Differential operators

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Differential operators. Let us write the linear differential equation

in a different notation

where D stands for d/dx and Dy for dy/dx. Thus

and, in general,

Thus the symbol D represents the operation of taking the derivative with respect to x of whatever follows it. The symbol D is called an operator because it defines an operation to be performed (in Dy, D performs an operation on y). Using this notation, the equation

can be written as

where 4) is defined as being equivalent to 3) i.e. the meaning of 4) is 3). Now note that the quantity within the brackets of 4) is a polynomial in D. And note that that polynomial in D is also an operator (just more complicated than the operator D) which, in this case, operates on y.

Def. Differential operator. A differential operator is a polynomial in the operator D where D stands for d/dx and Dy for dy/dx.

Example.

The polynomial in D

can be defined as that operator which, when applied to any function y, yields the result

(where y is assumed to possess as many derivatives as necessary to allow the operations to take place).

Def. Equality of operators. An operator A is said to be equal to an operator B if and only if the same result is produced when A acts upon a function y as when B acts upon y. In other words, A = B if and only if Ay = By for all functions y possessing the derivatives necessary for the operations involved.

Def. Product of two operators. The product AB of two operators A and B is defined as that operator which produces the same result as is obtained by using the operator B followed by the operator A. Thus ABy = A(By). The product of two differential operators always exists and is a differential operator. For operators with constant coefficients AB = BA. For operators with variable coefficients this is not generally true.

Example 1. Let A = D + 3 and B = 2D - 1. Compute ABy.

= (2D² + 5D - 3) y

Thus AB = (D +3)(2D - 1) = 2D² + 5D - 3

Let us now compute BAy.

= (2D² + 5D - 3) y

Thus BA = 2D² + 5D - 3 = AB

The above example involved operators with constant coefficients. Now let us consider a similar example in which one of the operators has variable coefficients.

Example 2. Let A = xD + 3 and B = D - 2. Compute ABy.

= (xD² - 2xD + 3D - 6) y

= [xD² - (2x - 3)D - 6] y

Thus AB = (xD + 3)(D - 2) = xD² - (2x - 3)D - 6

Now let us compute B(Ay).

= [xD² + (4 - 2x)D- 6] y

BA = xD² + (4 - 2x)D - 6

Thus we see that AB BA.

In general, in the case of operators with variable coefficients, the product is dependent on the order of the factors.

Def. Sum of operators. The sum of two differential operators is obtained by expressing each in the form

and adding corresponding coefficients.

Example. If

A = 2D² + 5D + x - 3

and

B = 3x²D² + D + 3x - 1

A + B = (3x² + 2)D² + 6D + 4x - 4

Differential operators are linear operators. A differential operator is a linear operator. In other words, if A is any differential operator, if c₁ and c₂ are arbitrary constants, and if f₁ and f₂ are any functions of x possessing the required number of derivatives, then

A(c₁f₁ + c₂f₂) = c₁Af₁ + c₂Af₂

Differential operators with constant coefficients. Let A, B and C be differential operators of type

a₀Dⁿ + a₁D^n-1+ ....... + a_n-1D + a_n

in which the coefficients a₀, a₁, ...., a_n are constants. With the above definitions of addition and multiplication the following algebraic laws hold:

1) A + B = B + A commutative law of addition

2) (A + B) + C = A+ (B + C) associative law of addition

3) (AB)C = A(BC) associative law of multiplication

4) A(B+C) = AB + AC distributive law of multiplication over addition

5) AB = BA commutative law of multiplication

In addition, if m and n are any two integers

6) D^mDⁿ = D^{m + n}

Thus differential operators with constant coefficients satisfy all the laws of ordinary algebra with regard to addition and multiplication and consequently can be manipulated algebraically in the same way we manipulate regular algebraic polynomials as far as the operations of addition and multiplication are concerned. Also, we can factor operators with constant coefficients using synthetic division.

In algebra if we multiply both sides of some equation such as

5x² + 3x = x³ + 9

by some expression such as

2x² + x + 1

the solution set of the new equation will contain the solution set of the old equation as a subset i.e. none of the solutions of the original equation are lost in doing this. Equations involving differential operators with constant coefficients can be manipulated in this same way. For example, if we multiply both sides of the equation

(2) (D² + 2D - 1)(3D³ - 4)y = 5x² + 2

by the operator

(D² + 3)

the solution set of the new equation will contain the solution set of the old as a subset i.e. the solution set of the old equation is not affected by doing this. Thus this type of operation is valid. To understand why this is true one must realize that “multiplication” by a factor such as (D² + 3) simply means that a certain operation is to be performed on both sides of the equation. For example, (D² + 3)y means (1) Take the second derivative of y. (2) Multiply y by 3. (3) Add the results of (1) and (2).

Some theorems involving e^mx and x^ke^mx. The following theorems are important in regard to solving linear differential equations with constant coefficients.

Theorem 1. For any constant m and integer k

D^ke^mx = m^ke^mx

Theorem 2. Let f(D) be a polynomial in D

f(D) = a₀Dⁿ + a₁D^n-1+ ....... + a_n-1D + a_n

Then, from Theorem 1,

f(D)e^mx = a₀mⁿe^mx + a₁m^{n -1}e^mx + ........ + a_{n -1}me^mx + a₀e^mx

1) f(D)e^mx = e^mxf(m)

From 1) we see that if m is a root of the equation f(m) = 0, then f(D)e^mx = 0.

Example. Let f(D) = 2D² + D - 6. Then the equation for f(m) = 0 is

2m² + m - 6 = 0

(m + 2)(2m - 3) = 0

which has the roots m₁ = -2 and m₂ = 3/2.

From 1) we can see that

(2D² + D - 6)e^-2x = 0

and

(2D² + D - 6)e^3/2x = 0

In other words, y₁ = e^-2x and y₂ = e^3/2x are solutions to the differential equation

(2D² + D - 6)y = 0

The following theorems show the effect of the operator (D - m)ⁿ on the quantity x^ke^mx.

Theorem 3

(D - m)^k x^ke^mx. = k!e^mx

Proof.

(D - m)x^ke^mx. = k x^k-1e^mx + mx^ke^mx - mx^ke^mx

= k x^k-1e^mx

(D - m)² x^ke^mx. = k(D - m)x^k-1e^mx

= k(k - 1)x^k-2e^mx

Repeating this process we get

(D - m)^k x^ke^mx. = k(k-1) ...... 2 ·1x⁰e^mx

= k!e^mx

Theorem 4.

(D - m)ⁿ x^ke^mx. = 0 for k = 0, 1, 2, .... , (n - 1) i.e. for all k < n

Proof. We wish to show that for all n > k,

(D - m)ⁿ x^ke^mx. = 0

Suppose n = k + 1. Then

(D - m)^k+1 x^ke^mx = (D - m)(D - m)^k x^ke^mx

= (D - m)k!e^mx (from Theorem 3)

= mk!e^mx - mk!e^mx

= 0

For n = k + 2 we have

(D - m)^k+2 x^ke^mx = (D - m) [(D - m)^k+1 x^ke^mx] = (D - m) [0] = 0,

etc.

References

1. Earl D. Rainville. Elementary Differential Equations