 0, if Y ≥ Y0.
2. VY Y (0) < 0.
3. There exists ε > 0 such that VY Y < 0, if Y ∈ (Y0 − ε; Y0).
Proof. The even function which is positive if Y ∈ [0; Y0), equal to the zero if Y ∈ [Y0; +∞)
and is the solution of problem (4) for all Y ∈ [0; Y0) is named the generalized solution
of problem (4)( [2, p.173-174]). As V − (γ + 1)δ|V |γ−1V is a continuous function, then,
according to [4, p.89], there exists the solution of problem (4). If Ṽ = V −1, then
ṼY Y = 2Ṽ −1
(
ṼY
)2
+ δ(γ + 1)Ṽ 2−γ − Ṽ , Y ≥ 0, Ṽ (0) = V −1
0 , ṼY (0) = 0.
The right part of this equation is a local Lipschitz function on Ṽ ∈ [V −1
0 ; +∞), ṼY ∈
(−∞; +∞). So, the solution of this problem is unique and the solution of problem (4) is
unique too. This solution is a continuous function and V (0) > 0. This means that either
V (Y ) > 0 ∀Y ≥ 0, or there exists a point Y0 > 0 such that Y0 = inf {Y ≥ 0 : V (Y ) ≤ 0} <
+∞. In the last case, we define V (Y ) = 0 if Y ≥ Y0. In the first case, we obtain a
nonnegative classical solution of problem (4). In the second case, we obtain a nonnegative
generalized solution of problem (4). The solution of problem (4) is not a constant because
if V (Y ) ≡ C, then C − (γ + 1)δCγ = 0 and either C = 0, or C = ((γ + 1)δ)
1
1−γ . But
0 < V (0) < ((γ + 1)δ)
1
1−γ . This means that V (Y ) is not a constant. If we reduce the
order of Eq. (4), we obtain
(5) VY (Y ) = −
√
C0 + V 2(Y ) − 2δV γ+1(Y ),
where C0 = 2δV γ+1
0 − V 2
0 > 0.
On the right part of this equation, we write a single ”-” as, according to Remark 2, the
space form must be a decreasing function on Y ≥ 0. According to (5), VY → −√
C0 < 0,
if V → 0. This means that there exists a point Y0 ∈ (0; +∞) such, that V (Y0) = 0.
Hence, there exists a generalized solution of problem (4) which is a nonnegative compactly
supported function.
Let us determine the type of convexity for V (Y ) in neighborhoods of the points Y = 0
and Y = Y0. Let us rewrite Eq. (4) in the next form: VY Y (Y ) = V (Y )− (γ + 1)δV γ(Y ).
Then VY Y (0) = V γ
0
(
V 1−γ
0 − (γ + 1)δ
)
< 0, because V (0) < ((γ+1)δ)
1
1−γ and γ ∈ (0; 1).
In a neighborhood of the point Y0, the values of the function V (Y ) are near zero. There-
fore, in this neighborhood, the inequality VY Y (Y ) = V (Y )γ
(
V 1−γ(Y ) − (γ + 1)δ
)
< 0 is
correct. Lemma 1 is proved.
58 SERGEY MELNIK
Lemma 2. If 0 < V0 < ((γ +1)δ)
1
1−γ , then the solution of problem (4) is bounded above
by
V̂ (Y ) =
{
V0 cos L̂Y, 0 ≤ Y ≤ π
2L̂
,
0, Y > π
2L̂
,
here L̂ =
√
δ(γ + 1)V γ−1
0 − 1.
Proof. The positive part of V̂ (y) is the solution of the problem:
(6) V̂Y (Y ) = −L̂
√
V 2
0 − V̂ 2(Y ), Y ≥ 0, V̂ (0) = V0.
We will prove that L̂2(V 2
0 − V 2) − C0 − V 2 + 2δV γ+1 ≤ 0 if V ∈ [0; V0]. If V = 0, then
δ(γ +1)V γ+1
0 −V 2
0 ≤ 2δV γ+1
0 −V 2
0 , because 0 < γ < 1. If V = V0, then both parts of the
inequality are equal to zero. The derivative of the left part of the inequality is equal to
2δ(γ + 1)V γ
(
1 −
(
V
V0
)1−γ
)
≥ 0. Hence, the left part of the inequality is an increasing
function, and its maximum is in V = V0. But, in this case, the parts of the inequality
are equal to each other. Therefore, the left part of the inequality is less or equal to its
right part, and the right part of Eq. (6) is more or equal to the right part of Eq. (5).
