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Inverse Laplace transform of $K_0 left(r sqrts^2-1right)$
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This question is about inverse Laplace transform $mathscrL^-1:srightarrow t$. Although I was not able to find appropriate contour to invert $K_0 left(r sright)$, I somehow know that
$$mathscrL^-1K_0 left(r sright)=fractheta(t-r)sqrtt^2-r^2.$$
How do one show that statement rigorously? The main part of my question is: Is it possible, similarly, to express a similar inverse fourier transform
$$mathscrL^-1leftK_0 left(r sqrts^2-1right)right$$
in terms of elementary functions? Thank you for suggestions. ($K_0$ is the modified Bessel function of second kind, $theta$ is just Heaviside theta, $r>0$ is a positive real parameter).
Important note: The function $K_0$ in the second laplace transform is ill-defined for $sin (0,1)$ and is ment to represent only its real part, equivalently, using common identities for Bessel functions,
$$K_0 left(r sqrts^2-1right) = -fracpi2Y_0left(rsqrt1 - s^2right)$$
which extends the domain of the original function to $sin (0,1)$.
integration laplace-transform bessel-functions
asked Jul 25 at 18:32
Machinato
1,0921018
add a comment |Â
up vote
3
down vote
favorite
This question is about inverse Laplace transform $mathscrL^-1:srightarrow t$. Although I was not able to find appropriate contour to invert $K_0 left(r sright)$, I somehow know that
$$mathscrL^-1K_0 left(r sright)=fractheta(t-r)sqrtt^2-r^2.$$
How do one show that statement rigorously? The main part of my question is: Is it possible, similarly, to express a similar inverse fourier transform
$$mathscrL^-1leftK_0 left(r sqrts^2-1right)right$$
in terms of elementary functions? Thank you for suggestions. ($K_0$ is the modified Bessel function of second kind, $theta$ is just Heaviside theta, $r>0$ is a positive real parameter).
Important note: The function $K_0$ in the second laplace transform is ill-defined for $sin (0,1)$ and is ment to represent only its real part, equivalently, using common identities for Bessel functions,
$$K_0 left(r sqrts^2-1right) = -fracpi2Y_0left(rsqrt1 - s^2right)$$
which extends the domain of the original function to $sin (0,1)$.
integration laplace-transform bessel-functions
asked Jul 25 at 18:32
Machinato
1,0921018
I am not getting a closed form, but something like $$fracpi2 int_-1^1 dx , J_0left ( r sqrt1-x^2 right ) , e^-t x $$
– Ron Gordon
Jul 25 at 19:15
This looks very nice. However, it cannot be true since if you do the Laplace transform on it and compare with the original function, the graphs looks differently for $s>1$ - and even for $s<1$ they do not coincide.
– Machinato
Jul 25 at 19:39
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This question is about inverse Laplace transform $mathscrL^-1:srightarrow t$. Although I was not able to find appropriate contour to invert $K_0 left(r sright)$, I somehow know that
$$mathscrL^-1K_0 left(r sright)=fractheta(t-r)sqrtt^2-r^2.$$
How do one show that statement rigorously? The main part of my question is: Is it possible, similarly, to express a similar inverse fourier transform
$$mathscrL^-1leftK_0 left(r sqrts^2-1right)right$$
in terms of elementary functions? Thank you for suggestions. ($K_0$ is the modified Bessel function of second kind, $theta$ is just Heaviside theta, $r>0$ is a positive real parameter).
Important note: The function $K_0$ in the second laplace transform is ill-defined for $sin (0,1)$ and is ment to represent only its real part, equivalently, using common identities for Bessel functions,
$$K_0 left(r sqrts^2-1right) = -fracpi2Y_0left(rsqrt1 - s^2right)$$
which extends the domain of the original function to $sin (0,1)$.
integration laplace-transform bessel-functions
asked Jul 25 at 18:32
Machinato
1,0921018
This question is about inverse Laplace transform $mathscrL^-1:srightarrow t$. Although I was not able to find appropriate contour to invert $K_0 left(r sright)$, I somehow know that
$$mathscrL^-1K_0 left(r sright)=fractheta(t-r)sqrtt^2-r^2.$$
How do one show that statement rigorously? The main part of my question is: Is it possible, similarly, to express a similar inverse fourier transform
$$mathscrL^-1leftK_0 left(r sqrts^2-1right)right$$
in terms of elementary functions? Thank you for suggestions. ($K_0$ is the modified Bessel function of second kind, $theta$ is just Heaviside theta, $r>0$ is a positive real parameter).
