This week, I studied two problems involving differential equations: "Missile Tracking" and the "Solar Furnace" problem. Through these, I have deepened my understanding of differential equations and familiarized myself with related operations. I am recording this here as a modest contribution to spark further discussion.
The Essence of Differential Equations
Many readers know that since Newton and Leibniz invented calculus, it has rapidly permeated almost all academic disciplines, later developing into many remarkable branches such as the calculus of variations and differential equations. It is well known that differential equations are tools for solving many important problems. I wonder what the readers’ understanding of differentials and differential equations is?
In fact, for ordinary differential equations, their essence is the same as the algebraic equations we have already studied, except that the corresponding operational relationships between them include not only the conventional addition, subtraction, multiplication, division, and powers, but also two additional relationships: differentiation and integration. For example, for a first-order differential equation \dot{y}=f(x,y), one might think it is an equation with two variables. In fact, it is not; it is a system of equations consisting of "four unknowns and three equations", which we can write as:
dy=f(x,y)dx, \quad y=\int dy, \quad x=\int dx
The unknowns are x, y, dx, and dy. One point must be noted here: although dx and dy tend toward zero, they are variables rather than constants (please refer to books on rigorous calculus analysis). You might wonder what the benefit of describing it this way is. Actually, this helps us reduce the sense of mystery (not necessarily the difficulty) surrounding differential equations. At the same time, when solving systems of differential equations, it helps guide us in the direction of eliminating variables. Once this is understood, we can purposefully use common algebraic operations and differentiation to eliminate unknowns until only x and dx (or y and dy) remain.
The Calculation of Differentials
This week, I explored the relationship between \frac{d^2 y}{dx^2} and \frac{d^2 x}{dy^2}. Let \frac{dy}{dx}=P, then:
\frac{d^2 x}{dy^2}=\frac{d(1/P)}{dy}=-\frac{dP}{P^2 dy}=-\frac{dP dx}{P^2 dx dy}
Since \frac{dP}{dx}=\frac{d^2 y}{dx^2}, we have:
\frac{d^2 x}{dy^2}=-\frac{\frac{d^2 y}{dx^2}}{(\frac{dy}{dx})^3}
After simplification, it is found that:
\frac{dy}{dx}=-\frac{d^2 y}{d^2 x}
Readers might find this conclusion quite nice, as it links high-order differentials with first-order differentials. However, although this conclusion is not "wrong" in its own context, it is imprecise and perhaps even meaningless. This is because, regarding differentials, I ignored an important issue: generally speaking, the d^2 y in \frac{d^2 y}{dx^2} is not equal to the d^2 y in \frac{d^2 y}{dt^2}. That is to say, even for the same functional expression, its second-order differential is not fixed and depends on the choice of the independent variable. So, what is the significance of the expression above?
BoJone on 2010.11.03
When reprinting, please include the original address of this article: https://kexue.fm/archives/1045
For more detailed information regarding reprinting, please refer to: "Scientific Space FAQ"