Convolution

Transforms of Derivatives: Solving ODEs

Pop Quiz: Which of the following statements is false? $$ \begin{array}{ll} &\\ \mbox{a.}&\displaystyle \mathscr{L}\{f(t)g(t)\}= \mathscr{L}\{f(t)\}\mathscr{L}\{g(t)\}\\ &\\ \mbox{b.}&\displaystyle \mathscr{L}\{f(t)+g(t)\}= \mathscr{L}\{f(t)\}+\mathscr{L}\{g(t)\}\\ &\\ \mbox{c.}&\displaystyle \mathscr{L}\{cf(t)\}= c\mathscr{L}\{f(t)\}\\ \end{array} $$

A New Kind of Product

We now introduce a new kind of product of functions that plays nice with $\mathscr{L}.$

Convolution

Let $f(t)$ and $g(t)$ be functions of exponential order.

We define the convolution of $f$ and $g$, denoted by $f*g,$ as follows $$ (f*g)(t)=\int_{0}^{t} f(\tau)g(t-\tau)\,d\tau $$

Example

For $f(t)=t$ and $g(t)=t^2,$ find $f*g.$

$$ \begin{array}{lll} \displaystyle (f*g)(t)&\displaystyle=\int_0^t f(\tau)g(t-\tau)\,d\tau &\mbox{}\\ \displaystyle &\displaystyle=\int_0^t \tau (t-\tau)^2\,d\tau &\mbox{}\\ \displaystyle &\displaystyle=\int_0^t \tau (t^2-2\tau t +\tau^2)\,d\tau &\mbox{}\\ \displaystyle &\displaystyle=\int_0^t \tau t^2-2\tau^2 t +\tau^3\,d\tau &\mbox{}\\ \displaystyle &\displaystyle=\left[ \frac{1}{2}\tau^2 t^2 -\frac{2}{3}\tau^3 t +\frac{1}{4}\tau^4 \right]_0^t &\mbox{}\\ \displaystyle &\displaystyle= \frac{1}{2}t^4 -\frac{2}{3}t^4 +\frac{1}{4}t^4 &\mbox{}\\ \displaystyle &\displaystyle= \frac{1}{12}t^4 &\mbox{}\\ \end{array} $$

Properties of Convolution

Although convolution is quite different from ordinary multiplication, (for example, $1*f\neq f$) it does share some properties.

$$ \begin{array}{lll} \displaystyle &\displaystyle f*g=g*f &\mbox{convolution is commutative}\\ &&\\ \displaystyle &\displaystyle f*(g+h)=f*g+f*h &\mbox{convolution distributes over addition}\\ &&\\ \displaystyle &\displaystyle f*(g*h)=(f*g)*h &\mbox{convolution is associative}\\ &&\\ \displaystyle &\displaystyle f*0=0 &\mbox{}\\ &&\\ \displaystyle &\displaystyle (cf)*g=f*(cg)=c(f*g) &\mbox{}\\ \end{array} $$

Convolution is Commutative

Above we computed $(f*g)(t)=t * t^2=\displaystyle \int_0^t \tau (t-\tau)^2\,d\tau=\frac{1}{12}t^4.$

We note that $$ \begin{array}{lll} \displaystyle (g*f)(t)&\displaystyle=\displaystyle t^2*t &\mbox{}\\ \displaystyle &\displaystyle=\displaystyle \int_0^t g(\tau)f(t-\tau)\,d\tau &\mbox{}\\ \displaystyle &\displaystyle=\displaystyle \int_0^t \tau^2 (t-\tau)\,d\tau &\mbox{}\\ \displaystyle &\displaystyle=\displaystyle \int_0^t \tau^2 t-\tau^3\,d\tau &\mbox{}\\ \displaystyle &\displaystyle=\displaystyle \left[ \frac{1}{3}\tau^3 t-\frac{1}{4}\tau^4\right]_0^t&\mbox{}\\ \displaystyle &\displaystyle=\displaystyle \frac{1}{3}t^4-\frac{1}{4}t^4&\mbox{}\\ \displaystyle &\displaystyle=\displaystyle \frac{1}{12}t^4&\mbox{}\\ \end{array} $$

There is one property of convolution which we will be making extensive use of...

Theorem

Let $f(t)$ and $g(t)$ be of exponential order, then $$ \mathscr{L}\{f(t)*g(t)\}= \mathscr{L}\{f(t)\}\mathscr{L}\{g(t)\} $$ "The transform of the convolution is the product of the transforms."

