Integral equation- solution $u in L^2(RR)$
Determine the solution $u in L^2(RR)$ of the integral equation :
$ u(x)-1/4 int_(-oo)^(+oo) e^(-|x-y|) u(y)dy =xe^(-|x|) $ .
$ u(x)-1/4 int_(-oo)^(+oo) e^(-|x-y|) u(y)dy =xe^(-|x|) $ .
Risposte
"elgiovo":
Of course you're right. I forgot $1/(2pi)$ before the integral. What hard work!
Of course you're right. I forgot $1/(2pi)$ before the integral. What hard work!
I get $u(x)= -4(e^(-|x|)- e^(-|x|/sqrt(2))) sign(x) $ .
I made a mistake in solving for $U(omega)$, so I get your result, too. 
With a little bit of simplifications, $u(x)=-8pi*(e^(-|x|)-e^(-(|x|)/sqrt2))*"sign"(x)$.
With a little bit of simplifications, $u(x)=-8pi*(e^(-|x|)-e^(-(|x|)/sqrt2))*"sign"(x)$.
I didn't check your calculations but arrive to a different formula : $U(omega)= (-8iomega)/((1+omega^2)(1+2omega^2))$ and consequently to a different , more simple expression for $ u(x) $.
I used the property that the transform of the convolution of 2 functions is the product of the transforms.
I used the property that the transform of the convolution of 2 functions is the product of the transforms.
The specification $u in L^2(RR)$ seems to suggest the use of the Fourier transform:
$U(omega)-1/4int_(-oo)^(oo) int_(-oo)^(oo) e^(-|x-y|)*u(y)*e^(-i omega y)dydx=-(4i omega)/((omega^2+1)^2)$.
Let's evaluate the double integral:
$int_(-oo)^(oo) int_(-oo)^(oo) e^(-|x-y|)*u(y)dydx=int_(-oo)^(oo) u(y) int_(-oo)^(oo)e^(-|x-y|)*e^(-i omega x)dxdy$.
The inner integral happens to be the Fourier transform of $f(x)=e^(-|x-y|)$, i.e. $F(omega)=(2e^(-i omega y))/(omega^2+1)$,
and the factor $e^(-i omega y)$ is immediately absorbed in the outer integral, which happens to be (how many coincidences!) the Fourier transform of $u(x)$, i.e. $U(omega)$.
Solving for $U(omega)$, we get $U(omega)=-(16 i omega)/((omega^2+1)(4omega^2+3))$, whose inverse transform should be something like $u(x)=6pi(e^x-e^(sqrt3/2 x))*"H"(-x)+6pi(e^(-sqrt3/2 x)-"cosh"(x)+"sinh"(x))*"H"(x)$.
$U(omega)-1/4int_(-oo)^(oo) int_(-oo)^(oo) e^(-|x-y|)*u(y)*e^(-i omega y)dydx=-(4i omega)/((omega^2+1)^2)$.
Let's evaluate the double integral:
$int_(-oo)^(oo) int_(-oo)^(oo) e^(-|x-y|)*u(y)dydx=int_(-oo)^(oo) u(y) int_(-oo)^(oo)e^(-|x-y|)*e^(-i omega x)dxdy$.
The inner integral happens to be the Fourier transform of $f(x)=e^(-|x-y|)$, i.e. $F(omega)=(2e^(-i omega y))/(omega^2+1)$,
and the factor $e^(-i omega y)$ is immediately absorbed in the outer integral, which happens to be (how many coincidences!) the Fourier transform of $u(x)$, i.e. $U(omega)$.
Solving for $U(omega)$, we get $U(omega)=-(16 i omega)/((omega^2+1)(4omega^2+3))$, whose inverse transform should be something like $u(x)=6pi(e^x-e^(sqrt3/2 x))*"H"(-x)+6pi(e^(-sqrt3/2 x)-"cosh"(x)+"sinh"(x))*"H"(x)$.
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