Comment on ( Unilateral Global Bifurcation from Intervals for Fourth-Order Problems and Its Applications )

In the recent paperW. Shen and T. He and G. Dai and X. Han established unilateral global bifurcation result for a class of nonlinear fourth-order eigenvalue problems. They show the existence of two families of unbounded continua of nontrivial solutions of these problems bifurcating from the points and intervals of the line trivial solutions, corresponding to the positive or negative eigenvalues of the linear problem. As applications of this result, these authors study the existence of nodal solutions for a class of nonlinear fourth-order eigenvalue problems with sign-changing weight. Moreover, they also establish the Sturm type comparison theorem for fourth-order problems with sign-changing weight. In the present comment, we show that these papers of above authors contain serious errors and, therefore, unfortunately, the results of these works are not true. Note also that the authors used the results of the recent work by G. Dai which also contain gaps.

We want to point out that the assertions of the papers [1,2] cannot be true, as they contradict classical results of mathematical analysis, since nonlinear eigenvalue problems of fourth-order arise in many applications, (see [3,4] and the references therein).
Note that if  is positive on [0, 1] then problem (4) has one sequence of positive eigenvalues (see [1]).
Let  +  ,  ∈ N, denote the set of functions in  which have exactly  − 1 generalized simple zeros in  and are positive near  = 0 and set  −  = − +  , and   =  +  ∪  −  .Let  be the closure of the set of nontrivial solution of problem (1).
One of the main results of the work [2] is the following theorem which plays an essential role in the study of problems considered in [1].
The proof of Lemma A contain gaps.Now we demonstrate this fact.Let V fl   1 .It follows by ( 9) that In the proof of Lemma A the authors claim that (see [2, p. 9403, formula (3.2)]) "By simple computation, one has that for any constant ."Multiplying the equations in (10) by V  and   1 , respectively, and adding both sides we obtain Integrating this relation from 0 to , we have Formula (13) shows that the formula (11) is not true, except in the case  1 () ≡ const.We show that the statement of theorem A ([1, Lemma 3]) that the eigenfunction V ]  (),  ∈ N, corresponding to the eigenvalue  ]  , has exactly  − 1 generalized simple zeros in  is not true.The proof of this statement has been achieved as follows (see proof of Proposition 3.3 from [5]).Let  ] ,1 be the first zero of V ]  (),  > 1, in .Set  1 = (0,  ] ,1 ).For any  ∈  1 ≡  1,2  0 ( 1 ) ∩  2,2 ( 1 ), let φ be the extension by zero of  on .It is obvious that φ ∈  ≡  1,2  0 () ∩  2,2 ().By Definition 2.1 from [5], we have Hence the restriction of V ]  in  1 is a nonnegative solution of the following problem: In fact, the restriction of V ]  in  1 is a classical solution of problem (15).Indeed, Proposition 2.1 from [5] Substituting in ( 14) and integrating by parts we obtain Since  is arbitrary, it follows from last equality that (V ]  )  ( ] ,1 ) = 0. Note that this proof is not true.Indeed, if, for any  ∈  1 , one lets φ be the extension by zero of  on , then it does not follow that φ ∈ .By the embedding theorem (see [6]), we have  1 →  1, ( 1 ) with 0 <  < 1/2, and so  ∈  1, ( 1 ).If we take a function  such that   ( ] ,1 − 0) ̸ = 0, then by definition of φ it follows that φ ( ] ,1 − 0) ̸ = 0. But, then again, by definition of φ we have φ ( ] ,1 + 0) = 0. From this we conclude that the function φ has not a derivative at the point  ] ,1 .Hence, φ ∉  by the embedding  →  1, ().Thus the equality (14) is not true for all  ∈  1 .
[1, p. 2]  and[2, p. 9401]  the authors write that "Clearly, the sets    ,  ∈ N,  ∈ {+, −}, are disjoint and open in " and further they use this assertion for the prove of Theorem B. It is obvious that these sets are disjoint, but they are not open in .Hence, the statements of the Theorem B and [1, Theorems 10 and 19] are not true.Now we will show that for  > 1 the set    ,  ∈ N,  ∈ {+, −}, are not open in .For simplicity, consider the case  = 2 (this fact is shown for any  > 2 similarly).
() in (0, 1), although   is contained in  neighborhood of the function  in .This means that  + 2 is not open in .Hence  − 2 is also not open in .