A simple 2nd order ODE using Green's functions

[quasar_4]

Homework Statement

Solve $$\frac{d^2 u(x)}{dx^2}=x(1-x)$$ subject to the homogeneous boundary conditions u(0)=0=u(1), using Green's functions.

Homework Equations

Green's functions...

The Attempt at a Solution

There are three parts:

i - homogeneous eqn:

$$\frac{d^2 u(x)}{dx^2}=0$$ which has the obvious solutions a, bx. Given the BCs these can be modified to be u(x) = x or u(x) = 1-x; the first satisfies the first BC and the second satisfies the second.

ii - construction of Green's function:

$$G(x,z) = \left\{ \begin{array}{rcl} c_1(z) x, \hspace{3mm} 0 < z < x \\ c_2(z) (1-x), \hspace{3mm} x < z < 1 \end{array}\right$$

iii - application of continuity requirements

We demand that:

$$c_1(z) z - c_2(z) (1-z) = 0$$
and
$$c_1(z) + c_2(z) = -1$$

which tells us after some algebra that

$$c_1(z) = -(1+z), c_2(z) = z$$. Thus

$$G(x,z) = \left\{ \begin{array}{rcl} -(1+z) x, 0 < z < x \\ z (1-x), x < z < 1 \end{array}\right$$.

So we should just be able to solve for u(x) problem-free:

$$u(x) = \int_0^1 G(x,z) z(1-z) dz = \int_0^x -(1+z) x z(1-z) dz + \int_x^1 z (1-x) z(1-z) dz = (11/12)x^4-x^2-(1/12)x-(1/4)x^5+1/12-(1/3)x^3$$

But this u(x) does not satisfy the original ODE!

What am I doing wrong? I suspect that I approached part i wrong. It seemed suspicious - I thought that both my two homogeneous solns had to individually satisfy BOTH BC's, and each only satisfies one - is that where I'm going wrong?

 

[HallsofIvy]

Homework Statement

Solve $$\frac{d^2 u(x)}{dx^2}=x(1-x)$$ subject to the homogeneous boundary conditions u(0)=0=u(1), using Green's functions.

Homework Equations

Green's functions...

The Attempt at a Solution

There are three parts:

i - homogeneous eqn:

$$\frac{d^2 u(x)}{dx^2}=0$$ which has the obvious solutions a, bx. Given the BCs these can be modified to be u(x) = x or u(x) = 1-x; the first satisfies the first BC and the second satisfies the second.

ii - construction of Green's function:

$$G(x,z) = \left\{ \begin{array}{rcl} c_1(z) x, \hspace{3mm} 0 < z < x \\ c_2(z) (1-x), \hspace{3mm} x < z < 1 \end{array}\right$$

You have these reversed. Since the first satisfies G(0,z)= 0, you must have 0< x< z. Since the second satisfies G(1,z)= 0, it must be for z< x< 1.

iii - application of continuity requirements

We demand that:

$$c_1(z) z - c_2(z) (1-z) = 0$$
and
$$c_1(z) + c_2(z) = -1$$

which tells us after some algebra that

$$c_1(z) = -(1+z), c_2(z) = z$$. Thus

$$G(x,z) = \left\{ \begin{array}{rcl} -(1+z) x, 0 < z < x \\ z (1-x), x < z < 1 \end{array}\right$$.

So we should just be able to solve for u(x) problem-free:

$$u(x) = \int_0^1 G(x,z) z(1-z) dz = \int_0^x -(1+z) x z(1-z) dz + \int_x^1 z (1-x) z(1-z) dz = (11/12)x^4-x^2-(1/12)x-(1/4)x^5+1/12-(1/3)x^3$$

But this u(x) does not satisfy the original ODE!

What am I doing wrong? I suspect that I approached part i wrong. It seemed suspicious - I thought that both my two homogeneous solns had to individually satisfy BOTH BC's, and each only satisfies one - is that where I'm going wrong?