Unique Solution differential equation system/ linear limited












0












$begingroup$


I have the following differential equation system:



$$begin{align}
x'(t)&=sqrt{1+x^2} +y^3 sin x -x^7\
y'(t)&=x(1-y^2 sin x)
end{align}$$
with $ x(0)=x_0, y(0)=y_0$



I have to show, that $ forall t geq 0 $ the System has a unique solution:



I have to show, that the right hand side is linear limited:
$|f(t,x,y)|leq a(t) + b(t)|(x,y)| $
Can somebody explain me, how I manage to do that?






ordinary-differential-equations lipschitz-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Thank you for editing:)
    $endgroup$
    – Steven33
    Dec 2 '18 at 10:20


















0












$begingroup$


I have the following differential equation system:



$$begin{align}
x'(t)&=sqrt{1+x^2} +y^3 sin x -x^7\
y'(t)&=x(1-y^2 sin x)
end{align}$$
with $ x(0)=x_0, y(0)=y_0$



I have to show, that $ forall t geq 0 $ the System has a unique solution:



I have to show, that the right hand side is linear limited:
$|f(t,x,y)|leq a(t) + b(t)|(x,y)| $
Can somebody explain me, how I manage to do that?






ordinary-differential-equations lipschitz-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Thank you for editing:)
    $endgroup$
    – Steven33
    Dec 2 '18 at 10:20
















0












0








0





$begingroup$


I have the following differential equation system:



$$begin{align}
x'(t)&=sqrt{1+x^2} +y^3 sin x -x^7\
y'(t)&=x(1-y^2 sin x)
end{align}$$
with $ x(0)=x_0, y(0)=y_0$



I have to show, that $ forall t geq 0 $ the System has a unique solution:



I have to show, that the right hand side is linear limited:
$|f(t,x,y)|leq a(t) + b(t)|(x,y)| $
Can somebody explain me, how I manage to do that?






ordinary-differential-equations lipschitz-functions







share|cite|improve this question











$endgroup$




I have the following differential equation system:



$$begin{align}
x'(t)&=sqrt{1+x^2} +y^3 sin x -x^7\
y'(t)&=x(1-y^2 sin x)
end{align}$$
with $ x(0)=x_0, y(0)=y_0$



I have to show, that $ forall t geq 0 $ the System has a unique solution:



I have to show, that the right hand side is linear limited:
$|f(t,x,y)|leq a(t) + b(t)|(x,y)| $
Can somebody explain me, how I manage to do that?






ordinary-differential-equations lipschitz-functions




ordinary-differential-equations lipschitz-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 10:12









mrtaurho

5,46041237




5,46041237










asked Dec 2 '18 at 10:08









Steven33Steven33

295




295












  • $begingroup$
    Thank you for editing:)
    $endgroup$
    – Steven33
    Dec 2 '18 at 10:20




















  • $begingroup$
    Thank you for editing:)
    $endgroup$
    – Steven33
    Dec 2 '18 at 10:20


















$begingroup$
Thank you for editing:)
$endgroup$
– Steven33
Dec 2 '18 at 10:20






$begingroup$
Thank you for editing:)
$endgroup$
– Steven33
Dec 2 '18 at 10:20












1 Answer
1






active

oldest

votes


















0












$begingroup$

The solutions are indeed bounded, as the graphic below demonstrates. To get a theoretical proof of that, consider the radius square $V(x,y)=x^2+y^2$ so that along solutions you get for $v(t)=V(x(t),y(t))$
begin{align}
v'(t)=frac{d}{dt}(x(t)^2+y(t)^2)=2(xx'+yy')
&=2left(xsqrt{1+x^2}-x^8+xyright)\
&le (1+2x^2)+(x^2+y^2)le 1+3v(t),
end{align}
using $2able a^2+b^2$ twice. This differential inequality provides not exactly the boundedness of the solutions, but an exponential bound for the growth of solutions,
$$
v(t)le(tfrac13+v(0))e^{3t}-tfrac13.
$$
As the solutions can thus not diverge to infinity in finite time, they are defined for all times.



stream plot



def eqn(x,y): return (1+x**2)**0.5+y**3*np.sin(x)-x**7, x*(1-y**2*np.sin(x))



# generate streamplot

r = 5

X, Y = np.meshgrid(np.arange(-r, r+.01, 0.1), np.arange(-r, r+.05, 0.1))



dotX, dotY = eqn(X,Y)



plt.streamplot(X, Y, dotX, dotY, density=2.5)

plt.xlim(-r,r); plt.ylim(-r,r);

plt.grid(); plt.show()





share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your answer. V is the solution of my problem? Can I argument without knowing V, because at the moment we are not able to solve differential equation systems.
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:11












