Separation of the changeable

The method of separating the variables , separating the variables , separating method or separating the variables is a method from the theory of ordinary differential equations . It can be used to solve separable first-order differential equations . These are differential equations in which the first derivation is a product of one function that is only dependent on and one only dependent on: The term “separation of the changeable” goes back to Johann I Bernoulli , who used it in 1694 in a letter to Gottfried Wilhelm Leibniz . ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle y'=f(y)g(x).}$

A similar procedure for certain partial differential equations is the separation approach .

Solution of the initial value problem

We investigate the initial value problem

${\displaystyle y'(x)=f(y(x))g(x)\ ,\ y(x_{0})=y_{0}}$

for continuous (real) functions and . If so, this initial value problem is solved by the constant function . This solution does not have to be unique under the specified conditions. ${\displaystyle f}$ ${\displaystyle g}$ ${\displaystyle f(y_{0})=0}$ ${\displaystyle y(x):\equiv y_{0}}$

Formulation of the sentence

Let it be with . Then: ${\displaystyle x_{0},y_{0}\in \mathbb {R} }$ ${\displaystyle f(y_{0})\neq 0}$

• There is an extensive open interval with for everyone . Then the mapping to is well-defined and strictly monotonic. There is also an extensive open interval so that the mapping has values ​​in for all . ${\displaystyle y_{0}}$ ${\displaystyle U\subset \mathbb {R} }$ ${\displaystyle f(y)\neq 0}$ ${\displaystyle y\in U}$ ${\displaystyle \Phi (y):=\int _{y_{0}}^{y}{\frac {1}{f(s)}}{\rm {d}}s}$ ${\displaystyle U}$
  ${\displaystyle x_{0}}$ ${\displaystyle V\subset \mathbb {R} }$ ${\displaystyle x\mapsto \int _{x_{0}}^{x}g(s){\rm {d}}s}$ ${\displaystyle x\in V}$ ${\displaystyle \Phi (U)}$
• Be and like above. Then the unique solution to the initial value problem is on . ${\displaystyle U}$ ${\displaystyle V}$ ${\displaystyle u(x):=\Phi ^{-1}\left(\int _{x_{0}}^{x}g(s){\rm {d}}s\right)}$
  ${\displaystyle y'(x)=f(y(x))g(x)\ ,\ y(x_{0})=y_{0}}$
  ${\displaystyle V}$

The solution of the initial value problem in this case is the solution of the equation ${\displaystyle u}$

${\displaystyle \int _{y_{0}}^{u(x)}{\frac {1}{f(s)}}\,\mathrm {d} s=\int _{x_{0}}^{x}g(s)\,\mathrm {d} s}$.

Note that in the case of the concrete form of the separate variables actually local uniqueness at present, though , and no local Lipschitz condition need to be met. ${\displaystyle f(y_{0})\neq 0}$ ${\displaystyle f}$ ${\displaystyle g}$

proof

There and steadily, there is an extensive open interval , so that for everyone . In particular, on the same sign , so on is monotonic well defined and strict. is an open interval encompassing 0. So there is an extensive open interval so that applies to all . ${\displaystyle f(y_{0})\neq 0}$ ${\displaystyle f}$ ${\displaystyle y_{0}}$ ${\displaystyle U}$ ${\displaystyle f(y)\neq 0}$ ${\displaystyle y\in U}$ ${\displaystyle f}$ ${\displaystyle U}$ ${\displaystyle \Phi (y):=\int _{y_{0}}^{y}{\frac {1}{f(s)}}{\rm {d}}s}$ ${\displaystyle U}$ ${\displaystyle \Phi (U)}$ ${\displaystyle x_{0}}$ ${\displaystyle V\subset \mathbb {R} }$ ${\displaystyle \int _{x_{0}}^{x}g(s){\rm {d}}s\in \Phi (U)}$ ${\displaystyle x\in V}$

${\displaystyle u(x):=\Phi ^{-1}\left(\int _{x_{0}}^{x}g(s){\rm {d}}s\right)}$is on well-defined, and because all true ${\displaystyle V}$ ${\displaystyle \Phi '(y)={\frac {1}{f(y)}}\neq 0}$ ${\displaystyle y\in U}$

${\displaystyle u'(x)={\frac {g(x)}{\Phi '(\Phi ^{-1}(\int _{x_{0}}^{x}g(s){\rm {d}}s))}}=f\left(\Phi ^{-1}\left(\int _{x_{0}}^{x}g(s){\rm {d}}s\right)\right)g(x)=f(u(x))g(x)}$

on . The chain rule and the inverse rule were used in the derivation . Of course it is . This proves the existence of a solution to the given initial value problem. ${\displaystyle V}$ ${\displaystyle u'(x)}$ ${\displaystyle u(x_{0})=y_{0}}$

For uniqueness, assume that some solution to the initial value problem is on . It is now shown that on holds; the uniqueness to the left of is analogous. ${\displaystyle {\tilde {u}}}$ ${\displaystyle V}$ ${\displaystyle u={\tilde {u}}}$ ${\displaystyle \{x\in V\mid x>x_{0}\}}$ ${\displaystyle x_{0}}$

Assume that the uniqueness to the right of is violated. Because of the continuity of and there is a with such that ${\displaystyle x_{0}}$ ${\displaystyle u}$ ${\displaystyle {\tilde {u}}}$ ${\displaystyle M\in V}$ ${\displaystyle M\geq x_{0}}$

