Suppose $v:\mathbb{R}^{2}\rightarrow\mathbb{R}$ solves $\Delta v=0$ on $\mathcal{B}_{1}(0)$. Let $\theta\in(0,2\pi)$ and define $M:\mathcal{B}_{1}(0)\rightarrow\mathcal{B}_{1}(0)$ by $M(x,y)=(x\cos\theta-y\sin\theta,\,x\sin\theta+y\cos\theta).$ Define $u:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by $u(x,y)=v(M(x,y)).$ Show that $u$ solves $\Delta u=0$ on $\mathcal{B}_{1}(0).$
I am sure we have to use the chain rule to show that $\Delta u=0.$ I have that
$$\partial_{x}u=\cos\theta\,\cdot\partial_{x}v+\sin\theta\,\cdot\partial_{y}v,$$
$$\partial^{2}_{x}u=\cos\theta\left[ \cos\theta\,\cdot\partial^{2}_{x}v+\sin\theta\,\cdot\partial^{2}_{xy}v \right]+\sin\theta\left[ \cos\theta\,\cdot\partial^{2}_{xy}v+\sin\theta\,\cdot\partial^2_{y}v \right],$$
$$\partial_{y}u=-\sin\theta\,\cdot\partial_{x}v+\cos\theta\,\cdot\partial_{y}v$$
$$\partial^{2}_{y}u=-\sin\theta\left[ -\sin \theta\,\cdot \partial^{2}_{x}v + \cos \theta\, \cdot \partial^{2}_{xy}v \right] + \cos \theta \left[ -\sin \theta\, \cdot \partial^{2}_{xy}v + \cos \theta\, \cdot \partial^2_{y}v \right].$$
From the above it follows that $\Delta u=0.$
Is the above computation correct? Any feedback is much appreciated.