Scaling argument... again
Exercise:
Let [tex]$\Omega \subseteq\mathbb{R}^N$[/tex] be a bounded open domain and suppose there exists [tex]$\lambda >0$[/tex] s.t. the problem:
(A) [tex]$\begin{cases} -\Delta u(x) =\lambda u(x) &\text{, for } x\in \Omega \\ u(x)=0 &\text{, for } x\in \partial \Omega \end{cases}$[/tex]
has nonzero strong solutions [tex]$u\in C^2 (\Omega)\cap C(\overline{\Omega})$[/tex]; remember that [tex]\Delta u(x) =\text{div} (\text{D} u) =\sum_{n=1}^N u_{x_n x_n}[/tex].
1. Fix [tex]$\sigma >0$[/tex] and let [tex]$\sigma \Omega:=\{ \sigma x,\ x\in \Omega\}$[/tex] (for example, if [tex]$\Omega$[/tex] is the unit ball, then [tex]$\sigma \Omega$[/tex] is the ball of radius [tex]$\sigma$[/tex]).
Show that the problem:
(B) [tex]$\begin{cases} -\Delta v(y) =\mu v(y) &\text{, for } y\in \sigma \Omega \\ v(y)=0 &\text{, for } y\in \partial (\sigma \Omega) \end{cases}$[/tex]
has nonzero strong solutions [tex]$v\in C^2(\sigma \Omega) \cap C(\overline{\sigma \Omega})$[/tex] for some [tex]$\mu >0$[/tex].
2. Suppose that the set [tex]$\Lambda :=\{ \lambda >0:\ \text{problem ({\bf A}) has a nonzero strong solution}\}$[/tex] has a minimum and set [tex]$\lambda_1 (\Omega):=\min \Lambda$[/tex].
Prove that, for all [tex]$\sigma >0$[/tex], the set [tex]$M:=\{\mu >0:\ \text{problem ({\bf B}) has a nonzero strong solution}\}$[/tex] has minimum [tex]$\mu_1 (\sigma \Omega)$[/tex].
Moreover find the relation between [tex]$\lambda_1 (\Omega)$[/tex] and [tex]$\mu_1 (\sigma \Omega)$[/tex].
3. What happens if we replace the PDE in (A) and (B) with the following:
[tex]$-\text{div} (|\text{D} w|^{p-2}\ \text{D} w) =\rho\ |w|^{p-2} w$[/tex]
(with [tex]$w=u,\ \rho=\lambda$[/tex] in (A) and [tex]$w=v,\ \rho =\mu$[/tex] in (B), of course)?
That is, do claims 1 & 2 remain true when we consider problems (A) & (B) for the [tex]$p$[/tex]-Laplace operator [tex]$-\Delta_p u := \text{div} (|\text{D} u|^{p-2}\ \text{D} u)$[/tex]?
***
For the courious people outta here, problem (A) is called eigenvalue problem for the Laplace operator (i.e. the operator [tex]$-\Delta u(x) :=-\text{div} (\text{D} u(x))$[/tex]) under homogeneous Dirichlet boundary condition in strong form.
If [tex]$\Omega$[/tex] is smooth enough, such a problem always has a solution, that is there exists a denumerable family [tex]$\{ \lambda_k \}\in ]0,+\infty[$[/tex] s.t. (A) with [tex]$\lambda=\lambda_k$[/tex] is satisfied by some nonzero function in [tex]$C^2(\Omega)\cap C(\overline{\Omega})$[/tex].
Hence the set [tex]$\Lambda =\{ \lambda_k\}$[/tex] (which is the spectrum of [tex]$-\Delta$[/tex]) has minimum, which is called the first (or principal) eigenvalue of [tex]$-\Delta$[/tex] in [tex]$\Omega$[/tex]; it can be proved that the nonzero eigenfunctions related to [tex]$\lambda_1(\Omega)$[/tex] are continuous and have constant sign (that is, they are either positive or negative) in [tex]$\Omega$[/tex].
Let [tex]$\Omega \subseteq\mathbb{R}^N$[/tex] be a bounded open domain and suppose there exists [tex]$\lambda >0$[/tex] s.t. the problem:
(A) [tex]$\begin{cases} -\Delta u(x) =\lambda u(x) &\text{, for } x\in \Omega \\ u(x)=0 &\text{, for } x\in \partial \Omega \end{cases}$[/tex]
has nonzero strong solutions [tex]$u\in C^2 (\Omega)\cap C(\overline{\Omega})$[/tex]; remember that [tex]\Delta u(x) =\text{div} (\text{D} u) =\sum_{n=1}^N u_{x_n x_n}[/tex].
1. Fix [tex]$\sigma >0$[/tex] and let [tex]$\sigma \Omega:=\{ \sigma x,\ x\in \Omega\}$[/tex] (for example, if [tex]$\Omega$[/tex] is the unit ball, then [tex]$\sigma \Omega$[/tex] is the ball of radius [tex]$\sigma$[/tex]).
Show that the problem:
(B) [tex]$\begin{cases} -\Delta v(y) =\mu v(y) &\text{, for } y\in \sigma \Omega \\ v(y)=0 &\text{, for } y\in \partial (\sigma \Omega) \end{cases}$[/tex]
has nonzero strong solutions [tex]$v\in C^2(\sigma \Omega) \cap C(\overline{\sigma \Omega})$[/tex] for some [tex]$\mu >0$[/tex].
2. Suppose that the set [tex]$\Lambda :=\{ \lambda >0:\ \text{problem ({\bf A}) has a nonzero strong solution}\}$[/tex] has a minimum and set [tex]$\lambda_1 (\Omega):=\min \Lambda$[/tex].
