Please Make A Note: Integrating Exact Differentials

Often times I am faced with the integration of a system of differential equations of the form

$\bg_white \120dpi \bg_white \120dpi \begin{cases} \dfrac{\partial p}{\partial x} = S(x,y) \\[12pt] \dfrac{\partial p}{\partial y} = T(x,y) \end{cases}$

This system describes the behaviour of the primitive function "p". In fact, these equations stem from the total differential of p

$\bg_white \120dpi {\rm d}p = \frac{\partial p}{\partial x} {\rm d}x +\frac{\partial p}{\partial y} {\rm d}y$

Two cases arise in this situation. If "dp" is an exact differential, then the system is integrable independent of the path of integration. A differential equation "dp" is said to be an exact differential if the primitive function "p" exists. A necessary and sufficient condition for "dp" to be an exact differential if "p" is continuous and twice differentiable. In other words, when the mixed derivatives of "p" exist and are equal. For example, given

$\bg_white \120dpi {\rm d}p = \frac{\partial p}{\partial x} {\rm d}x + \frac{\partial p}{\partial y} {\rm d}y + \frac{\partial p}{\partial z} {\rm d}z$

then, "dp" is an exact differential if, and only if,

$\bg_white \120dpi \begin{cases} \dfrac{\partial^2 p}{\partial x \partial y} =\dfrac{\partial^2 p}{\partial y \partial x} \\[12pt] \dfrac{\partial^2 p}{\partial x \partial z} =\dfrac{\partial^2 p}{\partial z \partial x} \\[12pt] \dfrac{\partial^2 p}{\partial y \partial z} =\dfrac{\partial^2 p}{\partial z \partial y} \end{cases}$

If the total differential is not exact, then the integration can only be carried out along a specified path of the domain.

Integration of an Exact Differential

The typical approach to integrating the system of differential equations is to first integrate in one coordinate direction, add a function of the remaining coordinates, substitute into the remaining equations and so on.

However, there is another technique that only requires the following three steps:

Integrate each equation with respect to its coordinate without adding any integration constants
Add the results
Subtract the common part

Here's an example. Consider

$\bg_white \120dpi \bg_white \120dpi \bg_white \120dpi \begin{cases} \dfrac{\partial p}{\partial x} = 2xy + y^3 \cos x + \sin(x) \\[12pt] \dfrac{\partial p}{\partial y} = x^2 + 3y^2 \sin x \end{cases}$

Integrating in each direction, we get

$\bg_white \120dpi \bg_white \120dpi \bg_white \120dpi \begin{cases} p_1=\displaystyle{\int \dfrac{\partial p}{\partial x}{\rm d}x} = x^2y + y^3 \sin x - \cos x + c_1 \\[12pt] p_2=\displaystyle{\int \dfrac{\partial p}{\partial y}{\rm d}y} = x^2y + y^3 \sin x + c_2\end{cases}$

Then, the primitive function is

$\bg_white \120dpi p=p_1 + p_2 - p_{\rm commom}$

$\bg_white \120dpi \bg_white \120dpi p=x^2y + y^3 \sin x - \cos x + c$

While this methods works, I have not seen any proof of why it does. Therefore, in this post, I will provide an independent proof. To simplify the process, I will only present this in two dimensions. I will also provide two methods for proving this. The words integral and primitive will be used interchangeably. Furthermore, I will refer to the individual differentials of a function as its constitutive differentials. For example,

$\bg_white \120dpi \inline \dfrac{\partial p}{\partial x}; \quad \dfrac{\partial p}{\partial y}$

are called the constitutive differentials of "p".

