Laplace Transforms and Convolution

Laplace Transform

Laplace Transforms are a powerful tool for solving differential equations. A Laplace transform is an operator that can be applied to an equation that transforms the function domain from time to Laplace space. This transform allows for any differential equation to be solved algebraically. The method of transforming a function into Laplace space is the following.
\[\mathcal{L}[f(t)]=\int_0^\infty e^{-st}f(t)dt\]
To start finding solutions to differential equations using Laplace transforms, it is helpful to find Laplace transforms of common functions into Laplace space.

Laplace Transforms: Exponentials

\[f(t)=e^{at}\]
Applying the Laplace Transform operator
\[\mathcal{L}[e^{at}]=\int_0^\infty e^{(-st)}e^{at}dt]\]
\[\int_0^\infty e^{t(a-s)}dt\]
To solve this function we will assume the condition that \(s>a\). This will make the exponent negative for all values greater than 0.
\[\frac{e^{t(a-s)}}{a-s}\Bigg|_0^\infty=\frac{1}{s-a}\]

Laplace Transforms: Sinusoids

Normally, integration by parts would be used when solving for the Laplace Transform of a sinusoidal function. Instead, we can use an Euler's Formula substitution to solve for both the Sine and Cosine Laplace Transforms simultaneously.
\[\mathcal{L}[\cos(\omega t)+i\sin(\omega t)]=\mathcal{L}[e^{i\omega t}]\]
From this, we can use Laplace transform of \(e^{at}\) that we already solved.
\[\mathcal{L}[e^{i\omega t}]=\frac{1}{s-i\omega}\]
Multiply by the complex conjugate to rationalize the denominator
\[\frac{1}{s-i\omega}\frac{s+i\omega}{s+i\omega}=\frac{s}{s^2+\omega^2}+\frac{i\omega}{s^2+\omega^2}\]
We can infer that the real part of the solution is the Laplace transform of Cosine and the imaginary part is the Laplace Transform of Sine.
\[\mathcal{L}[\cos(\omega t)]=\frac{s}{s^2+\omega^2}\]
\[\mathcal{L}[\sin(\omega t)]=\frac{\omega}{s^2+\omega^2}\]

Laplace Transforms: Polynomial

\[f(t)=t\]
This integration will require the use of integration by parts to solve for the Laplace transform
\[\mathcal{L}[t]=\int_0^\infty te^{-st}dt\]
\[u=t\]\[du=1\]\[dv=e^{-st}\]\[v=-\frac{1}{s}e^{-st}\]
\[\frac{-t}{s}e^{-st}+\int\frac{1}{s}e^{-st}dt\]
\[\frac{-t}{s}e^{-st}-\frac{e^{-st}}{s^2}\Bigg|_0^\infty\]
As the term \(-\frac{t}{s}e^{-st}\) evaluated at \(\infty\) is \(\frac{\infty}{\infty}\), L'Hospital's rule will be used to evaluate the integral.
\[\lim_{t\to \infty}-\frac{t}{s}e^{-st}=0\]
The Laplace transform of \( \textit{t} \) can now be evaluated as
\[\mathcal{L}[t]=\frac{1}{s^2}\]

Laplace Transforms: Delta Function

The delta function's Laplace transform can be evaluated by the following expression
\[\mathcal{L}[\delta(t-\tau)]=\int_0^\infty e^{-st}\delta(t-\tau)dt\]
For any \(\tau>0\) the integral of the Delta function will always be 1. We can use this property to solve for the Laplace transform of the delta function
\[\mathcal{L}[\delta(t-\tau)]=e^{-s\tau}\]
if \(\tau=0\)
\[\mathcal{L}[\delta(t)]=1\]
Solving differential equations by Laplace transforms often requires a bank of known Laplace Transforms. We also need to know the Laplace transforms of any arbitrary function. Solving for a Laplace transform of any function requires the use of the above process. You can very quickly make your own bank of transforms.

Laplace Transforms: Arbitrary Functions

In differential equations, we are concerned with solving for solution functions based on certain known factors such as forcing functions and rates. The Laplace transform of arbitrary functions and their derivatives can also be derived to find solutions to a differential equation.
\[\mathcal{L}[y(t)]=\int_0^\infty ye^{-st}dt=Y\]
The Laplace transform of the first derivative can also be derived using integration by parts
\[\mathcal{L}[y']=\int_0^\infty y'e^{-st}dt\]
\[u=e^{-st}\]
\[du=-se^{-st}\]
\[dv=y'\]
\[v=y\]
\[ye^{-st}+s\int ye^{-st}dt\]
The first half of the equation is just the initial value of y(t). The second half of the equation is the Laplace transform of y multiplied by s. The Laplace transform of the first derivative of a function can then be written as
\[\mathcal{L}[y']=sY-y(0)\]
A similar process can be used to solve for the Laplace transform of the second derivative of a function.
\[\mathcal{L}[y'']=s^2Y-sy(0)-y'(0)\]

