Lila at 22:55, 25 October 2021

2021-10-25T22:55:40Z

Lila at 22:54, 25 October 2021

2021-10-25T22:54:23Z

Lila: Created page with "==Definition== Given a simply connected and open subset ''D'' of '''R'''² and two functions ''I'' and ''J'' which are continuous on ''D'', an implicit first-order o..."

2021-10-25T22:53:07Z

Created page with "==Definition== Given a simply connected and open subset ''D'' of '''R'''2 and two functions ''I'' and ''J'' which are continuous on ''D'', an implicit first-order o..."

New page

==Definition==
Given a simply connected and open subset ''D'' of '''R'''2 and two functions ''I'' and ''J'' which are continuous on ''D'', an implicit first-order ordinary differential equation of the form

: <math>I(x, y)\, dx + J(x, y)\, dy = 0,</math>

is called an '''exact differential equation''' if there exists a continuously differentiable function ''F'', called the '''potential function''', so that
:<math>\frac{\partial F}{\partial x} = I</math>
and
:<math>\frac{\partial F}{\partial y} = J.</math>

An exact equation may also be presented in the following form:
:<math>I(x, y) + J(x, y) \, y'(x) = 0</math>
where the same constraints on ''I'' and ''J'' apply for the differential equation to be exact.

The nomenclature of "exact differential equation" refers to the exact differential of a function. For a function <math>F(x_0, x_1,...,x_{n-1},x_n)</math>, the exact or total derivative with respect to <math>x_0</math> is given by
:<math>\frac{dF}{dx_0}=\frac{\partial F}{\partial x_0}+\sum_{i=1}^{n}\frac{\partial F}{\partial x_i}\frac{dx_i}{dx_0}.</math>

===Example===
The function <math>F:\mathbb{R}^{2}\to\mathbb{R}</math> given by

:<math>F(x,y) = \frac{1}{2}(x^2 + y^2)+c</math>

is a potential function for the differential equation

:<math>x\,dx + y\,dy = 0.\,</math>

==Existence of potential functions==
In physical applications the functions ''I'' and ''J'' are usually not only continuous but even continuously differentiable. Schwarz's Theorem then provides us with a necessary criterion for the existence of a potential function. For differential equations defined on simply connected sets the criterion is even sufficient and we get the following theorem:

Given a differential equation of the form (for example, when ''F'' has zero slope in the x and y direction at ''F''(''x'',''y'')):
: <math>I(x, y)\, dx + J(x, y)\, dy = 0,</math>
with ''I'' and ''J'' continuously differentiable on a simply connected and open subset ''D'' of '''R'''2 then a potential function ''F'' exists if and only if
:<math>\frac{\partial I}{\partial y}(x, y) = \frac{\partial J}{\partial x}(x, y).</math>

==Solutions to exact differential equations==
Given an exact differential equation defined on some simply connected and open subset ''D'' of '''R'''2 with potential function ''F'', a differentiable function ''f'' with (''x'', ''f''(''x'')) in ''D'' is a solution if and only if there exists real number ''c'' so that
:<math>F(x, f(x)) = c.</math>

For an initial value problem
:<math>y(x_0) = y_0</math>
we can locally find a potential function by
:<math>\begin{align}
F(x,y) &= \int_{x_0}^x I(t,y_0) dt + \int_{y_0}^y J(x,t) dt \\
&= \int_{x_0}^x I(t,y_0) dt + \int_{y_0}^y \left[ J(x_0,t) + \int_{x_0}^{x} \frac{\partial I}{\partial y}(u, t)\, du \right]dt.
\end{align}</math>

Solving
:<math>F(x,y) = c</math>
for ''y'', where ''c'' is a real number, we can then construct all solutions.

