A simple 2nd order ODE using Green's functions

In summary, the problem is to solve the differential equation \frac{d^2 u(x)}{dx^2}=x(1-x) with homogeneous boundary conditions u(0)=0=u(1) using Green's functions. The solution involves constructing Green's function, satisfying continuity requirements and solving for u(x). However, there is an error in the construction of Green's function which leads to a solution that does not satisfy the original ODE. The error is due to the incorrect approach in solving the homogeneous equation.
  • #1
quasar_4
290
0

Homework Statement



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.

Homework Equations



Green's functions...

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:
 
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  • #2
quasar_4 said:

Homework Statement



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.

Homework Equations



Green's functions...

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:
 

Related to A simple 2nd order ODE using Green's functions

1. What is a simple 2nd order ODE?

A simple 2nd order ODE (ordinary differential equation) is an equation that involves a function and its first and second derivatives. It can be written in the form: y'' + P(x)y' + Q(x)y = f(x), where P(x) and Q(x) are functions of x, and f(x) is a known function.

2. What is a Green's function?

A Green's function is a mathematical tool used to solve linear differential equations. It is defined as the solution to a specific type of boundary value problem, which involves a linear differential operator and a known forcing function. Green's functions can be used to find solutions to a wide range of differential equations.

3. How do you use Green's functions to solve a 2nd order ODE?

To use Green's functions to solve a 2nd order ODE, you first need to find the Green's function for the specific differential operator and boundary conditions involved in the equation. This can be done through a variety of methods, such as separation of variables or the method of variation of parameters. Once the Green's function is found, it can be used to find the particular solution to the ODE by integrating it with the given forcing function.

4. What are the advantages of using Green's functions to solve ODEs?

One advantage of using Green's functions is that they provide a systematic and efficient method for solving linear differential equations. They can also handle a wide range of boundary conditions and forcing functions. Additionally, Green's functions can be used to solve non-homogeneous equations, which may not have a general analytical solution.

5. Are there any limitations to using Green's functions to solve ODEs?

While Green's functions are a powerful tool for solving linear differential equations, they do have some limitations. They may not be suitable for solving non-linear or higher order ODEs. Additionally, finding the Green's function can be a complex and time-consuming process, especially for more complicated differential operators and boundary conditions. In some cases, it may be more efficient to use other methods for solving ODEs.

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