Tyler Clarke
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56 lines
3.5 KiB
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[=post-]
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<p>
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<i><a href="[^baseurl]/posts/differential-week-9.html">Next post in this series</a></i>
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</p>
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<p>
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Hello again! Our second midterm is getting close (you can read the review bit on it <a href="[^baseurl]/posts/differential-exam-2.html">here</a>), but we aren't talking about
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that now: the <i>sujet du jour</i> is week 8, and we're getting deep
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into Laplace transforms.
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</p>
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<p>
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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.
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</p>
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<h2>Impulse Functions</h2>
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<p>
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Often in physics we'll have a situation where a very quick <i>impulse</i> is applied to a system. This involves a piecewise function: at some time `t`, the function
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will go to a high value, and then shortly after it will go low. The integral of that function is the total <i>momentum</i> produced, in physics terms.
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</p>
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<p>
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Impulses are so useful there's actually a function specifically for representing them: the <i>Dirac delta function</i>, represented as `delta (t)`.
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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`.
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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
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is `1` despite only being nonzero over an <i>infinitely small interval</i>. `delta(0) = oo` doesn't satisfy! Integrating this would yield `oo`, not
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`1`.
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</p>
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<p>
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This property makes the delta function incredibly useful for modelling things that happen instantly. However, it also makes it hard to think about!
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Consider that we can actually take the Laplacian of `delta(t)`: `L(delta(t)) = 1`. This is actually fairly hard to prove; sufficiently so
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that I'm just going to accept it as a fact (yikes!).
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</p>
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<p>
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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
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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}`.
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</p>
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<p>
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Let's do an example, straight from the textbook. We're given `2y'' + y' + 2y = delta (t - 5), y(0) = 0, y'(0) = 0`:
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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}`,
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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
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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
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of the theorem `L(u_c(t) f(t - c)) = e^{-cs} F(s)`. In this case, `c` is clearly
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`5`, so we end up with `u_5(t) L^{-1} (frac 1 {(s + frac 1 4)^2 + frac 15 16})`.
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</p>
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<p>
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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
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`frac 4 { sqrt(15) }`. Using both of these in concert
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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!
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</p>
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<h2>Final Notes</h2>
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<p>
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This was pretty quick. More of a blurb than anything, really. See you next week and good luck on the exam!
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</p>
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[/]
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[=title "Differential Equations Week 8"]
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[=date "2025-7-7"]
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[=author "Tyler Clarke"]
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[=subject "Calculus"] |