  • $begingroup$
    I dont get it. How do you know that the solution is bounded by V? Where is the expression sin x, which was in my system?
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:16












  • $begingroup$
    No, $(x(t),y(t))$ is the solution, $v(t)=r(t)^2=x(t)^2+y(t)^2$ is the square of the distance to the origin. A bound on $v$ is thus also a bound on the size of the solution. The sine terms cancel in the sum $xx'+yy'$.
    $endgroup$
    – LutzL
    Dec 2 '18 at 12:18












  • $begingroup$
    I see. You showed: $ v'(t)leq 1+3 v(t)$ Can i apply Grönwall's inequality. Then i would get: v(t) is bounded on each compact set.
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:42












  • $begingroup$
    Yes, the next step is then $left[e^{-3t}(tfrac13+v(t))right]'le 0$, so that $e^{-3t}(frac13+v(t))le(frac13+v(0))$. I'll add the missing $1/3$.
    $endgroup$
    – LutzL
    Dec 2 '18 at 12:45













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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

The solutions are indeed bounded, as the graphic below demonstrates. To get a theoretical proof of that, consider the radius square $V(x,y)=x^2+y^2$ so that along solutions you get for $v(t)=V(x(t),y(t))$
begin{align}
v'(t)=frac{d}{dt}(x(t)^2+y(t)^2)=2(xx'+yy')
&=2left(xsqrt{1+x^2}-x^8+xyright)\
&le (1+2x^2)+(x^2+y^2)le 1+3v(t),
end{align}
using $2able a^2+b^2$ twice. This differential inequality provides not exactly the boundedness of the solutions, but an exponential bound for the growth of solutions,
$$
v(t)le(tfrac13+v(0))e^{3t}-tfrac13.
$$
As the solutions can thus not diverge to infinity in finite time, they are defined for all times.



stream plot



def eqn(x,y): return (1+x**2)**0.5+y**3*np.sin(x)-x**7, x*(1-y**2*np.sin(x))



# generate streamplot

r = 5

X, Y = np.meshgrid(np.arange(-r, r+.01, 0.1), np.arange(-r, r+.05, 0.1))



dotX, dotY = eqn(X,Y)



plt.streamplot(X, Y, dotX, dotY, density=2.5)

plt.xlim(-r,r); plt.ylim(-r,r);

plt.grid(); plt.show()





share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your answer. V is the solution of my problem? Can I argument without knowing V, because at the moment we are not able to solve differential equation systems.
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:11












  • $begingroup$
    I dont get it. How do you know that the solution is bounded by V? Where is the expression sin x, which was in my system?
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:16












  • $begingroup$
    No, $(x(t),y(t))$ is the solution, $v(t)=r(t)^2=x(t)^2+y(t)^2$ is the square of the distance to the origin. A bound on $v$ is thus also a bound on the size of the solution. The sine terms cancel in the sum $xx'+yy'$.
    $endgroup$
    – LutzL
    Dec 2 '18 at 12:18












  • $begingroup$
    I see. You showed: $ v'(t)leq 1+3 v(t)$ Can i apply Grönwall's inequality. Then i would get: v(t) is bounded on each compact set.
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:42












  • $begingroup$
    Yes, the next step is then $left[e^{-3t}(tfrac13+v(t))right]'le 0$, so that $e^{-3t}(frac13+v(t))le(frac13+v(0))$. I'll add the missing $1/3$.
    $endgroup$
    – LutzL
    Dec 2 '18 at 12:45


















0












$begingroup$

The solutions are indeed bounded, as the graphic below demonstrates. To get a theoretical proof of that, consider the radius square $V(x,y)=x^2+y^2$ so that along solutions you get for $v(t)=V(x(t),y(t))$
begin{align}
v'(t)=frac{d}{dt}(x(t)^2+y(t)^2)=2(xx'+yy')
&=2left(xsqrt{1+x^2}-x^8+xyright)\
&le (1+2x^2)+(x^2+y^2)le 1+3v(t),
end{align}
using $2able a^2+b^2$ twice. This differential inequality provides not exactly the boundedness of the solutions, but an exponential bound for the growth of solutions,
$$
v(t)le(tfrac13+v(0))e^{3t}-tfrac13.
$$
As the solutions can thus not diverge to infinity in finite time, they are defined for all times.



stream plot



def eqn(x,y): return (1+x**2)**0.5+y**3*np.sin(x)-x**7, x*(1-y**2*np.sin(x))



