Riccati DGL < gewöhnliche < Differentialgl. < Analysis < Hochschule < Mathe < Vorhilfe
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| Aufgabe | The Riccati equation is given by
[mm] $y'=P(x)\;y^2+Q(x)\;y+R(x)$ [/mm] .
(a)
Show that if one solution of this equation, say [mm] y_1(x), [/mm] is known, then the general solution can be found by using the transformation
$y = [mm] y_1 [/mm] + [mm] \frac{1}{u}$
[/mm]
where u is a new dependent variable.
(b)
Show that if two solutions are known, say [mm] y_1(x) [/mm] and [mm] y_2(x), [/mm] the general solution is
[mm] $\frac{y-y_1}{y-y_2}=ce^{\int P(x)*(y1-y2)dx}$
[/mm]
(c)
Show that if three solutions are known, say [mm] y_1(x), y_2(x) [/mm] and [mm] y_3(x), [/mm] then the general solution is
[mm] $\frac{(y-y_1)(y_2-y_3)}{(y-y_2)(y_1-y_3)}=c$ [/mm] . |
zu (a)
[mm] $y'=P(x)\;y^2+Q(x)\;y+R(x)$ [/mm]
[mm] $y_1'=P(x)\;y_1^2+Q(x)\;y_1+R(x)$ [/mm]
[mm] $(y'-y_1')=P(x)\;(y^2-y_1^2)+Q(x)\;(y-y_1)$ [/mm]
[mm] (y-y_1)=u
[/mm]
[mm] $u'=P(x)(y-y_1)(y+y_1)+Q(x)u$
[/mm]
[mm] $u'=P(x)u(u+2y_1)+Q(x)u$
[/mm]
[mm] $u'=(2y_1P(x)+Q(x))u+P(x)u^2$
[/mm]
Das ist eine Bernoulli DGL.
[mm] $-u'u^{-2}+(2y_1P(x)+Q(x))u^{-1}=-P(x)$
[/mm]
[mm] $[u^{-1}]'+(2y_1P(x)+Q(x))u^{-1}=-P(x)$
[/mm]
$z'+(2y_1P(x)+Q(x))z=-P(x)$
zu (b)
Ich habe probiert
[mm] $y=\frac{y_2-y_1}{ce^{\int P(x)*(y1-y2)dx}-1}+y_2$
[/mm]
[mm] $y'=\frac{ce^{\int P(x)*(y1-y2)dx}(y_2'-y_1')-(y_2-y_1)*P(x)*(y_1-y_2)*ce^{\int P(x)*(y1-y2)dx}}{[ce^{\int P(x)*(y1-y2)dx}-1]^2}+y_2'$
[/mm]
in die DGL einzusetzen; ohne Erfolg.
LG, Martinius
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Hallo Martinius,
> The Riccati equation is given by
>
> [mm]y'=P(x)\;y^2+Q(x)\;y+R(x)[/mm] .
>
> (a)
>
> Show that if one solution of this equation, say [mm]y_1(x),[/mm] is
> known, then the general solution can be found by using the
> transformation
>
> [mm]y = y_1 + \frac{1}{u}[/mm]
>
> where u is a new dependent variable.
>
>
> (b)
>
> Show that if two solutions are known, say [mm]y_1(x)[/mm] and
> [mm]y_2(x),[/mm] the general solution is
>
> [mm]\frac{y-y_1}{y-y_2}=ce^{\int P(x)*(y1-y2)dx}[/mm]
>
>
> (c)
>
> Show that if three solutions are known, say [mm]y_1(x), y_2(x)[/mm]
> and [mm]y_3(x),[/mm] then the general solution is
>
> [mm]\frac{(y-y_1)(y_2-y_3)}{(y-y_2)(y_1-y_3)}=c[/mm] .