Solving differential equations by Laplace transforms

\[x'+4x=e^{t}\]
\[x(0)=2\]
Solve for the Laplace Transform of every part
\[\mathcal{L}[x'+4x=e^{t}]\to (sX-2)+4X=\frac{1}{s-1}\]
Group Laplace transform of x(t) and solve for X
\[X(s+4)=\frac{1}{s-1}+2\]
\[X = \frac{1}{(s-1)(s+4)}+\frac{2}{s+4}\]
Separate the polynomial on the left by using partial fraction decomposition
\[\frac{A}{s-1}+\frac{B}{s+4}=\frac{1}{(s-1)(s+4)}\]
\[A(s+4)+B(s-1)=1\]
\[4A-B=1\]
\[A+B=0\]
\[A=\frac{1}{5}\]
\[B=-\frac{1}{5}\]
Now we have three distinct parts of the Laplace transform
\[X=\frac{1}{5}(\frac{1}{s-1})-\frac{1}{5}(\frac{1}{s+4})+\frac{2}{s+4}\]
We know the Laplace transform of \(e^{at}\). We will use that knowledge to solve for the inverse of our Laplace Transform of X
\[\mathcal{L}[e^{at}]=\frac{1}{s-a}\]
\[\mathcal{L}^{-1}\bigg[\frac{1}{5(s-1)}\bigg]=\frac{1}{5}e^{t}\]
\[\mathcal{L}^{-1}\bigg[-\frac{1}{5(s-4)}\bigg]=-\frac{1}{5}e^{-4t}\]
\[\mathcal{L}^{-1}\bigg[\frac{2}{s-1}\bigg]=2e^{-4t}\]
Adding these inverse Laplace transforms together gives the complete solution to the differential equation.
\[x(t)=\frac{9}{5}e^{-4t}+\frac{1}{5}e^{t}\]
When solving for a Laplace transform, we usually neglect the constant of integration as we use initial conditions. We naturally introduce that constant when solving for our Laplace transform of the arbitrary function. This is shown in the introduction of initial conditions. By solving with a known initial condition, the constant of integration is naturally added in. Sometimes, it is beneficial to add this constant into your solution when solving it generally.

General Laplace Transforms

Constant
\[\mathcal{L}[C]=\frac{C}{s}\]
General Polynomial
\[\mathcal{L}[t^n]=\frac{n!}{s^{n+1}}\]
Exponential Function
\[\mathcal{L}[e^{at}]=\frac{1}{s-a}\]
Sinusoidal Functions
\[\mathcal{L}[\cos(\omega t)]=\frac{s}{s^2+\omega^2}\]
\[\mathcal{L}[\sin(\omega t)]=\frac{\omega}{s^2+\omega^2}\]
Delta Function
\[\mathcal{L}[\delta(t-\tau)]=e^{-s\tau}\]
General Functions
\[\mathcal{L}[y(t)]=Y(s)\]
\[\mathcal{L}[y'(t)]=sY(s)-y(0)\]
\[\mathcal{L}[y''(t)]=s^2Y-sy(0)-y'(0)\]

Convolution

To discuss the convolution function, we will use Laplace transforms. When solving an inverse Laplace transform, it is common to run into two separate Laplace functions being multiplied together. As such we will need a method to solve for this multiplication as the functions cannot be inverted separately.
\[\mathcal{L}^{-1}[G(s)F(s)]\]
To start, we will consider the Laplace Transform of two functions multiplied together
\[\mathcal{L}[f(t)g(t)]=\int_0^\infty e^{-s\tau}f(\tau)d\tau \int_0^\infty e^{-s\nu}g(\nu)d\nu\]
As a necessity, the variable for time have been changed in both functions to show that they are distinct. We will now combine the two integrals while maintaining the different time variables.
\[\int_0^\infty \int_0^\infty e^{-s(\tau+\nu)}f(\tau)g(\nu)d\nu d\tau\]
To account for discreet time steps between our two arbitrary time variables, we will employ the following substitution.
\[\nu=t-\tau\]
\[\int_0^\infty \int_\tau^\infty e^{-st}f(\tau)g(t-\tau)d\tau dt\]
The substitution changes the bounds of integration. If the order of integration is changed as well, we can derive the definition of the Laplace transform of a product of functions.
\[\int_0^\infty e^{-st}\int_0^t g(t-\tau)f(\tau)d\tau dt\]
\[[f(t)\star g(t)]=\int_0^t g(t-\tau)f(\tau)d\tau\]
The outer integration is the definition of the Laplace Transform. The inner integration is known as the Convolution function. Notice as well the similarity between the convolution function and Green's function in solving differential equations. You can think of Green's function as the convolution of the homogeneous impulse response function to a forcing function. We can use this function to solve for the Inverse Laplace that was originally shown.
\[\mathcal{L}[f(t)g(t)]=\mathcal{L}[f(t)\star g(t)]=G(s)F(s)\]
\[\mathcal{L}^{-1}[G(s)F(s)]=[g(t)\star f(t)]\]
It is also important to know that Convolution is not order-dependent. You can choose either function to substitute \(t-\tau\) into. It is best to choose the function that requires the least amount of variable substitutions to simplify calculations.

Laplace Transforms and Convolution