# generate streamplot

r = 5

X, Y = np.meshgrid(np.arange(-r, r+.01, 0.1), np.arange(-r, r+.05, 0.1))



dotX, dotY = eqn(X,Y)



plt.streamplot(X, Y, dotX, dotY, density=2.5)

plt.xlim(-r,r); plt.ylim(-r,r);

plt.grid(); plt.show()





share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your answer. V is the solution of my problem? Can I argument without knowing V, because at the moment we are not able to solve differential equation systems.
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:11












  • $begingroup$
    I dont get it. How do you know that the solution is bounded by V? Where is the expression sin x, which was in my system?
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:16












  • $begingroup$
    No, $(x(t),y(t))$ is the solution, $v(t)=r(t)^2=x(t)^2+y(t)^2$ is the square of the distance to the origin. A bound on $v$ is thus also a bound on the size of the solution. The sine terms cancel in the sum $xx'+yy'$.
    $endgroup$
    – LutzL
    Dec 2 '18 at 12:18












  • $begingroup$
    I see. You showed: $ v'(t)leq 1+3 v(t)$ Can i apply Grönwall's inequality. Then i would get: v(t) is bounded on each compact set.
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:42












  • $begingroup$
    Yes, the next step is then $left[e^{-3t}(tfrac13+v(t))right]'le 0$, so that $e^{-3t}(frac13+v(t))le(frac13+v(0))$. I'll add the missing $1/3$.
    $endgroup$
    – LutzL
    Dec 2 '18 at 12:45
















0












0








0





$begingroup$

The solutions are indeed bounded, as the graphic below demonstrates. To get a theoretical proof of that, consider the radius square $V(x,y)=x^2+y^2$ so that along solutions you get for $v(t)=V(x(t),y(t))$
begin{align}
v'(t)=frac{d}{dt}(x(t)^2+y(t)^2)=2(xx'+yy')
&=2left(xsqrt{1+x^2}-x^8+xyright)\
&le (1+2x^2)+(x^2+y^2)le 1+3v(t),
end{align}
using $2able a^2+b^2$ twice. This differential inequality provides not exactly the boundedness of the solutions, but an exponential bound for the growth of solutions,
$$
v(t)le(tfrac13+v(0))e^{3t}-tfrac13.
$$
As the solutions can thus not diverge to infinity in finite time, they are defined for all times.



stream plot



def eqn(x,y): return (1+x**2)**0.5+y**3*np.sin(x)-x**7, x*(1-y**2*np.sin(x))



# generate streamplot

r = 5

X, Y = np.meshgrid(np.arange(-r, r+.01, 0.1), np.arange(-r, r+.05, 0.1))



dotX, dotY = eqn(X,Y)



plt.streamplot(X, Y, dotX, dotY, density=2.5)

plt.xlim(-r,r); plt.ylim(-r,r);

plt.grid(); plt.show()





share|cite|improve this answer











$endgroup$



The solutions are indeed bounded, as the graphic below demonstrates. To get a theoretical proof of that, consider the radius square $V(x,y)=x^2+y^2$ so that along solutions you get for $v(t)=V(x(t),y(t))$
begin{align}
v'(t)=frac{d}{dt}(x(t)^2+y(t)^2)=2(xx'+yy')
&=2left(xsqrt{1+x^2}-x^8+xyright)\
&le (1+2x^2)+(x^2+y^2)le 1+3v(t),
end{align}
using $2able a^2+b^2$ twice. This differential inequality provides not exactly the boundedness of the solutions, but an exponential bound for the growth of solutions,
$$
v(t)le(tfrac13+v(0))e^{3t}-tfrac13.
$$
As the solutions can thus not diverge to infinity in finite time, they are defined for all times.



stream plot



def eqn(x,y): return (1+x**2)**0.5+y**3*np.sin(x)-x**7, x*(1-y**2*np.sin(x))



# generate streamplot

r = 5

X, Y = np.meshgrid(np.arange(-r, r+.01, 0.1), np.arange(-r, r+.05, 0.1))



dotX, dotY = eqn(X,Y)



plt.streamplot(X, Y, dotX, dotY, density=2.5)

plt.xlim(-r,r); plt.ylim(-r,r);

plt.grid(); plt.show()






share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 2 '18 at 12:46

























answered Dec 2 '18 at 11:54









LutzLLutzL

58.7k42055




58.7k42055












  • $begingroup$
    Thank you for your answer. V is the solution of my problem? Can I argument without knowing V, because at the moment we are not able to solve differential equation systems.
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:11












  • $begingroup$
    I dont get it. How do you know that the solution is bounded by V? Where is the expression sin x, which was in my system?
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:16












