Math 3350 Supplementary Notes: Operators and Linearity Dr. Kevin Long Texas Tech University March 3, 2008 The textbook introduces linear operators in almost an offhand way. This concept is so important for differential equations (and for applied mathematics in general) that it needs a more extensive explanation than provided by the book. The notion of a linear operator may seem obvious, so much so that it’s sometimes hard at first to understand why mathematicians would make such a big deal out of such a simple concept. A short list of why linearity is a big deal would include: • Equations defined in terms of linear operators obey the principle of superposition. Superpo- sition is the foundation of two of the most powerful methods for solving practical problems: the method of Laplace transforms (which we will study in detail later in this course) and the method of Green’s functions (which you’ll see in a later course). • Any theorem proved for equations involving linear operators applies to any type of equation involving any type of linear operator. Theorems about properties such as superposition, and methods such as Laplace transforms and Green’s functions apply not only to linear ordinary differential equations but to linear partial differential equations, linear delay equations, linear integral equations, linear integrodifferential equations, and many other kinds of problems. The details of all of these problem types are very different, but the linear structure is the same; that common structure lets us reuse ideas and methods between fields from statics to quantum mechanics. I’ll repeat here what I told you in class: advanced mathematics is about using structure to sim- plify calculations. Linearity is an example of that concept: it is a structural property common to many types of problems; in particular, it is common to all linear ordinary differential equations. By referring to that abstract structure, we’ll be able to establish some very important theorems such as superposition and dependence of solutions on initial conditions in a very simple and general way, without going through a bunch of case-by-case calculations. These theorems then do two impor- tant practical things for us: first, they are used in developing solution methods such as Laplace transforms; second, they help us get physical insight that can be used to understand the behavior of an equation without doing complicated calculations. 1 Operators An operator transforms one function to another. For example, the derivative operator D trans- forms any function u to its derivative u0. We write this using the following notation: D[u] = u0. Operator notation is very similar to function notation; we will distinguish operators from functions by using square brackets [] rather than parentheses () to enclose the argument. Operator notation is similar to function notation because the operator concept is similar to the function concept: operators and functions both take “input” and produce “output.” The difference is that a function accepts one or more numbers as input and produces one or more numbers as output, whereas an operator accepts a function as input and produces a function as output. Examples Here are some simple examples: • The identity operator I returns the input argument unchanged: I[u] = u. • The derivative operator D returns the derivative of the input: D[u] = u0. • The zero operator Z returns zero times the input: Z[u] = 0. Here are some other examples. • Let’s represent as an operator the expression y00 + 2y0 + 5y. We can write this as D[D[y]] + 2D[y] + 5I[y], where D and I are the derivative and identity operators defined previously. Most of the time we won’t write the identity operator explicitly, so our operator will be D[D[y]] + 2D[y] + 5y. We’ll also usually use the shorthand D2[y] ≡ D[D[y]] 2 so that the expression y00 + 2y0 + 5y is, in operator notation, D2[y] + 2D[y] + 5y. Important: Notice that we use “power” notation D2[y] to represent repeated composition of an operator, not repeated multiplication of the results of an operator. The expression D3[u] means D[D[D[u]]], not (D[u])3. • We can represent the entire operation D2[y] + 2D[y] + 5y as an operator, by writing A = D2 + 2D + 5I so that A[y] = D2[y] + 2D[y] + 5y. Linear Operators In this course we will concentrate on an important subset of operators: those which are linear. Definition 0.1 Let u and v be arbitrary