The Inverse Transform of a Product

Since the transform of the convolution is the product of the transforms, $$ \mathscr{L}\{f(t)*g(t)\}= \mathscr{L}\{f(t)\}\mathscr{L}\{g(t)\}=F(s)G(s) $$ we also have $$ \mathscr{L}^{-1}\{F(s)G(s)\}=f(t)*g(t) $$ That is, the inverse transform of a product is a convolution.

Example

Find $$ \mathscr{L}^{-1}\left\{\frac{5}{s^4+s^2}\right\} $$

First, we note that $$ \frac{5}{s^4+s^2}=\frac{5}{s^2}\frac{1}{s^2+1}=\mathscr{L}\{5t\}\mathscr{L}\{\sin t\} $$ Then $$ \begin{array}{lll} \displaystyle \mathscr{L}^{-1}\left\{\frac{5}{s^4+s^2}\right\}&\displaystyle=\mathscr{L}^{-1}\left\{\frac{5}{s^2}\frac{1}{s^2+1}\right\} &\mbox{}\\ \displaystyle &\displaystyle=5t*\sin t &\mbox{}\\ \displaystyle &\displaystyle=\sin t * (5t) &\mbox{}\\ \displaystyle &\displaystyle=\int_0^t \sin \tau \cdot 5(t-\tau)\,d\tau &\mbox{}\\ \displaystyle &\displaystyle=5\int_0^t \sin \tau \cdot (t-\tau)\,d\tau &\mbox{}\\ \displaystyle &\displaystyle=5\int_0^t t\sin \tau-\tau\sin \tau\,d\tau &\mbox{}\\ \displaystyle &\displaystyle=5\left[-t\cos \tau-(\sin \tau-\tau \cos \tau)\right]_0^t &\mbox{by parts, or table of integrals}\\ \displaystyle &\displaystyle=5\left[-t\cos \tau-\sin \tau+\tau \cos \tau\right]_0^t &\mbox{}\\ \displaystyle &\displaystyle=5\left(\left[-t\cos t-\sin t+t \cos t\right]-\left[-t\cos 0-\sin 0+0 \cos 0\right]\right) &\mbox{}\\ \displaystyle &\displaystyle=5\left(\left[-\sin t\right]-\left[-t\right]\right) &\mbox{}\\ \displaystyle &\displaystyle=5\left(t-\sin t\right) &\mbox{}\\ \displaystyle &\displaystyle=5t-5\sin t &\mbox{}\\ \end{array} $$ Alternative Route

In contrast, we could have done this by partial fractions. $$ \begin{array}{lll} &\displaystyle \frac{5}{s^2(s^2+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+1}&\mbox{}\\ \implies &\displaystyle \frac{5}{s^2(s^2+1)}\color{magenta}{s^2(s^2+1)}=\frac{A}{s}\color{magenta}{s^2(s^2+1)}+\frac{B}{s^2}\color{magenta}{s^2(s^2+1)}+\frac{Cs+D}{s^2+1}\color{magenta}{s^2(s^2+1)}&\mbox{}\\ \implies &\displaystyle 5=As(s^2+1)+B(s^2+1)+(Cs+D)s^2&\mbox{}\\ \implies &\displaystyle 5=A(s^3+s)+B(s^2+1)+(Cs+D)s^2&\mbox{}\\ \implies &\displaystyle 5=As^3+As+Bs^2+B+Cs^3+Ds^2&\mbox{}\\ \implies &\displaystyle 5=(A+C)s^3+(B+D)s^2+As+B&\mbox{}\\ \implies &\displaystyle \begin{cases}A+C=0\\B+D=0\\A=0\\B=5\end{cases}&\mbox{}\\ \end{array} $$ By inspection, $A=0,$ $B=5,$ $C=0,$ and $D=-5.$

That is, $$ \frac{5}{s^2(s^2+1)}=\frac{5}{s^2}-\frac{5}{s^2+1} $$ from which it follows that $$ \mathscr{L}^{-1}\left\{\frac{5}{s^4+s^2}\right\}= \mathscr{L}^{-1}\left\{\frac{5}{s^2}-\frac{5}{s^2+1}\right\}=5t-5\sin t $$

The Big Idea

Among several benefits of the above ideas, we can now handle systems with an arbitrary input function $f(t).$

For example, let's consider a system modelled by $$ x''+x=f(t), \,\,\,\,x(0)=0,\,\,x'(0)=0 $$ Taking the transform of both sides we have $$ s^2X(s)+X(s)=F(s) $$ Solving for $X(s)$ we have $$ X(s)=F(s)\frac{1}{s^2+1} $$ $X(s)$ is a product of transforms! What does this mean???????????????????

