Green Function for Second Order Differential Equation with Initial Value Problem

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In summary, the student is trying to find a "green's function" for a problem with an equation of the form d^2X / dt^2 + 2 dX/dt + (1+k^2)X. The student has found that the general solution to the homogeneous equation is e^t(C_1cos(kt)+ C_2sin(kt)) which satisfies the first boundary condition and the "jump" condition on the derivative. However, since the coefficients A and B are both zero, the "green's function" is identically zero for all 0\le t\le \tau.
  • #1
mwg
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Homework Statement


need to find the green function for

d^2X / dt^2 + 2 dX/dt +(1+k^2)X= f(t)

with IV X(0)=dX/dt (=0) = 0

Homework Equations



eq. attached

The Attempt at a Solution


attached

whenever I try to solve the homogonous part, I get 0 due to the IV.
guessing my strategy is wrong.

what should it be??
 

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  • #2
A "Green's function", [itex]G(t, \tau)[/itex], for a problem like this (the coefficient of the second derivative is 1) must satisfy several properties:
1) It must satisfy the homogenous d.e. in t for every [itex]\tau[/itex].
2) It must satisfy the boundary conditions.
3) It must be continuous.
4) The "jump" in the first derivative at [itex]t= \tau[/itex] must be one.

Here, the general solution to the homogeneous equation is
[tex]e^t(C_1cos(kt)+ C_2sin(kt))[/tex]
which means that the Green's function must be of the form
[tex]G(t, \tau)= \begin{pmatrix}e^t(Acos(kt)+ Bsin(kt) & 0\le t\le \tau \\ e^t(Ccos(kt)+ Dsin(kt) & \tau\le t\le 1\end{pmatrix}[/tex]

To satisfy the first boundary condition, since [itex]0\le\tau[/itex], we must have
[tex]G(0,\tau)= A= 0[/tex]
To satisfy the second boundary conditon, we must have
[tex]G_t(0,\tau)= A+ B= 0[/tex]
which, since A= 0, give B= 0. That is, [itex]G(t, \tau)[/itex] is identically 0 for all [itex]0\le t\le \tau[/itex].

But that does NOT say that [itex]G(t,\tau)[/itex] must be 0 for [itex]\tau\le t\le 1[/itex].

The continuity condition gives that
[tex]\lim_{t\to\tau^+}G(t,\tau)= e^{\tau}(C cos(k\tau)+ Dsin(k\tau))= \lim_{t\to\tau^-} G(t,\tau)= 0[/tex]
so we must have
[tex]e^{\tau}(C cos(k\tau)+ D sin(k\tau))= 0[/tex]
which leads to
[tex]D= -\frac{cos(k\tau)}{sin(k\tau)}C[/tex]

The "jump" condition on the derivative says that
[tex]\lim_{t\to\tau^-}G_t(t,\tau)- \lim_{t\to\tau^+}G_t(t,\tau)= 1[/tex]
which gives
[tex]e^\tau(Ck cos(kt)+ Dk sin(kt))+ e^\tau(-Ck sin(kt)+ Dk cos(kt))= 1[/tex]

those two equations give non-zero values for C and D.
 
  • #3
Thank you so much!

one more question though, do I set sum lambda G = f(x) and solve? (sorry, couldn't figure the Latex here)
 

1. What is a Green's function for an equation?

A Green's function for an equation is a mathematical function that is used to solve a particular type of differential equation known as a boundary value problem. It is also sometimes referred to as the fundamental solution to the differential equation.

2. How is a Green's function used in scientific research?

A Green's function is used in scientific research to solve boundary value problems in a variety of fields, such as physics, engineering, and mathematics. It allows scientists to find solutions to complex equations that would otherwise be difficult or impossible to solve analytically.

3. How is a Green's function related to the concept of causality?

In the context of differential equations, a Green's function is often seen as a representation of causality. This means that the solution to a differential equation can be thought of as a response to an input, with the Green's function representing the influence of the input on the solution.

4. Can a Green's function be used to solve any type of differential equation?

No, a Green's function is specifically designed to solve boundary value problems, which are a type of differential equation with specific conditions at the boundaries of the problem. It cannot be used to solve other types of differential equations, such as initial value problems.

5. Are there any limitations to using a Green's function to solve equations?

While a Green's function can be a powerful tool for solving boundary value problems, it does have some limitations. It may not always be possible to find a closed-form solution using a Green's function, and in some cases, numerical methods may need to be used instead. Additionally, the use of a Green's function may not be appropriate for all types of differential equations.

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