Then, according to Theorem 3 [5, p.39], the solution of problem (4) is bounded above by
the solution of problem (6). Lemma 2 is proved.
Lemma 3. If 0 < V0 < ((γ +1)δ)
1
1−γ , then the solution of problem (4) is bounded below
by
V̌ (Y ) =
{
V0 cos ĽY, 0 ≤ Y ≤ π
2Ľ
,
0, Y > π
2Ľ
,
here Ľ =
√
2δV γ−1
0 − 1.
Proof. The positive part of V̌ (y) is the solution of the problem:
(7) V̌Y (Y ) = −Ľ
√
V 2
0 − V̌ 2(Y ), Y ≥ 0, V̌ (0) = V0.
We will prove that C0 + V 2 − 2δV γ+1 − Ľ2(V 2
0 − V 2) ≤ 0, if V ∈ [0; V0]. If V = V0 or
V = 0, then both parts of the inequality are equal to zero.
(
C0 + V 2 − 2δV γ+1 − Ľ2(V 2
0 − V 2)
)
V
= 2δV γ
(
2
(
V
V0
)1−γ
− (γ + 1)
)
= 0.
The solutions of this equation are: V1 = 0 and V2 =
(
2
γ+1
) 1
γ−1
V0 < V0. So, V2 is a point
of minimum. Then, V = V0 is a point of maximum. Hence, the right part of Eq. (7) is
less or equal to the right part of Eq. (5). Therefore, according to Theorem 3 [5, p.39],
the solution of problem (4) is bounded below by the solution of problem (7). Lemma 3
is proved.
Corollary. If Y0 is a front point of the solution of problem (4), then
π
2Ľ
≤ Y0 ≤ π
2L̂
.
THE DECOMPOSITION OF A SOLUTION 59
Lemma 4. Let δ > 0. There exists a unique nonnegative generalized solution of the
problem:
(8) VY Y (Y ) − (γ + 1)δ|V (Y )|γ−1V (Y ) + V (Y ) = 0,
Y ≥ 0, VY (0) = 0, V (0) = V0 > (2δ)
1
1−γ .
This solution has the following properties.
1. There exists a front point Y0 = inf {Y ≥ 0 : V (Y ) ≤ 0} < +∞ such that V (Y ) > 0,
if Y ∈ [0; Y0) and V (Y ) = 0, if Y ≥ Y0.
2. VY Y (0) < 0.
3. There exists ε > 0 such that VY Y > 0, if Y ∈ (Y0 − ε; Y0).
Proof. The proof of the existence and uniqueness is the same as in Lemma 1. Let us
determine the type of convexity for V (Y ) in neighborhoods of the points Y = 0 and
Y = Y0. Let us rewrite the equation (8) in next form: VY Y (Y ) = (γ +1)δV γ(Y )−V (Y ).
Then, VY Y (0) = V γ
0
(
(γ + 1)δ − V 1−γ
0
)
< 0, because V (0) > (2δ)
1
1−γ > ((γ+1)δ)
1
1−γ . In
a neighborhood of the point Y0, the values of the function V (Y ) are near zero. Therefore,
in this neighborhood, VY Y (Y ) = V γ(Y )
(
(γ + 1)δ − V 1−γ(Y )
)
> 0. Lemma 4 is proved.
Lemma 5. If V0 > (2δ)
1
1−γ , then the solution of problem (8) is bounded above by
�
V(Y ) =
⎧⎨
⎩
V0 cos
�
L Y, 0 ≤ Y ≤ π
2
�
L
,
0, Y > π
2
�
L
,
where
�
L =
√
1 − 2δV γ−1
0 .
Proof. If we reduce the order of Eq. (8), we obtain
(9) VY (Y ) = −
√
−C0 − V 2(Y ) + 2δV γ+1(Y ).
The positive part of
�
V(Y ) is the solution of the problem:
(10)
�
VY (Y ) = −
�
L
√
V 2
0 −
�
V
2
(Y ), Y ≥ 0,
�
V(0) = V0.
Now the proof of this lemma is the same as that of Lemma 3.
