# How would you explain to a dummy how a diff. eq. is linear?

1. Jan 19, 2017

### Eclair_de_XII

1. The problem statement, all variables and given/known data
This is the definition I got from my book:

"A crucial classification of differential equations is whether they are linear or nonlinear. The ordinary differential equation

$F(t,y,y',...,y^{(n)})=0$

is said to be linear if F is a linear function of the variables $y, y',...,y^{n}$; thus the general linear ordinary differential equation of order $n$ is

$a_0(t)y^{(n)}+a_1(t)y^{(n-1)}+...+a_n(t)y=g(t)$

2. Relevant equations
I saw something similar in my Linear Algebra class used to prove linear independence that dictates the conditions needed, that none of the vectors [derivatives] can be written as linear combinations of each other. Basically a set of vectors $v_i$ is a linearly independent set if the below equation is satisfied for any non-zero constant $a_i$.

$a_1v_1+...+a_nv_n=0$
$a_0(t)y^{(n)}+a_1(t)y^{(n-1)}+...+a_n(t)y=g(t)=0$

I think it'll probably come up in my Linear Algebra/Differential Equations class I will eventually be taking, but this was the closest relevant equation I could relate the concept of linear differential equations to.

3. The attempt at a solution
Anyway, my conclusion is that the $a_i(t)$ are all constant relative to the independent variable $y$ (that is, they do not change as the value of $y$ or any of its derivatives change for any $t$. The $a_i(t)$ can still be non-linear in respect to $t$, but not to $y$. Can anyone provide for me a better explanation of why a coefficient that is a function of $t$ can be nonlinear?

2. Jan 19, 2017

### BvU

In your book, are the $y$ functions of $t$ or are they functions of yet another variable, e.g. $x$ ?

Check here

3. Jan 19, 2017

### Staff: Mentor

Actually the vectors $\{v_i\}$ are linear independent, if $a_i=0$ is the only solution of this equation. If you have a solution with some $a_i \neq 0$, then you have a linear dependency.
However, the $a_i(t)$ change if the changes of the $y^{(i)}(t)$ are due to changes of $t$, because they depend on $t$.
The coefficients - as you have noticed - depend only on the variable $t$. Nothing is said about the nature of this dependence. It could be constant, linear, polynomial, even exponential or none of these. So this is a linear differential equation, too:
$$-\cos(t) \cdot y^{(4)}(t) + 2\cdot y^{(3)}(t) + e^{t^2}\cdot y^{(2)}(t)+4t \cdot y'(t)+(t^3-2t+1)\cdot y(t) = g(t)$$
and $g(t)$ can be any function of $t$ as well, e.g. $g(t) = \sin (\log |t|)$. The property "linear" in "linear differential equation" only refers to the function $F$. One could also write $\\ F(t,y(t),y'(t),\ldots,y^{(n)}(t)) = F \cdot (1,y(t),y'(t),\ldots,y^{(n-1)}(t),y^{(n)}(t))^\tau \\ = (-g(t),a_n(t),\ldots,a_0(t)) \cdot (1,y(t),y'(t),\ldots,y^{(n-1)}(t),y^{(n)}(t))^\tau$
as a matrix multiplication of two vectors, hence a linear function. Of course things get quickly rather inconvenient the more complicate the coefficient functions $a_i(t)$ are, so you will mostly deal with constant, linear or quadratic $a_i(t)$. But periodic (harmonic oscillator) or exponential ones (economic growth) also appear in the description of real world processes.

4. Jan 19, 2017

### Eclair_de_XII

No, in my book, $y$ is strictly a function of $t$. That second-to-last paragraph in that link helped, in any case. Thanks.

Ah, right. Thanks for the refresher.

Hm, so what you're saying, is that the function $F$ can be written as a linear combination of its derivative variables and possible non-linear coefficients...?

5. Jan 19, 2017

### Staff: Mentor

Yes, because linearity of $F$ means a linear combination of $\{1,y,y',\ldots ,y^{(n)}\}$ with coefficients, that are themselves functions of $t$. And these don't have to be linear as functions of $t$. So one has to distinguish between the role of the $a_i$ as coefficients, which define $F$, and the role of the $a_i(t)$ as functions of the variable $t$.

I don't want to confuse you and the comparison isn't really one-to-one, but imagine a vector $\vec{F}$ over the real numbers, i.e. $\vec{F}=(a_1,\ldots,a_n)$. Then you can consider the $a_i$ as the real coefficients of $\vec{F}$ or as something else, e.g. limits of sequences of rational numbers, say $a_i = (r^{(i)}_n)_{n \in \mathbb{N}}$. Here it is obvious whether we talk about the coefficients or the limits.
In the case of differential equations, all the $a_i(t)$ are functions as the $y^{(i)}$ are. Thus it's not as obvious which is meant when terms like "linearity" occur; $F$ is linear, the $a_i$ not necessarily.

6. Jan 19, 2017

### Eclair_de_XII

I see. So can I just remember it as $a_i(t)$ being constant relative to $y_i(t)$? The former does not depend directly on the latter, and they both change in respect to the same variable, albeit their rate of change will not necessarily be the same. Or is that kind of thinking wrong?

This is why it confused me.

7. Jan 20, 2017

### Staff: Mentor

Yes, that's correct. Or in formulas
$$\frac{d}{d y^{(i)}(t)} a_j(t)=0\; \;,\; \;\frac{d}{d t}a_j(t) \neq_{i.g.} 0\; \;,\; \;\frac{d}{d t}y^{(i)}(t) = y^{(i+1)}(t)\neq_{i.g.} 0$$