> zu (a)
>
> [mm]y'=P(x)\;y^2+Q(x)\;y+R(x)[/mm]
>
> [mm]y_1'=P(x)\;y_1^2+Q(x)\;y_1+R(x)[/mm]
>
> [mm](y'-y_1')=P(x)\;(y^2-y_1^2)+Q(x)\;(y-y_1)[/mm]
>
> [mm](y-y_1)=u[/mm]
>
> [mm]u'=P(x)(y-y_1)(y+y_1)+Q(x)u[/mm]
>
> [mm]u'=P(x)u(u+2y_1)+Q(x)u[/mm]
>
> [mm]u'=(2y_1P(x)+Q(x))u+P(x)u^2[/mm]
>
> Das ist eine Bernoulli DGL.
>
> [mm]-u'u^{-2}+(2y_1P(x)+Q(x))u^{-1}=-P(x)[/mm]
>
> [mm][u^{-1}]'+(2y_1P(x)+Q(x))u^{-1}=-P(x)[/mm]
>
> [mm]z'+(2y_1P(x)+Q(x))z=-P(x)[/mm]
>
>
> zu (b)
>
> Ich habe probiert
>
> [mm]y=\frac{y_2-y_1}{ce^{\int P(x)*(y1-y2)dx}-1}+y_2[/mm]
>
> [mm]y'=\frac{ce^{\int P(x)*(y1-y2)dx}(y_2'-y_1')-(y_2-y_1)*P(x)*(y_1-y_2)*ce^{\int P(x)*(y1-y2)dx}}{[ce^{\int P(x)*(y1-y2)dx}-1]^2}+y_2'[/mm]
>
> in die DGL einzusetzen; ohne Erfolg.
>
Probiere doch mal irgendwie die DGL
[mm]z'+(2y_1P(x)+Q(x))z=-P(x)[/mm]
heranzuziehen.
Beachte hier daß
[mm]y-y_{1}=\bruch{1}{u_{1}}=z_{1}[/mm]
und
[mm]y-y_{2}=\bruch{1}{u_{2}}=z_{2}[/mm]
ist.
>
> LG, Martinius
Gruß
MathePower
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Hallo MathePower,
vielen Dank für deinen Hinweis.
> Hallo Martinius,
>
> > The Riccati equation is given by
> >
> > [mm]y'=P(x)\;y^2+Q(x)\;y+R(x)[/mm] .
> >
> > (a)
> >
> > Show that if one solution of this equation, say [mm]y_1(x),[/mm] is
> > known, then the general solution can be found by using the
> > transformation
> >
> > [mm]y = y_1 + \frac{1}{u}[/mm]
> >
> > where u is a new dependent variable.
> >
> >
> > (b)
> >
> > Show that if two solutions are known, say [mm]y_1(x)[/mm] and
> > [mm]y_2(x),[/mm] the general solution is
> >
> > [mm]\frac{y-y_1}{y-y_2}=ce^{\int P(x)*(y1-y2)dx}[/mm]
> >
> >
> > (c)
> >
> > Show that if three solutions are known, say [mm]y_1(x), y_2(x)[/mm]
> > and [mm]y_3(x),[/mm] then the general solution is
> >
> > [mm]\frac{(y-y_1)(y_2-y_3)}{(y-y_2)(y_1-y_3)}=c[/mm] .
> > zu (a)
> >
> > [mm]y'=P(x)\;y^2+Q(x)\;y+R(x)[/mm]
> >
> > [mm]y_1'=P(x)\;y_1^2+Q(x)\;y_1+R(x)[/mm]
> >
> > [mm](y'-y_1')=P(x)\;(y^2-y_1^2)+Q(x)\;(y-y_1)[/mm]
> >
> > [mm](y-y_1)=u[/mm]
> >
> > [mm]u'=P(x)(y-y_1)(y+y_1)+Q(x)u[/mm]
> >
> > [mm]u'=P(x)u(u+2y_1)+Q(x)u[/mm]
> >
> > [mm]u'=(2y_1P(x)+Q(x))u+P(x)u^2[/mm]
> >
> > Das ist eine Bernoulli DGL.
