Exact differential equation

An exact (or complete ) differential equation is an ordinary differential equation of form

${\displaystyle \ p(x,y(x))+q(x,y(x)){\frac {{\rm {d}}y(x)}{{\rm {d}}x}}=0}$,

for which there is a continuously differentiable function such that ${\displaystyle \Phi (x,y)}$

${\displaystyle {\frac {\partial \Phi (x,y)}{\partial x}}=p(x,y)}$   and   . ${\displaystyle {\frac {\partial \Phi (x,y)}{\partial y}}=q(x,y)}$

Such a function is then called the potential function of the vector field . ${\displaystyle \Phi }$ ${\displaystyle (p,q)}$

introduction

The differential equation is liked in the representation by the separation of the variables ${\displaystyle \textstyle p(x,y)+q(x,y){\frac {{\rm {d}}y}{{\rm {d}}x}}(x)=0}$

${\displaystyle p(x,y)\mathrm {d} x+q(x,y)\mathrm {d} y=0}$

specified. The advantage of this representation lies in the fact that the left side of the differential equation - i.e. - can be understood as a component of a total differential , with ${\displaystyle p(x,y)\mathrm {d} x+q(x,y)\mathrm {d} y}$

${\displaystyle \mathrm {d} \Phi (x,y)=p(x,y)\mathrm {d} x+q(x,y)\mathrm {d} y}$.

Here the function takes on the meaning of a scalar potential with the condition as well . Accordingly, there must be a vector field that can be formed from the gradient of the scalar potential, i.e. ${\displaystyle \Phi (x,y)}$ ${\displaystyle {\frac {\partial \Phi }{\partial x}}(x,y)=p(x,y)}$ ${\displaystyle {\frac {\partial \Phi }{\partial y}}(x,y)=q(x,y)}$

${\displaystyle {\begin{pmatrix}p(x,y)\\q(x,y)\end{pmatrix}}=\nabla \Phi (x,y)}$.

If and are continuously partially differentiable and if the domain of and is a simply connected area , then there is a scalar potential if and only if the so-called integrability condition ${\displaystyle p}$ ${\displaystyle q}$ ${\displaystyle p}$ ${\displaystyle q}$ ${\displaystyle U\subseteq \mathbb {R} ^{2}}$ ${\displaystyle \Phi }$

${\displaystyle {\frac {\partial q}{\partial x}}(x,y)={\frac {\partial p}{\partial y}}(x,y)}$

is satisfied. Therefore, for the twofold continuously partially differentiable function must apply: ${\displaystyle \Phi (x,y)}$

${\displaystyle {\frac {\partial ^{2}\Phi }{\partial x\partial y}}(x,y)={\frac {\partial p}{\partial y}}(x,y)={\frac {\partial q}{\partial x}}(x,y)={\frac {\partial ^{2}\Phi }{\partial y\partial x}}(x,y)}$.

The fact that is is dealt with in Schwarz's theorem . In addition, the stated integrability condition also means that if the rotation of the vector field disappears in a simply connected area, that is, a scalar potential exists. ${\displaystyle \textstyle {\frac {\partial ^{2}\Phi }{\partial x\partial y}}(x,y)={\frac {\partial ^{2}\Phi }{\partial y\partial x}}(x,y)}$ ${\displaystyle (p,q)}$ ${\displaystyle \textstyle {\frac {\partial q}{\partial x}}(x,y)-{\frac {\partial p}{\partial y}}(x,y)=0}$ ${\displaystyle \Phi }$

If, on the other hand, the right-hand side of the differential equation is combined with the total differential of the function , a Pfaffian form results in the representation and the equation follows after integration on both sides ${\displaystyle p(x,y)\mathrm {d} x+q(x,y)\mathrm {d} y=0}$ ${\displaystyle \Phi (x,y)}$ ${\displaystyle \mathrm {d} \Phi (x,y)=0}$

${\displaystyle \int \mathrm {d} \Phi (x,y)=\Phi (x,y)=C}$.

