Solve nonlinear ODE's System $x_1'(t)=-x_1(t)x_3(t)$, $x_2'(t)=-x_2(t)x_3(t)$, $x_3'(t)=x_1^2(t)+x_2^2(t)$

Question

I need to study the flow generated by the vector field $X(x,y,z)=(-xz,-yz,x^2+y^2)$. Therefore, I need to solve the system:

$$ \left\{ \begin{array}{ccc} x_1'(t)&=&-x_1(t)x_3(t)\\ x_2'(t) &=& -x_2(t)x_3(t)\\ x_3'(t) &=&x_1^2(t)+x_2^2(t) \end{array}\right. $$

I don't know how to solve this. I would appretiate any reference or help to solve this kind of equations.

Answer 1

Hint

This might help

$$ \left\{ \begin{array}{ccc} x_1'(t)&=&-x_1(t)x_3(t)\\ x_2'(t) &=& -x_2(t)x_3(t) \end{array}\right. \implies \frac {dx_1}{dx_2}=\frac {x_1}{x_2} \implies x_1=Kx_2$$ $$ \left\{ \begin{array}{ccc} x_1'(t)&=&-x_1(t)x_3(t)\\ x_2'(t) &=& -x_2(t)x_3(t)\\ x_3'(t) &=&x_1^2(t)+x_2^2(t) \end{array}\right.$$ $$x_3'(t) =x_1^2(t)+x_2^2(t)$$ $$\frac {dx_3}{dx_2} =\frac {x_1^2(t)+x_2^2(t)}{-x_2(t)x_3(t)}$$ $$\frac {dx_3}{dx_2} =\frac {K^2x_2(t)+x_2(t)}{-x_3(t)}$$ $$-\int x_3{dx_3} =(K^2+1)\int x_2dx_2$$ $$x^2_3=-(K^2+1)x^2_2+C \implies x_2=h(x_3)$$ You can substitute this in the third equation and try to solve it $$x_3'(t) =(K^2+1)x_2^2(t)$$ $$x_3'(t) =C-x_3^2(t)$$ $$\int \frac {dx_3}{x_3^2-C} =-t+R$$ $$....$$

Answer 2

Put it into cylindrical coordinates $$ x_{\,1} (t) = r(t)\cos \left( {\alpha (t)} \right)\quad x_{\,2} (t) = r(t)\sin \left( {\alpha (t)} \right) $$ so $$ \left\{ \matrix{ x_{\,3} '(t) = r(t)^{\,2} \quad \to \quad x_{\,3} (t) = \int_0^t {r(\tau )^{\,2} d\tau} + c_{\,3} = I(t) + c_{\,3} \hfill \cr x_{\,1} '(t) = r'(t)\cos \left( {\alpha (t)} \right) - r(t)\alpha '(t)\sin \left( {\alpha (t)} \right) = \hfill \cr = - \left( {I(t) + c_{\,3} } \right)r(t)\cos \left( {\alpha (t)} \right) \hfill \cr x_{\,2} '(t) = r'(t)\sin \left( {\alpha (t)} \right) + r(t)\alpha '(t)\cos \left( {\alpha (t)} \right) = \hfill \cr = - \left( {I(t) + c_{\,3} } \right)r(t)\sin \left( {\alpha (t)} \right) \hfill \cr} \right. $$ which, omitting the time dependency to get a clearer view, becomes $$ \left\{ \matrix{ r'\cos \alpha - r\alpha '\sin \alpha = - \left( {I + c_{\,3} } \right)r\cos \alpha \hfill \cr r'\sin \alpha + r\alpha '\cos \alpha = - \left( {I + c_{\,3} } \right)r\sin \alpha \hfill \cr} \right. $$

Multiplying the first by $\cos\alpha$ and the second by $\sin\alpha$ and summing gives $$ r' = - \left( {I + c_{\,3} } \right)r\quad \to \quad {{d^{\,2} } \over {dt^{\,2} }}\ln \left( r \right) = - r^{\,2} \quad \to \quad {{d^{\,2} } \over {dt^{\,2} }}\rho = e^{\, - \,2\rho } $$

while instead multiplying by $-\sin\alpha$ and $\cos\alpha$ and summing we get $$ r\alpha ' = 0 $$