# Problem computing inverse Laplace transform of the Bessel function

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I am trying to evaluate the inverse Laplace transform of a function that includes the modified Bessel function of the second kind, i.e., my function is f_s = sqrt(a*s)*besselk(1,2*sqrt(a*s)), where a is a constant (say a = 0.5) and 's' is defined as a symbol. To evaluate the inverse Laplace transform of this function, I used the built-in Matlab function 'ilaplace' as transV = ilaplace(f_s,s,t), where I want to evaluate my function at 't' = 2. However, the problem I am getting is that the inverse transformed expression transV still contains 's' variable, which I supposed to transform to 't' after the inverse Laplace transform.

Can anyone help me with what is the problem with this approach? I checked the Matlab documentation and still couldn't figure out the solution. I already tried using 'vpa' function as it was suggested in one of the Matlab forums, but that didn't help to get rid of 's'. I am stuck with this and would really appreciate any help on it.

My code is

a = 0.5; t = 2;

syms s;

f_s = 1/s*sqrt(a*s)*besselk(1,2*sqrt(a*s));

transV = ilaplace(f_s,t); % or transV = vpa(ilaplace(f_s,t));

With this, the final result transV still has an expression which is a function of 's' although it should be transformed after ilaplace.

##### 2 Comments

David Goodmanson
on 11 Nov 2017

Hi Kevin,

It appears that the result is just a slightly different inverse laplace transform that the toolbox can't do either. Might not exist in closed form.

### Accepted Answer

Walter Roberson
on 11 Nov 2017

Maple says that for a > 0, the result is (1/2)*exp(-a/t)

##### 3 Comments

Walter Roberson
on 11 Nov 2017

Under the assumption that a1, a2, a3, a4, s are all > 0, then

(1/2)*(BesselK(1, 2*a2^(1/2)*a3^(1/2)/t)*exp((-a3-a2)/t)*a2^(1/2)*a3^(1/2)+BesselK(1, 2*a1^(1/2)*a4^(1/2)/t)*exp((-a4-a1)/t)*a4^(1/2)*a1^(1/2)+BesselK(1, 2*a2^(1/2)*a4^(1/2)/t)*exp((-a4-a2)/t)*a4^(1/2)*a2^(1/2)+BesselK(1, 2*a1^(1/2)*a3^(1/2)/t)*exp((-a3-a1)/t)*a1^(1/2)*a3^(1/2))/t

### More Answers (1)

BeeTiaw
on 5 Aug 2020

##### 0 Comments

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