This makes it clear that there must be a constant that fulfills the function for all . The solution is therefore the initial condition of the differential equation and represents an equipotential line. ${\displaystyle C}$ ${\displaystyle x,y}$ ${\displaystyle \Phi (x,y)}$ ${\displaystyle \Phi (x,y)=C}$

${\displaystyle \Phi (x,y)}$is also referred to as the first integral in connection with the exact differential equation .

definition

In a simply connected domain an exact differential equation is given by ${\displaystyle U\subseteq \mathbb {R} ^{2}}$

${\displaystyle p(x,y)\,\mathrm {d} x+q(x,y)\,\mathrm {d} y=0}$

if the following conditions apply:

• The functions are continuously partially differentiable. ${\displaystyle p,q\colon U\to \mathbb {R} }$

• The integrability condition is fulfilled. ${\displaystyle {\frac {\partial p}{\partial y}}={\frac {\partial q}{\partial x}}}$

• There is a twofold, continuously partially differentiable scalar potential , so that and applies. ${\displaystyle \Phi (x,y)}$ ${\displaystyle {\frac {\partial \Phi }{\partial x}}(x,y)=p(x,y)}$ ${\displaystyle {\frac {\partial \Phi }{\partial y}}(x,y)=q(x,y)}$

• An initial value is given. ${\displaystyle \Phi (x_{0},y_{0})=C}$

Solution method

To solve the exact differential equation, it is necessary to find the scalar potential as follows: ${\displaystyle \Phi (x,y)}$

• Integrability condition: The differential equation is exact if the integrability condition

${\displaystyle {\frac {\partial p}{\partial y}}(x,y)={\frac {\partial q}{\partial x}}(x,y)}$

is satisfied. If this is not the case, the differential equation can possibly be solved using an integrating factor .

• First integral: If there is an exact differential equation, the relationship becomes through integration

${\displaystyle {\frac {\partial \Phi }{\partial x}}(x,y)=p(x,y)}$

the scalar potential increases

${\displaystyle \Phi (x,y)=\int p(x,y)\mathrm {d} x+\varphi (y)}$

certainly. Here is an integration constant that is independent of, but is variable. In this respect, the scalar potential is determined except for an unknown function . In order to determine the still unknown function , the integrability condition is used in the integral representation. By integrating ${\displaystyle \varphi (y)}$ ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle \varphi (y)}$ ${\displaystyle \varphi (y)}$

${\displaystyle {\frac {\partial p}{\partial y}}(x,y)={\frac {\partial q}{\partial x}}(x,y)}$

you get

${\displaystyle {\frac {\partial }{\partial y}}\int p(x,y)\mathrm {d} x+{\frac {\mathrm {d} \varphi (y)}{\mathrm {d} y}}={\frac {\partial }{\partial x}}\int q(x,y)\mathrm {d} x=q(x,y)\,,}$

where the right hand side of the equation yields. After forming follows ${\displaystyle q(x,y)}$

${\displaystyle {\frac {\mathrm {d} \varphi (y)}{\mathrm {d} y}}=q(x,y)-{\frac {\partial }{\partial y}}\int p(x,y)\mathrm {d} x\,.}$

Another integration results

${\displaystyle \varphi (y)=\int \mathrm {d} \varphi (y)=\int \left(q(x,y)-{\frac {\partial }{\partial y}}\int p(x,y)\mathrm {d} x\right)\mathrm {d} y}$

and thus a solution of the searched scalar potential reads

${\displaystyle {\begin{aligned}\Phi (x,y)&=\int p(x,y)\mathrm {d} x+\int \left(q(x,y)-{\frac {\partial }{\partial y}}\int p(x,y)\mathrm {d} x\right)\mathrm {d} y\\&=\int p(x,y)\mathrm {d} x+\varphi (y)\;.\end{aligned}}}$

The antiderivative is also called the first integral of the exact differential equation. ${\displaystyle \Phi (x,y)}$

• Initial condition: In all previously performed integrations, the integration constant was not taken into account because it is calculated from the initial value. Since, in addition to the exact differential equation, an initial value is required for the solution, the scalar potential can now be determined with. ${\displaystyle \Phi (x_{0},y_{0})=C}$ ${\displaystyle \Phi (x,y)}$

