Boundary value problem for certain classes of non-linear ordinary differential equations with free boundary
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Інститут прикладної математики і механіки НАН України
1999
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| Schriftenreihe: | Нелинейные граничные задачи |
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| Zitieren: | Boundary value problem for certain classes of non-linear ordinary differential equations with free boundary / N.E. Tovmasyan // Нелинейные граничные задачи: сб. науч. тр. — 1999. — Т. 9. — С. 185-189. — Бібліогр.: 3 назв. — англ. |
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irk-123456789-1692912020-06-10T01:26:22Z Boundary value problem for certain classes of non-linear ordinary differential equations with free boundary Tovmasyan, N.E. 1999 Article Boundary value problem for certain classes of non-linear ordinary differential equations with free boundary / N.E. Tovmasyan // Нелинейные граничные задачи: сб. науч. тр. — 1999. — Т. 9. — С. 185-189. — Бібліогр.: 3 назв. — англ. 0236-0497 http://dspace.nbuv.gov.ua/handle/123456789/169291 en Нелинейные граничные задачи Інститут прикладної математики і механіки НАН України |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine |
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English |
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Article |
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Tovmasyan, N.E. |
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Tovmasyan, N.E. Boundary value problem for certain classes of non-linear ordinary differential equations with free boundary Нелинейные граничные задачи |
| author_facet |
Tovmasyan, N.E. |
| author_sort |
Tovmasyan, N.E. |
| title |
Boundary value problem for certain classes of non-linear ordinary differential equations with free boundary |
| title_short |
Boundary value problem for certain classes of non-linear ordinary differential equations with free boundary |
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Boundary value problem for certain classes of non-linear ordinary differential equations with free boundary |
| title_fullStr |
Boundary value problem for certain classes of non-linear ordinary differential equations with free boundary |
| title_full_unstemmed |
Boundary value problem for certain classes of non-linear ordinary differential equations with free boundary |
| title_sort |
boundary value problem for certain classes of non-linear ordinary differential equations with free boundary |
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Інститут прикладної математики і механіки НАН України |
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1999 |
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http://dspace.nbuv.gov.ua/handle/123456789/169291 |
| citation_txt |
Boundary value problem for certain classes of non-linear ordinary differential equations with free boundary / N.E. Tovmasyan // Нелинейные граничные задачи: сб. науч. тр. — 1999. — Т. 9. — С. 185-189. — Бібліогр.: 3 назв. — англ. |
| series |
Нелинейные граничные задачи |
| work_keys_str_mv |
AT tovmasyanne boundaryvalueproblemforcertainclassesofnonlinearordinarydifferentialequationswithfreeboundary |
| first_indexed |
2025-07-15T04:02:50Z |
| last_indexed |
2025-07-15T04:02:50Z |
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1837684145587748864 |
| fulltext |
BOUNDARY VALUE PROBLEM FOR CERTAIN
CLASSES OF NON-LINEAR ORDINARY
DIFFERENTIAL EQUATIONS WITH FREE BOUNDARY
c© N.E. Tovmasyan
In the interval (0, t0) the following problem will be considered
m(t)
dν
dt
= k1F (t)− f1(ν), (1)
−k2
dm
dt
= F (t), (2)
f2(ν) + K3F (t) = m(t)Kn, (3)
m(0) = M0, V (0) = V0, (4)
m(t0) = m0, (5)
where f1(V ) and f2(V ) are given continuous on V functions at V ≥ 0, k1, k2, k3, k4,
M0, m0 and V are given positive constants, m(t), F (t) and V (t) unknown functions to
be found in the interval (0, t0), satisfying the conditions
0 < m0 < M0, f1(0) = 0, f2(V0) < k4M0, (6)
f ′1(V ) > 0, f ′2(V ) > 0 at V > 0, (7)
f ′1(V ) → +∞, f ′2(V ) → +∞ at V → +∞.
In this section we prove the following theorems.
Theorem 1. The problem (1) - (5) is uniquely solvable. To prove theorems 1 and 2
we need the following lemma.
