Determining of right-hand side of higher order ultraparabolic equation

Theorem 2.1

Suppose that the conditions (A), (C), (G), (L), (F), (U), (S) hold, and, besides:

1. Dαaαγ∈L∞(Ωx),∂ykc∈L∞(QT)(0<|α|=|γ|≤m0,k∈{1,...,l}),q1∗∈L2(Ωx),q2∗∈L2(0,T),∂ykfs∈L2(QT)(s∈{0,1,2},k∈{1,…,l});

2. |∂yig(x,y,t,ξ)|≤g1(i∈{1,...,l}),g(x,y,t,0)|ST1=0 for almost all (x, y, t) ∈ Q_T and all ξ ∈ ℝ¹, where g¹ is a positive function;

3. f0|ST1=0,f1|ST1=0,f2|ST1=0, then there exists a unique function u^* ∈ V₃(Q_T) ∩ C([0, T];L²(G)), that satisfies the condition (2) and the equality

   ∫QT[∂tu∗v+∑i=1lλj(x,y,t)∂yiu∗v+∑0<|α|=|γ|≤m0aαγ(x)Dαu∗Dγv+c(x,y,t)u∗v++g(x,y,t,u∗)v]dxdydt=∫QT[f1(x,y,t)q1∗(x)+f2(x,y,t)q2∗(t)+f0(x,y,t)]vdxdydt (6)

   for all functions ν ∈ V₁(Q_T).

Moreover, u^* ∈ V₄(Q_T) ∩ C([0, T];L²(G)), u^* satisfies the condition (2) and the Eq. (1) for almost all (x, y, t) ∈ Q_T (so, u^* is a solution to the problem (1) –(3)).

The proof is carried out according to the scheme of proof of Theorem 1, 2 [5].

Remark

The solution u^* of the problem (1) – (3) has such appraisals:

where the constants M₁, M₂, M₃ are independent of q1∗,q2∗ .

Definition

A triple of functions (u(x, y, t), q₁(x), q₂(t)) is a solution to the problem (1) – (5), if u ∈ V₄(Q_T) ∩ C([0, T];L²(G)), q₁ ∈ L²(Ω_x), q₂ ∈ L²(0, T), it satisfies Eq. (1) for almost all (x, y, t) ∈ Q_T and the conditions (2), (4), (5) hold.

Denote:

Let (u∗(x,y,t),q1∗(x),q2∗(t)) be a solution to the problem (1) — (5). Multiplication Eq. (1) on K₁(y, t) and its integration on Π imply the equality

for almost all x ∈ Ω_x. Since the equality

follows from (4), then from (10), hypothesis (K) and (11) we obtain

for almost all x ∈ Ω_x.

On the basis of (1) and (5), we receive

for almost all t ∈ [0, T]. Integrating (13) by parts and using the hypothesis (K), we obtain

for almost all t ∈ [0, T].

Then let us show that if functions u∗∈V4(QT)∩C([0,T];L2(G)),q1∗∈L2(Ωx),q2∗∈L2(0,T) satisfy (2), (12), (14) and Eq. (1) for almost all (x, y, t) ∈ Q_T, then they are the solution to the problem (1) — (5) with the right-hand side function f1(x,y,t)q1∗(x)+f2(x,y,t)q2∗(t)+f0(x,y,t) in the Eq. (1).

Let E1∗(x)=∫ΠK1(y,t)u∗(x,y,t)dydt,x∈Ωx,E2∗(t)=∫GK2(x,y)u∗(x,y,t)dxdy,t∈[0,T]. Then it is obvious, that

On the other hand, q1∗(x),q2∗(t) and u^*(x, y, t) satisfy the equalities (12), (14). Then from (12), (14) – (16) we have

Since u^*(x, y, t) satisfies the condition (3), we get D^α E1∗ (x) = 0 (|α| ≤ m₀ − 1), x ∈ ∂Ω_x, thus from (17) it follows that E1∗ (x) = E₁(x), x ∈ Ω_x. Integrating (18) by parts, we get

Besides, (5) implies the equality E2(0)=∫GK2(x,y)u0(x,y)dxdy. Thus, E2∗ (0) = E₂(0). So, E2∗ (t) = E₂(t), t ∈ [0, T] and u^*(x, y, t) satisfies the overdetermination conditions (4), (5). Thus, we proved the following lemma

Lemma 2.2

The triple offunctions u∗(x,y,t),q1∗(x),q2∗(t)),whereu∗∈V4(QT)∩C([0,T];L2(G)),∂tu∗∈L2(QT),q1∗∈L2(Ωx),q2∗∈L2(0,T), is a solution to the problem (1) – (5) if and only if it satisfies Eq. (1) for almost all (x, y, t) ∈ Q_T, and (2), (12), (14) hold.

Denote

Lemma 2.3

Let C₁ < 1 and Δ₁(x) ≠ 0 for all x ∈ Ω_x, Δ₂(t) ≠ 0 for all t ∈ (0, T). Suppose that the conditions (A), (C), (L), (U), (G), (E), (K), (F) hold, and Dαaαγ∈L∞(Ωx),∂ykc∈L∞(QT),∂ykfs∈L2(QT),fs|ST1=0(s∈{0,1,2},0<|α|=|γ|≤m0,k∈{1,…,l}), Then for each fixed function u^* ∈ V₃(Q_T) ∩ C([0, T];L²(G)) there exists a unique solution (q₁(x), q₂(t)) of the system of equations (12), (14), where q₁ ∈ L²(Ω_x), q₂ ∈ L²(0, T).

Proof

We shall use the method of successive approximations. We construct an approximation (q1m(x),q2m(t)) to the solution of system of equations (12), (14), where q1m(x),q2m(t),m∈N, are such that

Now we show that the sequences {q1m(x)}m=1∞,{q2m(t)}m=1∞, are fundamental in L²(Ω_x), L²(0, T) respectively and they converge to a solution of the system of equations (12), (14). Denote

According to (19), (20)

Raising both sides of (21), (22) to the second power and integrating them on x, t respectively, by the using of the Hölder’s inequality, we get