Fourth 3 week block

The Laplace transform

Ordinary Differential Equations, Tenenbaum & Pollard, Lesson 27, pages292-312

In this block we will apply the Laplace transform to the solution of linear differential equations involving higher order derivatives.

Laplace’s idea was to transform differential equations into algebraic equations, rearrange the algebra (sometimes easier said than done) and then look for the unique solution function that gave this algebraic answer. In other words, the differential equation gives us an algebraic gadget, and if we can find a function from a catalog that gives the same algebraic gadget then we have, thanks to the uniqueness of the algebraic gadget, found the solution to the differential equation.

The Laplace transform is an operator on functions, and typically those functions are solutions to linear differential equations.

What makes the Laplace transform work is that solutions to linear differential equations do not grow too fast. We will use the following basic fact, without going into any more details at this point:

if $x(t)$ satisfies a linear differential equation (of any order) then the improper integral $\int_{0}^{\infty}e^{-st}x(t)dt$ converges for some value of $s$ .

What it means for the improper integral $\int_{0}^{\infty}e^{-st}x(t)dt$ to converge, is that $\lim_{R\to \infty}\int_{0}^{R}e^{-st}x(t)dt$ exists, for some $s$ .

Here is an example where the improper integral does not converge for any value of $s$ :

Let $x(t)=e^{t^2}$ . Then the improper integral

$\int_{0}^{\infty}e^{-st}x(t)dt= \int_{0}^{\infty}e^{-st}e^{t^2}dt= \int_{0}^{\infty}e^{-s}e^tdt = e^{-s}\int_{0}^{\infty}e^tdt$

$=e^{-s}\lim_{R \to \infty}\int_{0}^Re^tdt = \lim_{R\to\infty}e^{-s}(e^R-1)$

does not converge for any value of $s$ .

For convenience, let’s call a function $x(t)$ for which the improper integral $\int_{0}^{\infty}e^{-st}x(t)dt$ converges for some $s$ , exponentially-bounded. The name is meant to imply that, in some loose sense, these functions do not increase too much faster than an exponential function. The calculation above shows that the function $x(t)=e^{t^2}$ is not exponentially-bounded. Solutions to linear differential equations are always exponentially-bounded.

For an exponentially-bounded function $x(t)$ the Laplace transform of $x(t)$ is $\mathscr{L}[x(t)]:=\int_{0}^{\infty}e^{-st}x(t)dt$ .

Here is a very simple example:

the function $x(t)=1$ is exponentially-bounded because it is constant, and its Laplace transform is $\mathscr{L}[1]=\int_{0}^{\infty}e^{-st}dt=\lim_{R\to\infty}\int_0^R e^{-st}dt=\lim_{R\to\infty}\frac{1}{s}(1-e^{-sR})=\frac{1}{s}$ for $s>0$ .

For more complicated functions (that is, just about everything else) the calculation of Laplace transforms becomes more complicated. Not so long ago people complied tables of Laplace transforms because calculation of the Laplace transform was relatively less easy, and because one wanted to read the answer backwards – knowing what Laplace transform we got from a solution to a linear differential equation, which known function also had this Laplace transform. The text for this course has short tables of Laplace transforms on pages 306 and 310.

Nowadays, there is some progress in this area. Because computer algebra systems have built-in versions fo the latest algorithms for calculating integals, it is no surprise that these systems can readily calculate Laplace transforms. So the hard work of calculating integrals is passed to the computer algebra system which does this not only as well as we can do it, but better than we can, and as well as it can possibly be done.

The MATLAB command for calculating the Laplace transform $\mathscr{L}[x(t)]$ of a function $x(t)$ is

>> syms s t; laplace(x(t) ,t,s)

Without the command “syms s t” MATLAB does not know how to handle the symbols “s” and “t”.

So, for example,

>> syms s t; laplace(1 ,t,s)

yields

1/s

The Mathematica command for the Laplace transform is

LaplaceTransform[x[t], t, s]

Laplace transforms and linear differential equations

It is a simple consequence of basic properties of integrals that for any two exponentially-bounded functions $x(t),y(t)$ and any two constant $a,b$ we have $\mathscr{L}[ax(t)+by(t)]= a\mathscr{L}[x(t)]+b\mathscr{L}[y(t)]$ . In other words, the Laplace transform is a linear operator on exponentially-bounded functions. It is this linearity property that allows us to usefully apply the Laplace transform to linear differential equations.

