Green functions

1. Jan 5, 2012

matematikuvol

Eq

$$u'(x)+p(x)u=f(x)$$

with initial condition $$u(0)=0$$

It's homogenous solution is

$$u_h=Ce^{-\int^x_0 p(s)ds}$$

Complete solution

$$u(x)=e^{-\int^x_0 p(s)ds}\int^x_0f(s)e^{\int^s_0 p(z)dz}ds=\int^x_0 f(s)g(x,s)ds$$

where $$g(x,s)=e^{-\int^{x}_s p(\xi)d \xi }$$

I didn't see that last step from here $$u(x)=e^{-\int^x_0 p(s)ds}\int^x_0f(s)e^{\int^s_0 p(z)dz}ds$$ to $$\int^x_0 f(s)g(x,s)ds$$ here.

Can you explain me this?

2. Jan 5, 2012

CompuChip

I'll give you some clues, and leave the details up to you to figure out (if you don't manage, feel free to ask again).

You can take the prefactor into the integral, as it is constant w.r.t. s:
$$\int_0^x f(s) \exp\left\{ \int_0^s p(z) dz - \int_0^x p(z) dz \right\} \, ds$$
(note I renamed the dummy integration variable to z).

Now you can rewrite the integral in the exponent to
$$\int_0^s p(z) dz - \int_0^x p(z) dz = \int_x^s p(z) dz$$
and just call the whole thing g(x, s).

3. Jan 5, 2012

matematikuvol

Tnx. Just to ask you how I get this boundaries in integration?

$$u_h'+p(x)u_h=0$$

$$\frac{du_h}{dx}=-p(x)u_h(x)$$

$$u_h=Ce^{-\int p(x)dx}$$

So

$$u(x)=C(x)e^{-\int p(x)dx}$$

$$u'=C'(x)e^{-\int p(x)dx}-pu$$

When I put this to Eq

$$u'+p(x)u=f(x)$$

I get

$$C'(x)e^{-\int p(x)dx}-pu+pu=f$$

and

$$C(x)=\int f(x)e^{\int p(x)dx}dx + C'$$

So solution is

$$u(x)=(\int f(x)e^{\int p(x)dx}dx + C')e^{-\int p(x)dx}$$

How to get integration border $$[a,b]$$, $$\int^{b}_a$$ in integrals. Tnx for your help.

4. Jan 7, 2012

matematikuvol

Can you help me?