2
$\begingroup$

I have a system of linear differential equation with an algebraic nonlinear constrain:

$$\frac{d^2x(t)}{dt^2}+x(t)=y(t)$$ $$\frac{d^2y(t)}{dt^2}+y(t)=x(t)$$

$x(t)^2+y(t)^2=\alpha$

My question is: is it possible to find an analytical solution of this DAE system or the only way is to solve it numerically?

Thanks in advance

  • 0
    This system is overdetermined2012-02-16

1 Answers 1

4

Obviously, this problem have solution if $\alpha\geq 0$. From now I assume it. Substitute formula for $y(t)$ from the first equation into the second. After simplifications we get $$ x^{(4)}(t)+2x^{(2)}(t)=0 $$ Using standard algorithm for solving linear differential equations with constant coefficients we get $$ x(t)=C_1 \cos(t\sqrt{2})+C_2 \sin(t\sqrt{2})+C_3 t + C_4 $$ So, $$ y(t)=x^{(2)}(t)+x(t)=-C_1 \cos(t\sqrt{2})-C_2 \sin(t\sqrt{2})+C_3 t + C_4 $$ Now we find $$ x^2(t)+y^2(t)=2\left((C_3 t+C_4)^2+C_1^2\cos^2(t\sqrt{2})+C_2^2\sin^2(t\sqrt{2})+C_1 C_2\cos(t\sqrt{2})\sin(t\sqrt{2})\right) $$ It is easy to see that, the condition $x^2(t)+y^2(t)=\alpha={\rm const}$ is satisfied iff $$ C_1=C_2=C_3=0 $$ So, $x(t)=y(t)=C_4=\sqrt{\frac{\alpha}{2}}$

  • 0
    Ah, our answers are morally equivalent, so I deleted mine.2012-02-16
  • 0
    This means x(t) and y(t) are constant.2012-02-16
  • 0
    Yes. Moreover this problem have solutions iff $\alpha\geq 0$2012-02-16