# Differential Equations, Part II

This post is an introduction to the Laplace transform and how it can be used to solve differential equations. Let ${f:[0,\infty)\rightarrow\mathbb R}$ be a function. We say that ${f}$ is locally integrable if for any compact ${K\subset[0,\infty)}$,

$\displaystyle \int_K |f|\mathsf dx < \infty.$

(Recall that a subset of ${\mathbb R^n}$ is compact if and only if it is closed and bounded.) For a locally integrable function ${f}$, we define the Laplace transform of ${f}$ by

$\displaystyle \mathcal L\{f(t)\}(s) := F(s) := \int_{[0,\infty)}e^{-st}f(t)\mathsf dt,$

for any complex ${s}$ for which the above integral exists. (For the rest of this post, assume that all functions are defined on ${[0,\infty)}$.) For the most basic example, let ${f(t)=1}$. Then

$\displaystyle F(s) = \int_0^\infty e^{-st}\mathsf dt = \frac1s.$

We can generalize the above with the following theorem:

Theorem 1 For any nonnegative integer ${n}$,

$\displaystyle \mathcal L\{t^n\}=\frac{n!}{s^{n+1}}.$

Proof: We proved the base case above. Now suppose the claim holds for some ${n\geqslant 0}$. Then

$\displaystyle \mathcal L\{t^{n+1}\} = \int_0^\infty e^{-st}t^{n+1}\mathsf dt.$

Using integration by parts, the above is equal to

$\displaystyle \lim_{t\rightarrow\infty} \left(-\frac1s t^{n+1}e^{-st}\right) + 0 - \left(\int_0^\infty -\frac1se^{-st}(n+1)t^n \right),$

and further computation yields

$\displaystyle \frac{n+1}s\int_0^\infty e^{-st}t^n\mathsf dt = \frac{n+1}s \mathcal L\{t^n\}=\left(\frac{n+1}s\right)\left(\frac{n!}{s^n}\right)=\frac{(n+1)!}{s^{n+1}}.$

$\Box$

Now suppose ${f}$ is an infinitely differentiable, locally integrable function on ${[0,\infty)}$. Then we have the following theorem:

Theorem 2 For any nonnegative integer ${n}$,

$\displaystyle \mathcal L\{f^{(n)\}(t)} = s^nF(s) - \sum_{k=0}^{n-1}f^{(k)}(0)s^{n-1-k},$

where ${F(s):=\mathcal L\{f(t)\}}$ and ${f^{(n)}}$ is the ${n^{\mathrm{th}}}$ derivative of ${f}$, with ${f^{(0)} := f}$.

Proof: For ${n=0}$, the claim is evident:

$\displaystyle \mathcal L\{f^{(0)}(t) \}=\mathcal L\{f(t)\}=F(s).$

Assume the claims holds for some nonnegative integer ${n}$, then we compute

\displaystyle \begin{aligned} \mathcal L\{f^{(n+1)}(t) \} &= \int_0^\infty e^{-st}t^{n+1}\mathsf dt\\ &= \lim_{t\rightarrow\infty}e^{-st}f^{(n)}(t) - f^{(n)}(0) +s\int_0^\infty e^{-st}f^{(n)}(t)\mathsf dt\\ &= s\mathcal L\{f^{(n)}(t) \} - f^{(n)}(0)\\ &= s^nF(s) - \sum_{k=0}^{n-1}f^{(k)}(0)s^{n-1-k} - f^{(n)}(0)\\ &= s^nF(s) - \sum_{k=0}^{n-1}f^{(k)}(0)s^{n-k}. \end{aligned}

$\Box$

Another example; suppose ${a\in\mathbb R}$ and ${f(t)=e^{at}}$ for ${t\geqslant 0}$. Then

\displaystyle \begin{aligned} \mathcal L\{f(t)\} &= \int_0^\infty e^{-st}e^{at}\mathsf dt\\ &= \int_0^\infty e^{-(s-a)t}\mathsf dt\\ &= \frac1{s-a}, \end{aligned}

for ${\mathrm{Re}(s)>a}$.

One more theorem:

Theorem 3 The Laplace transform is a linear operator. That is, if ${f}$ and ${g}$ are locally integrable functions, and ${c\in\mathbb R}$, then

$\displaystyle \mathcal L\{cf(t) \} = c\mathcal L\{f(t)\}$

and

$\displaystyle \mathcal L\{(f+g)(t)\} = \mathcal L\{f(t)\} + \mathcal L\{g(t)\}.$

Proof: This follows from linearity of the Lebesgue integral. Left as an exercise to the reader 🙂 $\Box$

Now for an example, suppose ${y}$ is a locally integrable, twice-differentiable function defined on ${[0,\infty)}$ satisfying the differential equation

$\displaystyle y'' + 3y' + 2y = 0,$

and further that

$\displaystyle y(0) = 0, y'(0) = 1.$

Applying the Laplace transform to the LHS, we have (letting ${Y(s):=\mathcal L\{y(t)\})}$)

\displaystyle \begin{aligned} \mathcal L\{y''(t) + 3y'(t) + 2y(t)\} &= \mathcal L\{y''(t)\} + 3\mathcal L\{y'(t) \} + 2\mathcal L\{y(t)\}.\\ &= s^2Y(s) - (sy(0) + y'(0) + 3(sY(s) - y(0)) + 2Y(s)\\ &= s^2Y(s) - 1 + 3sY(s) + 2Y(s). \end{aligned}

As ${\mathcal L\{0\}=0}$, we have

$\displaystyle Y(s)(s^2 + 3s + 2) - 1 = 0$

and hence

$\displaystyle Y(s) = \frac1{s^2+3s+2} = \frac1{(s+1)(s+2)} = \frac1{s+1} - \frac1{s+2}.$

Since

$\displaystyle \mathcal L\{e^{at}\} = \frac1{s-a},$

it follows that

$\displaystyle y(t) = e^{-t} - e^{-2t}.$