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Laplace transforms · Differential Equations | Math Infinitum

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Differential Equations
Laplace transforms

Differential Equations · Chapter 07

Laplace transforms

An integral that turns differentiation into multiplication. Transform a linear equation, solve the algebra, invert with a table: initial values enter by themselves, and forcing that switches on and off, strikes in an instant or arrives as data is handled by the same three steps.

An integral that turns differentiation into multiplication. Transform a linear equation, solve the algebra, invert with a table: initial values enter by themselves, and forcing that switches on and off, strikes in an instant or arrives as data is handled by the same three steps.

L {f} (s)

0 \int \infty e^{- s t} f (t) d t

L {y^{'}} = s Y - y (0)

e^{- a s}

u (t - a)

δ (t - a)

(f * g) (t)

Y (s)

Lessons

Build the foundation

01The Laplace transformThe Laplace transform turns a function of time into a function of a new variable, s, by weighting it with a decaying exponential and adding up the result over all positive time. What makes it worth doing is what it does to derivatives: differentiation in time becomes multiplication by s, so a linear differential equation with constant coefficients becomes an algebraic equation, with its initial conditions built in. Solve the algebra, transform back, and the problem is done.Read lesson →
02A table of transformsA handful of transforms, computed once, solve almost every problem of this chapter: powers of t, exponentials, sines and cosines, and the rule that multiplying by an exponential shifts the transform. Each entry is one integration, usually by parts. After that the table is used the way a multiplication table is: forwards to transform, and backwards to invert.Read lesson →
03Transforms of derivativesThe whole method rests on one computation. Integrating by parts, the transform of a derivative is s times the transform of the function, minus the value of the function at zero. Differentiation in time becomes multiplication by s, and the initial value appears on its own, without being asked for. Applied twice, the rule gives the transform of a second derivative, which carries both the initial value and the initial slope.Read lesson →
04Solving initial value problemsThe method has three steps. Transform the equation, turning derivatives into powers of s and bringing the initial values in; solve the resulting algebraic equation for the transform of the solution; and invert, usually after partial fractions, reading the answer off the table. The same three steps work for every linear equation with constant coefficients, forced or not: the forcing simply adds its own transform to the algebra.

Read lesson →

05Step functionsReal forcing is often switched: a voltage turned on at one moment and off at another, a load placed on a beam and later removed. The unit step function, zero before a given time and one after it, is the switch. Multiplying a function by a step turns it on at that time, and differences of steps build pulses. The transform of a step is simple, and so is the rule that switching a function on later multiplies its transform by a decaying exponential.Read lesson →

06Discontinuous forcingWith the step function, a forcing that switches on and off becomes one formula, and the transform handles it without solving on each piece separately and matching the pieces up. The solution comes out as a sum of delayed responses, each switched on when its piece of forcing starts. Even when the forcing jumps, the solution of a second-order equation stays smooth: it is the acceleration that jumps, not the position or the velocity.Read lesson →

07Impulses and the delta functionA hammer blow, a kick, a sudden charge: a force so short and so large that only its total effect matters. Model it as a pulse of fixed area squeezed into a shorter and shorter time; in the limit it becomes the delta function, zero except at one instant and with area one. It is not a function in the ordinary sense, but its transform is simply an exponential, and in a mechanical system an impulse simply changes the velocity at one instant.Read lesson →

08ConvolutionA linear system answers an impulse with its impulse response, and any forcing can be thought of as a stream of impulses, one at every instant, each as large as the forcing is then. The response to all of them is the sum of the impulse responses they set off, each delayed to its own instant. That sum is an integral, the convolution of the forcing with the impulse response, and its transform is simply the product of the two transforms.Read lesson →

09Chapter practiceBring the chapter together: compute and invert transforms with the table and the shift rules, solve an initial value problem in three steps, write a switched forcing with steps, respond to an impulse, and read a product of transforms as a convolution. No worked example sits above the answers.Read lesson →