${\displaystyle u(x)={\tilde {u}}(x)}$ for all ${\displaystyle x\in [x_{0},M]}$

is true, for which however the statement

${\displaystyle u(x)={\tilde {u}}(x)}$ on ${\displaystyle [x_{0},M+\epsilon ]}$

for everyone with is wrong. In the following it is shown that there is still a positive one for which the above statement is true, which implies the desired contradiction. ${\displaystyle \epsilon >0}$ ${\displaystyle M+\epsilon \in V}$ ${\displaystyle \epsilon >0}$

Because there is a with , so that applies to everyone . In particular, is on well-defined, and it applies ${\displaystyle {\tilde {u}}(M)=u(M)\in U}$ ${\displaystyle \epsilon _{0}>0}$ ${\displaystyle M+\epsilon _{0}\in V}$ ${\displaystyle {\tilde {u}}(x)\in U}$ ${\displaystyle x\in [M,M+\epsilon _{0})}$ ${\displaystyle \Phi ({\tilde {u}})}$ ${\displaystyle [x_{0},M+\epsilon _{0})}$

${\displaystyle (\Phi ({\tilde {u}}))'(x)=\Phi '({\tilde {u}}(x)){\tilde {u}}'(x)={\frac {1}{f({\tilde {u}}(x))}}f({\tilde {u}}(x))g(x)=g(x)}$for everyone . ${\displaystyle x\in [x_{0},M+\epsilon _{0})}$

This implies , therefore for everyone , what corresponds to the definition of . This contradicts the ambiguity assumption. ${\displaystyle \Phi ({\tilde {u}}(x))=\int _{x_{0}}^{x}g(s){\rm {d}}s}$ ${\displaystyle {\tilde {u}}(x)=\Phi ^{-1}\left(\int _{x_{0}}^{x}g(s){\rm {d}}s\right)}$ ${\displaystyle x\in [x_{0},M+\epsilon _{0})}$ ${\displaystyle u}$

example

We are looking for the solution to the initial value problem ${\displaystyle y}$

${\displaystyle y'=xy^{2}+x\ ,\ y(0)=1}$.

This is a differential equation with separate variables:

${\displaystyle y'=x(y^{2}+1)}$.

So sit down

${\displaystyle \Phi (y):=\int _{1}^{y}{\frac {1}{1+s^{2}}}{\rm {d}}s=\arctan y-\arctan 1=\arctan y-{\frac {\pi }{4}}}$.

The inverse function is

${\displaystyle \Phi ^{-1}(y)=\tan \left(y+{\frac {\pi }{4}}\right)}$.

So the solution of the initial value problem is given by

${\displaystyle y(x)=\tan \left(\int _{0}^{x}s{\rm {d}}s+{\frac {\pi }{4}}\right)=\tan \left({\frac {x^{2}}{2}}+{\frac {\pi }{4}}\right)}$.

Differentials as a clear calculation aid

The principle of the separation of the changeable clearly states that the following procedure is permitted, i. H. leads to correct results (although the differentials and are actually only symbols that, strictly speaking, cannot be calculated): ${\displaystyle \mathrm {d} x}$ ${\displaystyle \mathrm {d} y}$

• Consistently write the derivative as . ${\displaystyle {\tfrac {\mathrm {d} y}{\mathrm {d} x}}}$
• Use common fractions to move all terms that contain a - including the - to the right and all others - including the - to the left. ${\displaystyle x}$ ${\displaystyle \mathrm {d} x}$ ${\displaystyle \mathrm {d} y}$
• It should then be on the left of the counter and on on the right of the counter . ${\displaystyle \mathrm {d} y}$ ${\displaystyle \mathrm {d} x}$
• Just put an integral symbol in front of both sides and integrate.
• If necessary, solve the equation for. ${\displaystyle y}$
• Find the constant of integration using the initial condition. ${\displaystyle C}$

The calculation for the above example would then proceed as follows:

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}=xy^{2}+x\;\Longrightarrow \,\int {\frac {\mathrm {d} y}{1+y^{2}}}=\int x\,\mathrm {d} x\;\Longrightarrow \,\arctan(y)={\frac {x^{2}}{2}}+C\;\Longrightarrow \,y=\tan \left({\frac {x^{2}}{2}}+C\right)}$

with , well . ${\displaystyle 1=y(0)=\tan(C)}$ ${\displaystyle C=\arctan(1)={\frac {\pi }{4}}}$

literature

• Wolfgang Walter: Ordinary differential equations .4. revised edition. Springer, 1990, ISBN 3-540-52017-1 , pp. 13-20
• Kurt Endl, Wolfgang Luh: Analysis I . 9th edition. Aula-Verlag, Wiesbaden 1989, ISBN 3-89104-498-4 , pp. 316–333
• Harro Heuser : Ordinary differential equations. Introduction to teaching and use . 6th updated edition. Vieweg + Teubner, 2009, ISBN 978-3-8348-0705-2 , pp. 102-122

Web links

• Jochen Merker: Differential Equations (PDF) script, summer semester 2011, University of Rostock, in particular pp. 12-14
• Eric W. Weisstein : Separation of Variables . In: MathWorld (English).
• Separation of variables . Paul's Online Math Notes, Lamar University
• Ron Larson: Separation of Variables . (PDF) (free book chapter from Calculus: Applied approach )

Individual evidence

1. ↑ Harro Heuser: Ordinary differential equations. Introduction to teaching and use. 2nd Edition. Teubner, Stuttgart 1991, ISBN 3-519-12227-8 , p. 128