> >
> > [mm]-u'u^{-2}+(2y_1P(x)+Q(x))u^{-1}=-P(x)[/mm]
> >
> > [mm][u^{-1}]'+(2y_1P(x)+Q(x))u^{-1}=-P(x)[/mm]
> >
> > [mm]z'+(2y_1P(x)+Q(x))z=-P(x)[/mm]
> >
> >
> > zu (b)
> >
> > Ich habe probiert
> >
> > [mm]y=\frac{y_2-y_1}{ce^{\int P(x)*(y1-y2)dx}-1}+y_2[/mm]
> >
> > [mm]y'=\frac{ce^{\int P(x)*(y1-y2)dx}(y_2'-y_1')-(y_2-y_1)*P(x)*(y_1-y_2)*ce^{\int P(x)*(y1-y2)dx}}{[ce^{\int P(x)*(y1-y2)dx}-1]^2}+y_2'[/mm]
>
> >
> > in die DGL einzusetzen; ohne Erfolg.
> >
>
>
> Probiere doch mal irgendwie die DGL
>
> [mm]z'+(2y_1P(x)+Q(x))z=-P(x)[/mm]
>
> heranzuziehen.
>
> Beachte hier daß
>
> [mm]y-y_{1}=\bruch{1}{u_{1}}=z_{1}[/mm]
>
> und
>
> [mm]y-y_{2}=\bruch{1}{u_{2}}=z_{2}[/mm]
>
> ist.
>
>
> >
> > LG, Martinius
>
>
> Gruß
> MathePower
Dann wäre das
[mm]z_1'+(2y_1P(x)+Q(x))z_1=-P(x)[/mm]
[mm]z_2'+(2y_1P(x)+Q(x))z_2=-P(x)[/mm]
[mm] $z_1=e^{-\int (2y1P(x)+Q(x))dx}*\int -P(x)*e^{\int (2y1P(x)+Q(x))dx}\;dx=\frac{1}{y-y_1}$
[/mm]
[mm] $z_2=e^{-\int (2y2P(x)+Q(x))dx}*\int -P(x)*e^{\int (2y2P(x)+Q(x))dx}\;dx=\frac{1}{y-y_2}$
[/mm]
[mm] $\frac{z_2}{z_1}=\frac{y-y_1}{y-y_2}=e^{\int 2y1P(x)\;dx-\int 2y2P(x)\;dx}*\frac{\int P(x)*e^{\int (2y2P(x)+Q(x))dx}\;dx}{\int P(x)*e^{\int (2y1P(x)+Q(x))dx}\;dx}$
[/mm]
Das sieht schon viel besser aus. Kann man das auch noch vereinfachen?
LG, Martinius
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Hallo Martinius,
> Hallo MathePower,
>
> vielen Dank für deinen Hinweis.
>
> > Hallo Martinius,
> >
> > > The Riccati equation is given by
> > >
> > > [mm]y'=P(x)\;y^2+Q(x)\;y+R(x)[/mm] .
> > >
> > > (a)
> > >
> > > Show that if one solution of this equation, say [mm]y_1(x),[/mm] is
> > > known, then the general solution can be found by using the
> > > transformation
> > >
> > > [mm]y = y_1 + \frac{1}{u}[/mm]
> > >
> > > where u is a new dependent variable.
> > >
> > >
> > > (b)
> > >
> > > Show that if two solutions are known, say [mm]y_1(x)[/mm] and
> > > [mm]y_2(x),[/mm] the general solution is
> > >
> > > [mm]\frac{y-y_1}{y-y_2}=ce^{\int P(x)*(y1-y2)dx}[/mm]
> > >
> > >
> > > (c)
> > >
> > > Show that if three solutions are known, say [mm]y_1(x), y_2(x)[/mm]
> > > and [mm]y_3(x),[/mm] then the general solution is
> > >
> > > [mm]\frac{(y-y_1)(y_2-y_3)}{(y-y_2)(y_1-y_3)}=c[/mm] .