• Without an initial value: If the initial value is not known, the differential equation gives the solution . This initial condition then, inserted into the first integral, delivers the desired solution of the exact differential equation ${\displaystyle \Phi (x_{0},y_{0})}$ ${\displaystyle {\frac {\mathrm {d} \Phi }{\mathrm {d} y}}(x,y)=0}$ ${\displaystyle \Phi (x,y)=C}$

${\displaystyle C=\int p(x,y)\mathrm {d} x+\varphi (y)\;.}$

• With initial value: If an initial value is specified, the equation must be fulfilled. This initial value then, inserted into the first integral, provides the desired solution of the exact differential equation ${\displaystyle \Phi (x_{0},y_{0})}$ ${\displaystyle \Phi (x,y)=\Phi (x_{0},y_{0})}$

${\displaystyle \Phi (x_{0},y_{0})=\int p(x,y)\mathrm {d} x+\varphi (y)\;.}$

• simply connected area: Finally, it has to be checked whether the solution covers a simply connected area. If this is not the case, it must be checked whether the solution can be reduced to a simply connected area through suitable restrictions.

example

Lemniscate from Gerono: solution set of ${\displaystyle x^{4}-x^{2}+y^{2}=0}$

The aim is to compute the exact differential equation of Gerono's lemniscate . So it becomes the differential equation

${\displaystyle (4x^{3}-2x)\mathrm {d} x+2y\,\mathrm {d} y=0}$

considered with the initial value . So is ${\displaystyle \Phi (1,0)=0}$

${\displaystyle p(x,y)=4x^{3}-2x\qquad q(x,y)=2y}$

and yields the integrability condition

${\displaystyle {\frac {\partial p}{\partial y}}=0\qquad \;\,{\frac {\partial q}{\partial x}}=0}$.

The differential equation is therefore exact and the first integral can be determined immediately. To do this, it is first calculated ${\displaystyle \varphi (y)}$

${\displaystyle {\begin{aligned}\varphi (y)&=\int \left(q-{\frac {\partial }{\partial y}}\int p\,\mathrm {d} x\right)\mathrm {d} y\\&=\int 2y\,\mathrm {d} y-\int {\frac {\partial }{\partial y}}\int \left(4x^{3}-2x\right)\mathrm {d} x\,\mathrm {d} y\\\\&=\int 2y\,\mathrm {d} y-\int \underbrace {{\frac {\partial }{\partial y}}\left(x^{4}-x^{2}\right)} _{=0}\mathrm {d} y\\&=y^{2}\;.\end{aligned}}}$

Thus, and the second integral vanishes, since the integrand is not dependent on. As explained above, the constants of integration are not taken into account. The first integral can be determined under this condition ${\displaystyle \varphi (y)=y^{2}}$ ${\displaystyle y}$

${\displaystyle {\begin{aligned}\Phi (x,y)&=\int p\,\mathrm {d} x+\varphi (y)\\&=\int \left(4x^{3}-2x\right)\mathrm {d} x+y^{2}\\&=x^{4}-x^{2}+y^{2}\;.\end{aligned}}}$

With and the initial value , the solution of the implicit curve results ${\displaystyle \Phi (x,y)=C=\Phi (1,0)}$ ${\displaystyle \Phi (1,0)=0}$

${\displaystyle x^{4}-x^{2}+y^{2}=0}$.

Integrating factors

For an ordinary differential equation of the form that does not meet the integrability condition, a zero-point-free, continuously differentiable function can occasionally be determined in such a way that ${\displaystyle \ p(x,y)+q(x,y){\tfrac {{\rm {d}}y}{{\rm {d}}x}}=0}$ ${\displaystyle {\tfrac {\partial p}{\partial y}}={\tfrac {\partial q}{\partial x}}}$ ${\displaystyle \mu (x,y)\neq 0}$

${\displaystyle \mu (x,y)p(x,y)+\mu (x,y)q(x,y){\frac {{\rm {d}}y}{{\rm {d}}x}}=0}$

becomes an exact differential equation. In this case it is called an integrating factor or Euler's multiplier . Since, by definition, it never becomes zero, the exact differential equation has the same solutions as before multiplication with where is an integrating factor if and only if the integrability condition in the representation ${\displaystyle \mu }$ ${\displaystyle \mu }$ ${\displaystyle \mu .}$ ${\displaystyle \mu (x,y)}$

${\displaystyle {\frac {\partial \mu p}{\partial y}}={\frac {\partial \mu q}{\partial x}}}$

is fulfilled.

