Based in Maryland, is REVOLUTIONIZINg the world, one post at a time. With topics ranging from Biology to Multivariable Calculus, this website covers the entire academic spectrum.

Laplace Transforms


MathJax TeX Test Page Before we discuss it at all, let's just define a Laplace transform. $$\mathscr{L}(f(t)) = F(s) = \int_0^{\infty}e^{-st}f(t)\, \mathrm{d}t$$


This makes it really easy to do many differential equations, so laplace transforms become really useful. Ignore the actual formula for now, it seems arbitrarily complex, but it is very helpful in doing many things. For example, the last post, The Eigenvalue Method, discussed solving systems of differential equations, and Laplace Transforms can do this. One of the reasons they are so good is they can turn an n'th degree equation into a first degree equation. What do I mean? 

MathJax TeX Test Page $$f(t) + f'(t) + f''(t) = g(t)$$ $$\mathscr{L}(f(t)) = F(s)$$ $$\mathscr{L}(f'(t)) = sF(s)-f(0)$$ $$\mathscr{L}(f''(t)) = s^2F(s)-sf(0) - f'(0)$$ Both $f(0)$ and $f'(0)$ are both constants, so on the left, you just get an expression of $F(s)$ and $s$


Many functions have easy laplace functions: e^at, t, 1, sin(t), etc. We're not expected to compute the integral each time, so we have table, called the L'Chart.

Definition-Like Theorems

MathJax TeX Test Page In order for a laplace transform to exist, the function must be of exponential order. That is to say, $$\boxed{\lim_{t \to \infty}\dfrac{f(t)}{e^{ct}} = 0}$$ This is because once $f(t)$ is too big, the integral is infinite, so there's no Laplace Transform.
$$\text{Corrollary: The Laplace Transform is of } s^{-1} \text{ order.}$$ $$\boxed{\lim_{s \to \infty}\mathscr{L}(f(t)) = 0}$$ Finally, Laplace transforms are unique. That is to say, $$\boxed{\text{If } \mathscr{L}(f(t)) = \mathscr{L}(g(t)) \to f(t) = g(t)}$$

Theorem: 1/s

MathJax TeX Test Page $$\boxed{\mathscr{L}^{-1}\left(\dfrac{F(s)}{s}\right) = \int_0^{t}f(\tau)\, \mathrm{d}\tau \to \dfrac{F(s)}{s} = \mathscr{L}\left(\int_0^{t}f(\tau)\, \mathrm{d}\tau\right)}$$ Mainly, only the inverse laplace operation theorem is useful.


MathJax TeX Test Page Convolution is useful because it allows us to perform operations on products. Here is what convolution $f * g$ is. $$f(t) * g(t) \text{ (NOT multiplication) } = \int_0^{t}f(\tau)g(t-\tau)\,\mathrm{d}\tau$$ In addition, $$\mathscr{L}(f(t) * g(t)) = \mathscr{L}(f(t)) \cdot \mathscr{L}(g(t))$$ $$\mathscr{L}^{-1}(F(s) \cdot G(s)) = f(t) * g(t)$$

Factor of t

MathJax TeX Test Page $$\mathscr{L}(t^nf(t)) = (-1)^n\mathscr{L}^{(n)}(f(t))$$ If you take the inverse laplace of both sides, you get $$t^nf(t) = (-1)^n\mathscr{L}^{-1}\left(\mathscr{L}^{(n)}(f(t))\right)$$ For t = 1, we get $$tf(t) = -\mathscr{L}^{-1}\left(\mathscr{L}'(f(t))\right)$$


MathJax TeX Test Page $$\mathscr{L}^{-1}(\arctan(s)) = -\dfrac{1}{t}\mathscr{L}^{-1}\left(\frac{d}{dx}\arctan(s)\right)$$ $$-\dfrac{1}{t}\mathscr{L}^{-1}\left(\frac{d}{dx}\arctan(s)\right) = -\dfrac{1}{t}\mathscr{L}^{-1}\left(\dfrac{1}{1 + s^2}\right) = \boxed{-\dfrac{\sin(t)}{t}}$$
David Witten

The Eigenvalue Method