> > > zu (a)
> > >
> > > [mm]y'=P(x)\;y^2+Q(x)\;y+R(x)[/mm]
> > >
> > > [mm]y_1'=P(x)\;y_1^2+Q(x)\;y_1+R(x)[/mm]
> > >
> > > [mm](y'-y_1')=P(x)\;(y^2-y_1^2)+Q(x)\;(y-y_1)[/mm]
> > >
> > > [mm](y-y_1)=u[/mm]
> > >
> > > [mm]u'=P(x)(y-y_1)(y+y_1)+Q(x)u[/mm]
> > >
> > > [mm]u'=P(x)u(u+2y_1)+Q(x)u[/mm]
> > >
> > > [mm]u'=(2y_1P(x)+Q(x))u+P(x)u^2[/mm]
> > >
> > > Das ist eine Bernoulli DGL.
> > >
> > > [mm]-u'u^{-2}+(2y_1P(x)+Q(x))u^{-1}=-P(x)[/mm]
> > >
> > > [mm][u^{-1}]'+(2y_1P(x)+Q(x))u^{-1}=-P(x)[/mm]
> > >
> > > [mm]z'+(2y_1P(x)+Q(x))z=-P(x)[/mm]
> > >
> > >
> > > zu (b)
> > >
> > > Ich habe probiert
> > >
> > > [mm]y=\frac{y_2-y_1}{ce^{\int P(x)*(y1-y2)dx}-1}+y_2[/mm]
> > >
> > > [mm]y'=\frac{ce^{\int P(x)*(y1-y2)dx}(y_2'-y_1')-(y_2-y_1)*P(x)*(y_1-y_2)*ce^{\int P(x)*(y1-y2)dx}}{[ce^{\int P(x)*(y1-y2)dx}-1]^2}+y_2'[/mm]
>
> >
> > >
> > > in die DGL einzusetzen; ohne Erfolg.
> > >
> >
> >
> > Probiere doch mal irgendwie die DGL
> >
> > [mm]z'+(2y_1P(x)+Q(x))z=-P(x)[/mm]
> >
> > heranzuziehen.
> >
> > Beachte hier daß
> >
> > [mm]y-y_{1}=\bruch{1}{u_{1}}=z_{1}[/mm]
> >
> > und
> >
> > [mm]y-y_{2}=\bruch{1}{u_{2}}=z_{2}[/mm]
> >
> > ist.
> >
> >
> > >
> > > LG, Martinius
> >
> >
> > Gruß
> > MathePower
>
>
> Dann wäre das
>
> [mm]z_1'+(2y_1P(x)+Q(x))z_1=-P(x)[/mm]
>
> [mm]z_2'+(2y_1P(x)+Q(x))z_2=-P(x)[/mm]
>
>
> [mm]z_1=e^{-\int (2y1P(x)+Q(x))dx}*\int -P(x)*e^{\int (2y1P(x)+Q(x))dx}\;dx=\frac{1}{y-y_1}[/mm]
>
> [mm]z_2=e^{-\int (2y2P(x)+Q(x))dx}*\int -P(x)*e^{\int (2y2P(x)+Q(x))dx}\;dx=\frac{1}{y-y_2}[/mm]
>
>
> [mm]\frac{z_2}{z_1}=\frac{y-y_1}{y-y_2}=e^{\int 2y1P(x)\;dx-\int 2y2P(x)\;dx}*\frac{\int P(x)*e^{\int (2y2P(x)+Q(x))dx}\;dx}{\int P(x)*e^{\int (2y1P(x)+Q(x))dx}\;dx}[/mm]
>
>
> Das sieht schon viel besser aus. Kann man das auch noch
> vereinfachen?
>
Beim ersten Faktor kann man den Exponenten noch etwas zusammenfassen.
Jetzt mußt Du wohl noch zeigen, daß
[mm]\frac{\int P(x)*e^{\int (2y2P(x)+Q(x))dx}\;dx}{\int P(x)*e^{\int (2y1P(x)+Q(x))dx}\;dx}[/mm]
eine Konstante ist.
>
> LG, Martinius
Gruß
MathePower
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