Lemma 6. If V0 > (2δ)
1
1−γ , then the solution of problem (8) is bounded below by
�
V(Y ) =
⎧⎪⎨
⎪⎩
V0 − 0.25
�
L
2
Y 2, 0 ≤ Y ≤ 2
√
V0
�
L
,
0, Y > 2
√
V0
�
L
,
where
�
L =
√
2V γ
0
(
V 1−γ
0 − (γ + 1)δ
)
.
Proof. The positive part of
�
V(y) is the solution of the problem:
(11)
�
VY (Y ) = −
�
L
√
V0 −
�
V(Y ), Y ≥ 0,
�
V(0) = V0.
We will prove such an inequality: H(V ) ≥ 0, if V ∈ [0; V0]. Here H(V ) =
�
L
2
(V0 −
V ) + C0 − 2δV γ+1 + V 2. Notice that H(0) = V γ+1
0
(
V 1−γ
0 − 2γδ
)
> 0 and H(V0) = 0.
Let us demonstrate that H(V ) is a decreasing function on [0;V0]. HV (V ) = 2(γ +
1)δ (V γ
0 − V γ) + 2(V − V0). Let us prove that HV (V ) ≤ 0 on V ∈ [0; V0]. HV (0) =
60 SERGEY MELNIK
2V γ
0
(
(γ + 1)δ − V 1−γ
0
)
< 0, HV (V0) = 0, HV V (V ) = 2V γ−1(V 1−γ − γ(γ + 1)δ). If
V ∈ (0; (γ(γ + 1)δ)
1
1−γ ), then HV V (V ) ≤ 0. Hence, HV (V ) is a decreasing function on
V ∈ (0; (γ(γ + 1)δ)
1
1−γ ) and HV ((γ(γ + 1)δ)
1
1−γ ) < 0. If V ∈ ((γ(γ + 1)δ)
1
1−γ ); V0], then
HV V (V ) > 0. Hence, HV (V ) is an increasing function on V ∈ ((γ(γ + 1)δ)
1
1−γ ); V0]
and H(V ) ≤ H(V0) = 0. Ultimately, HV (V ) ≤ 0 on V ∈ [0; V0]. Therefore, H(V ) is
a decreasing function on V ∈ [0; V0] and H(V ) ≥ H(V0) = 0 for all V ∈ [0; V0). It
means that the right part of Eq. (11) is less or equal to the right part of Eq. (9). Then,
according to Theorem 3 [5, p.39], the solution of problem (8) is bounded below by the
solution of problem (11). Lemma 6 is proved.
Corollary. If Y0 is a front point of the solution of problem (8), then
2
√
V0
�
L
≤ Y0 ≤ π
2
�
L
.
Lemma 7. For some v0 > 0, there exists a unique generalized solution of the problem:
(12) μ
(
p
q
)2
vyy(y) + (γ + 1)
p
q
vγ(y) − v(y) = 0,
y ≥ 0, vy(0) = 0, v(0) = v0.
This solution is a nonnegative function with bounded support.
Proof. Let N < 0, then μ > 0. Denote: p̌ = ‖V̌ ‖2, q̌ = |‖V̌ ‖|γ+1, p = ‖V ‖2, q =
|‖V ‖|γ+1, p̂ = ‖V̂ ‖2, q̂ = |‖V̂ ‖|γ+1. According to Lemmas 2 and 3, such an inequal-
ity is correct: p̌
q̂ ≤ p
q ≤ p̂
q̌ . For V̌ (Y ) and V̂ (Y ), we obtain p̌ = πV 2
0
2Ľ
, p̂ = πV 2
0
2L̂
,
q̌ =
√
πV γ+1
0 Γ(0.5γ+1)
ĽΓ(0.5γ+1.5)
, q̂ =
√
πV γ+1
0 Γ(0.5γ+1)
L̂Γ(0.5γ+1.5)
. Here, Γ is the gamma-function. Then the
inequality √
πL̂Γ(0.5γ + 1.5)V 1−γ
0
2ĽΓ(0.5γ + 1)
≤ p
q
≤
√
πĽΓ(0.5γ + 1.5)V 1−γ
0
2L̂Γ(0.5γ + 1)
is correct. If V0 runs from (2δ)
1
1−γ to +∞, then the right part runs from 0 to +∞ and
the left part runs from Z to +∞. Here, Z is the minimal value of the left part of this
inequality. This means that, for some δ > 0, there exists V0 such that p
q = δ. Then, Eq.
(4) transforms into
VY Y (Y ) + (γ + 1)
p
q
V γ(Y ) − V (Y ) = 0.
