7.5: Determining Whether an Expression is an Exact Differential
- Page ID
- 151694
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Since exact differentials have these important characteristics, it is valuable to know whether a given differential expression is exact or not. That is, given a differential expression of the form
\[df=M\left(x,y\right)dx+\ N\left(x,y\right)dy, \label{eq1} \]
we would like to be able to determine whether \(df\) is exact or inexact. It turns out that there is a simple test for exactness:
The differential in the form of Equation \ref{eq1} is exact if and only if
\[\dfrac{\partial M}{ \partial y} = \dfrac{\partial N}{ \partial x}. \label{eq2} \]
That is, this condition is necessary and sufficient for the existence of a function, \(f\left(x,y\right)\), for which \(M\left(x,y\right)=f_x\left(x,y\right)\) and \(N\left(x,y\right)=f_y\left(x,y\right)\).
In §4 we demonstrate that the condition is necessary. Now we want to show that it is sufficient. That is, we want to demonstrate: If Equation \ref{eq2} hold, then there exists a \(f\left(x,y\right)\), such that \(M\left(x,y\right)=f_x\left(x,y\right)\) and \(N\left(x,y\right)=f_y\left(x,y\right)\). To do this, we show how to find a function, \(f\left(x,y\right)\), that satisfies the given differential relationship. If we integrate \(M\left(x,y\right)\) with respect to \(x\), we have \[f\left(x,y\right)=\int{M\left(x,y\right)dx+h\left(y\right)} \nonumber \]
where \(h\left(y\right)\) is a function only of \(y\); it is the arbitrary constant in the integration with respect to \(x\), which we carry out with \(y\) held constant.
To complete the proof, we must find a function \(h\left(y\right)\) such that this \(f\left(x,y\right)\) satisfies the conditions:
\[\begin{align} \label{GrindEQ_1} M\left(x,y\right) &=f_x\left(x,y\right)\Leftrightarrow \\[4pt] &=\frac{\partial }{\partial x}\left[\int{M\left(x,y\right)dx+h\left(y\right)}\right] \end{align} \]
\[\begin{align}\label{GrindEQ_2} N\left(x,y\right) &=f_y\left(x,y\right) \Leftrightarrow \\[4pt] &=\frac{\partial }{\partial y}\left[\int{M\left(x,y\right)dx+h\left(y\right)}\right] \end{align} \]
The validity of condition in Equation \ref{GrindEQ_1} follows immediately from the facts that the order of differentiation and integration can be interchanged for a continuous function and that \(h\left(y\right)\) is a function only of \(y\), so that \({\partial h}/{\partial x=0}\).
To find \(h\left(y\right)\) such that condition in Equation \ref{GrindEQ_2} is satisfied, we observe that
\[\frac{\partial }{\partial y}\left[\int{M\left(x,y\right)dx+h\left(y\right)}\right]=\int{\left(\frac{\partial M\left(x,y\right)}{\partial y}\right)}dx+\frac{dh\left(y\right)}{dy} \nonumber \]
But since \[\frac{\partial M\left(x,y\right)}{\partial y} = \frac{\partial N\left(x,y\right)}{\partial x} \nonumber \]
this becomes
\[\frac{\partial }{\partial y}\left[\int{M\left(x,y\right)dx+h\left(y\right)}\right] = \int{\left(\frac{\partial N\left(x,y\right)}{\partial x}\right)}dx+\frac{dh\left(y\right)}dy =N\left(x,y\right)+\frac{dh\left(y\right)}{dy} \nonumber \]
Hence, condition in Equation \ref{GrindEQ_2} is satisfied if and only if \({dh\left(y\right)}/{dy}=0\), so that \(h\left(y\right)\) is simply an arbitrary constant.