Hello again! Our second midterm is getting close (you can read the review bit on it here), but we aren't talking about that now: the sujet du jour is week 8, and we're getting deep into Laplace transforms.
Last week was pretty reduced compared to our normal lecture load; about 66% less material than we'd typically cover. Hence, this will be a short post.
Often in physics we'll have a situation where a very quick impulse is applied to a system. This involves a piecewise function: at some time `t`, the function will go to a high value, and then shortly after it will go low. The integral of that function is the total momentum produced, in physics terms.
Impulses are so useful there's actually a function specifically for representing them: the Dirac delta function, represented as `delta (t)`. This function is not explicitly defined: instead, we define it `delta(t) = 0, t != 0` with the constraint that `int_{-oo}^{oo} delta (t) dt = 1`. There aren't any actual functions we can define that satisfy this: if you aren't convinced, take a minute to think about a function whose integral is `1` despite only being nonzero over an infinitely small interval. `delta(0) = oo` doesn't satisfy! Integrating this would yield `oo`, not `1`.
This property makes the delta function incredibly useful for modelling things that happen instantly. However, it also makes it hard to think about! Consider that we can actually take the Laplacian of `delta(t)`: `L(delta(t)) = 1`. This is actually fairly hard to prove; sufficiently so that I'm just going to accept it as a fact (yikes!).
Sometimes things don't happen at `t = 0` \[citation needed], so we also sometimes want to translate the delta function; that is, we want it to do its thing at `t = t_0` rather than `t = 0`. We do this the normal way: `delta(t - t_0)`. The Laplace transform is just `L(delta(t - t_0)) = e^{-s t_0}`.
Let's do an example, straight from the textbook. We're given `2y'' + y' + 2y = delta (t - 5), y(0) = 0, y'(0) = 0`: to solve this, we take the Laplace transform of both sides. I won't bore you with the details; we end up with `L(y) (2s^2 + s + 2) + y(0) (-1 - 2s) - 2y'(0) = e^{-5st}`, which substitutes to `L(y) (2s^2 + s + 2) = e^{-5st}` because both IVs are `0`. `L(y) = frac {e^{-5s}} {2s^2 + s + 2}`. Consider that we can use some square-completion magic to rewrite this as `L(y) = frac {e^{-5s}} 2 frac 1 {(s + frac 1 4)^2 + frac 15 16}`. Why did we do this? Because of the theorem `L(u_c(t) f(t - c)) = e^{-cs} F(s)`. In this case, `c` is clearly `5`, so we end up with `u_5(t) L^{-1} (frac 1 {(s + frac 1 4)^2 + frac 15 16})`.
Now we apply the second theorem: `L(e^{at} sin(bt)) = frac b {(s - a)^2 + b^2}`. `b = frac {sqrt(15)} 4`, and `a = frac 1 4`. We also have a constant multiplier `frac 4 { sqrt(15) }`. Using both of these in concert gives us our final result: `f(t) = u_5(t) frac 2 {sqrt(15)} e^{-frac (t - 5) 4} sin(frac {sqrt(15)} 4 (t - 5))`. Easy enough!
This was pretty quick. More of a blurb than anything, really. See you next week and good luck on the exam!
[/] [=title "Differential Equations Week 8"] [=date "2025-7-7"] [=author "Tyler Clarke"] [=subject "Calculus"]