Method 1:

A differential system of equations can be written in general as

$\bg_white \120dpi \bg_white \120dpi \begin{cases} \dfrac{\partial p}{\partial x} = f(x) + g(y) + h(x,y) \\[12pt] \dfrac{\partial p}{\partial y} = s(x) + l(y) + q(x,y)\end{cases}$

and let "p1" and "p2" denote the integrals of the above equations without the addition of an integration constant. Thus

$\bg_white \120dpi \begin{cases} p_1=\displaystyle\int \dfrac{\partial p}{\partial x} {\rm d}x = \int f(x){\rm d}x + \int g(y){\rm d}x + \int h(x,y) {\rm d}x \\[12pt] p_2=\displaystyle\int \dfrac{\partial p}{\partial y} {\rm d}y = \int s(x){\rm d}y + \int l(y){\rm d}y + \int q(x,y) {\rm d}y\end{cases}$

Now let us first integrate in the "x" direction

$\bg_white \120dpi \bg_white \120dpi p(x,y)=\int\dfrac{\partial p}{\partial x}{\rm d}x = \int f(x){\rm d}x + \int g(y) {\rm d}x + \int h(x,y) {\rm d}x + N(y)$

$\bg_white \120dpi \bg_white \120dpi \bg_white \120dpi p(x,y) = p_1 + N(y)$

Note the addition of the integration constant "N(y)". Substituting into the second differential, we get

$\bg_white \120dpi \bg_white \120dpi \dfrac{\partial p}{\partial y} = s(x) + l(y) + q(x,y) =\dfrac{\partial }{\partial y}\left[ \int g(y) {\rm d}x \right ]+ \dfrac{\partial }{\partial y}\left[\int h(x,y) {\rm d}x \right ] +\dfrac{{\rm d}N}{{\rm d}y}$

$\bg_white \120dpi \bg_white \120dpi \dfrac{{\rm d}N}{{\rm d}y} = s(x) + l(y) + q(x,y)-\dfrac{\partial }{\partial y}\left[ \int g(y) {\rm d}x \right ] - \dfrac{\partial }{\partial y}\left[\int h(x,y) {\rm d}x \right ]$

(Note that it is assumed here that dN/dy is a function of y ONLY, so that the right hand side of the above equation will evaluate of a function of y and only y). Integrating with respect to "y", we recover

$\bg_white \120dpi \bg_white \120dpi N(y) = \int s(x){\rm d}y + \int l(y){\rm d}y + \int q(x,y){\rm d}y - \int g(y) {\rm d}x - \int h(x,y){\rm d}x + c$

$\bg_white \120dpi \bg_white \120dpi \bg_white \120dpi N(y) = p_2 - \int g(y) {\rm d}x - \int h(x,y){\rm d}x + c$

substituting "N(y)" into the equation for "p(x,y)", we have

$\bg_white \120dpi p(x,y) = p_1 + p_2 \underbrace{-\int g(y) {\rm d}x - \int h(x,y){\rm d}x}_{-M(x,y)} + c$

So far, we have shown that the primitive function is the sum of the integrals in each direction (i.e. p1 and p2) minus some term denoted by M(x,y). It remains to show that M(x,y) represents a part that is common to both "p1" and "p2". To arrive at this, we invoke the fact that the this system's total differential is exat, i.e.

$\bg_white \120dpi \dfrac{\partial^2 p}{\partial x \partial y} =\dfrac{\partial^2 p}{\partial y \partial x}$

$\bg_white \120dpi \dfrac{\partial g}{\partial y} + \dfrac{\partial h}{\partial y} = \dfrac{\partial s}{\partial x} + \dfrac{\partial q}{\partial x}$

collecting terms, we get

$\bg_white \120dpi \dfrac{\partial (g + h)}{\partial y} = \dfrac{\partial (s + q)}{\partial x}$

This is true if

$\bg_white \120dpi \begin{cases} g(y)+h(x,y) = \dfrac{\partial T(x,y)}{\partial x}\\[12pt] s(x)+q(x,y) = \dfrac{\partial T(x,y)}{\partial y} \end{cases}$

with T(x,y) being continuous and twice differentiable. Given this, we rewrite M(x,y) as

