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matematikuvol
Jan5-12, 04:28 AM
If we have Green function

g(x,s)=exp[-\int^x_s p(z)dz] we want to think about that as distribution so we multiply it with Heaviside step function

g(x,s)=H(x-s)exp[-\int^x_s p(z)dz]

Why we can just multiply it with step function and tell that the function is the same. Tnx for the answer.

matematikuvol
Jan5-12, 04:29 PM
To be more precise.

If I say solution to eq is

u(x)=\int^{x}_0g(x,s)f(s)ds

where g=e^{-\int^x_yp(z)dz}

Then if I define

g=H(x-s)e^{-\int^x_sp(z)dz}

is then

u(x)=\int^{\infty}_0g(x,s)f(s)ds

and what is Green function this g=e^{-\int^x_yp(z)dz} or this g=H(x-s)e^{-\int^x_sp(z)dz}?

matematikuvol
Jan7-12, 05:14 PM
Can you help me?