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Solution to Nonlinear System of Differential Equations

Solution of system of linearly dependent equations.Solve numerical system of nonlinear equations?Finding a Lyapunov function for the differential system $x_1'=-8x_1^3-x_2$, $x_2'=-4x_2-4x_1^3$Solving a system of partial differential equations consist of 6 equations on 9 variables by using MatlabBendixson's condition for existence of limit cycle for a nonlinear systemNumerically solving a system of linear 2nd order differential equationsNonlinear equations systemsOrdinary Differential Equation with 3 unknownsTrouble with numerically solving this system of nonlinear ODEsHelp solving a system of differential equations

0

$begingroup$

I am working on a optimal control problem, specifically the design of a hypersonic aircraft nose. When minimizing the drag coefficient, I am led to the following system of differential equations:
$$
x_1' = -x_2
$$
$$
x_3' = frac-4x_2^31+x_2^2
$$
$$
x_3 = frac4x_1x_2^2(3+x_2^2)(1+x_2^2)^2
$$
with the boundary/initial conditions
$$
x_1(0) = a, qquad 4x_1(l) = x_3(l),
$$
where $l$ is just the endpoint, i.e. $xin[0,l]$. Is it possible to solve this system analytically, or will I have to resort to a numerical solution?

---

If it helps, the source of these equations comes from the Hamiltonian
$$
H(r,u,p) = frac4ru^31+u^2-pu,
$$
with the necessary conditions
$$
fracpartial Hpartial u = 0, qquad p' = fracdpdx = -fracpartial Hpartial r
$$
I just relabeled the variables as $x_1 equiv r, x_2 equiv u, x_3 equiv p$.

ordinary-differential-equations nonlinear-system optimal-control nonlinear-analysis

share|cite|improve this question

edited Mar 30 at 2:40

Josh Pilipovsky

asked Mar 30 at 2:22

Josh PilipovskyJosh Pilipovsky

1033

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Is there (supposed to be) an equation for $x_2'$? Or, how does $x_2$ evolve? Cheers!
  $endgroup$
  – Robert Lewis
  Mar 30 at 2:29
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No equation for $x_2'$ unfortunately. $x_2 = u(x)$ would be the control input for this system, and the necessary conditions give this system of equations.
  $endgroup$
  – Josh Pilipovsky
  Mar 30 at 2:33
  

  
  
  
  

add a comment | 

0

$begingroup$

I am working on a optimal control problem, specifically the design of a hypersonic aircraft nose. When minimizing the drag coefficient, I am led to the following system of differential equations:
$$
x_1' = -x_2
$$
$$
x_3' = frac-4x_2^31+x_2^2
$$
$$
x_3 = frac4x_1x_2^2(3+x_2^2)(1+x_2^2)^2
$$
with the boundary/initial conditions
$$
x_1(0) = a, qquad 4x_1(l) = x_3(l),
$$
where $l$ is just the endpoint, i.e. $xin[0,l]$. Is it possible to solve this system analytically, or will I have to resort to a numerical solution?

---

If it helps, the source of these equations comes from the Hamiltonian
$$
H(r,u,p) = frac4ru^31+u^2-pu,
$$
with the necessary conditions
$$
fracpartial Hpartial u = 0, qquad p' = fracdpdx = -fracpartial Hpartial r
$$
I just relabeled the variables as $x_1 equiv r, x_2 equiv u, x_3 equiv p$.

ordinary-differential-equations nonlinear-system optimal-control nonlinear-analysis

share|cite|improve this question

edited Mar 30 at 2:40

Josh Pilipovsky

asked Mar 30 at 2:22

Josh PilipovskyJosh Pilipovsky

1033

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Is there (supposed to be) an equation for $x_2'$? Or, how does $x_2$ evolve? Cheers!
  $endgroup$
  – Robert Lewis
  Mar 30 at 2:29
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No equation for $x_2'$ unfortunately. $x_2 = u(x)$ would be the control input for this system, and the necessary conditions give this system of equations.
  $endgroup$
  – Josh Pilipovsky
  Mar 30 at 2:33
  

  
  
  
  

add a comment | 

$begingroup$

I am working on a optimal control problem, specifically the design of a hypersonic aircraft nose. When minimizing the drag coefficient, I am led to the following system of differential equations:
$$
x_1' = -x_2
$$
$$
x_3' = frac-4x_2^31+x_2^2
$$
$$
x_3 = frac4x_1x_2^2(3+x_2^2)(1+x_2^2)^2
$$
with the boundary/initial conditions
$$
x_1(0) = a, qquad 4x_1(l) = x_3(l),
$$
where $l$ is just the endpoint, i.e. $xin[0,l]$. Is it possible to solve this system analytically, or will I have to resort to a numerical solution?

