Linearity, and What It Does and Does Not Let You Do
Chapter Forty-Seven
Syllabus topic Module 1, "1.2 The Laplace Transform"
Pages 104 to 105 of 303
In one line
The transform of a sum is the sum of the transforms, and a constant passes straight through, but nothing else is respected.
The property
This is the first of MU's "Theorems on Important Properties of Laplace Transformation", and the one used in every single question. Her label covers seven results in all, and the table at the end of the chapter on multiplying by t gathers the properties of the Laplace transform in one place, each with its own chapter: linearity, the two shifting theorems, multiplication by a power of t, division by t, change of scale, the transform of a derivative and the transform of an integral.
L{a f(t) + b g(t)} = a L{f(t)} + b L{g(t)}
In words: the Laplace transform is a linear operator.
Why it holds
Because integration is linear, and the transform is an integral. The integral of a sum is the sum of the integrals, and a constant factor comes out in front. There is nothing more to the proof than that, and an examination answer can be three lines long.
Write the transform of the combination as its defining integral, split the integral at the plus sign, and take the constants out. Each remaining integral is one of the two transforms by definition.
Worked
L{3 + 4t} = 3/s + 4/s^2
L{2 e^(3t) - 5 sin(2t)} = 2/(s - 3) - 10/(s^2 + 4)
L{t^2 - 3 cos(4t) + e^(-t)} = 2/s^3 - 3 s/(s^2 + 16) + 1/(s + 1)
L{(2 + t)^2} = 4/s + 4/s^2 + 2/s^3
The last line needs the bracket expanded before linearity can be used: 4 plus 4t plus t squared. Linearity handles sums, and a square is not a sum until you make it one.
What linearity does NOT give you
This is the real content of the chapter, and it is the source of more wrong answers than any other single thing in the module.
L{f(t) g(t)} = L{f(t)} L{g(t)}
L{f(t)/g(t)} = L{f(t)}/L{g(t)}
L{f(t)^2} = L{f(t)}^2
None of those is true. Here is the counterexample, on the simplest possible functions. Take f(t) = g(t) = t.
L{t} = 1/s^2
L{t t} = 2/s^3
The product of the two transforms would be one over s to the fourth. The transform of the product is 2 over s cubed. They are not equal and they are not even the same shape.
So whenever a product appears, one of three things has to happen.
Turn the product into a sum first, with an algebraic or trigonometric identity. This is what to try first, and it works surprisingly often.
L{sin(t) cos(t)} = 1/(s^2 + 4)
L{t(t + 1)} = 2/s^3 + 1/s^2
Use a shifting theorem, if one factor is an exponential. The next chapter.
Linearity, and What It Does and Does Not Let You Do
Use the convolution theorem, which is what actually replaces the false product rule, and which says that a product of transforms corresponds to a convolution and not a product.
Where linearity is used without being noticed
Three places, so that you see how much rests on one small theorem.
Solving a differential equation. The whole method depends on being able to transform each term of the equation separately. Without linearity there would be no method at all.
Partial fractions. Splitting F(s) into simpler fractions and inverting each one is linearity applied to the inverse transform, which is linear for exactly the same reason.
L⁻¹{1/(s - 1) + 1/(s + 1)} = e^t + e^(-t)
Building the table. Every compound entry, such as the transform of a polynomial, is linearity applied to the elementary rows.
Linearity of the inverse transform
Worth stating separately, because it is used constantly and because a question sometimes asks for it.
L⁻¹{2/s + 3/s^2} = 2 + 3t
L⁻¹{1/(s - 2) - 1/(s - 3)} = e^(2t) - e^(3t)
The reason: if F is the transform of f and G of g, then by the linearity above aF plus bG is the transform of af plus bg, so the inverse of aF plus bG is af plus bg.
Check yourself
L{5 - 2t + t^2} = 5/s - 2/s^2 + 2/s^3
L{3 cos(t) + 4 sin(t)} = 3 s/(s^2 + 1) + 4/(s^2 + 1)
L{e^(t) + e^(-t)} = 1/(s - 1) + 1/(s + 1)
L{2 cosh(t)} = 2 s/(s^2 - 1)
The last two lines are the same function, since e to the t plus e to the minus t is twice cosh t, so the two answers must agree. Adding the two fractions of the third line gives 2s over s squared minus 1, which is the fourth. Checks like that cost ten seconds and catch a slip.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.