Charlie Conner

2022-09-30

Differential equations notation

I've always wondered why does the differential equation notation for linear equations differ from the standard terminology of vector spaces.

We all know that the equation ${y}^{\u2033}+p(x){y}^{\prime}+q(x)y=g(x)$ for some function g is called linear and that the associated equation ${y}^{\u2033}+p(x){y}^{\prime}+q(x)y=0$ is called homogeneous. But why is that? WHY should mathematicians explicitly cause confusion with the rest of the theory of vector spaces?

What I mean by that is : Why not call the equation ${y}^{\u2033}+p(x){y}^{\prime}+q(x)y=g(x)$ an affine equation and call ${y}^{\prime}+p(x){y}^{\prime}+q(x)y=0$ a linear equation? Because linear equations (in the sense of differential equations) are not linear in the sense of vector spaces unless they're homogeneous ; and linear equations (in the sense of differential equations) remind me more of a linear system of the form $Ax=b$ (which is called an affine equation in vector space theory) than of a linear equation at all.

Just so that I made myself clear ; I perfectly know the difference between linear equations in linear algebra and linear equations in differential equations theory ; I'm asking for some reason of "why the name".

I've always wondered why does the differential equation notation for linear equations differ from the standard terminology of vector spaces.

We all know that the equation ${y}^{\u2033}+p(x){y}^{\prime}+q(x)y=g(x)$ for some function g is called linear and that the associated equation ${y}^{\u2033}+p(x){y}^{\prime}+q(x)y=0$ is called homogeneous. But why is that? WHY should mathematicians explicitly cause confusion with the rest of the theory of vector spaces?

What I mean by that is : Why not call the equation ${y}^{\u2033}+p(x){y}^{\prime}+q(x)y=g(x)$ an affine equation and call ${y}^{\prime}+p(x){y}^{\prime}+q(x)y=0$ a linear equation? Because linear equations (in the sense of differential equations) are not linear in the sense of vector spaces unless they're homogeneous ; and linear equations (in the sense of differential equations) remind me more of a linear system of the form $Ax=b$ (which is called an affine equation in vector space theory) than of a linear equation at all.

Just so that I made myself clear ; I perfectly know the difference between linear equations in linear algebra and linear equations in differential equations theory ; I'm asking for some reason of "why the name".

Clare Acosta

Beginner2022-10-01Added 7 answers

Step 1

There is no confusion at all: these are in fact the same concepts when viewed in the right light.

First of all, we need to find a vector space to put all these functions in. Let ${C}^{\mathrm{\infty}}(\mathbb{R})$ be the space of all smooth functions $\mathbb{R}\to \mathbb{R}$; this is naturally a R-vector space, albeit of infinite dimension. Consider the operator $D:{C}^{\mathrm{\infty}}(\mathbb{R})\to {C}^{\mathrm{\infty}}(\mathbb{R})$ defined by

$$Df=\sum _{k=0}^{n}{a}_{k}{f}^{(k)}$$

where ${a}_{k}:\mathbb{R}\to \mathbb{R}$ are some smooth functions and ${f}^{(k)}$ is the k-th derivative of f. D is easily seen to be a linear operator, and a differential equation of the form

$$Df=g$$

is precisely a linear ODE of order n. That is to say, the word ‘linear’ refers to the linearity of D as an operator! In this light, solving a linear ODE consists of two steps:

1. Finding the kernel of D, i.e. solving the homogeneous linear ODE $Df=0$, and

2. Finding a ‘particular integral’ ${f}_{0}$ such that $D{f}_{0}=g$.

Step 2

This is exactly the same as solving a system of (possibly inhomogeneous) linear equations in ordinary linear algebra!

Now, one might be tempted to find an analogue of Gaussian elimination to work with linear ODEs, but the fact that ${C}^{\mathrm{\infty}}(\mathbb{R})$ has infinite dimension and no natural basis tends to screw things up a little...

There is no confusion at all: these are in fact the same concepts when viewed in the right light.

First of all, we need to find a vector space to put all these functions in. Let ${C}^{\mathrm{\infty}}(\mathbb{R})$ be the space of all smooth functions $\mathbb{R}\to \mathbb{R}$; this is naturally a R-vector space, albeit of infinite dimension. Consider the operator $D:{C}^{\mathrm{\infty}}(\mathbb{R})\to {C}^{\mathrm{\infty}}(\mathbb{R})$ defined by

$$Df=\sum _{k=0}^{n}{a}_{k}{f}^{(k)}$$

where ${a}_{k}:\mathbb{R}\to \mathbb{R}$ are some smooth functions and ${f}^{(k)}$ is the k-th derivative of f. D is easily seen to be a linear operator, and a differential equation of the form

$$Df=g$$

is precisely a linear ODE of order n. That is to say, the word ‘linear’ refers to the linearity of D as an operator! In this light, solving a linear ODE consists of two steps:

1. Finding the kernel of D, i.e. solving the homogeneous linear ODE $Df=0$, and

2. Finding a ‘particular integral’ ${f}_{0}$ such that $D{f}_{0}=g$.

Step 2

This is exactly the same as solving a system of (possibly inhomogeneous) linear equations in ordinary linear algebra!

Now, one might be tempted to find an analogue of Gaussian elimination to work with linear ODEs, but the fact that ${C}^{\mathrm{\infty}}(\mathbb{R})$ has infinite dimension and no natural basis tends to screw things up a little...

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