Tempered solutions of $-u''+u =0 $
Prove that the equation $-u'' +u=0 $ has no tempered solutions except the null function.
Risposte
El giovo solution is simple and correct.
But I like complicated matters.
I start from his equation, after application of FT : $ (1+omega^2) hat u =0 $ and look for solutions of equation $1+omega^2=0 $ in $CC$.
Let’s see if it works.
We have now a problem of division for a distribution $(hat u )$ multiplied by a function $(1+omega^2) in C^(oo)$.
The roots are $omega=+-i $ and are both simple.
The division theorem says that the general solution of the division is $hat u(x)= c_1delta(x-i)+c_2delta(x+i) .
It is convenient to put : $d_(1)=c_1/(2pi) ;d_(2)=c_2/(2pi) $ .
Consequently we get $hatu(x)=2pi d_(1)delta(x-i) +2pi d_(2)delta(x+i) .
Antitrasforming ( remember that $hat1 = 2pi delta$ etc.) we arrive to :
$u(x)= d_(1) e^(i^2x) +d_(2) e^(-i^2x) =d_(1)e^(-x)+d_(2) e^x $.
This is the most general solution of the ODE.
We note that neither $e^(-x)$ nor $e^x $ are tempered solutions ; but we are looking for them. The only way to get one is to fix $d_1=d_2=0 $ and so the only tempered solution of the ODE is the null function.
Question
The result is correct, but is it allowed to use division theorem looking for roots of $1+omega^2$ in $CC$ instead that in $RR$ only?
I recall the theorem :
Let $I$ be an open interval and $Psi $ a function of class $C^(oo) $ in $I $, with all zeroes isolated and of finite order.
The set of all solutions of the equation : $Psi u =0$ in $I $ is given by the distributions of the type :
$u(x) = sum_(z in ZZ) sum_(i=0)^(nu(z)-1) c_(z,i) delta^(i)(x-z) $ where $ZZ$ is the set of the zeroes of $psi$, $nu(z) $
is the order of the generic zero $z$ and $c_(z,i) $ are arbitrary constants.
But I like complicated matters.
I start from his equation, after application of FT : $ (1+omega^2) hat u =0 $ and look for solutions of equation $1+omega^2=0 $ in $CC$.
Let’s see if it works.
We have now a problem of division for a distribution $(hat u )$ multiplied by a function $(1+omega^2) in C^(oo)$.
The roots are $omega=+-i $ and are both simple.
The division theorem says that the general solution of the division is $hat u(x)= c_1delta(x-i)+c_2delta(x+i) .
It is convenient to put : $d_(1)=c_1/(2pi) ;d_(2)=c_2/(2pi) $ .
Consequently we get $hatu(x)=2pi d_(1)delta(x-i) +2pi d_(2)delta(x+i) .
Antitrasforming ( remember that $hat1 = 2pi delta$ etc.) we arrive to :
$u(x)= d_(1) e^(i^2x) +d_(2) e^(-i^2x) =d_(1)e^(-x)+d_(2) e^x $.
This is the most general solution of the ODE.
We note that neither $e^(-x)$ nor $e^x $ are tempered solutions ; but we are looking for them. The only way to get one is to fix $d_1=d_2=0 $ and so the only tempered solution of the ODE is the null function.
Question
The result is correct, but is it allowed to use division theorem looking for roots of $1+omega^2$ in $CC$ instead that in $RR$ only?
I recall the theorem :
Let $I$ be an open interval and $Psi $ a function of class $C^(oo) $ in $I $, with all zeroes isolated and of finite order.
The set of all solutions of the equation : $Psi u =0$ in $I $ is given by the distributions of the type :
$u(x) = sum_(z in ZZ) sum_(i=0)^(nu(z)-1) c_(z,i) delta^(i)(x-z) $ where $ZZ$ is the set of the zeroes of $psi$, $nu(z) $
is the order of the generic zero $z$ and $c_(z,i) $ are arbitrary constants.
By FT, $U(1+omega^2)=0$. But $1+omega^2 ne 0$, $forall omega in RR$. Then $U=0 rightarrow u=0$.
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