Exact differential

This article is about the concept from elementary differential calculus. For the generalized advanced mathematical concept from differential topology and differential geometry, see closed and exact differential forms.

In multivariate calculus, a differential or differential form is said to be exact or perfect (so called an exact differential), as contrasted with an inexact differential, if it is equal to the general differential dQ for some differentiable function Q.

An exact differential is sometimes also called a total differential, or a full differential, or, in the study of differential geometry, it is termed an exact form.

The integral of an exact differential over any integral path is integral path-independent, and this fact is used to identity state functions in thermodynamics.

Overview

Definition

Even if we work in three dimensions here, the definitions of exact differentials for other dimensions are similar to the three dimensional definition. In three dimensions, a form of the type

$A(x,y,z)\,dx+B(x,y,z)\,dy+C(x,y,z)\,dz$

is called a differential form. This form is called exact on a domain $D\subset \mathbb {R} ^{3}$ in space if there exists some differentiable scalar function $Q=Q(x,y,z)$ defined on $D$ such that

$dQ\equiv \left({\frac {\partial Q}{\partial x}}\right)_{y,z}\,dx+\left({\frac {\partial Q}{\partial y}}\right)_{x,z}\,dy+\left({\frac {\partial Q}{\partial z}}\right)_{x,y}\,dz,$   $dQ=A\,dx+B\,dy+C\,dz$

throughout $D$, where $x,y,z$ are orthogonal coordinates (e.g., Cartesian, cylindrical, or spherical coordinates). In other words, in some domain of three dimensional space, a differential form is an exact differential if it is equal to the general differential of a differentiable function.

Note: In this mathematical expression, the subscripts outside the parenthesis indicate which variables are being held constant during differentiation. Due to the definition of the partial derivative, these subscripts are not required, but they are included as a reminder.

Integral path independence

The exact differential for a differentiable function $Q$ is equal to $dQ=\nabla Q\cdot d\mathbf {r}$, that is the scalar product between the conservative vector field $(A,B,C)=\nabla Q$ (where the right hand side is the gradient of $Q$) for the corresponding potential $Q$ and the general differential displacement vector $(dx,dy,dz)=d\mathbf {r}$.

The gradient theorem states

$\int _{i}^{f}dQ=\int _{i}^{f}\nabla Q(\mathbf {r} )\cdot d\mathbf {r} =Q\left(f\right)-Q\left(i\right)$

that does not depend on which integral path between the given path endpoints $i$ and $f$ is chosen. So it is concluded that the integral of an exact differential is independent of the choice of an integral path between given path endpoints (path independence).

For three dimensional spaces, this integral path independence can also be proved by using the vector calculus identity $\nabla \times (\nabla Q)=\mathbf {0}$ and the Stokes' theorem.

$\oint _{\partial \Sigma }\nabla Q\cdot d\mathbf {r} =\iint _{\Sigma }(\nabla \times \nabla Q)\cdot d\mathbf {a} =0$

for a closed loop $\partial \Sigma$ with the smooth oriented surface $\Sigma$ in it.

Thermodynamic state function

In thermodynamics, when $dQ$ is exact, the function $Q$ is a state function (a mathematical function depending only on the current equilibrium state, not depending on the system path taken to reach the equilibrium state) of the system. The thermodynamic functions $U$ (internal energy), $S$ (entropy), $H$ (enthalpy), $A$ (Helmholtz free energy), and $G$ (Gibbs free energy) are state functions. Generally, neither work nor heat is a state function.

One dimension

In one dimension, a differential form

$A(x)\,dx$

is exact if and only if $A$ has an antiderivative (but not necessarily one in terms of elementary functions). If $A$ has an antiderivative and let $Q$ be an antiderivative of $A$ so ${\frac {dQ}{dx}}=A$, then $A(x)\,dx$ obviously satisfies the condition for exactness. If $A$ does not have an antiderivative, then we cannot write $dQ={\frac {dQ}{dx}}dx$ with $A={\frac {dQ}{dx}}$ for a differentiable function $Q$ so $A(x)\,dx$ is inexact.