== Second order exact differential equations ==
The concept of exact differential equations can be extended to second order equations.<ref>{{Cite book|title=Ordinary Differential Equations: An Elementary Textbook for Students of Mathematics, Engineering and the Sciences|url=https://archive.org/details/ordinarydifferen00tene_850|url-access=limited|last1=Tenenbaum|first1=Morris|last2=Pollard|first2=Harry|publisher=Dover|year=1963|isbn=0-486-64940-7|location=New York|pages=[https://archive.org/details/ordinarydifferen00tene_850/page/n131 248]|chapter=Solution of the Linear Differential Equation with Nonconstant Coefficients. Reduction of Order Method.}}</ref> Consider starting with the first-order exact equation:

:<math>I\left(x,y\right)+J\left(x,y\right){dy \over dx}=0</math>

Since both functions {{nowrap|<math>I\left(x,y\right)</math>,}} <math>J\left(x,y\right)</math> are functions of two variables, implicitly differentiating the multivariate function yields

:<math>{dI \over dx} +\left({ dJ\over dx}\right){dy \over dx}+{d^2y \over dx^2}\left(J\left(x,y\right)\right)=0</math>

Expanding the total derivatives gives that

:<math>{dI \over dx}={\partial I\over\partial x}+{\partial I\over\partial y}{dy \over dx}</math>

and that

:<math>{dJ \over dx}={\partial J\over\partial x}+{\partial J\over\partial y}{dy \over dx}</math>

Combining the <math display="inline">{dy \over dx}</math> terms gives

:<math>{\partial I\over\partial x}+{dy \over dx}\left({\partial I\over\partial y}+{\partial J\over\partial x}+{\partial J\over\partial y}{dy \over dx}\right)+{d^2y \over dx^2}\left(J\left(x,y\right)\right)=0</math>

If the equation is exact, then {{nowrap|<math display="inline">{\partial J\over\partial x}={\partial I\over\partial y}</math>.}} Additionally, the total derivative of <math>J\left(x,y\right)</math> is equal to its implicit ordinary derivative <math display="inline">{dJ \over dx}</math>. This leads to the rewritten equation

:<math>{\partial I\over\partial x}+{dy \over dx}\left({\partial J\over\partial x}+{dJ \over dx}\right)+{d^2y \over dx^2}\left(J\left(x,y\right)\right)=0</math>

Now, let there be some second-order differential equation

:<math>f\left(x,y\right)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^2y \over dx^2}\left(J\left(x,y\right)\right)=0</math>

If <math>{\partial J\over\partial x}={\partial I\over\partial y}</math> for exact differential equations, then

:<math>\int \left({\partial I\over\partial y}\right)dy=\int \left({\partial J\over\partial x}\right)dy</math>

and

:<math>\int \left({\partial I\over\partial y}\right)dy=\int \left({\partial J\over\partial x}\right)dy=I\left(x,y\right)-h\left(x\right)</math>

where <math>h\left(x\right)</math> is some arbitrary function only of <math>x</math> that was differentiated away to zero upon taking the partial derivative of <math>I\left(x,y\right)</math> with respect to <math>y</math>. Although the sign on <math>h\left(x\right)</math> could be positive, it is more intuitive to think of the integral's result as <math>I\left(x,y\right)</math> that is missing some original extra function <math>h\left(x\right)</math> that was partially differentiated to zero.

Next, if

:<math>{dI\over dx}={\partial I\over\partial x}+{\partial I\over\partial y}{dy \over dx}</math>

then the term <math>{\partial I\over\partial x}</math> should be a function only of <math>x</math> and <math>y</math>, since partial differentiation with respect to <math>x</math> will hold <math>y</math> constant and not produce any derivatives of <math>y</math>. In the second order equation

:<math>f\left(x,y\right)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^2y \over dx^2}\left(J\left(x,y\right)\right)=0</math>

only the term <math>f\left(x,y\right)</math> is a term purely of <math>x</math> and <math>y</math>. Let <math>{\partial I\over\partial x}=f\left(x,y\right)</math>. If <math>{\partial I\over\partial x}=f\left(x,y\right)</math>, then

:<math>f\left(x,y\right)={ dI\over dx}-{\partial I\over\partial y}{dy \over dx}</math>

Since the total derivative of <math>I\left(x,y\right)</math> with respect to <math>x</math> is equivalent to the implicit ordinary derivative <math>{dI \over dx}</math> , then