$\bg_white \120dpi M(x,y) = \int \left[g(y)+h(x,y) \right ]{\rm d}x=\int \dfrac{\partial T}{\partial x}{\rm d}x=T(x,y)$

Finally, we now rewrite "p1" and "p2" using T(x,y)

$\bg_white \120dpi \begin{cases} \displaystyle p_1= \int f(x){\rm d}x + T(x,y) \\[12pt] \displaystyle p_2= \int l(y){\rm d}y + T(x,y) \end{cases}$

It is therefore clear that M(x,y) is actually a common term for "p1" and "p2" and the primitive is given by

$\bg_white \120dpi p(x,y) = p_1 + p_2 - T(x,y) + c$

Q.E.D.

Note: One can now verify that the integration constant N(y) is a function of y, and only y. Reconsider the equation for dN/dy

writing in terms of T(x,y), we have

$\bg_white \120dpi \dfrac{{\rm d}N}{{\rm d}y} = l(y) + \dfrac{\partial T}{\partial y} -\dfrac{\partial }{\partial y}\underbrace{\left\{ \int \left[ g(y) + h(x,y)\right ] {\rm d}x \right\}}_{T(x,y)}$

$\bg_white \120dpi \dfrac{{\rm d}N}{{\rm d}y} = l(y) + \dfrac{\partial T}{\partial y} -\dfrac{\partial T}{\partial y}=l(y)$

Method 2:

Consider a real continuous and twice differentiable function "p(x,y)". In general, this function can be written as

$\bg_white \120dpi \bg_white \120dpi p(x,y)=f(x) + g(y) + q(x,y) + c$

where "c" is a constant and the component functions are also real, continuous, and twice differentiable. Now form the total differential of "p"

$\bg_white \120dpi {\rm d}p = \frac{\partial p}{\partial x} {\rm d}x +\frac{\partial p}{\partial y} {\rm d}y$

where

$\bg_white \120dpi \begin{cases} \dfrac{\partial p}{\partial x} = \dfrac{{\rm d}f}{{\rm d}x} + \dfrac{\partial q}{\partial x} \\[12pt] \dfrac{\partial p}{\partial y} = \dfrac{{\rm d}g}{{\rm d}y} + \dfrac{\partial q}{\partial y} \end{cases}$

The key component of this derivation are the derivatives of "q" in both differentials. We now define the integrals of the above system without adding an integration constant. We get

$\bg_white \120dpi \begin{cases} p_1=\displaystyle\int \dfrac{\partial p}{\partial x} {\rm d}x = f(x) + q(x,y) + c_1\\[12pt] p_2=\displaystyle\int \dfrac{\partial p}{\partial y} {\rm d}y = g(y){\rm d}y + q(x,y) + c_2 \end{cases}$

but we know that

$\bg_white \120dpi \bg_white \120dpi p(x,y)=f(x) + g(y) + q(x,y) + c$

When written in terms of "p1" and "p2", one recovers

$\bg_white \120dpi p(x,y)= p_1 + p_2 - q(x,y)$

where q(x,y) is the common part.

Q.E.D.

This shows that the primitive of an exact differential is equal to the sum of primitives of its constitutive functions minus the common part. Integration of the constitutive differentials is carried out without adding integration constants.

Summary:

While I have shown that the method of separate integrations works, one must remember that this technique is applicable ONLY to exact differentials. Therefore, the mixed derivatives test must be first applied before carrying out this type of integration. In my next post, I will discuss how to handle inexact differentials.

Cite as:
Saad, T. "Integrating Exact Differentials". Weblog entry from Please Make A Note. https://pleasemakeanote.blogspot.com/2010/07/integrating-totalexact-differentials.html?m=0

Please Make A Note

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Saturday, July 3, 2010

Integrating Exact Differentials

Integration of an Exact Differential

Method 1:

A differential system of equations can be written in general as

Method 2:

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