Two and three dimensions

By symmetry of second derivatives, for any "well-behaved" (non-pathological) function $Q$, we have

${\frac {\partial ^{2}Q}{\partial x\,\partial y}}={\frac {\partial ^{2}Q}{\partial y\,\partial x}}.$

Hence, in a simply-connected region R of the xy-plane, a differential form

$A(x,y)\,dx+B(x,y)\,dy$

is an exact differential if and only if the equation

$\left({\frac {\partial A}{\partial y}}\right)_{x}=\left({\frac {\partial B}{\partial x}}\right)_{y}$

holds. If it is an exact differential so $A={\frac {\partial Q}{\partial x}}$ and $B={\frac {\partial Q}{\partial y}}$, then $Q$ is a differentiable (smoothly continuous) function along $x$ and $y$, so $\left({\frac {\partial A}{\partial y}}\right)_{x}={\frac {\partial ^{2}Q}{\partial y\partial x}}={\frac {\partial ^{2}Q}{\partial x\partial y}}=\left({\frac {\partial B}{\partial x}}\right)_{y}$. If $\left({\frac {\partial A}{\partial y}}\right)_{x}=\left({\frac {\partial B}{\partial x}}\right)_{y}$ holds, then $A$ and $B$ are differentiable (again, smoothly continuous) functions along $y$ and $x$ respectively, and $\left({\frac {\partial A}{\partial y}}\right)_{x}={\frac {\partial ^{2}Q}{\partial y\partial x}}={\frac {\partial ^{2}Q}{\partial x\partial y}}=\left({\frac {\partial B}{\partial x}}\right)_{y}$ is only the case.

For three dimensions, in a simply-connected region R of the xyz-coordinate system, by a similar reason, a differential

$dQ=A(x,y,z)\,dx+B(x,y,z)\,dy+C(x,y,z)\,dz$

is an exact differential if and only if between the functions A, B and C there exist the relations

$\left({\frac {\partial A}{\partial y}}\right)_{x,z}\!\!\!=\left({\frac {\partial B}{\partial x}}\right)_{y,z}$; $\left({\frac {\partial A}{\partial z}}\right)_{x,y}\!\!\!=\left({\frac {\partial C}{\partial x}}\right)_{y,z}$; $\left({\frac {\partial B}{\partial z}}\right)_{x,y}\!\!\!=\left({\frac {\partial C}{\partial y}}\right)_{x,z}.$

These conditions are equivalent to the following sentence: If G is the graph of this vector valued function then for all tangent vectors X,Y of the surface G then s(X, Y) = 0 with s the symplectic form.

These conditions, which are easy to generalize, arise from the independence of the order of differentiations in the calculation of the second derivatives. So, in order for a differential dQ, that is a function of four variables, to be an exact differential, there are six conditions (the combination $C(4,2)=6$) to satisfy.

Partial differential relations

If a differentiable function $z(x,y)$ is one-to-one (injective) for each independent variable, e.g., $z(x,y)$ is one-to-one for $x$ at a fixed $y$ while it is not necessarily one-to-one for $(x,y)$, then the following total differentials exist because each independent variable is a differentiable function for the other variables, e.g., $x(y,z)$.

$dx={\left({\frac {\partial x}{\partial y}}\right)}_{z}\,dy+{\left({\frac {\partial x}{\partial z}}\right)}_{y}\,dz$

$dz={\left({\frac {\partial z}{\partial x}}\right)}_{y}\,dx+{\left({\frac {\partial z}{\partial y}}\right)}_{x}\,dy.$

Substituting the first equation into the second and rearranging, we obtain

$dz={\left({\frac {\partial z}{\partial x}}\right)}_{y}\left[{\left({\frac {\partial x}{\partial y}}\right)}_{z}dy+{\left({\frac {\partial x}{\partial z}}\right)}_{y}dz\right]+{\left({\frac {\partial z}{\partial y}}\right)}_{x}dy,$

$dz=\left[{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial y}}\right)}_{z}+{\left({\frac {\partial z}{\partial y}}\right)}_{x}\right]dy+{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial z}}\right)}_{y}dz,$

$\left[1-{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial z}}\right)}_{y}\right]dz=\left[{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial y}}\right)}_{z}+{\left({\frac {\partial z}{\partial y}}\right)}_{x}\right]dy.$

Since $y$ and $z$ are independent variables, $dy$ and $dz$ may be chosen without restriction. For this last equation to generally hold, the bracketed terms must be equal to zero. The left bracket equal to zero leads to the reciprocity relation while the right bracket equal to zero goes to the cyclic relation as shown below.

Reciprocity relation

Setting the first term in brackets equal to zero yields

${\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial z}}\right)}_{y}=1.$

A slight rearrangement gives a reciprocity relation,

${\left({\frac {\partial z}{\partial x}}\right)}_{y}={\frac {1}{{\left({\frac {\partial x}{\partial z}}\right)}_{y}}}.$

There are two more permutations of the foregoing derivation that give a total of three reciprocity relations between $x$, $y$ and $z$.