It is usually difficult to solve this partial differential equation in general. But since you only need a special solution , you will try to find a solution with special approaches . Such approaches could be, for example: ${\displaystyle \mu (x,y)}$ ${\displaystyle \mu (x,y)}$

${\displaystyle \mu =\mu (x),\quad \mu =\mu (y),\quad \mu =\mu (x+y),\quad \mu =\mu (xy)}$.

Integrating factor and ${\displaystyle \mu (x)}$ ${\displaystyle \mu (y)}$

A simple example of an integrating factor is given when it only depends on one variable or . ${\displaystyle \mu }$ ${\displaystyle x}$ ${\displaystyle y}$

First, the case is considered in which the integrating factor is only dependent on and is as a result of it. With this assumption, the integrability condition results ${\displaystyle x}$ ${\displaystyle {\tfrac {\partial \mu }{\partial y}}=0}$

${\displaystyle {\frac {\partial \mu p}{\partial y}}={\frac {\partial \mu q}{\partial x}}}$

the following illustration in connection with the product rule

${\displaystyle \mu {\frac {\partial p}{\partial y}}=\mu {\frac {\partial q}{\partial x}}+q{\frac {\partial \mu }{\partial x}}}$

and after forming follows

${\displaystyle q{\frac {\partial \mu }{\partial x}}=\mu \left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)\,,}$

which can also be written as

${\displaystyle {\frac {1}{\mu }}{\frac {\partial \mu }{\partial x}}={\frac {1}{q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)\,.}$

The chain rule for the logarithmic derivative finally yields

${\displaystyle {\frac {\partial \ln \mu }{\partial x}}={\frac {1}{q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)\,.}$

Integration of this equation on both sides results in omitting the integration constants

${\displaystyle \ln \mu =\int {\frac {1}{q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)\mathrm {d} x}$

or

${\displaystyle \mu (x)=\exp \left(\int {\frac {1}{q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)\mathrm {d} x\right)=\exp \left(\int f(x)\,\mathrm {d} x\right)\,.}$

Accordingly, the integrating factor is only dependent on if the following expression is only a function of : ${\displaystyle \mu (x)}$ ${\displaystyle x}$ ${\displaystyle x}$

${\displaystyle {\frac {1}{q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)=f(x).}$

In the same way it can be shown that the integrating factor only depends on if ${\displaystyle \mu (y)}$ ${\displaystyle y}$

${\displaystyle -{\frac {1}{p}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)=f(y)}$

has only one dependency and the integrating factor is then ${\displaystyle y}$

${\displaystyle \mu (y)=\exp \left(-\int {\frac {1}{p}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)\mathrm {d} y\right)=\exp \left(\int f(y)\,\mathrm {d} y\right)\,.}$

example

Starting from the differential equation

${\displaystyle 2y^{2}\mathrm {d} x+2xy\mathrm {d} y=0}$

With

${\displaystyle p(x,y)=2y^{2}\qquad q(x,y)=2xy}$

and

${\displaystyle {\frac {\partial p}{\partial y}}=4y\qquad \quad \;\;{\frac {\partial q}{\partial x}}=2y}$

it becomes apparent that the integrability condition is not met. Since it only depends on, it makes sense to choose the integrating factor in such a way that it only depends on and thus ${\displaystyle p}$ ${\displaystyle y}$ ${\displaystyle \mu (x)}$ ${\displaystyle x}$

${\displaystyle f(x)={\frac {1}{q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)={\frac {1}{2xy}}\left(4y-2y\right)={\frac {1}{x}}\;.}$

