System of differential equations (non-linear)

[xyzt]

I am trying to solve the system of ODEs:

 

[math]\frac{d}{dt}(\frac{u}{\sqrt{1-u^2-v^2}})=k_1[/math]

[math]\frac{d}{dt}(\frac{v}{\sqrt{1-u^2-v^2}})=k_2[/math]

 

where [math]k_1,k_2[/math] are two constants and [math]u(0)=u_0[/math], [math]v(0)=v_0[/math].

 

I tried using the symmetry by transforming into polar coordinates:

 

[math]u=\rho(t) cos \theta (t)[/math]

 

[math]v=\rho(t) sin \theta (t)[/math]

 

but it did not help. Any ideas? This is not homework.

 

PS: Wolfram Alpha couldn't solve it even after I made [math]k_1=k_2[/math]

Edited April 26, 2015 by xyzt

[mathematic]

You can integrate both sides with respect to t and get 2 algebraic equations in u and v.

[xyzt]

You can integrate both sides with respect to t and get 2 algebraic equations in u and v.

Thank you,

 

Both equations reduce to :

 

[math]\sqrt{1-u^2-v^2}=k_1t+c_1[/math]

[math]\sqrt{1-u^2-v^2}=k_2t+c_2[/math]

This forces [math]k_1=k_2, c_1=c_2[/math], so, the system has a solution only for [math]k_1=k_2[/math]. It doesn't seem right, I don't think that it is allowable to integrate the equations as if v is a constant in the first equation while u is a constant for the second equation.

Edited April 26, 2015 by xyzt

[mathematic]

Thank you,

 

Both equations reduce to :

 

[math]\sqrt{1-u^2-v^2}=k_1t+c_1[/math]

[math]\sqrt{1-u^2-v^2}=k_2t+c_2[/math]

This forces [math]k_1=k_2, c_1=c_2[/math], so, the system has a solution only for [math]k_1=k_2[/math]. It doesn't seem right, I don't think that it is allowable to integrate the equations as if v is a constant in the first equation while u is a constant for the second equation.

The left sides are wrong. When you integrate [math] \frac{df}{dt}[/math] You get f + c. For your problem, for each equation, the left sides should be [math] \frac {u}{\sqrt{1-u^2-v^2}} \ and\ \frac {v}{\sqrt{1-u^2-v^2}}[/math].

[xyzt]

The left sides are wrong. When you integrate [math] \frac{df}{dt}[/math] You get f + c. For your problem, for each equation, the left sides should be [math] \frac {u}{\sqrt{1-u^2-v^2}} \ and\ \frac {v}{\sqrt{1-u^2-v^2}}[/math].

[math]\frac{d}{dt}(\frac{u}{\sqrt{1-u^2-v^2}})=k_1[/math]

 

produces

 

[math]\frac{du}{\sqrt{1-u^2-v^2}}=k_1dt[/math]

 

practically a variable separation which gives , by integrating wrt u in the LHS and t in the RHS the answers that I posted earlier. I think that you are claiming that one should get:

 

[math]d\frac{u}{\sqrt{1-u^2-v^2}}=k_1dt[/math]

 

resulting into

 

[math]\frac{u}{\sqrt{1-u^2-v^2}}=\int{k_1dt}[/math]

 

This doesn't seem right. Neither of them is right

 

The right thing is to observe that

 

[math]d(\frac{1}{\sqrt{1-u^2-v^2}})=\frac{udu+vdv}{(\sqrt{1-u^2-v^2})^3}[/math]

 

So, the equation is [math]\frac{udu+vdv}{(\sqrt{1-u^2-v^2})^3}=k_1dt[/math]

The problem is that the second equation produces:

[math]\frac{udu+vdv}{(\sqrt{1-u^2-v^2})^3}=k_2dt[/math]

 

so, the system is unsolvable if [math]k_1 \ne k_2[/math] and indeterminate if [math]k_1 = k_2[/math]. I think this is why Wolfram gave up on this problem. Unfortunately this is not an arbitrary exercise, it is rooted in particle physics.

Edited April 28, 2015 by xyzt

[andrewcellini]

xyzt,

 

[math]\frac{u}{\sqrt{1-u^2-v^2}}=k_1dt[/math]

 

 

shouldn't it be equal to kt, not kdt, after the integration with respect to dt?

 

integrating both sides with respect to dt, the dt's will cancel on the LHS, leaving the integral of df on the LHS, and the integral kdt on RHS

 

but i could be mistaken.

Edited April 27, 2015 by andrewcellini

[xyzt]

xyzt,

 

 

shouldn't it be equal to kt, not kdt, after the integration with respect to dt?

 

integrating both sides with respect to dt, the dt's will cancel on the LHS, leaving the integral of df on the LHS, and the integral kdt on RHS

 

but i could be mistaken.

I think you are mistaken because both u and v are functions of t, so one cannot separate the variables as you suggest. Think about it, Wolfram Alpha couldn't solve this system , the solution cannot be that simple.