Lemma 1. Let F (t), V (t) and m(t) be the solution of the problem (1) - (4) in the
interval 0 < t < t0 and m(t) satisfy the condition
m0 ≤ m(t) ≤ M0 (8)
then V (t) > 0, F (t) > 0 at 0 ≤ t < t, t0 < ∞.
Proof of lemma 1. Let V (t) be equal to zero at some point τ, 0 < τ < t0 and τ be
the smallest number such that V (τ) = 0. As V (0) = V0 > 0 then V (t) > 0 at 0 ≤ t < τ ,
so
V ′(τ) ≤ 0. (9)
Substituting into (3) t = τ and recalling the equality f1(0) = 0 and inequality (8),
we get
F (τ) ≥ m0k4
k3
. (10)
Putting into (1) t = τ , we get
m(τ)V ′(τ) = F (τ)k1. (11)
From (9), (10) and (11) we have
V ′(τ) > 0. (12)
The condition (12) is a contradiction with (9). This means that the supposition
V (τ) = 0 at some point τ ∈ (0, t0) is not true. Hence
V (τ) > 0 at 0 ≤ t < t0.
We prove now that F (t) > 0 at 0 ≤ t < t0. From the condition f2(V0) < M0k4 and
equality (3) at t = 0 it follows that F (0) > 0. Let F (t) be equal to zero at some point
of the interval (0, t0). Then, as for (9), we can show that there exists a point τ ∈ (0, t0)
where
F (τ) = 0, F ′(τ) ≤ 0. (13)
Putting in (2) t = τ , we get
m′(τ) = 0. (14)
Differentating the both sides of (3) with respect to t and substituting t = τ , we get
V ′(τ)f ′2(V (τ)) + F ′(τ)k3 = 0. (15)
As V (τ) > 0, the condition (7) implies f ′2(V (τ)) > 0. So from (13) and (15) we have
V ′(τ) ≥ 0. (16)
Putting into (1) t = τ and recalling the condition F (τ) = 0 we obtain
m(τ)V ′(τ) = −f1(V (τ)). (17)
As f1(0) = 0 and f ′(V ) > 0 at V > 0, then f1(V ) > 0 at V > 0. So from (17) it follows
that
V ′(τ) < 0. (18)
The condition (18) is a contradiction with (16). This means that F (t) is not equal to
zero in the interval (0, t0). As F (0) > 0 we get F (t) > 0 at 0 ≤ t < t0.
We prove now that t0 < ∞. Integrating the inequality (2) from zero to t, t ∈ (0, t0)
and using the inequality (8) we have
t0∫
0
F (t)dt ≤ k2(M0 −m0). (19)
As F (t) > 0 the inequality (19) implies that the left - hand side has a limit at t → t0.
Passing in (19) to the limit at t → t0, we get
t0∫
0
F (t)dt ≤ k2(M0 −m0). (20)
Let ω and Ω be two subsets of the interval (0, t0) satisfying the conditions
F (t) ≤ m0k4
2k3
at t ∈ ω, (21)
F (t) ≤ m0k4
2k3
at t ∈ Ω, (22)
respectively. Denote by ω0 and Ω0 the Lebesgue measure of the sets ω and Ω,
0 ≤ ω0 ≤ t0, Ω0 = t0 − ω0.
