📂Odinary Differential Equations

Laplace Transform Convolution

Definitions¹

Let’s call $\mathcal{L}$ the Laplace transform. The function $f*g$ that satisfies the following expression is called the convolution with respect to the Laplace transforms of $f$ and $g$.

$$ \mathcal{L}(f*g) = \mathcal{L}(f) \cdot \mathcal{L}(g) $$

Theorem

The convolution of $f$ and $g$ with respect to their Laplace transforms $h=f*g$ is as follows.

$$ h(t) = f*g(t) = \int_{0}^t f(t-\tau)g(\tau)d\tau = \int_{0}^t f(\tau) g(t-\tau)d\tau $$

Proof

$$ \begin{align*} \mathcal{L} \left\{ h(t) \right\} &= \int_{0}^{\infty} e^{-st} \left( \int_{0}^t f(t-\tau)g(\tau)d\tau \right)dt \\ &= \int_{0}^{\infty}\int_{0}^t e^{-st} f(t-\tau)g(\tau)d\tau dt \end{align*} $$

By changing the order of integration (refer to appendix)

$$ \int_{0}^{\infty}\int_{0}^t d\tau dt\quad \implies \quad \int_{0}^{\infty}\int_\tau^{\infty} dt d\tau $$

Thus, the above expression becomes as follows.

$$ \int_{0}^{\infty}\int_\tau^{\infty} e^{-st} f(t-\tau)g(\tau)dt d\tau $$

If we let $u=t-\tau$

$$ \int_{0}^{\infty}\int_{0}^{\infty} e^{-su}e^{-s\tau} f(u)g(\tau)du d\tau $$

And separate the integrals with respect to $u$ and $\tau$ respectively

$$ \int_{0}^{\infty}e^{-s\tau}g(\tau)d\tau\int_{0}^{\infty} e^{-su} f(u)du=G(s)F(s)
$$

Hence

$$ \mathcal{L}(f*g)(s) = \mathcal{L}h(s) = H(s) = G(s) F(s) = \mathcal{L}g \cdot \mathcal{L}f $$

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Examples

1

Find the inverse Laplace transform of $H(s)=\dfrac{a}{s^2(s^2+a^2)}$.

Since $\mathcal{L} \left\{ t \right\}=\dfrac{1}{s^2}$ and $\mathcal{L} \left\{ \sin (at)\right\}=\dfrac{a}{s^2+a^2}$, $h(t)$ is the convolution of $t$ and $\sin (at)$.

$$ h(t) = \int_{0}^t (t-\tau)\sin (a\tau) d\tau = \dfrac{at-\sin(at)}{a^2} $$

Reference: Table of Laplace Transforms

2

Solve the following initial value problem.

$$ \begin{cases} y^{\prime \prime}+4y=g(t) \\ y(0)=3, \quad y^{\prime}(0)=-1 \end{cases} $$

Assuming the given 2nd order linear inhomogeneous differential equation is as follows.

$$ ay^{\prime \prime} + by^{\prime} + cy = g(t) $$

And let $\mathcal{L} \left\{ y \right\} =Y(s)$, $\mathcal{L} \left\{ g(t) \right\}=G(s)$. Then

$$ Y(s) = \dfrac{ (as + b)y(0) + ay^{\prime}(0) } {as^2+bs+c} + \dfrac{G(s) }{as^2+bs+c} $$

Applying the formula for solving 2nd order differential equations using the Laplace transform

$$ Y(s)=\dfrac{3s-1}{s^2+4}+\dfrac{G(s)}{s^2+4} $$

Organizing it we get

$$ Y(s)=3\dfrac{s}{s^2+4}-\dfrac{1}{2}\dfrac{2}{s^2+4}+\dfrac{1}{2}\dfrac{2}{s^2+4}G(s) $$

Then, since $\mathcal{L} \left\{ \cos (2t) \right\}=\dfrac{s}{s^2+4}$ and $\mathcal{L} \left\{ \sin (2t) \right\}=\dfrac{2}{s^2+4}$, $y(t)$ is

$$ y(t)=3\cos (2t) -\dfrac{1}{2}\sin (2t) +\dfrac{1}{2} \int_{0}^t \sin (2(t-\tau))g(\tau)d\tau $$

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Appendix