Oftentimes we, as engineers, have to write energy/mass/momentum balances to derive the governing equations for a quantity $\phi(x,t)$ of a physical system. One of many derivations of such balances involves writing a net sum of fluxes $\Phi(x,t)$ over the system's surface and summing them up to the accumulation term $\partial\phi/\partial t$. In such a method, it is possible to select simplified systems (e.g. a cube or cylinder, et similia) and write each balance for a specific dimension. In the case of a cube, we could write for the $x$ direction over the surface $A_x$ as the following Eulerian spec balance $$ \left.\frac{\partial\phi}{\partial t}\right|_x=A_x\left.\Phi\right|_x-A_x\left.\Phi\right|_{x+dx}+Vf(x,t) $$ where $f(x,t)$ is a source volume term. Then, we almost always assume that $\Phi(x,t)$ is smooth around $x$, and the following relation holds: $$ \Phi_{x+dx}=\Phi_x+\left.\frac{\partial\Phi}{\partial x}\right|_xdx+o(dx^2)\tag{1}\label{fp} $$

Substituting the first order expansion of the fluxes, one obtains the well known balance $$ \frac{1}{V}\left.\frac{\partial\phi}{\partial t}\right|_x=-\left.\frac{\partial\Phi}{\partial x}\right|_x+f(x,t)\quad\to\quad\frac{1}{V}\frac{\partial\vec{\phi}}{\partial t}=-\nabla_x\Phi+f(x,t) $$

I am aware that the above derivation can be substituted with any smooth and closed system $\Omega$, by writing the balances in surface/volume integral form, yielding the same result. However, even in such generalized derivation, step $\eqref{fp}$ is performed with the same first-order approx.

My question is

why stop at the first derivative in approximating $\Phi_x$ functional form?

Under the assumptions of the above derivation, why even the most prominent reference texts report that fluxes at the system's boundary have to be treated at first order, are there other cases in which this approximation fails or it is not sufficient? I've tried to look for relevant literature but failed.

  • $\begingroup$ ... are there other cases in which this approximation fails or it is not sufficient? Yes. Double and triple integrals can be a mess. If the integral doesn't evaluate identically to a non-zero amount when the order of integration is changed, then the integral isn't valid. Typically, you get a 0 with one order and a non-zero with a different order. This usually happens when the boundaries are at infinity. Setting non-infinite boundaries makes these problems very hard to solve. But the solutions may have features that are entirely lacking from the bogus solutions at infinity. $\endgroup$
    – Phil Sweet
    Commented Aug 2, 2022 at 23:11
  • $\begingroup$ Fluxes are, in effect, velocities. It turns out that in the physical universe we inhabit, all of it's dynamics can be described by two differentials in time. So position, velocity, and acceleration handle all spacial descripions. We don't need a jounce term. In your example, the boundary is in x, and the flux is x', so you only need one flux differential to handle dynamics in this universe. See also en.wikipedia.org/wiki/Ostrogradsky_instability, and the section Higher derivatives of generalized coordinates $\endgroup$
    – Phil Sweet
    Commented Aug 3, 2022 at 3:03

3 Answers 3


Great question, ultimately related to the definitions of derivatives and integrals using limits and the fundamental theorem of calculus. Not a mathematician, so things might be informal and a bit chaotic, but overall I hope my answer helps the intuition about these concepts.

Short answer:

Because the fundamental theorem of calculus makes this exact in the limit $dx\to0$.

Long but hopefully more intuitive answer:

Looking at the derivative

We start with your equation (1): $$ \Phi_{x+dx}=\Phi_x+\left.\frac{\partial\Phi}{\partial x}\right|_xdx+o(dx^2). $$ This relationship becomes exact in first order sense as we take the limit $dx\to0$: $$ \lim_{dx\to0}\Phi_{x+dx}=\Phi_x+\lim_{dx\to0}\left.\frac{\partial\Phi}{\partial x}\right|_xdx. $$ This might become clearer if we plug in the definition of the derivative: $$ \lim_{dx\to0}\Phi_{x+dx}=\Phi_x+\lim_{dx\to0}\frac{\Phi_{x + dx}-\Phi_{x}}{dx}dx. $$ In fact, we can lose the limit entirely here! $$ \Phi_{x+dx}=\Phi_x+\frac{\Phi_{x + dx}-\Phi_{x}}{dx}dx=\Phi_x+\Phi_{x + dx}-\Phi_{x}=\Phi_{x + dx}. $$

First- and second order expansions

For non-zero dx, a first order expansion approaches the exact function for smaller $dx$, and a higher-order expansion approaches that function faster (in a small region). In a sense, the "region" of confidence is smaller for a first order approximation. You already mention smoothness and texts on Taylor expansions elaborate on the region of convergence, requirements of the function, etc.