One more ingredient we will need is a formula to relate the Laplace transform of the derivative $\frac{dx}{dt}$ of a function $x(t)$ to the Lapalce transform of $x(t)$ .

The formula we want comes about by a straightforward use of integration by parts:

$\int_0^{\infty}e^{-st}\frac{dx(t)}{dt}dt=\lim_{R\to\infty}(e^{-sR}x(R)-x(0))+s\int_0^{\infty}e^{-st}x(t)dt=\mathscr{L}[x(t)]-x(0)$

First-order homogeneous linear differential equations

These are differential equations of the form $\frac{dx}{dt}+Bx =0$ .

By applying a Laplace transfrom to both sides of this differential equation we see that we must have $s\mathscr{L}[x(t)]-x(0)+B\mathscr{L}[x(t)]]=0$ for a solution $x(t)$ to the differential equation. That is, $\mathscr{L}[x(t)]=\frac{x(0)}{B+s}$ for $s>-B$ .

So, if we can find a function $x(t)$ with $\mathscr{L}[x(t)]=\frac{1}{B+s}$ then , by the uniqueness of Laplace transforms for exponentially-bounded functions, we will have found, to within a constant, the solution to the differential equation.

Fortunately for us, both MATLAB and Mathematica have built-in commands for finding the inverse of the Laplace transform. For MATLAB the command is

>> syms A B s t; ilaplace(A/(B+s),s,t)

with the result

>> A*exp(-B*t)

For Mathematica the command is

Inverse LaplaceTransform[A/(B+s),s,t]

with the result

$\bf{A e^{-Bt}}$

In our case, $A=x(0)$ so the solution to the differential equation is $x(t) = x(0)e^{-Bt}$ .

First-order non-homogeneous linear differential equations

These are differential equations of the form $\frac{dx}{dt}+Bx =C$ where $C\neq0$ . A similar approach to the homogeneous case shows us that the Laplace transform $\mathscr{L}[x(t)]$ of a solution $x(t)$ must satisfy $\mathscr{L}[x(t)]=\frac{x(0)}{B+s}+\frac{C}{s(B+s)}$ .

Using inverse Laplace transforms we get $x(t) = \frac{C}{B}+\frac{AB-C}{B}e^{-Bt}$ where $A= x(0)$ .

Second-order linear differential equations

These are linear differential equations of the form $\frac{d^2x}{dt^2}+B\frac{dx}{dt}+Cx(t)=D$ .

We apply the Laplace transform to $\frac{d^2x}{dt^2}$ to get

$\mathscr{L}[\frac{d^2x}{dt^2}]$

$=s\mathscr{L}[\frac{dx}{dt}]-\frac{dx}{dt}\vert_{t=0}$

$=s(s\mathscr{L}[x(t)]-x(0))-\frac{dx}{dt}\vert_{t=0}$

$=s^2\mathscr{L}[x(t)]-sx(0)-\frac{dx}{dt}\vert_{t=0}$

Applying this to the second-order differential equation above gives

$s^2\mathscr{L}[x(t)]-sx(0)-\frac{dx}{dt}\vert_{t=0} +B(s\mathscr{L}[x(t)]-x(0))+C\mathscr{L}[x(t)]=\frac{D}{s}$ .

This is an algebraic expression which we solve for $\mathscr{L}[x(t)]$ . We then apply the inverse Laplace transform as before to find the solution of the linear differential equation.

Details for doing this are given on pages 185-187 of differential Equations with MATLAB, Hunt and others, or in the “Applications” section for “LaplaceTransform” in the Mathematica help system.

Note that in the second-order differential equation above, we have assumed $D=0$ , but there is no reason to do this: we can replace D by any function for which we can find a Laplace transform, that is, by any exponentially-bounded function.

Second-order equations and systems of first order equations

The second-order linear equation $\frac{d^2x}{dt^2}+B\frac{dx}{dt}+Cx(t)=D$ can be written as a system of two first-order linear equations by the simple device of writing $y(t)=\frac{dx}{dt}$ . By this device, we can re-write the second-order linear differential equation as the following system of first-order linear equations:

$\frac{dx}{dt}+By(t)+Cx(t)=D$

$layex \frac{dx}{dt}=y(t)$.

In the other direction we can, in general, write a system of linear-first-order equations as a single second-order linear differential equation. The exception is when the two first-order equations are completely decoupled.

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Fourth 3 week block

The Laplace transform

Laplace transforms and linear differential equations

First-order homogeneous linear differential equations

First-order non-homogeneous linear differential equations

Second-order linear differential equations

Second-order equations and systems of first order equations

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