Cyclic relation

The cyclic relation is also known as the cyclic rule or the Triple product rule. Setting the second term in brackets equal to zero yields

${\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial y}}\right)}_{z}=-{\left({\frac {\partial z}{\partial y}}\right)}_{x}.$

Using a reciprocity relation for ${\tfrac {\partial z}{\partial y}}$ on this equation and reordering gives a cyclic relation (the triple product rule),

${\left({\frac {\partial x}{\partial y}}\right)}_{z}{\left({\frac {\partial y}{\partial z}}\right)}_{x}{\left({\frac {\partial z}{\partial x}}\right)}_{y}=-1.$

If, instead, reciprocity relations for ${\tfrac {\partial x}{\partial y}}$ and ${\tfrac {\partial y}{\partial z}}$ are used with subsequent rearrangement, a standard form for implicit differentiation is obtained:

${\left({\frac {\partial y}{\partial x}}\right)}_{z}=-{\frac {{\left({\frac {\partial z}{\partial x}}\right)}_{y}}{{\left({\frac {\partial z}{\partial y}}\right)}_{x}}}.$

Some useful equations derived from exact differentials in two dimensions

(See also Bridgman's thermodynamic equations for the use of exact differentials in the theory of thermodynamic equations)

Suppose we have five state functions $z,x,y,u$, and $v$. Suppose that the state space is two-dimensional and any of the five quantities are differentiable. Then by the chain rule

$dz=\left({\frac {\partial z}{\partial x}}\right)_{y}dx+\left({\frac {\partial z}{\partial y}}\right)_{x}dy=\left({\frac {\partial z}{\partial u}}\right)_{v}du+\left({\frac {\partial z}{\partial v}}\right)_{u}dv$ 1

but also by the chain rule:

$dx=\left({\frac {\partial x}{\partial u}}\right)_{v}du+\left({\frac {\partial x}{\partial v}}\right)_{u}dv$ 2

and

$dy=\left({\frac {\partial y}{\partial u}}\right)_{v}du+\left({\frac {\partial y}{\partial v}}\right)_{u}dv$ 3

so that (by substituting (2) and (3) into (1)):

${\begin{aligned}dz=&\left[\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial u}}\right)_{v}+\left({\frac {\partial z}{\partial y}}\right)_{x}\left({\frac {\partial y}{\partial u}}\right)_{v}\right]du\\+&\left[\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial v}}\right)_{u}+\left({\frac {\partial z}{\partial y}}\right)_{x}\left({\frac {\partial y}{\partial v}}\right)_{u}\right]dv\end{aligned}}$ 4

which implies that (by comparing (4) with (1)):

$\left({\frac {\partial z}{\partial u}}\right)_{v}=\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial u}}\right)_{v}+\left({\frac {\partial z}{\partial y}}\right)_{x}\left({\frac {\partial y}{\partial u}}\right)_{v}$ 5

Letting $v=y$ in (5) gives:

$\left({\frac {\partial z}{\partial u}}\right)_{y}=\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial u}}\right)_{y}$ 6

Letting $u=y$ in (5) gives:

$\left({\frac {\partial z}{\partial y}}\right)_{v}=\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial y}}\right)_{v}+\left({\frac {\partial z}{\partial y}}\right)_{x}$ 7

Letting $u=y$ and $v=z$ in (5) gives:

$\left({\frac {\partial z}{\partial y}}\right)_{x}=-\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial y}}\right)_{z}$ 8

using ($\partial a/\partial b)_{c}=1/(\partial b/\partial a)_{c}$ gives the triple product rule:

$\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}=-1$ 9

See also

• Closed and exact differential forms for a higher-level treatment
• Differential (mathematics)
• Inexact differential
• Integrating factor for solving non-exact differential equations by making them exact
• Exact differential equation

References

• Perrot, P. (1998). A to Z of Thermodynamics. New York: Oxford University Press.
• Dennis G. Zill (14 May 2008). A First Course in Differential Equations. Cengage Learning. ISBN 978-0-495-10824-5.

External links

• Inexact Differential – from Wolfram MathWorld
• Exact and Inexact Differentials – University of Arizona
• Exact and Inexact Differentials – University of Texas
• Exact Differential – from Wolfram MathWorld