Not sure whether this is the sub-forum for this or not but since Laplace Transforms can be used quite a bit in the solutions for differential equations, thought I would post it here.
The question, which I'll get to in a moment here, arose because of another thread here asking about partial fraction decomposition for a particular Laplace transform. Because the equation had no real roots, partial fraction decomposition really wasn't the way to go formally since the transforms are considered to be for f(t) where t and f(t) are real valued as opposed to, for example, complex valued. However in certain (general?) circumstances one can ignore that 'requirement' that t and f be real and just 'plug in the numbers'. For example, consider the two inverse Laplace transforms
\(\displaystyle L^{-1}(\frac{s}{s^2+a^2}) = cos(at)\)
and
\(\displaystyle L^{-1}(\frac{s}{s^2-a^2}) = cosh(at)\)
where a is real valued.
Now let's 'play around' and allow complex values in the function space (as opposed to the transform space which already assumes s can be complex). If we allow a to be complex, can we still say
\(\displaystyle L^{-1}(\frac{s}{s^2+a^2}) = cos(at)\)?
Certainly this works for the
\(\displaystyle L^{-1}(\frac{s}{s^2-a^2}) = cosh(at)\)
since replacing a by i b we get
\(\displaystyle L^{-1}(\frac{s}{s^2+(ib)^2}) = L^{-1}(\frac{s}{s^2-b^2}) = cos(ibt) = cosh(bt)\)?
Now if one didn't realize cos(ibt) = cosh(bt), the answer might look a little strange, i.e. I thought Laplace transforms were only for real functions kind of puzzlement, but still the answer is correct. So, second question: Does this only work when, in fact, the inverse transform gives a real valued function (or, see EDIT, a 'nice' complex function) no matter what it appears to be?
EDIT: Certainly if f is a function of a real variable and is itself complex in a 'nice way', i.e. f(t) = f1(t) + i f2(t) where f1 and f2 are real and suitably integrable we would have Laplace transforms work for these kinds of complex functions because of the linearity , etc. of the Laplace transform. Note that cos(ibt) (and the sine function) are linear in this way.
The question, which I'll get to in a moment here, arose because of another thread here asking about partial fraction decomposition for a particular Laplace transform. Because the equation had no real roots, partial fraction decomposition really wasn't the way to go formally since the transforms are considered to be for f(t) where t and f(t) are real valued as opposed to, for example, complex valued. However in certain (general?) circumstances one can ignore that 'requirement' that t and f be real and just 'plug in the numbers'. For example, consider the two inverse Laplace transforms
\(\displaystyle L^{-1}(\frac{s}{s^2+a^2}) = cos(at)\)
and
\(\displaystyle L^{-1}(\frac{s}{s^2-a^2}) = cosh(at)\)
where a is real valued.
Now let's 'play around' and allow complex values in the function space (as opposed to the transform space which already assumes s can be complex). If we allow a to be complex, can we still say
\(\displaystyle L^{-1}(\frac{s}{s^2+a^2}) = cos(at)\)?
Certainly this works for the
\(\displaystyle L^{-1}(\frac{s}{s^2-a^2}) = cosh(at)\)
since replacing a by i b we get
\(\displaystyle L^{-1}(\frac{s}{s^2+(ib)^2}) = L^{-1}(\frac{s}{s^2-b^2}) = cos(ibt) = cosh(bt)\)?
Now if one didn't realize cos(ibt) = cosh(bt), the answer might look a little strange, i.e. I thought Laplace transforms were only for real functions kind of puzzlement, but still the answer is correct. So, second question: Does this only work when, in fact, the inverse transform gives a real valued function (or, see EDIT, a 'nice' complex function) no matter what it appears to be?
EDIT: Certainly if f is a function of a real variable and is itself complex in a 'nice way', i.e. f(t) = f1(t) + i f2(t) where f1 and f2 are real and suitably integrable we would have Laplace transforms work for these kinds of complex functions because of the linearity , etc. of the Laplace transform. Note that cos(ibt) (and the sine function) are linear in this way.
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