It is clear that
t0∫
0
F (t)dt ≥
∫
F (t)dt ≥ ω0
m0k4
2k3
. (23)
From (20) and (23) follows the inequality
ω0 ≤ 2k2k3(M0 −m0)
k4m0
. (24)
From the inequalities (10), (22) and the equality (3) we have
f2(V (t)) ≥ m0k4
2
(25)
Dividing the both sides of (1) by m(t) we obtain
f1(V (t))
m(t)
=
k1F (t)
m(t)
− V ′(t). (26)
Integrating (26) from zero to t, (t ∈ (0, t0)) we get
t∫
0
f1(V (t))
m(t)
dt = k1
t∫
0
F (t)
m(t)
dt− V ′(t) + V. (27)
As V (t) > 0 and F (t) > 0 at t ∈ (0, t0) and m0 ≤ m(t) ≤ M0, (29) implies
t∫
0
f1(V (τ))dτ
m(τ)
≤ k1
m0
t∫
0
F (τ)dτ + V0. (28)
The last inequality and (20) yield
t∫
0
f1(V (τ))dτ
m(τ)
≤ k1k2
m0
(M0 −m0) + V0, t ∈ (0, t0). (29)
As V (τ) > 0, f1(V (τ)) > 0 and m(τ) > 0, from the inequality (29) we deduce that
the left - hand side has a limit at t → t0 and this limit satisfies the inequality
t∫
0
f1(V (τ))dτ
m(τ)
≤ V0 +
k1k2
m0
(M0 −m0). (30)
Let c0 be a positive solution of the equation
f2(c0) =
m0k4
2
. (31)
As f1(V ) and f2(V ) are increasing functions, from (25) and (31) it follows that
V (t) ≥ c0 at t ∈ Ω, (32)
f1(V (t)) ≥ f1(c0) at t ∈ Ω. (33)
From (10) and (33) we have
t0∫
0
f1(V (τ))dτ
m(τ)
≤
∫
Ω
f1(V (τ))dτ
m(τ)
≥ f1(c0)
M0
Ω0. (34)
From the inequalities (30) and (34) we obtain
Ω0 ≤ M0
f1(c0)
[
V0 +
k1k2
m0
(M0 −m0)
]
. (35)
From the relations (4) and (35) we get t0 < ∞. Lemma 1 is proved.
Proof of theorem 1. It is known that the problem (1) - (4) possesses a solution in
a sufficiently small neighbourhood (0, ε) (. [1], [2]).
As F (0) > 0, from (2) deduce that m(t) is a monotone decreasing in this neighbour-
hood function an
m0 < m(t) < M0 at t ∈ (0, ε). (36)
Let (0, t0) be the maximal neighbourhood where the solution of the problem (1) -
(4), satisfying the inequality
m0 < m(t) < M0 at t ∈ (0, t0) (37)
exists.
According to lemma 1, the interval (0, t0) is bounded.
As m(t) is decreasing in interval (0, t0) and satisfies the inequality (37) the limit of
m(t) at t → t0 exists. Denote this limit by m(t0). It is clear that m0 < m(t) < M0. We
show that m(t0) = m0.
According to lemma 1 V (t) > 0 and F (t) > 0 at t ∈ (0, t0). Hence f2(V ) is also
positive. Thus from (3) and (37) we get
f2(V ) ≤ M0k4, k3F (t) ≤ M0k4. (38)
So
0 ≤ V ≤ c1, F (t) ≤ M0k4
k3
at 0 ≤ t < t0, (39)
where c1 is a positive solution of the equation f2(c1) = M0k4. The inequalities (37)
and (39) imply that the right - hand side of (1) is bounded in the interval (0, t0). This
means that is also bounded in this interval. Hence there exists the limit of V (t) at
t → t0. Denote this limit by V (t0). Therefore, from (1) - (3) we may conclude that
F (t),m(t) and V (t) are continously derivable on the segment [0, t0] functions. As in the
case of lemma 1, we can show that
F (t0) > 0, V (t0) > 0. (40)
Let m(t0) 6= m0. Then
m0 < m(t0) < M0. (41)
Denote V (t0) = V1, m(t0) = m1. Consider Cauchy problem for the system of equa-
tions (1) - (3) in the interval (t0, t0 + ε) with boundary conditions
V (t0) = V1,m(t0) = m1. (42)
As it is know this problem admits a solution for sufficiently small ε. The double
inequality m0 < m(t0) < M0 shows that letting ε to be small one might ensure the
inequality m0 < m(t0) < M0 at t0 ≤ t ≤ t0 + ε. So the solution of the problem (1)-(4),
satisfying the inequality (37) in the interval (0, t0 + ε) exists. But this is not possible
as (0, t0) is the maximal neighbourhood, where such a solution exists. This means that
our supposition, namely m(t0) 6= m0, is not true, hence m(t0) = m0 . Theorem 1 is
proved.
The above results are used in mathematical modelling of the flight of winged aircraft
along a given trajectory.
References
1. Petrovsky I.G., Lectures on the theory of ordinar01.y differential equations (Russian), M., Nauka,
1970.
2. Tricomi F., Differential equations (Russian), M., IL, 1962.
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