Given the system of ODE $$x_1'=x_2$$ $$x_2'=-a \sin x_1- b x_2$$ Where $a,b >0$ real numbers. Using the Lyapunov function $$v(x_1,x_2)=\int_0^{x_1} a \sin{s}\,ds+\frac{1}{2} X^TPX$$ where $X=(x_1,x_2)$ and P some matrix, prove the system is asymptotically stable.
I need to find an appropriate matrix $P$ so that $v$ satisfies $v(0,0)=0$, $v(x_1,x_2)>0$ and $\frac{dv(0,0)}{dt}=0$, $\frac{dv}{dt}<0$,
I tried many values for $P$ but couldn't couldn't come up with anything. Can you help me prove this?
Let $$ v=Ax_1^2+2Bx_1x_2+Cx_2^2+a\int_0^{x_1}\sin sds. $$ Then \begin{eqnarray} \frac{dv}{dt}&=&2Ax_1x_1'+2B(x_1'x_2+x_1x_2')+2Cx_2x_2'+a\sin x_1\cdot x_1'\\ &=&2Ax_1x_2+2B(x_2^2-ax_1\sin x_1-bx_1x_2)+2C(-ax_2\sin x_1-bx_2^2)+ax_2\sin x_1\\ &=&2(A-bB)x_1x_2+a(1-2C)x_2\sin x_1-2bBx_1\sin x_1+2(B-bC)x_2^2. \end{eqnarray} Let $$ A-2bB=0, 1-2C=0$$ which implies $A=2bB,C=\frac12$. The positive definiteness of $Ax_1^2+2Bx_1x_2+Cx_2^2$ requires $$ B^2-AC<0 $$ or $$B^2-bB<0 \text{ or }B<b.$$ Choose $B=\frac14b, A=\frac12b^2$ and then $$ \frac{dv}{dt}=-\frac12b^2x_1\sin x_1-\frac14bx_2^2\le0. $$