Let us transform the variables: Y = q√
μpy, v(y) = V ( q√
μpy). We consider that
p =
q√
μp
p, q =
q√
μp
q,
p
q
=
p
q
.
Then, the function v(y) is the solution of Eq. (12). As V (Y ) ≥ 0, VY (0) = 0, V (0) = V0,
then v(y) ≥ 0, vy(0) = 0, v(0) = V0. Since V (Y ) has bounded support, v(y) has bounded
support and, for its front point, the inequality
π
√
μp
2qĽ
≤ y0 ≤ π
√
μp
2qL̂
is correct. Let N > 0, then μ < 0. Using Lemmas 5 and 6, we obtain the similar results.
Lemma 7 is proved.
THE DECOMPOSITION OF A SOLUTION 61
Corollary. The even function, which is the same as the solution of problem (12) on
y ≥ 0, is the solution of problem (3).
Remark 4. As the functions r(t), v(y), and V (Y ) are nonnegative, we will pass to the
notations of their moduli.
4. Proof of Theorem 1.
The functional f(t; u) is defined for (Ft)t≥0-adapted u ∈ L2([0; T ) × Ω;W). We will
prove that f(t ∧ τ ; u) has the Gateaux differential on the subspace W in mean square,
and this differential has the form
Df(t ∧ τ ; u) =
∫
u(t ∧ τ, x)g(x)dx −
∫
u(0, x)g(x)dx+
+a
t∧τ∫
0
∫
ux(s, x)gx(x)dxds − b
t∧τ∫
0
∫
|u(s, x)|γ−1u(s, x)g(x)dxdw(s).(13)
Here, g ∈ W.
We will calculate the Gateaux differential on the subspace according to [6, p.118]. It
is known that the Lebesgue integrals are Gateaux differentiable. Let us prove that the
stochastic integral is also Gateaux differentiable on the subspace W in mean square, and
its differential is equal to (γ + 1)
t∧τ∫
0
∫ |u(s, x)|γ−1u(s, x)g(x)dxdw(s). We must prove
that
(14)
lim
h→0
M
⎛
⎝ t∧τ∫
0
∫ |u + hg|γ+1 − |u|γ+1
h
dxdw(s) − (γ + 1)
t∧τ∫
0
∫
|u|γ−1ugdxdw(s)
⎞
⎠
2
= 0.
Using the mean value theorem, we obtain
M
⎛
⎝ t∧τ∫
0
∫ ( |u + gh|γ+1 − |u|γ+1
h
− (γ + 1)|u|γ−1ug
)
dxdw(s)
⎞
⎠
2
=
= (γ + 1)2M
t∧τ∫
0
⎛
⎝∫ 1∫
0
(|u + θhg|γ+1(u + θhg) − |u|γ−1u
)
dθgdx
⎞
⎠
2
ds ≤
≤ (γ + 1)2||g||2M
t∧τ∫
0
∫ 1∫
0
(|u + θhg|γ−1(u + θhg) − |u|γ−1u
)2
dθdxds.(15)
If |h| ≤ 1, then
(|u + θhg|γ−1(u + θhg) − |u|γ−1u
)2 ≤ 2(|u| + |g|)2γ + 2|u|2γ . According
to the Lebesgue theorem [7, p.284], we obtain (14).
Thus, a stationary point for f(t∧ τ ; u) is a generalized solution of problem (1). Let us
prove that there exists a stationary point for f(t∧ τ ; u) and this stationary point admits
a decomposition. By substituting u(t, x) = r(t)φ(p, q)v(y) into f(t ∧ τ ; u), we obtain
f(t ∧ τ ; u) =
1
2
pφ3−γ(p, q)
(
r3−γ(t ∧ τ) − r3−γ(0)
)
+
+
az
2
φγ+1(p, q)
t∧τ∫
0
rγ+1(s)ds − bq
γ + 1
φ2(p, q)
t∧τ∫
0
r2(s)dw(s).
62 SERGEY MELNIK
Here, r(t) is the solution of problem (2) with some real A and positive r0 and B,
φ is some nonnegative differentiable function, v(y) is some function in W, and y =
xrγ−1(t)φγ−1(p, q). Let φ(p, q) satisfy the equality
1
2
pφ3−γ(p, q) =
bq
(γ + 1)(3 − γ)B
φ2(p, q).