Important note: The function $K_0$ in the second laplace transform is ill-defined for $sin (0,1)$ and is ment to represent only its real part, equivalently, using common identities for Bessel functions,
$$K_0 left(r sqrts^2-1right) = -fracpi2Y_0left(rsqrt1 - s^2right)$$
which extends the domain of the original function to $sin (0,1)$.
integration laplace-transform bessel-functions
asked Jul 25 at 18:32
Machinato
1,0921018
asked Jul 25 at 18:32
Machinato
1,0921018
asked Jul 25 at 18:32
Machinato
1,0921018
asked Jul 25 at 18:32
Machinato
1,0921018
1,0921018
I am not getting a closed form, but something like $$fracpi2 int_-1^1 dx , J_0left ( r sqrt1-x^2 right ) , e^-t x $$
– Ron Gordon
Jul 25 at 19:15
This looks very nice. However, it cannot be true since if you do the Laplace transform on it and compare with the original function, the graphs looks differently for $s>1$ - and even for $s<1$ they do not coincide.
– Machinato
Jul 25 at 19:39
add a comment |Â
I am not getting a closed form, but something like $$fracpi2 int_-1^1 dx , J_0left ( r sqrt1-x^2 right ) , e^-t x $$
– Ron Gordon
Jul 25 at 19:15
This looks very nice. However, it cannot be true since if you do the Laplace transform on it and compare with the original function, the graphs looks differently for $s>1$ - and even for $s<1$ they do not coincide.
– Machinato
Jul 25 at 19:39
I am not getting a closed form, but something like $$fracpi2 int_-1^1 dx , J_0left ( r sqrt1-x^2 right ) , e^-t x $$
– Ron Gordon
Jul 25 at 19:15
I am not getting a closed form, but something like $$fracpi2 int_-1^1 dx , J_0left ( r sqrt1-x^2 right ) , e^-t x $$
– Ron Gordon
Jul 25 at 19:15
This looks very nice. However, it cannot be true since if you do the Laplace transform on it and compare with the original function, the graphs looks differently for $s>1$ - and even for $s<1$ they do not coincide.
– Machinato
Jul 25 at 19:39
This looks very nice. However, it cannot be true since if you do the Laplace transform on it and compare with the original function, the graphs looks differently for $s>1$ - and even for $s<1$ they do not coincide.
– Machinato
Jul 25 at 19:39
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The result given by Mariusz can be verified as follows. The substitution $t = r cosh tau$ gives
$$F(s) = int_0^infty
frac cosh sqrt t^2 - r^2 sqrt t^2 - r^2
theta(t - r) e^-s t dt =
int_0^infty e^-r s cosh tau cosh(r sinh tau) dtau, \
operatornameRe s > 1.$$
Converting $cosh(r sinh tau)$ to $e^pm r sinh tau$ and writing $a sinh tau + b cosh tau$ as $A cosh(tau + tau_0)$ gives
$$F(s) = frac 1 2 int_0^infty e^A cosh(tau - tau_0) dtau +
frac 1 2 int_0^infty e^A cosh(tau + tau_0), \
A = -r sqrt s^2 - 1, ;
tau_0 = operatornamearcsinh frac 1 sqrt s^2 - 1.$$
Since $cosh$ is even,
$$F(s) = frac 1 2 int_-tau_0^infty e^A cosh tau dtau +
frac 1 2 int_tau_0^infty e^A cosh tau dtau =
int_0^infty e^A cosh tau dtau = \
K_0(-A).$$
answered Jul 26 at 2:32
Maxim
2,035113
add a comment |Â
up vote
1
down vote
With CAS help:
$$mathcalL_s^-1left[K_0left(r sqrts^2-1right)right](t)=fractheta (t-r) cosh left(sqrtt^2-r^2right)sqrtt^2-r^2$$
for $r>0$
Maple code:
`assuming`([inttrans:-invlaplace(BesselK(0, r*sqrt(s^2-1)), s, t)], [r > 0])
#Heaviside(t-r)*cosh(sqrt(-r^2+t^2))/sqrt(-r^2+t^2)
answered Jul 25 at 22:21
Mariusz Iwaniuk
1,6511615
The square root inside $cosh$ is missing in the MathJax formula.
– Maxim
Jul 26 at 2:34
@Maxim. Thanks,it was a typo.
– Mariusz Iwaniuk
Jul 26 at 7:24
This is miraculous! Thank you, this is what I was looking for.
