Sentence examples for a supersolution of the problem from inspiring English sources

Thus is a supersolution of the problem (3.7), and.

This shows that w1 is a supersolution of the problem P λ 1 f.

Let u -, u+ S1,2(G, loc) ∩ C G) be respectively a subsolution and a supersolution of the problem L u + f ( ξ, u ) = 0, (4.4).

Lemma 1 leads to a comparison principle for a subsolution and a supersolution of the problem left { textstylebegin{array}l@{quad}l} -varDelta _{p} u+lambda_{1} |u|^{p-2}u-mu varDelta _{q}u+mulambda_{2} |u|^{q-2}u=g u)& mbox{in } D, u=0& mbox{on } partial D, end{array}displaystyle right.

Noting v'(r)le0,quad r>0, one further sees that u̅ is a supersolution to the problem (1 - 3) if (u_{0}) satisfies (27).

The main idea is to construct a supersolution of the equation.

We can show that ((alpha _{1},alpha_{2})), ((beta_{1},betare2})) are a subsolution and a supersolution of the system (1), respectively.

It is easy to show that ((alpha_{1},alpha_{2})), ((beta_{1},betare2})) are a subsolution and a supersolution of the system (1), respectively.

Proof: Define υ t, x) := u t, x + ct) then (1) may be reformulated as Let M′ > max{ M, sup x u0 x)} and let z = z t, x) be the solution of Since M′ is a supersolution of the elliptic operator at the right-hand side of the differential equation, we know that z t < 0 (see, for instance Sattinger, 1973, p. 33).

Lemmas 3.1 and 3.2 guarantee that is a subsolution of problem (1.1) and is a supersolution of problem (1.1).

Clearly, ( u ¯, v ¯ ) = ( A, B ) is a supersolution of problem (1.1), then by comparison principle, the solution of (1.1) should be global.