Proving that $\lVert u \rVert_{L^2} \leq Ce^{-\nu t}$ for certain pde Given that $$\frac{\partial u}{\partial t}+\sin(y)\frac{\partial u}{\partial x}=\nu\Bigl(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\Bigr)$$
With the following periodic boundary conditions:
$$u(-\pi,y,t)=u(\pi,y,t) \\ u(x,-\pi,t)=u(x,\pi,t) \\u_x(-\pi,y,t)=u_x(\pi,y,t)\\
u_y(x,-\pi,t)=u_y(x,\pi,t)\\
u(x,y,0)=F(x,y)$$
Prove that $$\lVert u \rVert_{L^2} \leq Ce^{-\nu t}$$
I have used the finite Fourier transform to get that 
$$\frac{du_{mn}}{dt}=-\nu (n^2+m^2)u_{mn} -\int_{-\pi}^{\pi}\sin(y) u_ne^{-imy}dy$$
Where $$u_{mn}=\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} u(x,y,t)e^{-imx} *e^{-iny} dxdy$$
Second I tried Energy method Multiply by u and then integrate, still I didn't get the required result.
How to get the required result ?
Any Hint ?
 A: We have $$\frac{\partial u}{\partial t}+\sin(y)\frac{\partial u}{\partial x}=\nu\Bigl(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\Bigr) \ \ *$$
Multiplying * with u and integrating 
$$\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} u \frac{\partial u}{\partial t} dx\ dy+ \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} u \sin(y)\frac{\partial u}{\partial x} dx \ dy=\int_{-\pi}^{\pi} \int_{-\pi}^{\pi}\nu u \Bigl(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\Bigr)dx \ dy $$
Now integrating by parts and using the boundary conditions we get :
$$\frac{1}{2}\frac{d}{dt} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} u^2 \ dx \ dy =-\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} (\frac{\partial u}{\partial x})^2 +(\frac{\partial u}{\partial y})^2 dx \ dy$$
Or Equivalently $$\frac{1}{2}\frac{d}{dt}\lVert u \rVert_{L^2}^2 =-\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} |\nabla u|^2 dx \ dy$$
Now using Poincare's inequality we get $$\frac{1}{2}\frac{d}{dt}\lVert u \rVert_{L^2}^2 \leq -\nu C \lVert u \rVert_{L^2}^2$$
Now Let $z=e^{2 \nu Ct} \lVert u \rVert_{L^2}^2$
Which implies that $$\frac{dz}{dt} \leq 0$$
So $z(t) \leq z(0)$
Hence the result $$\lVert u \rVert_{L^2}^2 \leq C_1 e^{- \nu C t}$$
