Construct the Green s function for the equation

Question

Construct the Green s function for the equation y^''+ 2y^'+2y=0 Which boundary conditions y(0)=0 , y(π/2)=0 Is this Green s function symmetric? What is the Green s function, if the differential equation is e^2x y^''+2e^2x y^'+2e^2x y=0 Is this Green s function symmetric?

Answer 1

1. Solve the characteristic equation $\lambda^2+2\lambda+2=0$
2. Write down the general solution $y=e^{-x}(A\cos x+B\sin x)$.
3. Find a solution $u$ such that $u(a)=0$ and $u'(a)=1$, where $a$ is the pole of Green's function. Knowing the phase form of 2 will help: $u(x)=e^{a-x}\sin(x-a)$.
4. By virtue of 3, the function $u(x)H(x-a)$, where $H$ is the Heaviside function, produces $\delta_a$ when plugged into your equation.
5. Find a solution $y$ such that $y(x)+u(x)H (x-a)$ satisfies the boundary condition. This amounts to asking $y(0)=0$ and $y(\pi/2)=-e^{a-\pi/2} \cos a$. Hence, $y(x)=-e^{a-x}\sin x\cos a$.
6. You have Green's function: $$g(x,a) = y(x)+u(x)H (x-a)= e^{a-x}(-\sin x\cos a+\sin(x-a) H(x-a)) $$
7. Is $g$ symmetric? No, of course not. Whatever happens with the trigonometric part, the exponential $e^{a-x}$ is not going to magically turn into $e^{x-a}$. You may want to observe that the first derivative $y'$ in the ODE is responsible for the presence of exponential. The first derivative operator is not symmetric: it distinguishes left and right directions.

Your other equation is just the same as the first one, with irrelevant factor of $e^{2x}$.

Answer 2

∴ G(x ,η)= {( e^(η-x) cosη sinx 0≤x ≤η e^(η-x) sinη cosx η≤x ≤π/2 )┤ This is not symmetric due to the trigonometric part, the exponential e^(η-x) is not going to turn into e^(x-η)