:<math>f\left(x,y\right)+{\partial I\over\partial y}{dy \over dx}={dI \over dx}={d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)+{dh\left(x\right) \over dx}</math>

So,

:<math>{dh\left(x\right) \over dx}=f\left(x,y\right)+{\partial I\over\partial y}{dy \over dx}-{d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)</math>

and

:<math>h\left(x\right)=\int\left(f\left(x,y\right)+{\partial I\over\partial y}{dy \over dx}-{d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)\right)dx</math>

Thus, the second order differential equation

:<math>f\left(x,y\right)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^2y \over dx^2}\left(J\left(x,y\right)\right)=0</math>

is exact only if <math>g\left(x,y,{dy \over dx}\right)={ dJ\over dx}+{\partial J\over\partial x}={dJ \over dx}+{\partial J\over\partial x}</math> and only if the below expression

:<math>\int\left(f\left(x,y\right)+{\partial I\over\partial y}{dy \over dx}-{d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)\right)dx=\int \left(f\left(x,y\right)-{\partial \left(I\left(x,y\right)-h\left(x\right)\right)\over\partial x}\right)dx</math>

is a function solely of <math>x</math>. Once <math>h\left(x\right)</math> is calculated with its arbitrary constant, it is added to <math>I\left(x,y\right)-h\left(x\right)</math> to make <math>I\left(x,y\right)</math>. If the equation is exact, then we can reduce to the first order exact form which is solvable by the usual method for first-order exact equations.

:<math>I\left(x,y\right)+J\left(x,y\right){dy \over dx}=0</math>

Now, however, in the final implicit solution there will be a <math>C_1x</math> term from integration of <math>h\left(x\right)</math> with respect to <math>x</math> twice as well as a <math>C_2</math>, two arbitrary constants as expected from a second-order equation.

=== Example ===
Given the differential equation

:<math>\left(1-x^2\right)y''-4xy'-2y=0</math>

one can always easily check for exactness by examining the <math>y''</math> term. In this case, both the partial and total derivative of <math>1-x^2</math> with respect to <math>x</math> are <math>-2x</math>, so their sum is <math>-4x</math>, which is exactly the term in front of <math>y'</math>. With one of the conditions for exactness met, one can calculate that

:<math>\int \left(-2x\right)dy=I\left(x,y\right)-h\left(x\right)=-2xy</math>

Letting <math>f\left(x,y\right)=-2y</math>, then

:<math>\int \left(-2y-2xy'-{d \over dx}\left(-2xy \right)\right)dx=\int \left(-2y-2xy'+2xy'+2y\right)dx=\int \left(0\right)dx=h\left(x\right)</math>

So, <math>h\left(x\right)</math> is indeed a function only of <math>x</math> and the second order differential equation is exact. Therefore, <math>h\left(x\right)=C_1</math> and <math>I\left(x,y\right)=-2xy+C_1</math>. Reduction to a first-order exact equation yields

:<math>-2xy+C_1+\left(1-x^2\right)y'=0</math>

Integrating <math>I\left(x,y\right)</math> with respect to <math>x</math> yields

:<math>-x^2y+C_1x+i\left(y\right)=0</math>

where <math>i\left(y\right)</math> is some arbitrary function of <math>y</math>. Differentiating with respect to <math>y</math> gives an equation correlating the derivative and the <math>y'</math> term.

:<math>-x^2+i'\left(y\right)=1-x^2</math>

So, <math>i\left(y\right)=y+C_2</math> and the full implicit solution becomes

:<math>C_1x+C_2+y-x^2y=0</math>

Solving explicitly for <math>y</math> yields

:<math>y= \frac{C_1x+C_2}{1-x^2}</math>

== Higher order exact differential equations ==
The concepts of exact differential equations can be extended to any order. Starting with the exact second order equation

