- #1

#### member 428835

Hi PF!

The ODE $$g''(x) + (1-k^2)g(x) = f(x)\\ g(0) = y(\pi/3) = 0$$

where ##f(x)## is a forcing function and ##k \in \mathbb N## is a constant has a Green's function via variation of parameters as

$$

G_L = \frac{L(y)R(x)}{W} : 0<x<y<\pi/3\\

G_R = \frac{L(x)R(y)}{W} : 0<y<x<\pi/3

$$

with solutions $$L(x) = \frac{\sin(\beta x)}{\beta}\\

R(x) = \sin\beta(x-\pi/3)\\

W = \sin (\beta \pi/3)\\

\beta = \sqrt{1-k^2}$$

where ##L(x)## is the left sided solution and ##R(x)## is the right sided solution (obviously). I can verify this Green's function is correct for the given ODE. However, despite being correct, the inequalities look wrong. Why isn't the Green's function instead defined as

$$

G_L = \frac{L(y)R(x)}{W} : 0<y<x<\pi/3\\

G_R = \frac{L(x)R(y)}{W} : 0<x<y<\pi/3

$$

since ##L## is the solution at ##x=0## and ##R## is the solution at ##x=\pi/3##?

If We look at a similar problem, $$d_x^2 g = f(x)\\g(0)=g(1) = 0$$ we see its Green's function is $$G_L = l(x) r(y):0<x<y<\pi/3\\G_R=l(y)r(x):0<y<x<\pi/3$$

where $$l(x) = -x\\ r(x) = (1-x)$$ which makes sense, since ##l(0)=0## and ##r(1)=0##.

So why is there a difference (or am I not seeing something) in the two approaches?

The ODE $$g''(x) + (1-k^2)g(x) = f(x)\\ g(0) = y(\pi/3) = 0$$

where ##f(x)## is a forcing function and ##k \in \mathbb N## is a constant has a Green's function via variation of parameters as

$$

G_L = \frac{L(y)R(x)}{W} : 0<x<y<\pi/3\\

G_R = \frac{L(x)R(y)}{W} : 0<y<x<\pi/3

$$

with solutions $$L(x) = \frac{\sin(\beta x)}{\beta}\\

R(x) = \sin\beta(x-\pi/3)\\

W = \sin (\beta \pi/3)\\

\beta = \sqrt{1-k^2}$$

where ##L(x)## is the left sided solution and ##R(x)## is the right sided solution (obviously). I can verify this Green's function is correct for the given ODE. However, despite being correct, the inequalities look wrong. Why isn't the Green's function instead defined as

$$

G_L = \frac{L(y)R(x)}{W} : 0<y<x<\pi/3\\

G_R = \frac{L(x)R(y)}{W} : 0<x<y<\pi/3

$$

since ##L## is the solution at ##x=0## and ##R## is the solution at ##x=\pi/3##?

If We look at a similar problem, $$d_x^2 g = f(x)\\g(0)=g(1) = 0$$ we see its Green's function is $$G_L = l(x) r(y):0<x<y<\pi/3\\G_R=l(y)r(x):0<y<x<\pi/3$$

where $$l(x) = -x\\ r(x) = (1-x)$$ which makes sense, since ##l(0)=0## and ##r(1)=0##.

So why is there a difference (or am I not seeing something) in the two approaches?

Last edited by a moderator: