Struggling to understand the reasoning behind the "D operator" method in solving linear ODEs.

Question

In my ODE course at school, we were introduced to various techniques for solving differential equations, but without any proofs or explanations—just formulas and methods to memorize. I’m struggling to understand the reasoning behind one technique in particular: the "D operator" method.

Here’s what I know about this method:

• Suppose we have a linear differential equation with constant coefficients, in the form the form $y''+ay' +b=P(x)$ (it could be of any order really ) where P(x) is a polynomial (also works for other functions). The D operator method involves rewriting the equation as follows:

$$y = \frac{P(x)}{D^2 + aD + b}$$

then to find the particular solution we set $y =\frac{P(x)}{D^2+aD+b}$ then expand $\frac{1}{D^2+aD+b}$ as a series and then differentiate. For example $1-y''= x^6 $ we have that $y =\frac{x^6 }{1-D^2}=(x^6 )(1+D^2 +D^4+D^6+D^8+....) =x^6+30x^4+360x^2+720$

• It also works for exponential function $e^{ax}$, Assume that we want to solve $ay''+by'+cy=e^{ax}$ we will find the particular solution by substituting $D=a$ and this will be the the particular solution unless the denominator is $0$ then we multiply the numerator with $x$ and differentiate the denominator For example $$y''-2y'-3y=4e^{5x}$$ $$y_p= \frac{4e^{5x}}{D^2-2D-3} = \frac{4e^{5x}}{5^2-2*5-3}=\frac{e^{5x}}{3}$$ However here I have no idea what is going on and why do we differentiate and multiply by $x$? and how we even differentiate a differential operator? for example $$y''-2y'-3y=4e^{3x}$$ $$y_p= \frac{4e^{3x}}{D^2-2D-3}=\frac{4e^{3x}}{3^2-2*3-3}\text{we had differentiate }$$ $$y_p= \frac{4xe^{3x}}{2D-2}=xe^{3x}$$
• It also works for $\sin(ax), \cos(ax)$ the difference that is $D^2=-a$, Also it works for $\sinh(ax), \cosh(ax)$ but $D^2=a^2$

My concern is how to rigorously prove all of these steps or how to justify it.

1. What exactly is an operator? How does it work in the context of differential equations? How to rigorously define it?
2. How can we justify expanding $\frac{1}{1-D}$ as a geometric series, I know how to prove that it converges for $|z|<1 , z\in \mathbb{C}$ and why it works for complex but why does it work with $D$?
3. How is the fraction $\frac{1}{1-D}$ defined with an operator like $D$? In particular, what does this fraction mean, and how do we interpret expressions like $D^2$ (I know it is the second derivative but how did the expansion in the geometric series turns the nth power to the nth derivative? )
4. How does multiplying the series by the function lead to derivatives?
5. In case of $f(x)= e^{ax}, \sin(ax), $ etc, how to rigorously prove all its steps ?
6. What does it mean to differentiate $D$? Why do we do it? Why do we multiply with $x$ ?
7. Is there a field or course that studies these concepts rigorously? I’ve studied real analysis mainly form Rudin's books and haven’t found any mention of this operator approach or how to justify or prove any of these steps. Would functional analysis cover this, or is there another area of mathematics that treats operators in this way?

Answer 1

The 'operator' $D$ is a function that acts on functions. Specifically, we have $D(f) = f'$ whenever $f$ is differentiable. It's also possible to interpret a scalar like $2$ as an operator -- a function that acts on functions -- where $2:f \mapsto 2f$.

We can do arithmetic with functions. You might remember in algebra calculating things like $(f+g)(x)$ or $(fg)(x)$ given two functions $f(x)$ and $g(x)$. Since $D$ is a function, you can do things like this with it too. So for instance we can consider a new operator $D+2$, which does the following to a function $f$: $$(D+2)(f) = D(f) + 2(f) = f'+2f $$ Writing $D^2$ means applying $D$ twice, so $D^2(f) = D(f')=f''$ and so on for higher powers.

Operator notation for a DE is a way to think of the left-hand-side of the DE itself as an operator that acts on the solution to the DE. For example $$y-y'' = (1-D^2)y$$ The benefit of this approach is exactly that you can now do algebra with this "LHS operator", like you have learned to do in your examples. Getting into the details of proving that it works is beyond the scope of this answer I think, but some good relevant other questions are linked in the comments.

Answer 2

$$\text{Wrong}:\require{enclose}\enclose{updiagonalstrike,downdiagonalstrike}{y = \frac{P(x)}{D^2 + aD + b}}$$ $$\text{Correct}: y= \frac{1}{D^2 + aD + b}P(x),\,\text{or better:}\,\, y=(D^2+aD+b)^{-1}P(x) $$

If $h(x)=\sum_{n=0}^\infty a_n x^n$ then $h(D)=\sum_{n=0}^\infty a_n D^n$.

Example 1: Find the particular solution of the ODE, $(D+a)y=x^2.$

Solution: Since $(x+a)^{-1}=\sum_{n=0}^\infty\frac{(-1)^nx^n}{a^{n+1}}$, the particular solution is $y=(D+a)^{-1}x^2=\sum_{n=0}^\infty\frac{(-1)^nD^nx^2}{a^{n+1}}=\frac{x^2}a-\frac {2x}{a^2}+\frac2{a^3}.$

Example 2: Find the particular solution of the ODE, $(D^2+(a+b)D+ab)y=x^2.$

Solution: By using example 1, $y=(D+b)^{-1}(D+a)^{-1}x^2=(D+b)^{-1}(\frac{x^2}a-\frac {2x}{a^2}+\frac2{a^3})=\sum_{n=0}^\infty\frac{(-1)^nD^n}{b^{n+1}}(\frac{x^2}a-\frac {2x}{a^2}+\frac2{a^3})\\=\frac{x^2}{ab}-\frac{2x}{ab^2}+\frac2{ab^3}-\frac{2x}{a^2b}+\frac{2}{a^2b^2}+\frac2{a^3b}=\frac{x^2}{ab}-\frac{2(a+b)x}{a^2b^2}+\frac{2(a^2+ab+b^2)}{a^3b^3}.$

More examples can be given. Note that "D" is the differentiation operator defined by $Dy=y'$.

Answer 3

The simplest approach is the best: With differential equations, it's It's always hard to prove that a particular method will yield a solution. But, once you have a candidate for solution, it's always easy to prove that a particular function $y(x)$ is (or is not) - just differentiate. So don't try to prove that a method will work. Just use, and prove that it did work.

Think this is amateurish? It's not. It's used by the greatest mathematicians.

So, instead of proving that the $D$ operator will produce meaningful results, just use it, and prove the candidate you get is correct (which is as easy as taking a deriviative).

Answer 4

A first order ordinary differential equation with constant coefficient has the form

$$f'(x)+af(x)=g(x),$$

or using the notation $D$ as a differentiation operator, $$(D+a)f(x)=g(x).$$

The equation can be solved by multiplying by $e^{at}$, which gives

$$e^{ax}f'(x)+ae^{ax}f(x)=(e^{ax}f(x))'=e^{ax}g(x)$$

and

$$f(x)=e^{-ax}\int e^{ax}g(x)\,dx$$

which we symbolically denote

$$f(x)=\frac1{D+a}g(x).$$

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We can show that the operator $D$ follows some rules of arithmetic. In particular, the product of binomials $D+c$ is equivalent to the successive application of these binomials.

$$(D+a)(D+b)=D^2+aD+bD+ab$$ yields

$$(D+a)(D+b)f(x)=(D+a)(f'(x)+bf(x))=f''(x)+af'(x)+bf'(x)+abf(x).$$

Similarly, one can show that the solution to the second order equation

$$f''(x)+(a+b)f'(x)+abf(x)=g(x)$$

is

$$f(x)=e^{-ax}\int e^{ax}\left(e^{-bx}\int e^{bx}g(x)\,dx\right)dx\\=\frac1{D+a}\frac1{D+b}g(x)$$ which can be written symbolically

$$\dfrac1{D^2+(a+b)D+ab}g(x).$$

In particular, for an exponential RHS, it is easy to show that

$$\frac1{P(D)}e^{cx}=\frac1{P(c)}e^{cx}.$$

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Now given the identity

$$(D+a)(D^2-aD+a^2)=D^3+a^3,$$

we can write that, provided that the third derivative of $g(x)$ vanishes ($f(x)$ is a polynomial at most quadratic), then

$$\frac1{D+a}g(x)=\frac1{a^3}(D^2-aD+a^2)g(x).$$

This generalizes to other degrees.

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As you can see, all these rules are rigorous, and the notation with the $D$ operator can be seen as just a shorthand. It is no coincidence that it behaves similarly to an ordinary variable, and this sets the basis of a very convenient operational calculus.

Anyway, any other arithmetic calculation using the $D$ operator must be validated in the frame of ordinary calculus. For instance, one must be cautious about expressing $\dfrac1{P(D)}$ in terms of a Taylor expansion, as this raises convergence issues.

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PS: the indefinite integrals must be understood with the usual integration constant.