---

If it helps, the source of these equations comes from the Hamiltonian
$$
H(r,u,p) = frac4ru^31+u^2-pu,
$$
with the necessary conditions
$$
fracpartial Hpartial u = 0, qquad p' = fracdpdx = -fracpartial Hpartial r
$$
I just relabeled the variables as $x_1 equiv r, x_2 equiv u, x_3 equiv p$.

ordinary-differential-equations nonlinear-system optimal-control nonlinear-analysis

share|cite|improve this question

edited Mar 30 at 2:40

Josh Pilipovsky

asked Mar 30 at 2:22

Josh PilipovskyJosh Pilipovsky

1033

$endgroup$

share|cite|improve this question

edited Mar 30 at 2:40

Josh Pilipovsky

asked Mar 30 at 2:22

Josh PilipovskyJosh Pilipovsky

1033

share|cite|improve this question

edited Mar 30 at 2:40

Josh Pilipovsky

asked Mar 30 at 2:22

Josh PilipovskyJosh Pilipovsky

1033

asked Mar 30 at 2:22

Josh PilipovskyJosh Pilipovsky

1033

asked Mar 30 at 2:22

Josh PilipovskyJosh Pilipovsky

1033

1 Answer
1

active

oldest

votes

0

$begingroup$

$$begincases
fracdx_1dx = -x_2 \
fracdx_3dx= frac-4x_2^31+x_2^2 \
x_3 = frac4x_1x_2^2(3+x_2^2)(1+x_2^2)^2
endcases$$
$$fracdx_1dx_3= frac1+x_2^24x_2^2= frac14x_2^2+ frac14 quadimpliesquad frac1x_2^2 =4fracdx_1dx_3-1$$
$$x_3 = frac4x_1(3 frac1x_2^2+1)( frac1x_2^2+1)^2=
frac4x_1(3 (4fracdx_1dx_3-1 )+1)( 4fracdx_1dx_3-1 +1)^2=
frac12 x_1frac6fracdx_1dx_3-1(fracdx_1dx_3)^2$$

$$x_1left(fracdx_3dx_1right)^2 -6x_1fracdx_3dx_1+2x_3=0$$
Or, on a more usual form, with $x_1=X$ and $x_3=Y$ :
$$(Y')^2-6Y'+frac2XY=0$$
This kind of non-linear ODE has no simple general solution.

There is an obvious particular solution : $quad Y=4Xquadimpliesquad x_3=4x_1$ . As a consequence a particular solution of the problem can be found :

$x_3=4x_1=frac4x_1x_2^2(3+x_2^2)(1+x_2^2)^2
quadimpliesquad x_2=pm 1$

$fracdx_1dx = -x_2=mp 1quad$ and with condition $x_1(0)=a$ :
$$begincasesx_1= mp x+a \
x_2:=pm 1 \
x_3=mp 4x+4a
endcases$$

This an exact solution of the problem if $quad 4x_1(l)=x_3(l)=4amp 4(l)$ .

If $quad 4x_1(l)=x_3(l)neq 4amp 4(l) quad$ I am afraid that there is no closed form solution (made of a finite number of standard functions). Then numerical calculus is required.

share|cite|improve this answer

answered Mar 30 at 9:19

JJacquelinJJacquelin

45.5k21857

$endgroup$

add a comment | 

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0

$begingroup$

$$begincases
fracdx_1dx = -x_2 \
fracdx_3dx= frac-4x_2^31+x_2^2 \
x_3 = frac4x_1x_2^2(3+x_2^2)(1+x_2^2)^2
endcases$$
$$fracdx_1dx_3= frac1+x_2^24x_2^2= frac14x_2^2+ frac14 quadimpliesquad frac1x_2^2 =4fracdx_1dx_3-1$$
$$x_3 = frac4x_1(3 frac1x_2^2+1)( frac1x_2^2+1)^2=
frac4x_1(3 (4fracdx_1dx_3-1 )+1)( 4fracdx_1dx_3-1 +1)^2=
frac12 x_1frac6fracdx_1dx_3-1(fracdx_1dx_3)^2$$

$$x_1left(fracdx_3dx_1right)^2 -6x_1fracdx_3dx_1+2x_3=0$$
Or, on a more usual form, with $x_1=X$ and $x_3=Y$ :
$$(Y')^2-6Y'+frac2XY=0$$
This kind of non-linear ODE has no simple general solution.

There is an obvious particular solution : $quad Y=4Xquadimpliesquad x_3=4x_1$ . As a consequence a particular solution of the problem can be found :

$x_3=4x_1=frac4x_1x_2^2(3+x_2^2)(1+x_2^2)^2
quadimpliesquad x_2=pm 1$

$fracdx_1dx = -x_2=mp 1quad$ and with condition $x_1(0)=a$ :
$$begincasesx_1= mp x+a \
x_2:=pm 1 \
x_3=mp 4x+4a
endcases$$

This an exact solution of the problem if $quad 4x_1(l)=x_3(l)=4amp 4(l)$ .

