A simple 2nd order ODE using Green's functions

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Green's functions...

3. The attempt at a solution

There are three parts:

i - homogeneous eqn:

[tex] \frac{d^2 u(x)}{dx^2}=0 [/tex] 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:

[tex] 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 [/tex]

iii - application of continuity requirements

We demand that:

[tex] c_1(z) z - c_2(z) (1-z) = 0 [/tex]
and
[tex] c_1(z) + c_2(z) = -1 [/tex]

which tells us after some algebra that

[tex] c_1(z) = -(1+z), c_2(z) = z[/tex]. Thus

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

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

[tex] 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 [/tex]

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? :confused:
 

HallsofIvy

Science Advisor
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1. The problem statement, all variables and given/known data

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

2. Relevant equations

Green's functions...

3. The attempt at a solution

There are three parts:

i - homogeneous eqn:

[tex] \frac{d^2 u(x)}{dx^2}=0 [/tex] 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:

[tex] 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 [/tex]
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:

[tex] c_1(z) z - c_2(z) (1-z) = 0 [/tex]
and
[tex] c_1(z) + c_2(z) = -1 [/tex]

which tells us after some algebra that

[tex] c_1(z) = -(1+z), c_2(z) = z[/tex]. Thus

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

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

[tex] 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 [/tex]

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? :confused:
 

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