Fourier Transform - Laplace Transform - Which variable transform?

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I need to know when do I have to transform $x$ and when $y$ in a PDE in Fourier Transform and Laplace Transform.

In an exercise of Fourier Transform, I have to solve a Laplace Equation, where $y>0$ and $x$ can be any value. In this case, I have to transform the $x$ variable

In an exercise of Fourier Cosine Transform, I have $0<x<2$ and $y>0$ and I have to transform the $y$ variable

In an exercise of Laplace Transform, I have to solve a PDE where $0<x<2$ and $t>0$. This time, I have to transform the $t$ variable.

So, what can you tell me about which variable do I have to transform? Do I have to transform the "more infinite" variable of both?

Thanks!

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Firstly, remember what I said on your other post yesterday about domains.. Here, for the Fourier Transform of the Laplace Equation, only $x$ has "infinite" domain $(-\infty, \infty)$, where as $y$ has "semi infinite" domain i.e. $y \in (0, \infty)$. So you take the Fourier Transform with respect to $x$. So

1. Largest domain of any variable is infinite $\implies$ Complex (Normal) Fourier Transform that variable.

Secondly, you use the Fourier Sine or Cosine Transforms when dealing with a "semi infinite" domain. Here, $y \in (0, \infty)$ where as $x \in (0, 2)$, so obviously $x$ can be dealt with discretely and you will need to transform the $y$ variable. So

2. Largest domain of any variable is semi infinite $\implies$ Sine/Cosine Fourier Transform that variable.

Lastly, I think you are getting confused between Fourier Transforms and Laplace Transforms. Laplace Transforms are always taken with respect to time $t$ (just look at the integral representation of it)

$$F(s) = \int_{0}^{\infty} f(t)e^{-st} dt$$

Notice we are integrating over $t$? That is because a Laplace Transform "transforms a function of time into a function of complex frequency" (quote taken from Wikipedia), with $s = x + iy$ the complex part. So

3. Laplace Transform $\implies$ Transform time, $t$.

Note, this is a general guide only.