1340 CHAPTER 39. WEAK SOLUTIONS

where p> 1. Thus this is positive for a while and then when r is larger, it becomes negative.Thus there is r0 > 0 where 1

2 r20−CArp+1

0 ≡ a > 0. Hence when ∥u∥= r0, you have I (u)≥a > 0. This is part of the mountain pass conditions. Now consider the other part. Letting∥u∥= 1 be fixed, the estimates imply

I (ru)≤ 12

r2−∫

Uα |u|p+1 rp+1dx≤ 1

2r2− rp+1C

Hence, for r large enough, the right side becomes negative because p+1 > 2. Therefore,r→ I (ru) is positive for small r and is eventually negative as r gets larger. hence thereis some value of r where this equals 0. Then v = ru. This verifies the conditions for themountain pass theorem.

conclusions

It follows from the mountain pass theorem that there is some u ̸= 0 such that I′ (u) = 0.From the above computations,

u−R−1 f (u) = 0

Now R =−∆ the Laplacian. In terms of weak derivatives,

⟨−∆u,v⟩H−1,H10=∫

U∇u ·∇vdx = (u,v)H1

0 (U)

and so in terms of weak derivatives,

−∆u = f (u) in H−1 (U) , u ∈ H10 (U) so u = 0 on ∂U .

This proves the following theorem.

Theorem 39.2.2 Suppose the conditions 39.2.8 - 39.2.10 hold. Then there exists a nonzerou ∈ H1

0 (U) such that−∆u = f (u)

One can verify that an example of such a function f (u) is

f (u) = |u|p−2 u

This is very exciting to a large number of people because it gives an interesting example ofnon uniqueness of a boundary value problem. It is clear that u = 0 works.