– Machinato
Jul 26 at 8:01
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The result given by Mariusz can be verified as follows. The substitution $t = r cosh tau$ gives
$$F(s) = int_0^infty
frac cosh sqrt t^2 - r^2 sqrt t^2 - r^2
theta(t - r) e^-s t dt =
int_0^infty e^-r s cosh tau cosh(r sinh tau) dtau, \
operatornameRe s > 1.$$
Converting $cosh(r sinh tau)$ to $e^pm r sinh tau$ and writing $a sinh tau + b cosh tau$ as $A cosh(tau + tau_0)$ gives
$$F(s) = frac 1 2 int_0^infty e^A cosh(tau - tau_0) dtau +
frac 1 2 int_0^infty e^A cosh(tau + tau_0), \
A = -r sqrt s^2 - 1, ;
tau_0 = operatornamearcsinh frac 1 sqrt s^2 - 1.$$
Since $cosh$ is even,
$$F(s) = frac 1 2 int_-tau_0^infty e^A cosh tau dtau +
frac 1 2 int_tau_0^infty e^A cosh tau dtau =
int_0^infty e^A cosh tau dtau = \
K_0(-A).$$
answered Jul 26 at 2:32
Maxim
2,035113
add a comment |Â
up vote
2
down vote
accepted
The result given by Mariusz can be verified as follows. The substitution $t = r cosh tau$ gives
$$F(s) = int_0^infty
frac cosh sqrt t^2 - r^2 sqrt t^2 - r^2
theta(t - r) e^-s t dt =
int_0^infty e^-r s cosh tau cosh(r sinh tau) dtau, \
operatornameRe s > 1.$$
Converting $cosh(r sinh tau)$ to $e^pm r sinh tau$ and writing $a sinh tau + b cosh tau$ as $A cosh(tau + tau_0)$ gives
$$F(s) = frac 1 2 int_0^infty e^A cosh(tau - tau_0) dtau +
frac 1 2 int_0^infty e^A cosh(tau + tau_0), \
A = -r sqrt s^2 - 1, ;
tau_0 = operatornamearcsinh frac 1 sqrt s^2 - 1.$$
Since $cosh$ is even,
$$F(s) = frac 1 2 int_-tau_0^infty e^A cosh tau dtau +
frac 1 2 int_tau_0^infty e^A cosh tau dtau =
int_0^infty e^A cosh tau dtau = \
K_0(-A).$$
answered Jul 26 at 2:32
Maxim
2,035113
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The result given by Mariusz can be verified as follows. The substitution $t = r cosh tau$ gives
$$F(s) = int_0^infty
frac cosh sqrt t^2 - r^2 sqrt t^2 - r^2
theta(t - r) e^-s t dt =
int_0^infty e^-r s cosh tau cosh(r sinh tau) dtau, \
operatornameRe s > 1.$$
Converting $cosh(r sinh tau)$ to $e^pm r sinh tau$ and writing $a sinh tau + b cosh tau$ as $A cosh(tau + tau_0)$ gives
$$F(s) = frac 1 2 int_0^infty e^A cosh(tau - tau_0) dtau +
frac 1 2 int_0^infty e^A cosh(tau + tau_0), \
A = -r sqrt s^2 - 1, ;
tau_0 = operatornamearcsinh frac 1 sqrt s^2 - 1.$$
Since $cosh$ is even,
$$F(s) = frac 1 2 int_-tau_0^infty e^A cosh tau dtau +
frac 1 2 int_tau_0^infty e^A cosh tau dtau =
int_0^infty e^A cosh tau dtau = \
K_0(-A).$$
answered Jul 26 at 2:32
Maxim
2,035113
The result given by Mariusz can be verified as follows. The substitution $t = r cosh tau$ gives
$$F(s) = int_0^infty
frac cosh sqrt t^2 - r^2 sqrt t^2 - r^2
theta(t - r) e^-s t dt =
int_0^infty e^-r s cosh tau cosh(r sinh tau) dtau, \
operatornameRe s > 1.$$
Converting $cosh(r sinh tau)$ to $e^pm r sinh tau$ and writing $a sinh tau + b cosh tau$ as $A cosh(tau + tau_0)$ gives
$$F(s) = frac 1 2 int_0^infty e^A cosh(tau - tau_0) dtau +
frac 1 2 int_0^infty e^A cosh(tau + tau_0), \
A = -r sqrt s^2 - 1, ;
tau_0 = operatornamearcsinh frac 1 sqrt s^2 - 1.$$
Since $cosh$ is even,
$$F(s) = frac 1 2 int_-tau_0^infty e^A cosh tau dtau +
frac 1 2 int_tau_0^infty e^A cosh tau dtau =
int_0^infty e^A cosh tau dtau = \
K_0(-A).$$
answered Jul 26 at 2:32
Maxim
2,035113
edited Jul 26 at 3:01
answered Jul 26 at 2:32
Maxim
2,035113
answered Jul 26 at 2:32
Maxim
2,035113
answered Jul 26 at 2:32
Maxim
2,035113
2,035113
add a comment |Â
add a comment |Â
up vote
1
down vote
With CAS help:
$$mathcalL_s^-1left[K_0left(r sqrts^2-1right)right](t)=fractheta (t-r) cosh left(sqrtt^2-r^2right)sqrtt^2-r^2$$
for $r>0$
Maple code:
`assuming`([inttrans:-invlaplace(BesselK(0, r*sqrt(s^2-1)), s, t)], [r > 0])
#Heaviside(t-r)*cosh(sqrt(-r^2+t^2))/sqrt(-r^2+t^2)
answered Jul 25 at 22:21
Mariusz Iwaniuk
1,6511615
The square root inside $cosh$ is missing in the MathJax formula.