The Big Idea

It means that $$ X(s)=F(s)\frac{1}{s^2+1}=\mathscr{L}\{f(t)\}\mathscr{L}\{\sin t\} $$ so that $$ x(t)=\mathscr{L}^{-1}\left\{F(s)\frac{1}{s^2+1}\right\}=f(t)*\sin t=\int_{0}^{t}f(\tau)\sin(t-\tau)\,d\tau $$

In General...

For undamped systems starting at rest, $$ x''+\omega_0^2x=f(t), \,\,\,\,x(0)=0,\,\,x'(0)=0 $$ we can apply the above technique to get $$ x(t)=f(t)*\sin(\omega_0 t)=\int_{0}^{t}f(\tau)\sin(\omega_0(t-\tau))\,dt $$ or $$ x(t)=\sin(\omega_0 t)*f(t)=\int_{0}^{t}\sin(\omega_0\tau)f(t-\tau)\,dt $$

Even More Generally

For any system starting at rest, $$ \begin{array}{lll} &\displaystyle mx''+bx'+kx=f(t), \,\,\,\,x(0)=0,\,\,x'(0)=0&\mbox{}\\ \implies &\displaystyle ms^2X(s)+bsX(s)+kX(s)=F(s)&\mbox{}\\ \implies &\displaystyle (ms^2+bs+k)X(s)=F(s)&\mbox{}\\ \implies &\displaystyle X(s)=\frac{1}{ms^2+bs+k}F(s)&\mbox{}\\ \implies &\displaystyle X(s)=H(s)F(s)&\mbox{}\\ \end{array} $$ where $H(s)$ is the transfer function. Letting $h(t)=\mathscr{L}^{-1}\{H(s)\},$ the solution to the system is $$ x(t)=(h*f)(t)=\int_0^t h(\tau)f(t-\tau)\,d\tau $$

The Big Deal

We can now state the solution to a system $$mx''+bx'+kx=f(t), \,\,\,\,x(0)=x_0,\,\,x'(0)=x_1$$ in terms of an arbitrary input $f(t)!$

Example

Solve the initial value problem $$ x''-x=f(t), \,\,\,\,x(0)=1,\,\,x'(0)=1 $$ where $f(t)$ is any suitable function.

$$ \begin{array}{lll} &\displaystyle x''-x=f(t) &\mbox{}\\ \implies &\displaystyle s^2X(s)-sx(0)-x'(0)-X(s)=F(s)&\mbox{}\\ \implies &\displaystyle s^2X(s)-s-1-X(s)=F(s)&\mbox{}\\ \implies &\displaystyle s^2X(s)-X(s)=s+1+F(s)&\mbox{}\\ \implies &\displaystyle (s^2-1)X(s)=s+1+F(s)&\mbox{}\\ \implies &\displaystyle X(s)=\frac{s+1}{s^2-1}+\frac{1}{s^2-1}F(s)&\mbox{}\\ \implies &\displaystyle X(s)=\frac{1}{s-1}+\frac{1}{s^2-1}F(s)&\mbox{}\\ \end{array} $$ Since $\displaystyle \mathscr{L}^{-1}\left\{\frac{1}{s-1}\right\}=e^{t}$ and $\displaystyle \mathscr{L}^{-1}\left\{\frac{1}{s^2-1}\right\}=\sinh t,$ we have $$ x(t)=e^t+\sinh t * f(t)=e^t+\int_0^t \sinh(\tau) f(t-\tau)\, d\tau $$ or $$ x(t)=e^t+f(t)*\sinh t=e^t+\int_0^t f(\tau)\sinh(t-\tau) \, d\tau $$

Example

Recall that last time we solved the IVP $$ y''+2y'+5y=1,\,\,\,\,y(0)=0,\,\,y'(0)=0 $$ with transfer function $\displaystyle H(s)=\frac{1}{s^2+2s+5}.$

Solve the IVP stating the solution as a convolution.

