# Green's function for a BVP

• Kate2010
In summary: I'll look into that.In summary, I solved the homogeneous equation, looking for 2 linearly independent solutions, and found A (constant) and exp(-x). I am struggling with the boundary conditions though.

## Homework Statement

Find a green's function G(x,t) for the BVP y'' + y' = f(x), y(0) = 0, y'(1) = 0.

## The Attempt at a Solution

I solved the homogeneous equation, looking for 2 linearly independent solutions, and found A (constant) and exp(-x). I am struggling with the boundary conditions though. My solution y1=A satisfies y1'(1) = 0 but I can't find a solution to satisfy y(0) = 0. If I were to find this, my method would be to write y(x) = c1(x)y1(x) + c2(x)y2(x), then find integral expressions for c1 and c2. However, as I can't find another linearly independent solution to the homogeneous equation do I need to use a different method?

Thank you.

y(x) = A + B e-x

It already consists of two linearly independent solutions. You just have to set the constants so they satisfy the boundary condition, as you usually do. In this case, you get A=-B, so the solution you want is

y(x) = A(1 - e-x)

Sorry if I am being dumb, but surely when I use y(x) = A(1-exp(-x)) and try to satisfy the boundary condition y'(1) = 0, I get y'(x) = Aexp(-x) so y'(1) = Aexp(-1) so A = 0?

After I posted, I thought that's what might be confusing you. The Green's function satisfies

$$y''(x)+y'(x) = -\delta(x-t)$$

So you actually have two solutions, u(x) for when x<t and v(x) for when x>t, where 0<t<1. The solution you found would be v(x)=c2. Since it's only valid for x>t, it only has to satisfy the boundary condition at x=1. Similarly, u(x)=c1(1-e-x) is valid for x<t, so it only has to satisfy the boundary condition at x=0. So you have

$$G(x,t) = \left\{\begin{array}{lr}c_1(1-e^{-x}) & \mathrm{when}~x<t \\ c_2 & \mathrm{when}~x>t\end{array}\right.$$

The idea now is to figure out what c1 and c2 need to be so that G(x,t) satisfies the requirements of a Green's function.

Do you have a textbook? It probably shows how to construct the Green's function once you have the two solutions u(x) and v(x).

Last edited:
Ah yes :) thanks!

## 1. What is a Green's function for a BVP?

A Green's function for a boundary value problem (BVP) is a mathematical tool used to solve a differential equation with specified boundary conditions. It represents the response of the system to a unit impulse at a specific point.

## 2. How is a Green's function used to solve a BVP?

The Green's function is used as a fundamental solution to the differential equation and is convolved with the inhomogeneous term to obtain the particular solution. The complete solution is then obtained by adding the homogeneous solution to the particular solution.

## 3. What are the properties of a Green's function for a BVP?

A Green's function for a BVP must satisfy the boundary conditions of the problem, be symmetric, and have a singularity at the same point as the inhomogeneous term. It also has the property of linearity, which means it can be scaled and added together to obtain a solution for a more complex problem.

## 4. How does a Green's function relate to the concept of a Green's theorem?

Green's theorem is a generalization of the fundamental theorem of calculus and relates line integrals to double integrals. In the context of a BVP, the Green's function can be seen as the kernel of the double integral in Green's theorem.

## 5. Can a Green's function be used to solve any type of BVP?

No, a Green's function can only be used for linear, constant coefficient differential equations with specified boundary conditions. It cannot be used for non-linear equations or equations with variable coefficients.