So the integrating factor is

${\displaystyle \mu (x)=\exp \left(\int f(x)\,\mathrm {d} x\right)=\exp \left(\int {\frac {1}{x}}\,\mathrm {d} x\right)=\exp \left(\log(x)\right)=x\;.}$

Integrating factor ${\displaystyle \mu (x+y)}$

Depends on from, so is the integrating factor ${\displaystyle f\;{\stackrel {\mathrm {def} }{=}}{\frac {1}{p-q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)}$ ${\displaystyle x+y}$

${\displaystyle \mu (x+y)=\exp \left(-\int {\frac {1}{p-q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)\mathrm {d} x-\int {\frac {1}{p-q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)\mathrm {d} y\right)\;{\stackrel {\mathrm {def} }{=}}\exp \left(-\int f(t)\mathrm {d} t\right){\bigg |}_{t=x+y}\,.}$

proof

It is

${\displaystyle {\begin{aligned}{\frac {\partial \mu p}{\partial y}}&=\mu {\frac {\partial p}{\partial y}}+p{\frac {\partial \mu }{\partial y}}=\exp \left(-\int f(t)\mathrm {d} t\right){\bigg |}_{t=x+y}{\frac {\partial p}{\partial y}}-f(x+y)\exp \left(-\int f(t)\mathrm {d} t\right){\bigg |}_{t=x+y}p\\\\&=\exp \left(-\int f(t)\mathrm {d} t\right){\bigg |}_{t=x+y}\left({\frac {\partial p}{\partial y}}-{\frac {p}{p-q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)\right)\end{aligned}}}$

and in the same way arises

${\displaystyle {\begin{aligned}{\frac {\partial \mu q}{\partial x}}&=\mu {\frac {\partial q}{\partial x}}+q{\frac {\partial \mu }{\partial x}}=\exp \left(-\int f(t)\mathrm {d} t\right){\bigg |}_{t=x+y}{\frac {\partial q}{\partial x}}-f(x+y)\exp \left(-\int f(t)\mathrm {d} t\right){\bigg |}_{t=x+y}q\\\\&=\exp \left(-\int f(t)\mathrm {d} t\right){\bigg |}_{t=x+y}\left({\frac {\partial q}{\partial x}}-{\frac {q}{p-q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)\right)\;.\end{aligned}}}$

If the integrability condition is now brought into the representation , it follows ${\displaystyle \textstyle {\frac {\partial \mu p}{\partial y}}-{\frac {\partial \mu q}{\partial x}}=0}$

${\displaystyle {\begin{aligned}{\frac {\partial \mu p}{\partial y}}-{\frac {\partial \mu q}{\partial x}}&=\exp \left(-\int f(t)\mathrm {d} t\right){\bigg |}_{t=x+y}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}-{\frac {p}{p-q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)+{\frac {q}{p-q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)\right)\\\\&=\exp \left(-\int f(t)\mathrm {d} t\right){\bigg |}_{t=x+y}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}-{\frac {p-q}{p-q}}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}\right)\right)\\\\&=\exp \left(-\int f(t)\mathrm {d} t\right){\bigg |}_{t=x+y}\left({\frac {\partial p}{\partial y}}-{\frac {\partial q}{\partial x}}-{\frac {\partial p}{\partial y}}+{\frac {\partial q}{\partial x}}\right)\\\\&=0\end{aligned}}}$

${\displaystyle \Box }$

literature

• Ludwig Bieberbach : Theory of differential equations Springer, reprint Berlin Heidelberg New York 1979, page 15-21 (scanned page 31: 15-37: 21), uni-goettingen.de
• Harro Heuser : Ordinary differential equations , Vieweg + Teubner 2009 (6th edition), pages 91-102, ISBN 978-3-8348-0705-2
• Wolfgang Walter : Ordinary differential equations , Springer-Verlag 2000, page 37-47, ISBN 3540676422
• Jochen Merker: Differential equations , script, summer semester 2011, University of Rostock, pages 19–21

Individual evidence

1. ↑ Harro Heuser : Ordinary differential equations , Vieweg + Teubner 2009 (6th edition), pages 100-102, ISBN 978-3-8348-0705-2