Consider the equation $y'' + 4y = f(t), y(0) = 1, y'(0) = 0$. Use the Laplace Transform to compute the Green’s function for this equation.

Question

Consider the equation $y'' + 4y = f(t), y(0) = 1, y'(0) = 0$. Use the Laplace Transform to compute the Green’s function for this equation.

$y'' + 4y = f(t) \rightarrow L\{y'' + 4y = f(t)\}=L\{f(t)\}$

$(s^2+4)Y-s=L\{f(t)\}$

then how we processed

Answer 1

You just want to have a function that satisfies zero boundary conditions and the differential equation $y^{\prime\prime}+4y=\delta(t-\tau)$. In Laplace transform domain this is $-(0)-s(0)+s^2Y(s)+4Y(s)=e^{-s\tau}$. Then $$Y(s)=\frac{e^{-s\tau}}{s^2+4}$$ The inverse transform is $$G(t,\tau)=\frac12\sin(2(t-\tau))u(t-\tau)$$ You need to add $\cos(2t)$ to the solution eventually to satisfy the nonzero boundary conditions. Check: $$\begin{align}y(t)&=\int_0^{\infty}\frac12\sin2(t-\tau)u(t-\tau)f(\tau)d\tau+\cos2t\\ &=\int_0^t\frac12\sin2(t-\tau)f(\tau)d\tau+\cos2t\end{align}$$ $$\begin{align}y^{\prime}(t)&=\frac12\sin2(t-t)f(t)+\int_0^t\cos2(t-\tau)f(\tau)d\tau-2\sin2t\\ &=\int_0^t\cos2(t-\tau)f(\tau)d\tau-2\sin2t\end{align}$$ $$\begin{align}y^{\prime\prime}(t)&=\cos2(t-t)f(t)-2\int_0^t\sin2(t-\tau)f(\tau)d\tau-4\cos2t\\ &=f(t)-2\int_0^t\sin2(t-\tau)f(\tau)d\tau-4\cos2t\\ &=f(t)-4y(t)\end{align}$$ And $$y(0)=\int_0^0\frac12\sin2(t-\tau)f(\tau)d\tau+\cos0=0+1=1$$ $$y^{\prime}(0)=\int_0^0\cos2(t-\tau)f(\tau)d\tau-2\sin0=0-0=0$$