Then φ(p, q) = M
(
p
q
) 1
γ−1
and
f(t ∧ τ ; u) =
p
2
M3−γ
(
p
q
) 3−γ
γ−1
⎛
⎝r3−γ(t ∧ τ) − r3−γ(0) − (3 − γ)B
t∧τ∫
0
r2(s)dw(s)
⎞
⎠+
+
az
2
Mγ+1
(
p
q
) γ+1
γ−1
t∧τ∫
0
rγ+1(s)ds.
According to the Itô formula,
r3−γ(t ∧ τ) − r3−γ(0) =
(3 − γ)N
2
t∧τ∫
0
rγ+1(s)ds + (3 − γ)B
t∧τ∫
0
r2(s)dw(s)
and
f(t ∧ τ ; u) =
M3−γ(3 − γ)N
4
t∧τ∫
0
rγ+1(s)ds
[
p
2
γ−1 q
3−γ
1−γ − μzp
γ+1
γ−1 q
γ+1
1−γ
]
.
As
‖u(s, •)‖2 = r3−γ(s)M3−γp
2
γ−1 q
γ−3
γ−1 , ‖ux(s, •)‖2 = rγ+1(s)Mγ+1p
γ+1
γ−1 q
γ+1
1−γ z,
(16) |‖u(s, •)‖|γ+1 = r2(s)M2p
2
γ−1 q
γ−3
γ−1 ,
then
f(t ∧ τ ; u) =
(3 − γ)N
4
M2(1−γ)
t∧τ∫
0
[ |‖u(s, •)‖|2(γ+1)
‖u(s, •)‖2
− μ‖ux(s, •)‖2
]
ds.
Let us calculate the Gateaux differential on the subspace W 1
2 (R1) ∩ Lγ+1(R1):
Df(t ∧ τ ; u) =
(3 − γ)N
2
M2(1−γ)×
×
t∧τ∫
0
∫ [ |‖u(s, •)‖|γ+1
‖u(s, •)‖2
(γ + 1)uγ(s, x) − |‖u(s, •)‖|2(γ+1)
‖u(s, •)‖4
u(s, x) + μuxx
]
g(x)dxds.
Using (16), we obtain
Df(t ∧ τ ; u) =
(3 − γ)N
4
(
p
q
) 1
γ−1
t∧τ∫
0
r2γ−1(s)ds×
×
∫ [
(γ + 1)
p
q
vγ(y) − v(y) + μ
(
p
q
)2
vyy(y)
]
g(x)dx.
THE DECOMPOSITION OF A SOLUTION 63
If v(y) is the solution of problem (3), then Df(t ∧ τ ; u) = 0 and the function
u(t, x) = r(t)M
(
p
q
) 1
γ−1
v
(
xrγ−1(t)Mγ−1 p
q
)
is a stationary point for f(t∧ τ ; u). Thus, u(t, x) is a generalized solution of problem (1)
which admits a decomposition. Theorem 1 is proved.
5. Dynamics of the solution.
If
u0(x) = r(0)M
(
p
q
) 1
γ−1
v
(
xrγ−1(0)Mγ−1 p
q
)
,
then
u(t, x) = r(t)M
(
p
q
) 1
γ−1
v
(
xrγ−1(t)Mγ−1 p
q
)
.
Consider the dynamics for r(t).
Theorem 2. Let r(t) be the solution of problem (2).
If 2A < B2, then P
{
lim
t→τ(ω)
r(t) = 0
}
= 1.
If 2A > B2, then P
{
lim
t→τ(ω)
r(t) = +∞
}
= 1.
Proof. Denote λ = 1 − 2AB−2. Let P (r, ε, β, α) be the solution of the problem
0.5B2r2γPrr + Ar2γ−1Pr = 0,
0 < ε < r < β, P (ε, ε, β, α) = α, P (β, ε, β, α) = 1 − α, 0 ≤ α ≤ 1.
If α = 1, then P is the probability of such an event: the process r(t) leaves the interval
(ε; β) over the left endpoint. If α = 0, then P is the probability of such an event: the
process r(t) leaves the interval (ε; β) over the right endpoint.
Let 2A < B2, then P (r, ε, β, 1) = rλ−βλ
ελ−βλ and lim
β→+∞
P (r, ε, β, 1) = 1, ∀r > ε ≥ 0
because λ > 0. This means that r(t) reaches the zero level with probability 1.