functions of a variable t, and let α be an arbitrary complex constant. A linear operator is any operator L having both of the following properties: 1. Distributivity over addition: L[u + v] = L[u] + L[v] 2. Commutativity with multiplication by a constant: αL[u] = L[αu] Examples 1. The derivative operator D is a linear operator. To prove this, we simply check that D has both properties required for an operator to be linear. • D distributes over addition. By the definition of D, d D[u + v] = (u + v) dt and then by the sum rule for derivatives, d du dv (u + v) = + . dt dt dt Putting the previous steps together demonstrates that D[u + v] = D[u] + D[v], so that D has property 1, distribution over addition. 3 • D commutes with multiplication by a constant. By the definition of D, d D[αu] = (αu) dt which in turn equals (by the product rule) du α , dt or αD[u]. Because D[αu] = αD[u], D has property 2. Because D has both properties required of a linear operator, it is linear. 2. If A and B are linear operators, their sum A + B is a linear operator. The sum of two operators is defined so that the sum of the operators returns the sum of their outputs: (A + B)[u] ≡ A[u] + B[u]. We prove linearity by checking that A+B has the two properties required of a linear operator. • A + B distributes over addition. From the definition of A + B we know (A + B)[u + v] = A[u + v] + B[u + v]. We have specified that A and B are linear, so (A + B)[u + v] = A[u] + A[v] + B[u] + B[v], which after reordering the sum gives (A + B)[u + v] = A[u] + B[u] + A[v] + B[v] or (A + B)[u + v] = (A + B)[u] + (A + B)[v] which means A + B distributes over addition. A + B therefore has property 1. • A + B commutes with multiplication by a constant. We have (A + B)[αu] = A[αu] + B[αu] and then because A and B are assumed to be linear, (A + B)[αu] = αA[u] + αB[u]. Factoring out α on the right-hand side gives (A + B)[αu] = α(A[u] + B[u]) or (A + B)[αu] = α(A + B)[u] 4 3. Let’s next see an example of an operator that is not linear. Define the exponential operator E[u] = eu. We test the two properties required for linearity: • Property 1: E[u + v] = e(u + v) = E[u]E[v] which is not equal to E[u] + E[v]. Therefore the exponential operator does not have the first property required for linearity, and is therefore not a linear operator. • Property 2: E[αu] = eαu = (eu)α = E[u]α which is not equal to αE[u]. Therefore the exponential operator does not have the second property required for linearity, and is therefore not a linear operator. 4. Here’s another nonlinear operator: Z x 1/3 A[u] = (u(t))3 dt . 0 We’ll find that this operator has property 2 (commutativity with multiplication by a constant) but not property 1 (distributivity over addition). • Property 1: Z x 1/3 A[u + v] = (u(t) + v(t))3 dt 0 Z x 1/3 = (u(t))3 + 3u(t)(v(t))2 + 3v(t)(u(t))2 + (v(t))3 + dt 0 which is not equal to Z x 1/3 Z x 1/3 A[u] + A[v] = (u(t))3 dt + (v(t))3 dt 0 0 Therefore A does not have property 1. • Property 2: Z x 1/3 A[αu] = (αu(t))3 dt 0 Z x 1/3 = α3 (u(t))3 dt 0 Z x 1/3 = α (u(t))3 dt 0 = αA[u]. Therefore A has property 2. The operator A has property 2 but because it must have both properties required for linearity, it is not a linear operator. 5 Worked Examples Here are some worked examples relating to problems in basic statics. Imagine a 3D object V , with fixed boundary surface and known density distribution function ρ(x). In statics, we would need to compute a number of properties of this body, such as its total mass and center of mass. All of these computations can be expressed in terms of operators acting on the density function ρ. These operators will be integral operators, meaning that they each involve the calculation of an integral involving the density function ρ. Later in the semester we’ll see another integral operator useful in solving differential equations: the Laplace transform. Operators for mass and center of mass The mass M of an object V is computed by integrating the density over volume: Z M = ρ(x) dV. V The integral is a volume integral, which you will remember from calculus 3; if you’re not good with volume integrals, don’t worry, we won’t do any actual computations with them here. We’re just interested in the structure of the operations. We can regard the computation of mass as an operator acting on the density function ρ: it takes ρ as an input, and produces a constant function M as output.