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} <br /> c_1(z) x, \hspace{3mm} 0 &lt; z &lt; x \\<br /> c_2(z) (1-x), \hspace{3mm} x &lt; z &lt; 1<br /> \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} <br /> -(1+z) x, 0 &lt; z &lt; x \\<br /> z (1-x), x &lt; z &lt; 1<br /> \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} <br /> c_1(z) x, \hspace{3mm} 0 &lt; z &lt; x \\<br /> c_2(z) (1-x), \hspace{3mm} x &lt; z &lt; 1<br /> \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} <br /> -(1+z) x, 0 &lt; z &lt; x \\<br /> z (1-x), x &lt; z &lt; 1<br /> \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?