Let 2A > B2, then P (r, ε, β, 0) = rλ−ελ
βλ−ελ and lim
ε→0
P (r, ε, β, 0) = 1, ∀r < β. This means
that r(t) tends to +∞ with probability 1 if t → τ(ω). Theorem 2 is proved.
Remark 5. If A = γ
2B2, then the process
r(t) =
(
r1−γ(0) + (1 − γ)Bw(t)
) 1
1−γ
is the solution of problem (2).
Now we can describe the dynamics of u(t, x).
Let 2A < B2. Then, according to Theorem 2, the amplitude and the front point tend
to zero. The configuration of the space form depends on the sign of the parameter N .
If 2A < (γ − 2)B2, then μ > 0 and, for V0 <
(
(γ + 1)p
q
) 1
1−γ
, we obtain the first space
form for v(y), according to Lemma 1. If (γ − 2)B2 < 2A < B2, then μ < 0 and, for
V0 <
(
2 p
q
) 1
1−γ
, we obtain the second space form for v(y), according to Lemma 4.
Let 2A > B2. Then, according to Theorem 2, the amplitude and the front point tend
to infinity. As γ ∈ (0; 1), then γ − 2 < 1, μ < 0, and the space form has the second type,
according to Lemma 4.
64 SERGEY MELNIK
6. Conclusion.
Let us compare the dynamics of the solution of the stochastic problem (1) and the
solution of a similar deterministic problem. According to [8, p.221], if the deterministic
equation has the source term and γ ∈ (0; 1), then its solution tends to infinity on the
whole space. The sign of the last term of Eq. (1) is variable, and it is not possible to
classify this term neither as the source nor as the absorber. If the drift coefficient of
problem (2) is large comparatively to the diffusion coefficient, then the process u(t, x)
develops with intensification. If the drift coefficient of problem (2) is small comparatively
to the diffusion coefficient, then the process u(t, x) develops in the peaking regime.
Bibliography
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tions for Quasilinear Parabolic Equations, Nauka, Moscow, 1987, pp. 475.
3. J.I. Gikhman, A.V. Skorokhod, Stochastic Differential Equations and Its Applications, Naukova
Dumka, Kyiv, 1982, pp. 536.
4. S.A. Melnik, The group analysis of stochastic differential equations, Ann. Uni. Sci. Budapest
21 (2002), 69–79.
5. E. Kamke, Differentialgleichungen. Lösungsmethoden und Lösungen, Leipzig, 1959.
6. H.-H. Kuo, Gaussian Measures in Banach Spaces, Springer, Berlin, 1975.
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E-mail : melnik@ iamm.ac.donetsk.ua
|
| id | nasplib_isofts_kiev_ua-123456789-4507 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 0321-3900 |
| language | English |
| last_indexed | 2025-12-07T15:24:03Z |
| publishDate | 2007 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Melnik, S. 2009-11-19T13:59:42Z 2009-11-19T13:59:42Z 2007 The decomposition of a solution of the quasilinear stochastic parabolic equation with weak source / S. Melnik // Theory of Stochastic Processes. — 2007. — Т. 13 (29), № 3. — С. 55–64. — Бібліогр.: 8 назв.— англ. 0321-3900 https://nasplib.isofts.kiev.ua/handle/123456789/4507 519.21 We obtain conditions which guarantee the existence of a decomposition of a solution of the quasilinear stochastic parabolic equation with a weak source. en Інститут математики НАН України The decomposition of a solution of the quasilinear stochastic parabolic equation with weak source Article published earlier |
| spellingShingle | The decomposition of a solution of the quasilinear stochastic parabolic equation with weak source Melnik, S. |
| title | The decomposition of a solution of the quasilinear stochastic parabolic equation with weak source |
| title_full | The decomposition of a solution of the quasilinear stochastic parabolic equation with weak source |
| title_fullStr | The decomposition of a solution of the quasilinear stochastic parabolic equation with weak source |
| title_full_unstemmed | The decomposition of a solution of the quasilinear stochastic parabolic equation with weak source |
| title_short | The decomposition of a solution of the quasilinear stochastic parabolic equation with weak source |
| title_sort | decomposition of a solution of the quasilinear stochastic parabolic equation with weak source |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/4507 |
| work_keys_str_mv | AT melniks thedecompositionofasolutionofthequasilinearstochasticparabolicequationwithweaksource AT melniks decompositionofasolutionofthequasilinearstochasticparabolicequationwithweaksource |