Solving a stochastic differential equation with trigonometric functions
Consider the stochastic differential equations $$dX_t = -\frac{1}{2} X_t dt - Y_t d W_t,$$ $$dY_t = -\frac{1}{2} Y_t dt + X_t d W_t,$$ with $X_0 = 1$ and $Y_0 = 0.$ My goal is to transform this into polar coordinates and solve using $R_t$ and $\Theta_t.$ After about 12 pages of work (front and back), I got this: $$d R_t = 0,$$ $$d \Theta_t = 2R_t \tan \Theta _t dt + 2R_t \sec^2 \Theta _t d W_t,$$ with $R_0 = 1$ and $\Theta _0 = 0.$ Now, I know the $dR_t = 0$ part is correct, but like I said, I had about 12 pages of work both front and back to get all of this, and it was very messy. Luckily, it simplified surprisingly nicely, so I am hoping this is correct. But now, I have no idea how to solve this. Can anyone help?
Thanks.
From the two initial SDE, by multiply the first one with $X_t$ and the second one with $Y_t$, we have $$X_tdX_t = -\frac{1}{2}X_t^2dt-X_tY_tdW_t\tag{1}$$ $$Y_tdY_t = -\frac{1}{2}Y_t^2dt+Y_tX_tdW_t\tag{2}$$ Take $(1)+(2)$, we obtain $$X_tdX_t+Y_tdY_t=-\frac{1}{2}\left( X_t^2+Y_t^2\right)dt \tag{3}$$ Let us denote $$X_t = R_t\sin(\Theta_t)$$ $$Y_t = R_t\cos(\Theta_t)$$ With Ito lemma and $(3)$, we have $$\begin{align} d(R_t^2) &= d(X_t^2+Y_t^2) \\ &=2X_tdX_t+\frac{1}{2}\cdot2 \langle dX_t,dX_t\rangle +2Y_tdY_t+\frac{1}{2}\cdot \langle dY_t,dY_t\rangle\\ &=2(X_tdX_t+Y_tdY_t)+(\langle dX_t,dX_t\rangle+\langle dY_t,dY_t\rangle)\\ &=-( X_t^2+Y_t^2)dt+(Y_t^2+X_t^2)dt\\ &=0 \end{align} $$ Then, $$R_t = R_0 = 1 \tag{4}$$
Besides, from the two intial SDE, by multiply the first one with $Y_t$ and the second one with $X_t$, we have also $$Y_tdX_t = -\frac{1}{2}X_tY_tdt-Y_t^2dW_t\tag{5}$$ $$X_tdY_t = -\frac{1}{2}X_tY_tdt+X_t^2dW_t\tag{6}$$
Take $(6)-(5)$ and by knowing $(4)$ we obtain $$X_tdY_t-Y_tdX_t = dW_t $$ $$ \iff \sin(\Theta_t) d\cos(\Theta_t) - \cos(\Theta_t) d\sin(\Theta_t) = dW_t \tag{7}\\ $$ By Ito lemma, we know that $$d\cos(\Theta_t) = -\sin(\Theta_t) d\Theta_t+\frac{1}{2}(-\cos(\Theta_t))\langle d\Theta_t,d\Theta_t \rangle$$ $$d\sin(\Theta_t) = \cos(\Theta_t) d\Theta_t+\frac{1}{2}(-\sin(\Theta_t))\langle d\Theta_t,d\Theta_t\rangle$$
Then, from $(7)$, we have $$ \begin{align} dW_t &= \sin(\Theta_t) d\cos(\Theta_t) - \cos(\Theta_t) d\sin(\Theta_t) \\ &= -\sin^2(\Theta_t)d\Theta_t - \frac{1}{2}\sin(\Theta_t)\cos(\Theta_t)\langle d\Theta_t,d\Theta_t\rangle -\cos^2(\Theta_t)d\Theta_t+\frac{1}{2}\sin(\Theta_t)\cos(\Theta_t)\langle d\Theta_t,d\Theta_t\rangle \\ &=d\Theta_t \tag{8} \end{align} $$
From $(8)$, we deduce easily that, with $c\in \Bbb R$ $$\Theta_t = W_t +c$$ Then $$X_t = \sin(W_t+c)$$ $$Y_t = \cos(W_t+c)$$
Q.E.D
At least part of what you have is correct: $dR_t = 0$. Unfortunately, I suspect that your equation for $d\Theta_t$ is wrong, but without seeing the work it would be hard to say why.
The key here is to note that since $dR_t = 0$, we know $R_t = R_0 = X_0^2 + Y_0^2 = 1$. The fact that $X_t^2 + Y_t^2 = 1$ suggests we should try guessing that the solution is $X_t = \cos(W_t)$, $Y_t = \sin(W_t)$. Then Ito's formula gives \begin{align*} dX_t &= -\sin(W_t) dW_t + \frac 12 \cos(W_t) dt = -Y_t dW_t + \frac 12 X_t dt \\ dY_t &= \cos(W_t) dW_t - \frac 12 \sin(W_t) dt = X_t dW_t - \frac 12 Y_t dt \end{align*} so this is indeed the solution.