– Maxim
Jul 26 at 2:34
@Maxim. Thanks,it was a typo.
– Mariusz Iwaniuk
Jul 26 at 7:24
This is miraculous! Thank you, this is what I was looking for.
– Machinato
Jul 26 at 8:01
add a comment |Â
up vote
1
down vote
With CAS help:
$$mathcalL_s^-1left[K_0left(r sqrts^2-1right)right](t)=fractheta (t-r) cosh left(sqrtt^2-r^2right)sqrtt^2-r^2$$
for $r>0$
Maple code:
`assuming`([inttrans:-invlaplace(BesselK(0, r*sqrt(s^2-1)), s, t)], [r > 0])
#Heaviside(t-r)*cosh(sqrt(-r^2+t^2))/sqrt(-r^2+t^2)
answered Jul 25 at 22:21
Mariusz Iwaniuk
1,6511615
The square root inside $cosh$ is missing in the MathJax formula.
– Maxim
Jul 26 at 2:34
@Maxim. Thanks,it was a typo.
– Mariusz Iwaniuk
Jul 26 at 7:24
This is miraculous! Thank you, this is what I was looking for.
– Machinato
Jul 26 at 8:01
add a comment |Â
up vote
1
down vote
up vote
1
down vote
With CAS help:
$$mathcalL_s^-1left[K_0left(r sqrts^2-1right)right](t)=fractheta (t-r) cosh left(sqrtt^2-r^2right)sqrtt^2-r^2$$
for $r>0$
Maple code:
`assuming`([inttrans:-invlaplace(BesselK(0, r*sqrt(s^2-1)), s, t)], [r > 0])
#Heaviside(t-r)*cosh(sqrt(-r^2+t^2))/sqrt(-r^2+t^2)
answered Jul 25 at 22:21
Mariusz Iwaniuk
1,6511615
With CAS help:
$$mathcalL_s^-1left[K_0left(r sqrts^2-1right)right](t)=fractheta (t-r) cosh left(sqrtt^2-r^2right)sqrtt^2-r^2$$
for $r>0$
Maple code:
`assuming`([inttrans:-invlaplace(BesselK(0, r*sqrt(s^2-1)), s, t)], [r > 0])
#Heaviside(t-r)*cosh(sqrt(-r^2+t^2))/sqrt(-r^2+t^2)
answered Jul 25 at 22:21
Mariusz Iwaniuk
1,6511615
edited Jul 26 at 7:22
answered Jul 25 at 22:21
Mariusz Iwaniuk
1,6511615
answered Jul 25 at 22:21
Mariusz Iwaniuk
1,6511615
answered Jul 25 at 22:21
Mariusz Iwaniuk
1,6511615
1,6511615
The square root inside $cosh$ is missing in the MathJax formula.
– Maxim
Jul 26 at 2:34
@Maxim. Thanks,it was a typo.
– Mariusz Iwaniuk
Jul 26 at 7:24
This is miraculous! Thank you, this is what I was looking for.
– Machinato
Jul 26 at 8:01
add a comment |Â
The square root inside $cosh$ is missing in the MathJax formula.
– Maxim
Jul 26 at 2:34
@Maxim. Thanks,it was a typo.
– Mariusz Iwaniuk
Jul 26 at 7:24
This is miraculous! Thank you, this is what I was looking for.
– Machinato
Jul 26 at 8:01
The square root inside $cosh$ is missing in the MathJax formula.
– Maxim
Jul 26 at 2:34
The square root inside $cosh$ is missing in the MathJax formula.
– Maxim
Jul 26 at 2:34
@Maxim. Thanks,it was a typo.
– Mariusz Iwaniuk
Jul 26 at 7:24
@Maxim. Thanks,it was a typo.
– Mariusz Iwaniuk
Jul 26 at 7:24
This is miraculous! Thank you, this is what I was looking for.
– Machinato
Jul 26 at 8:01
This is miraculous! Thank you, this is what I was looking for.
– Machinato
Jul 26 at 8:01
add a comment |Â
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I am not getting a closed form, but something like $$fracpi2 int_-1^1 dx , J_0left ( r sqrt1-x^2 right ) , e^-t x $$
– Ron Gordon
Jul 25 at 19:15
This looks very nice. However, it cannot be true since if you do the Laplace transform on it and compare with the original function, the graphs looks differently for $s>1$ - and even for $s<1$ they do not coincide.
– Machinato
Jul 25 at 19:39