Uniqueness of Radial Solutions for Elliptic Equation Involving the Pucci Operator ()
1. Introduction
We study the nonlinear elliptic equation
(1)
where is Pucci maximal operator, the potential f is super linear with some further constraints. Using to denote the eigenvalues of then explicitly, the Pucci operator is given by
For more detailed discussion, see for example [1,2]. This equation has been extensively studied, see [3-5], etc. and the references therein.
Normalize to be for simplicity. We will in this paper investigate the uniqueness of positive radial solution of (1) in the annulus
with Dirichlet boundary condition. In this case, Equation (1) reduces to
(2)
where
Throughout the paper, we assume Note that Now we could state our main results.
Theorem 1. Suppose is small enough and
Then (2) has at most one positive solution with Dirichlet boundary condition.
If instead of the smallness of we assume further growing condition on then we have the following Theorem 2. Suppose that for,
where
Then (2) has at most one positive solution with Dirichlet boundary condition.
In the case the Pucci operator reduces to the usual Laplace operator, and the corresponding unique results are proved by Ni and Nussbaum in [6].
We also remark that the above theorems could be generalized to nonlinearities which also depends on We will not pursue this further in this paper.
2. Lane-Emden Transformation and Uniqueness of the Radial Solutions
2.1. Proof of Theorem 1
We shall perform a Lane-Emden type transformation to Equation (2). Let us introduce a new function
where with
Then satisfies
(3)
where we have denoted
and Note that m may not be continuous at the points where or Additionally, if and then
Lemma 3. Let w be a positive solution of (3) with Then there exists such that and
Proof. If for some then
The conclusion of the lemma follows immediately from this inequality. ■
Given the solution of (3) with and will be denoted by. Let
By standard argument, we know that positive solution of (3) with Dirichlet boundary condition is unique if we could show that
whenever is a positive solution to (3) with
The functions and satisfy the following equations:
The initial condition satisfied by is:,.
Now let be a positive constant such that is a positive solution to (3) with. To show that, let us first prove that must vanish at some point in the interval In the following, we write simply as
Lemma 4. There exists such that.
Proof. Let us consider the function
We have
We remark that is indeed not everywhere differentiable, since m is not continuous. It however could be shown that the jump points of m are isolated. Here by, we mean the derivative of at the point where it is differentiable. The same remark applies to the functions and below.
Now if for then
Since we infer that
It follows that
This is a contradiction, since and . ■
With the above lemma at hand, we wish to show that in the interval vanishes at only one point ξ. For this purpose, let us define functions and Put
and
Lemma 5. We have
(4)
(5)
Proof. Differentiate the Equation (3) with respect to s gives us
(6)
Hence
As to the function h, there holds
Combining this with (3) and (6) we get
It follows that
■
Now we are ready to prove Theorem 1.
Proof of Theorem 1. We need to show that.
We first of all claim that the first zero of in must stay in the interval where is given by Lemma 3. Suppose to the contrary that
By (5) using the fact that we find that if is small enough, then in the interval
Since we find that
Therefore
This is a contradiction, since and
Now the first zero of lies in If then the second zero of lies in Note that in Therefore, by identity (4)
This together with
implies that
but this contradicts with, , and This finishes the proof. ■
2.2. Proof of Theorem 2
Similar arguments as that of Theorem 1 could be used to prove Theorem 2. In this case, we shall make the following transform:
where
and Then
(7)
With this transformation, in the interval , w satisfies
(8)
where
By the definition of one could verify that Note that and are step functions and not continous.
Let be the solution of (8) with and. Now similar as in the proof of Theorem 1, we suppose is a positive solution with Dirichlet boundary condition and. We have the following lemma, whose proof will be omitted.
Lemma 6. There exists such that , and
With this lemma at hand, we observe that by (8)
This combined with (7) tells us that Then it is not difficult to show that for and while for
Recall that satisfies
Consider the function then
From this we infer that the function must change sign in the interval similar as that of Theorem 1.
Now let us define
and
where and Moreover, denote
Lemma 7. There holds
Proof. Direct calculation shows
and
This then leads to the desired identity. ■
Now with the help of this lemma, we could prove Theorem 2.
Proof of Theorem 2. First we show the first zero of is in the interval Otherwise, since
one could then use the fact that in and to deduce that in
But this contradicts with and.
Now if the second zero of is in Then since
one could use in to deduce that in which contradicts with and ■
3. Acknowledgements
The author would like to thank Prof. P. Felmer for useful discussion.
NOTES