Exactness of first order expansion when taking the limit

Letting $dx$ approach zero, the higher order terms vanish quickly. In fact, $dx=10^{-3}$ gives $dx^2 = 10^{-6}$, $dx^3 = 10^{-9}$, etc., i.e. the order of magnitude of the difference between first and second (or third) order terms increases quadratically (or cubically, respectively). Therefore, we only need the first order term to determine $\Delta \Phi$ in a small region. If we weren't interested in changes, we could also use a constant value. Letting taking the limit of $dx$ to zero could perhaps be more intuitively written as: $$ \lim_{dx\to 0} dx = \lim_{n\to \infty} 10^{-n} $$ and the difference in magnitude between first and second order terms scales with $dx$: $$ \lim_{dx\to 0} \frac{dx}{dx^2} = \lim_{dx\to 0} \frac{1}{dx} \to \infty. $$ (Technically, the limit doesn't exist I guess.)

Use in a derivative balance equation

The value of $\lim_{dx\to0}\Phi_{x+dx}$ is never used "raw". It is simply \Phi_{x} for a "smooth" function. Your resulting balance equation uses the gradients, or tangent line in the single variable case, which involves the derivative and thus dividing by the volume $V$. Somewhere along the way by doing that you implicitly divided by $dx$, resulting in the gradient/derivative of $\Phi$. Again, implicitly, you also used the limit $dx\to 0$ in your substitution, which allows you to strike out second order terms and remain exact.

Use in an integral equation

As an exercise, look at the integral form of the balance equation and the relation between the definition of the derivative and the integral as a Riemann sum. For simplicity, let's take a function $\Phi(x)$ on an interval $[a,b]$ and look at the arc length $L(\Phi)$ using the well-known arc-length formula:

The length of a small piece of the curve may be given as: $$ L_{dx}(\Phi) = \sqrt{dx^2 + dy^2} = \sqrt{dx^2 + (\Phi(x+dx)-\Phi(x))^2} = dx\sqrt{1 + \left(\frac{\Phi(x+dx)-\Phi(x)}{dx}\right)^2} $$ For a nonzero $dx$, this is a crude approximation, which could benefit from a second or higher order term. We can sum these segments up to get the length: $$ L_{[a,b]}(\Phi) = dx\sum_{i=1}^n \sqrt{1 + \left(\frac{\Phi(x+dx)-\Phi(x)}{dx}\right)^2}, $$ for some appropriate $n$ and with $x=a+i*dx$. This sum approaches the true value as $dx$ approaches zero. Taking the limit transforms the finite difference into the derivative and the sum into the integral and results in the exact arc length:

$$ L_{[a,b]}(\Phi) = \int_a^b \sqrt{1 + \left(\frac{d\Phi(x)}{dx}\right)^2}dx, $$

No need for a second order term!


When you have several variables, you need to check that all the second order terms in fact converge faster than all of the first order terms. If not, you have to change your units of some of the variables until they do. The concept of natural units is that all are of order 1 in any problem, and this guarantees all second order terms can be dropped.

There are often coefficients attached to the terms that can be recursively derived, and may be heavily divergent. This can be another problem where you need to ensure the growth of the coefficient is covered as well. This is a nontrivial problem for discrete marching algorithms.

Yet another problem is that there are often areas of interest where all the first order terms cancel each other out. This is usually a problem. Second order terms can rescue some numerical methods some of the time.

  • $\begingroup$ "When you have several variables, you need to check that all the second order terms in fact converge faster than all of the first order terms. If not, you have to change your units of some of the variables until they do." That's the whole point of the question, why is that? Why must that happen in order to write the governing equation? $\endgroup$
    – TheVal
    Commented Aug 2, 2022 at 19:27
  • 1
    $\begingroup$ Basically, because 400 years of problem-solving methods are based on that idea, ie. Lagrangian dynamics, Hamiltonian dynamics, The idea is to get ODE systems that are solvable - perhaps ugly as sin - but solvable. All the examples you are shown are ones that work! That isn't to say that it works all the time - it just works all the time in the classroom. Study the development of the theory of variations. Mathematicians will do nearly anything to reduce the order of the equation set $\endgroup$
    – Phil Sweet
    Commented Aug 2, 2022 at 22:21

By definition smooth is a function that changes by a curve with a radius, $r$ which is inversely proportional to the second derivative of the function at $x$.

we approximate the r as the distance of the intersection of two perpendicular lines to two tangents to the curve at $x$ and $x+dx$ to the chord $x$, $dx$.

As the dx gets smaller the tangents intersect farther and at the limit, $r$ becomes infinite and the second derivative becomes zero.

2nd derivative limit


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