关于数值数学的一个典型问题

<正> Collatz L.在综述性文章1]和2]中就数值数学的典型问题归纳为五类,第一类是方程Tu=φ或Tu=u的解.关于这类问题主要是寻找解的存在性定理和解的存在区间以及唯一性定理等等. 如所周知,由初始元u_o出发,经过迭代

ON A TYPICAL PROBLEM OF NUMERICAL MATHEMATICS

Let R be a partially ordered linear space (See 3]) and operator T be defined on R. We shall find conditions in order that the equation Tv+r=v has a solution for a given r∈R. We formulate the results as follows:Theorem 1.If u_o,ω_o∈R(u_o≤ω_o) are two given elements and Tu_o≥u_o, Tω_o≤ω_o, furthermore if there exists a constant N<1 such that for u, ω(u≤ω) in u_o, ω_o] Tω-Tu≥N(ω-u),then and the equation Tv=v has a solution in u_n, ω_n] .Theorem 2. If u_o,ω_o∈R(u_o≤ω_o) are two giren elements and Tu_o≤u_o,Tω_o≥ω_o, furthermore if there exists a constant M>1 such that for u, ω(u≤ω) in u_o,ω_o] Tω-Tu≤M(ω-u), then u_o≤u_1≤u_2≤…≤u_n≤…≤ω_n≤…≤ω_2≤ω_1≤ω_o and the equation Tv=v has a solution in u_n, ω_n].Let operator T_1 be increasing and operator T_2 be decreasing. Define T=T_1+T_2. We haveTheorem 3. If u_o, ω_o ∈R(u_o≤ω_o) are two given elements and T_1u_o+T_2ω_o+γ≥ u_o, T_1ω_o + T_2u_o + r≤ω_o, furthermore if T is additive,then u_o≤u_1≤u_2≤…≤u_n≤…≤ω_n≤…≤ω_2≤ω_1≤ω_o and the equation Tv+γ=v has a solution in u_n, ω_n].