  • $begingroup$
    No, $(x(t),y(t))$ is the solution, $v(t)=r(t)^2=x(t)^2+y(t)^2$ is the square of the distance to the origin. A bound on $v$ is thus also a bound on the size of the solution. The sine terms cancel in the sum $xx'+yy'$.
    $endgroup$
    – LutzL
    Dec 2 '18 at 12:18












  • $begingroup$
    I see. You showed: $ v'(t)leq 1+3 v(t)$ Can i apply Grönwall's inequality. Then i would get: v(t) is bounded on each compact set.
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:42












  • $begingroup$
    Yes, the next step is then $left[e^{-3t}(tfrac13+v(t))right]'le 0$, so that $e^{-3t}(frac13+v(t))le(frac13+v(0))$. I'll add the missing $1/3$.
    $endgroup$
    – LutzL
    Dec 2 '18 at 12:45




















  • $begingroup$
    Thank you for your answer. V is the solution of my problem? Can I argument without knowing V, because at the moment we are not able to solve differential equation systems.
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:11












  • $begingroup$
    I dont get it. How do you know that the solution is bounded by V? Where is the expression sin x, which was in my system?
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:16












  • $begingroup$
    No, $(x(t),y(t))$ is the solution, $v(t)=r(t)^2=x(t)^2+y(t)^2$ is the square of the distance to the origin. A bound on $v$ is thus also a bound on the size of the solution. The sine terms cancel in the sum $xx'+yy'$.
    $endgroup$
    – LutzL
    Dec 2 '18 at 12:18












  • $begingroup$
    I see. You showed: $ v'(t)leq 1+3 v(t)$ Can i apply Grönwall's inequality. Then i would get: v(t) is bounded on each compact set.
    $endgroup$
    – Steven33
    Dec 2 '18 at 12:42












  • $begingroup$
    Yes, the next step is then $left[e^{-3t}(tfrac13+v(t))right]'le 0$, so that $e^{-3t}(frac13+v(t))le(frac13+v(0))$. I'll add the missing $1/3$.
    $endgroup$
    – LutzL
    Dec 2 '18 at 12:45


















$begingroup$
Thank you for your answer. V is the solution of my problem? Can I argument without knowing V, because at the moment we are not able to solve differential equation systems.
$endgroup$
– Steven33
Dec 2 '18 at 12:11






$begingroup$
Thank you for your answer. V is the solution of my problem? Can I argument without knowing V, because at the moment we are not able to solve differential equation systems.
$endgroup$
– Steven33
Dec 2 '18 at 12:11














$begingroup$
I dont get it. How do you know that the solution is bounded by V? Where is the expression sin x, which was in my system?
$endgroup$
– Steven33
Dec 2 '18 at 12:16






$begingroup$
I dont get it. How do you know that the solution is bounded by V? Where is the expression sin x, which was in my system?
$endgroup$
– Steven33
Dec 2 '18 at 12:16














$begingroup$
No, $(x(t),y(t))$ is the solution, $v(t)=r(t)^2=x(t)^2+y(t)^2$ is the square of the distance to the origin. A bound on $v$ is thus also a bound on the size of the solution. The sine terms cancel in the sum $xx'+yy'$.
$endgroup$
– LutzL
Dec 2 '18 at 12:18






$begingroup$
No, $(x(t),y(t))$ is the solution, $v(t)=r(t)^2=x(t)^2+y(t)^2$ is the square of the distance to the origin. A bound on $v$ is thus also a bound on the size of the solution. The sine terms cancel in the sum $xx'+yy'$.
$endgroup$
– LutzL
Dec 2 '18 at 12:18














$begingroup$
I see. You showed: $ v'(t)leq 1+3 v(t)$ Can i apply Grönwall's inequality. Then i would get: v(t) is bounded on each compact set.
$endgroup$
– Steven33
Dec 2 '18 at 12:42






$begingroup$
I see. You showed: $ v'(t)leq 1+3 v(t)$ Can i apply Grönwall's inequality. Then i would get: v(t) is bounded on each compact set.
$endgroup$
– Steven33
Dec 2 '18 at 12:42














$begingroup$
Yes, the next step is then $left[e^{-3t}(tfrac13+v(t))right]'le 0$, so that $e^{-3t}(frac13+v(t))le(frac13+v(0))$. I'll add the missing $1/3$.
$endgroup$
– LutzL
Dec 2 '18 at 12:45






$begingroup$
Yes, the next step is then $left[e^{-3t}(tfrac13+v(t))right]'le 0$, so that $e^{-3t}(frac13+v(t))le(frac13+v(0))$. I'll add the missing $1/3$.
$endgroup$
– LutzL
Dec 2 '18 at 12:45




















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