Prove that, for all [tex]$\sigma >0$[/tex], the set [tex]$M:=\{\mu >0:\ \text{problem ({\bf B}) has a nonzero strong solution}\}$[/tex] has minimum [tex]$\mu_1 (\sigma \Omega)$[/tex].
Moreover find the relation between [tex]$\lambda_1 (\Omega)$[/tex] and [tex]$\mu_1 (\sigma \Omega)$[/tex].
3. What happens if we replace the PDE in (A) and (B) with the following:
[tex]$-\text{div} (|\text{D} w|^{p-2}\ \text{D} w) =\rho\ |w|^{p-2} w$[/tex]
(with [tex]$w=u,\ \rho=\lambda$[/tex] in (A) and [tex]$w=v,\ \rho =\mu$[/tex] in (B), of course)?
That is, do claims 1 & 2 remain true when we consider problems (A) & (B) for the [tex]$p$[/tex]-Laplace operator [tex]$-\Delta_p u := \text{div} (|\text{D} u|^{p-2}\ \text{D} u)$[/tex]?
***
For the courious people outta here, problem (A) is called eigenvalue problem for the Laplace operator (i.e. the operator [tex]$-\Delta u(x) :=-\text{div} (\text{D} u(x))$[/tex]) under homogeneous Dirichlet boundary condition in strong form.
If [tex]$\Omega$[/tex] is smooth enough, such a problem always has a solution, that is there exists a denumerable family [tex]$\{ \lambda_k \}\in ]0,+\infty[$[/tex] s.t. (A) with [tex]$\lambda=\lambda_k$[/tex] is satisfied by some nonzero function in [tex]$C^2(\Omega)\cap C(\overline{\Omega})$[/tex].
Hence the set [tex]$\Lambda =\{ \lambda_k\}$[/tex] (which is the spectrum of [tex]$-\Delta$[/tex]) has minimum, which is called the first (or principal) eigenvalue of [tex]$-\Delta$[/tex] in [tex]$\Omega$[/tex]; it can be proved that the nonzero eigenfunctions related to [tex]$\lambda_1(\Omega)$[/tex] are continuous and have constant sign (that is, they are either positive or negative) in [tex]$\Omega$[/tex].
Risposte
Correct! 
1 and 2.
Everything boils down to this:
if $r>0$ and $v(y)=u(r y)$ then $Deltav(y)=r^2Deltau(ry)$.
This said, take one $u$ solving (A). Let $v(y)=u(y/sigma)$. Clearly $v \in C^2(sigmaOmega)nnC(bar{sigmaOmega})$ and $v(y)=0$ if $y \in partial(sigmaOmega)$. Also, $-Deltav(y)= -sigma^{-2}Deltau(y/sigma)=-lambda sigma^{-2} u(y/sigma)$ if $y in sigma Omega$; that is, $v$ solves B with $mu=sigma^{-2}lambda$.
This procedure can be reversed. Doing so we get a one-one correspondence between the sets $Lambda$ and $M$:
$lambda \mapsto sigma^{-2}lambda$;
in particular [size=75](*)[/size], one of the sets has got a minimum if and only if the other one has got a minimum and if that is the case we have
$lambda_1(Omega)=sigma^{-2}mu_1(sigma Omega)$.
3.
Not surprisingly, this time the scaling factor is $r^p$, that is:
if $r>0$ and $v(y)=u(r y)$ then $Delta_pv(y)=r^pDelta_p u(ry)$.
Operating exactly as before we prove claims 1 and 2, only this time we'll have
$lambda_1(Omega)=sigma^{-p}mu_1(sigma Omega)$.
_____________
(*) This one sounds a lot like spaghetti English. How do you say "in particolare" in English?
Everything boils down to this:
if $r>0$ and $v(y)=u(r y)$ then $Deltav(y)=r^2Deltau(ry)$.
This said, take one $u$ solving (A). Let $v(y)=u(y/sigma)$. Clearly $v \in C^2(sigmaOmega)nnC(bar{sigmaOmega})$ and $v(y)=0$ if $y \in partial(sigmaOmega)$. Also, $-Deltav(y)= -sigma^{-2}Deltau(y/sigma)=-lambda sigma^{-2} u(y/sigma)$ if $y in sigma Omega$; that is, $v$ solves B with $mu=sigma^{-2}lambda$.
This procedure can be reversed. Doing so we get a one-one correspondence between the sets $Lambda$ and $M$:
$lambda \mapsto sigma^{-2}lambda$;
in particular [size=75](*)[/size], one of the sets has got a minimum if and only if the other one has got a minimum and if that is the case we have
$lambda_1(Omega)=sigma^{-2}mu_1(sigma Omega)$.
3.
Not surprisingly, this time the scaling factor is $r^p$, that is:
if $r>0$ and $v(y)=u(r y)$ then $Delta_pv(y)=r^pDelta_p u(ry)$.
Operating exactly as before we prove claims 1 and 2, only this time we'll have
$lambda_1(Omega)=sigma^{-p}mu_1(sigma Omega)$.
_____________
(*) This one sounds a lot like spaghetti English. How do you say "in particolare" in English?
Ok, dissonance. 
By the way, this is a useful (though easy) exercise to students interested in PDE; the argument is similar to the one Rigel used to solve this puzzle.
By the way, this is a useful (though easy) exercise to students interested in PDE; the argument is similar to the one Rigel used to solve this puzzle.
Gugo, please do not reveal the solution to this exercise, since I'm planning on working on it in my spare time (and this means it will be quite long before I can accomplish something good ... 
). Thanks!
). Thanks!
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