:<math>{d^2y \over dx^2}\left(J\left(x,y\right)\right)+{dy \over dx}\left({dJ \over dx}+{\partial J\over\partial x}\right)+f\left(x,y\right)=0</math>

it was previously shown that equation is defined such that

:<math>f\left(x,y\right)={dh\left(x\right) \over dx}+{d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)-{\partial J\over\partial x}{dy \over dx}</math>

Implicit differentiation of the exact second-order equation <math>n</math> times will yield an <math>\left(n+2\right)</math>th order differential equation with new conditions for exactness that can be readily deduced from the form of the equation produced. For example, differentiating the above second-order differential equation once to yield a third-order exact equation gives the following form

:<math>{d^3y \over dx^3}\left(J\left(x,y\right)\right)+{d^2y \over dx^2}{dJ \over dx}+{d^2y \over dx^2}\left({dJ \over dx}+{\partial J\over\partial x}\right)+{dy \over dx}\left({d^2J \over dx^2}+{d \over dx}\left({\partial J\over\partial x}\right)\right)+{df\left(x,y\right) \over dx}=0</math>

where

:<math>{df\left(x,y\right) \over dx}={d^2h\left(x\right) \over dx^2}+{d^2 \over dx^2}\left(I\left(x,y\right)-h\left(x\right)\right)-{d^2y \over dx^2}{\partial J\over\partial x}-{dy \over dx}{d \over dx}\left({\partial J\over\partial x}\right)=F\left(x,y,{dy \over dx}\right)</math>
and where <math>F\left(x,y,{dy \over dx}\right)</math>
is a function only of <math>x,y</math> and <math>{dy \over dx}</math>. Combining all <math>{dy \over dx}</math> and <math>{d^2y \over dx^2}</math> terms not coming from <math>F\left(x,y,{dy \over dx}\right)</math> gives

:<math>{d^3y \over dx^3}\left(J\left(x,y\right)\right)+{d^2y \over dx^2}\left(2{dJ \over dx}+{\partial J\over\partial x}\right)+{dy \over dx}\left({d^2J \over dx^2}+{d \over dx}\left({\partial J\over\partial x}\right)\right)+F\left(x,y,{dy \over dx}\right)=0</math>

Thus, the three conditions for exactness for a third-order differential equation are: the <math>{d^2y \over dx^2}</math> term must be <math>2{dJ \over dx}+{\partial J\over\partial x}</math>, the <math>{dy \over dx}</math> term must be <math>{d^2J \over dx^2}+{d \over dx}\left({\partial J\over\partial x}\right)</math> and

:<math>F\left(x,y,{dy \over dx}\right)-{d^2 \over dx^2}\left(I\left(x,y\right)-h\left(x\right)\right)+{d^2y \over dx^2}{\partial J\over\partial x}+{dy \over dx}{d \over dx}\left({\partial J\over\partial x}\right)</math>

must be a function solely of <math>x</math>.

=== Example ===
Consider the nonlinear third-order differential equation

:<math>yy'''+3y'y''+12x^2=0</math>

If <math>J\left(x,y\right)=y</math>, then <math>y''\left(2{dJ \over dx}+{\partial J\over\partial x}\right)</math> is <math>2y'y''</math> and <math>y'\left({d^2J \over dx^2}+{d \over dx}\left({\partial J\over\partial x}\right)\right)=y'y''</math>which together sum to <math>3y'y''</math>. Fortunately, this appears in our equation. For the last condition of exactness,

:<math>F\left(x,y,{dy \over dx}\right)-{d^2 \over dx^2}\left(I\left(x,y\right)-h\left(x\right)\right)+{d^2y \over dx^2}{\partial J\over\partial x}+{dy \over dx}{d \over dx}\left({\partial J\over\partial x}\right)=12x^2-0+0+0=12x^2</math>

which is indeed a function only of <math>x</math>. So, the differential equation is exact. Integrating twice yields that <math>h\left(x\right)=x^4+C_1x+C_2=I\left(x,y\right)</math>. Rewriting the equation as a first-order exact differential equation yields