If $quad 4x_1(l)=x_3(l)neq 4amp 4(l) quad$ I am afraid that there is no closed form solution (made of a finite number of standard functions). Then numerical calculus is required.

share|cite|improve this answer

answered Mar 30 at 9:19

JJacquelinJJacquelin

45.5k21857

$endgroup$

add a comment | 

0

$begingroup$

$$begincases
fracdx_1dx = -x_2 \
fracdx_3dx= frac-4x_2^31+x_2^2 \
x_3 = frac4x_1x_2^2(3+x_2^2)(1+x_2^2)^2
endcases$$
$$fracdx_1dx_3= frac1+x_2^24x_2^2= frac14x_2^2+ frac14 quadimpliesquad frac1x_2^2 =4fracdx_1dx_3-1$$
$$x_3 = frac4x_1(3 frac1x_2^2+1)( frac1x_2^2+1)^2=
frac4x_1(3 (4fracdx_1dx_3-1 )+1)( 4fracdx_1dx_3-1 +1)^2=
frac12 x_1frac6fracdx_1dx_3-1(fracdx_1dx_3)^2$$

$$x_1left(fracdx_3dx_1right)^2 -6x_1fracdx_3dx_1+2x_3=0$$
Or, on a more usual form, with $x_1=X$ and $x_3=Y$ :
$$(Y')^2-6Y'+frac2XY=0$$
This kind of non-linear ODE has no simple general solution.

There is an obvious particular solution : $quad Y=4Xquadimpliesquad x_3=4x_1$ . As a consequence a particular solution of the problem can be found :

$x_3=4x_1=frac4x_1x_2^2(3+x_2^2)(1+x_2^2)^2
quadimpliesquad x_2=pm 1$

$fracdx_1dx = -x_2=mp 1quad$ and with condition $x_1(0)=a$ :
$$begincasesx_1= mp x+a \
x_2:=pm 1 \
x_3=mp 4x+4a
endcases$$

This an exact solution of the problem if $quad 4x_1(l)=x_3(l)=4amp 4(l)$ .

If $quad 4x_1(l)=x_3(l)neq 4amp 4(l) quad$ I am afraid that there is no closed form solution (made of a finite number of standard functions). Then numerical calculus is required.

share|cite|improve this answer

answered Mar 30 at 9:19

JJacquelinJJacquelin

45.5k21857

$endgroup$

add a comment | 

$begingroup$

$$begincases
fracdx_1dx = -x_2 \
fracdx_3dx= frac-4x_2^31+x_2^2 \
x_3 = frac4x_1x_2^2(3+x_2^2)(1+x_2^2)^2
endcases$$
$$fracdx_1dx_3= frac1+x_2^24x_2^2= frac14x_2^2+ frac14 quadimpliesquad frac1x_2^2 =4fracdx_1dx_3-1$$
$$x_3 = frac4x_1(3 frac1x_2^2+1)( frac1x_2^2+1)^2=
frac4x_1(3 (4fracdx_1dx_3-1 )+1)( 4fracdx_1dx_3-1 +1)^2=
frac12 x_1frac6fracdx_1dx_3-1(fracdx_1dx_3)^2$$

$$x_1left(fracdx_3dx_1right)^2 -6x_1fracdx_3dx_1+2x_3=0$$
Or, on a more usual form, with $x_1=X$ and $x_3=Y$ :
$$(Y')^2-6Y'+frac2XY=0$$
This kind of non-linear ODE has no simple general solution.

There is an obvious particular solution : $quad Y=4Xquadimpliesquad x_3=4x_1$ . As a consequence a particular solution of the problem can be found :

$x_3=4x_1=frac4x_1x_2^2(3+x_2^2)(1+x_2^2)^2
quadimpliesquad x_2=pm 1$

$fracdx_1dx = -x_2=mp 1quad$ and with condition $x_1(0)=a$ :
$$begincasesx_1= mp x+a \
x_2:=pm 1 \
x_3=mp 4x+4a
endcases$$

This an exact solution of the problem if $quad 4x_1(l)=x_3(l)=4amp 4(l)$ .

If $quad 4x_1(l)=x_3(l)neq 4amp 4(l) quad$ I am afraid that there is no closed form solution (made of a finite number of standard functions). Then numerical calculus is required.

share|cite|improve this answer

answered Mar 30 at 9:19

JJacquelinJJacquelin

45.5k21857

$endgroup$

share|cite|improve this answer

answered Mar 30 at 9:19

JJacquelinJJacquelin

45.5k21857

answered Mar 30 at 9:19

JJacquelinJJacquelin

45.5k21857

answered Mar 30 at 9:19

JJacquelinJJacquelin

45.5k21857