Taking the transform of both sides $$ \begin{array}{lll} &\displaystyle y''+2y'+5y=1&\mbox{}\\ \implies &\displaystyle s^2X(s)+2sX(s)=\mathscr{L}\{1\}&\mbox{}\\ \implies &\displaystyle (s^2+2s+5)X(s)=\mathscr{L}\{1\}&\mbox{}\\ \implies &\displaystyle X(s)=\frac{1}{s^2+2s+5}\mathscr{L}\{1\}&\mbox{}\\ \end{array} $$ That is, $$ X(s)=H(s)\mathscr{L}\{1\} $$ Letting $h(t)=\mathscr{L}^{-1}\{H(s)\},$ the solution to can now be stated as a convolution. $$ x(t)=\mathscr{L}^{-1}\left\{H(s)\mathscr{L}\{1\}\right\}=(h*1)(t) $$ Now, $$ \begin{array}{lll} \displaystyle h(t)&\displaystyle=\mathscr{L}^{-1}\left\{H(s)\right\} &\mbox{}\\ \displaystyle &\displaystyle=\mathscr{L}^{-1}\left\{\frac{1}{s^2+2s+5}\right\} &\mbox{}\\ \displaystyle &\displaystyle=\mathscr{L}^{-1}\left\{\frac{1}{(s+1)^2+4}\right\} &\mbox{}\\ \displaystyle &\displaystyle=\mathscr{L}^{-1}\left\{\frac{1}{\color{blue}{2}}\frac{\color{blue}{2}}{(s+1)^2+4}\right\} &\mbox{}\\ \displaystyle &\displaystyle=\frac{1}{2}e^{-t}\sin(2t) &\mbox{by the first shifting property}\\ \end{array} $$ So, we have $$ \begin{array}{ll} x(t)&=(h*1)(t)\\ &=\displaystyle\int_0^t h(\tau)\,d\tau\\ &=\displaystyle \int_0^t \frac{1}{2}e^{-\tau}\sin(2\tau)\,d\tau\\ &=\displaystyle \frac{1}{5}-\frac{1}{5}e^{-t}\cos (2t)-\frac{1}{10}e^{-t}\sin(2t)\\ \end{array} $$ which is the same solution we obtained last time. (With less effort!)

Other Uses of Convolution: Integro-Differential Equations

An integro-differential equation is an equation involving an unknown function $x$ which contains both derivatives and integrals of $x.$

A common area of application of integro-differential equations is in modelling populations and other ecological phenomena.

Example

Use Laplace transform methods to solve the following integro-differential equation $$ x'(t)=1-\int_{0}^{t}x(t-\tau)e^{-2\tau}\,d\tau,\,\,\,\,x(0)=1 $$

We take the transform of both sides. $$ \begin{array}{lll} &\displaystyle x'(t)=1-\int_{0}^{t}x(t-\tau)e^{-2\tau}\,d\tau&\mbox{}\\ \implies &\displaystyle x'(t)=1-x(t)*e^{-2t}&\mbox{state integral as convolution}\\ \implies &\displaystyle sX(s)-x(0)=\frac{1}{s}-X(s)\mathscr{L}\{e^{-2t}\}&\mbox{apply transform}\\ \implies &\displaystyle sX(s)-1=\frac{1}{s}-X(s)\frac{1}{s+2}&\mbox{$x(0)=1$}\\ \implies &\displaystyle sX(s)+X(s)\frac{1}{s+2}=1+\frac{1}{s}&\mbox{}\\ \implies &\displaystyle \color{magenta}{s(s+2)}sX(s)+\color{magenta}{s(s+2)}X(s)\frac{1}{s+2}=\color{magenta}{s(s+2)}\cdot 1+\color{magenta}{s(s+2)}\frac{1}{s}&\mbox{}\\ \implies &\displaystyle (s^3+2s^2)X(s)+sX(s)=s^2+2s+s+2&\mbox{}\\ \implies &\displaystyle (s^3+2s^2+s)X(s)=s^2+3s+2&\mbox{}\\ \implies &\displaystyle X(s)=\frac{s^2+3s+2}{s^3+2s^2+s}&\mbox{}\\ \implies &\displaystyle X(s)=\frac{(s+1)(s+2)}{s(s+1)(s+1)}&\mbox{}\\ \implies &\displaystyle X(s)=\frac{s+2}{s(s+1)}&\mbox{}\\ \implies &\displaystyle X(s)=\frac{2}{s}-\frac{1}{s+1}&\mbox{by partial fractions}\\ \implies &\displaystyle x(t)=2-e^{-t}&\mbox{}\\ \end{array} $$

Volterra Integral Equation

The Volterra Integral Equation is $$ x(t)=f(t)+\int_{0}^{t}g(t-\tau)x(\tau)\,d\tau $$ where $f(t)$ and $g(t)$ are known functions.

