So, I'm reading the classical paper

  • Gidas, B., Ni, WM. & Nirenberg, L., Symmetry and Related Properties via the Maximum Principle, Commun. Math. Phys. 68 (1979) pp. 209–243, doi:10.1007/BF01221125, Project Euclid,

and I am in trouble with some passages. Since giving a complete description of the framework where the problem is posed would be unfeasible, I seek help from people familiar with the referred paper, and this is why I'm posting it here instead of posting at MathStackExchange.

First question

In the proof of Lemma 2.1, when the hypothesis that $f(0) \geq 0$ in $\Omega_\varepsilon$ is made, we obtain equation $\widehat{\text{(2.1)}}$: $$ \Delta u + b_1 u_1 + f(u) - f(0) \leq 0. $$ Then the authors claim that, by the Mean Value Theorem, $$ \Delta u + b_1 u_1 + c(x) u \leq 0, \quad (*) $$ for some function $c(x)$.

How was the Mean Value Theorem used to yield $(*)$?

Second question

In the proof of Lemma 2.2, the authors claim that $$ w(x) = v(x) - u(x) \leq 0, \quad w \not\equiv 0 $$ and $$ \Delta w + b_1(x) w_1 + c(x) w \geq 0, \quad (**) $$ by the integral form of the Mean Value Theorem. Again,

How was the Mean Value Theorem used?

Third question

Again in the proof of Lemma 2.2, the authors use the Maximum Principle for the equation $(**)$

Do we know if c(x) is negative in order to apply the Maximum Principle? Is there a reference for a Maximum Principle where $c$ is anything (which is what they use, as I understand)?

Thanks in advance.


For your first question: you just need that given a solution $u$ there exists a function $c(x)$ such that $f(u(x)) - f(0) = c(x) u(x)$. This is possible because by the Mean Value Theorem applied to $f$ there exists a function $v(x)$ between $u(x)$ and $0$ such that $f(u(x)) - f(0) = f'(v(x)) u(x)$.

For your second question, the MVT is applied to get $f(v(x)) - f(u(x)) = f'(z(x)) (v(x) - u(x))$ for some $z(x)$ between $v(x)$ and $u(x)$.

For your third question, the authors explain exactly this point on page 212 of the paper, in section 1.3. As they explained, the case $c(x) \leq 0$ is well-known, but the proof for the specific statement that they need also works for general-signed $c(x)$. They included a proof of the statement, which spans the bottom lines of page 212 and the top half of page 213.

  • $\begingroup$ Thanks a lot! I had indeed take a glimpse at pages 212 and 213, but I thought the proof was just for Hopf's Lemma. One last issue: why do they mention the "integral" form of the Mean Value Theorem, in the second time they use it? $\endgroup$ – Danilo Gregorin Afonso Feb 24 at 15:33
  • $\begingroup$ @DaniloGregorinAfonso: you are right, the proof they included is only for the Hopf Mx. Principle. But they also include a reference to [7] which contains the proof of the statement that they need. Note that their specific maximum principle only deals with the case of $u \leq 0$ with the maximum point at $u = 0$, which is probably what makes it work. (It's been a while since I looked at this paper.) // I have no idea why they specified the integral form of the MVT, since $f$ is $C^1$. $\endgroup$ – Willie Wong Feb 24 at 16:59
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    $\begingroup$ Actually, if you let $u = e^{\lambda\cdot x} v(x)$. Then $$ 0 \leq Lu =e^{\lambda\cdot x}( a \cdot \partial^2 v + b \cdot \partial v + 2a(\lambda)\cdot \partial v ) + e^{\lambda \cdot x} ( c + a(\lambda,\lambda) + b \cdot \lambda) v $$ For large $\lambda$, the final term is "positive" times $v$, which is non-positive, and hence $v$ is a non-positive function with $L' v \geq 0$ where $L'$ has no zero-order term, and so you can apply the standard strong maximum principle. $\endgroup$ – Willie Wong Feb 24 at 17:14

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