:<math>x^4+C_1x+C_2+yy'=0</math>

Integrating <math>I\left(x,y\right)</math> with respect to <math>x</math> gives that <math>{x^5\over 5}+C_1x^2+C_2x+i\left(y\right)=0</math>. Differentiating with respect to <math>y</math> and equating that to the term in front of <math>y'</math> in the first-order equation gives that
<math>i'\left(y\right)=y</math> and that <math>i\left(y\right)={y^2\over 2}+C_3</math>. The full implicit solution becomes

:<math>{x^5\over 5}+C_1x^2+C_2x+C_3+{y^2\over 2}=0</math>

The explicit solution, then, is

:<math>y=\pm\sqrt{C_1x^2+C_2x+C_3-\frac{2x^5}{5}}</math>

==Licensing==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Exact_differential_equation Exact differential equation, Wikipedia] under a CC BY-SA license

@@ Line 54: / Line 54: @@
 :<math>I\left(x,y\right)+J\left(x,y\right){dy \over dx}=0</math>
-Since both functions {{nowrap|<math>I\left(x,y\right)</math>,}} <math>J\left(x,y\right)</math> are functions of two variables, implicitly differentiating the multivariate function yields
+Since both functions <math>I\left(x,y\right)</math>, <math>J\left(x,y\right)</math> are functions of two variables, implicitly differentiating the multivariate function yields
 :<math>{dI \over dx} +\left({ dJ\over dx}\right){dy \over dx}+{d^2y \over dx^2}\left(J\left(x,y\right)\right)=0</math>
@@ Line 70: / Line 70: @@
 :<math>{\partial I\over\partial x}+{dy \over dx}\left({\partial I\over\partial y}+{\partial J\over\partial x}+{\partial J\over\partial y}{dy \over dx}\right)+{d^2y \over dx^2}\left(J\left(x,y\right)\right)=0</math>
-If the equation is exact, then {{nowrap|<math display="inline">{\partial J\over\partial x}={\partial I\over\partial y}</math>.}} Additionally, the total derivative of <math>J\left(x,y\right)</math> is equal to its implicit ordinary derivative <math display="inline">{dJ \over dx}</math>. This leads to the rewritten equation
+If the equation is exact, then <math display="inline">{\partial J\over\partial x}={\partial I\over\partial y}</math>. Additionally, the total derivative of <math>J\left(x,y\right)</math> is equal to its implicit ordinary derivative <math display="inline">{dJ \over dx}</math>. This leads to the rewritten equation
 :<math>{\partial I\over\partial x}+{dy \over dx}\left({\partial J\over\partial x}+{dJ \over dx}\right)+{d^2y \over dx^2}\left(J\left(x,y\right)\right)=0</math>

@@ Line 50: / Line 50: @@
 == Second order exact differential equations ==
-The concept of exact differential equations can be extended to second order equations.<ref>{{Cite book|title=Ordinary Differential Equations: An Elementary Textbook for Students of Mathematics, Engineering and the Sciences|url=https://archive.org/details/ordinarydifferen00tene_850|url-access=limited|last1=Tenenbaum|first1=Morris|last2=Pollard|first2=Harry|publisher=Dover|year=1963|isbn=0-486-64940-7|location=New York|pages=[https://archive.org/details/ordinarydifferen00tene_850/page/n131 248]|chapter=Solution of the Linear Differential Equation with Nonconstant Coefficients. Reduction of Order Method.}}</ref> Consider starting with the first-order exact equation:
+The concept of exact differential equations can be extended to second order equations. Consider starting with the first-order exact equation:
 :<math>I\left(x,y\right)+J\left(x,y\right){dy \over dx}=0</math>

Exact Differential Equations - Revision history

Lila at 22:55, 25 October 2021

Lila at 22:54, 25 October 2021

Lila: Created page with "==Definition== Given a simply connected and open subset ''D'' of '''R'''2 and two functions ''I'' and ''J'' which are continuous on ''D'', an implicit first-order o..."

Lila: Created page with "==Definition== Given a simply connected and open subset ''D'' of '''R'''² and two functions ''I'' and ''J'' which are continuous on ''D'', an implicit first-order o..."