Volterra Integral Equation

Taking the transform of both sides of $$ x(t)=f(t)+\int_{0}^{t}g(t-\tau)x(\tau)\,d\tau=f(t)+g(t)*x(t) $$ gives us $$ X(s)=F(s)+G(s)X(s) $$ so that $$ X(s)=F(s)\frac{1}{1-G(s)} $$ Then, $$ x(t)=\mathscr{L}^{-1}\left\{F(s)\frac{1}{1-G(s)}\right\}=f(t)*\mathscr{L}^{-1}\left\{\frac{1}{1-G(s)}\right\} $$

Example

Solve the integral equation $$ x(t)=e^{-t}+\int_{0}^{t}\sinh(t-\tau)x(\tau)\,d\tau $$

Taking the transform of both sides, $$ \begin{array}{lll} &\displaystyle x(t)=e^{-t}+\int_{0}^{t}\sinh(t-\tau)x(\tau)\,d\tau&\mbox{}\\ \implies &\displaystyle x(t)=e^{-t}+\sinh t * x(t)&\mbox{}\\ \implies &\displaystyle X(s)=\frac{1}{s+1}+\frac{1}{s^2-1}X(s)&\mbox{}\\ \implies &\displaystyle X(s)-\frac{1}{s^2-1}X(s)=\frac{1}{s+1}&\mbox{}\\ \implies &\displaystyle \color{magenta}{(s^2-1)}X(s)-\color{magenta}{(s^2-1)}\frac{1}{s^2-1}X(s)=\color{magenta}{(s^2-1)}\frac{1}{s+1}&\mbox{}\\ \implies &\displaystyle (s^2-1)X(s)-X(s)=s-1&\mbox{}\\ \implies &\displaystyle (s^2-2)X(s)=s-1&\mbox{}\\ \implies &\displaystyle X(s)=\frac{s-1}{s^2-2}&\mbox{}\\ \implies &\displaystyle X(s)=\frac{s}{s^2-2}-\frac{1}{s^2-2}&\mbox{}\\ \implies &\displaystyle X(s)=\frac{s}{s^2-2}-\color{blue}{\frac{1}{\sqrt{2}}}\frac{\color{blue}{\sqrt{2}}}{s^2-2}&\mbox{prep for Laplace table}\\ \implies &\displaystyle x(t)=\cosh\left(\sqrt{2}t\right)-\frac{1}{\sqrt{2}}\sinh\left(\sqrt{2}t\right)&\mbox{}\\ \end{array} $$

Recall: Transfer Functions

Transfer functions help us to understand how a system (mass-spring, electrical circuit, etc.) modelled by an equation responds to a forcing function $f(t)$ when all initial conditions set to $0.$

That is to say, we consider an equation $$ Lx=f(t) $$ with initial conditions all zero, and apply the Laplace transform to both sides $$ A(s)X(s)=F(s) $$ We then have $$ X(s)=\frac{1}{A(s)}F(s)=H(s)F(s) $$ The function $H(s)$ is called the transfer function.

Transfer Functions and Convolution

Suppose we have a constant-coefficient system satisfying the IVP $$ Lx=f(t),\,\,\,\,x(0)=0,\,\,x'(0)=0 $$ The transform of the solution $X(s)$ depends on the transfer function $H(s)$ as $$ X(s)=H(s)F(s) $$ so that $$ x(t)=h(t)*f(t) $$ where $h(t)$ is the inverse transform of $H(s).$

Prequel: Impulse Response

Suppose $H(s)$ is the transfer function of the the system $Lx=f(t).$

The function $$ h(t)=\mathscr{L}^{-1}\{H(s)\} $$ turns out to be what we will be calling the impulse response function of the system.

This function will turn out to be important when

1) handling situations when a lot of force is imparted to a system over a very short time, and

2) handling situations where large forces are applied over small regions of space.

Summary of Properties So Far

Suppose $\mathscr{L}\{f(t)\}=F(s)$ and $\mathscr{L}\{g(t)\}=G(s).$ Then $$ \begin{array}{lll} \hline &\displaystyle \phi(t) & \Phi(s)=\mathscr{L}\{\phi(t)\}&\\\hline &\displaystyle e^{-at}f(t)&F(s+a) \\ &\displaystyle f(t-a)u(t-a)& e^{-as}F(s)\\ &f'(t) & sF(s)-f(0)\\ &f''(t) & s^2F(s)-sf(0)-f'(0)\\ &f'''(t) & s^3F(s)-s^2f(0)-sf'(0)-f''(0)\\ &f^{(n)}(t) & s^nF(s)-s^{n-1}f(0)- \cdots -sf^{(n-2)}(t)-f^{(n-1)}(t)\\ &tf(t) & -F'(s)&\\ &t^nf(t) & (-1)^nF^{(n)}(s)&\\ &\displaystyle\frac{1}{t}f(t) & \displaystyle \int_{s}^{\infty}F(\sigma)\,d\sigma\\ &\displaystyle \int_{0}^{t}f(\tau) \,d\tau & \displaystyle \frac{1}{s}F(s) \\ &f(t)*g(t)&F(s)G(s)\\ \hline \end{array} $$