# How to get general solution via Green's function?

• psie
psie
Homework Statement
Find a fundamental solution of the equation ##x''-x=0##. Solve using this the equation ##x''-x=e^t\sin{t}##.
Relevant Equations
The definition of a fundamental solution is given in theorem below.
I'll start with a characterization of the Green's function as a fundamental solution to a differential operator. This theorem is given in Ordinary Differential Equations by Andersson and Böiers.

Theorem. Let $$L(t,\lambda)=\lambda^n+a_{n-1}(t)\lambda^{n-1}+\ldots+a_1(t)\lambda+a_0(t)\quad\text{and }D=\frac{d}{dt}.\tag1$$ Denote by ##E(t,\tau)## the uniquely determined solution ##u(t)## of the initial value problem
\begin{align}
&L(t,D)u=0 \tag2\\
\end{align}
Then
$$y(t)=\int_{t_0}^t E(t,\tau)g(\tau)d\tau\tag4$$
is the solution of the problem
\begin{align}
&L(t,D)y=g(t) \tag5\\
&y(t_0)=y'(t_0)=\ldots=y^{(n-1)}(t_0)=0. \tag6
\end{align}
##E(t,\tau)## is known as the fundamental solution to the differential operator ##L(t,D)##, also known as Green's function.

Now, consider the equation ##x''-x=e^t\sin{t}##. The homogeneous solution is ##x_h(t)=Ae^t+Be^{-t}##. To obtain ##E(t,\tau)##, note that ##x_h'(t)=Ae^t-Be^{-t}##. According to ##(3)##, ##x_h(\tau)=x_h'(\tau)=0##, so we have two equations and two unknowns. Solving this system, we obtain ##A=\frac{e^{-\tau}}{2}## and ##B=-\frac{e^\tau}{2}##, so $$E(t,\tau)=\frac12(e^{t-\tau}-e^{\tau-t})=\sinh{(t-\tau)}.$$

Here is where I'm stuck. ##(4)## seems to depend on an initial value ##t_0##, which I don't have. Moreover, I'm a little unsure whether or not ##(4)## is the general solution to ##L(t,D)y=g(t)## or simply a particular solution. I'm looking to obtain the general solution of ##x''-x=e^t\sin{t}## using Green's function, i.e. ##E(t,\tau)##. Grateful for any help.

Let's begin by rewriting your general theorem as it applies to your specific ##2^{nd}##-order ODE, using the same equation numbers as in the theorem. Your Green's function ##E\left(t,\tau\right)## must satisfy:
$$\frac{d^{2}E\left(t,\tau\right)}{dt^{2}}-E\left(t,\tau\right)=0\tag{2}$$with
$$E\left(\tau,\tau\right)=0,\quad\left.\frac{dE\left(t,\tau\right)}{dt}\right|_{t=\tau}=1\tag{3}$$In your post, you solve this system for the unique result:$$E\left(t,\tau\right)=\sinh\left(t-\tau\right)$$which is clearly correct. Turning then to eq.##(4)## of the theorem, you write:
psie said:
Here is where I'm stuck. ##(4)## seems to depend on an initial value ##t_{0}##, which I don't have.
True, we don't have a specific value for ##t_{0}##, just like we don't have a specific value for ##t##, so we just carry them both along. We make this manifest by altering the left of eq.##(4)## to make it clear that the ultimate solution ##x## can depend on both of the variables ##t## and ##t_{0}##:
$$x\left(t,t_{0}\right)=\intop_{t_{0}}^{t}E\left(t,\tau\right)g\left(\tau\right)d\tau=\intop_{t_{0}}^{t}\sinh\left(t-\tau\right)e^{\tau}\sin\left(\tau\right)d\tau\tag{4}$$So now you've arrived at the most tedious step: performing the integral on the right of ##(4)##. It's elementary, in that it boils down to integrating products of exponentials, but the final answer is a sum over several terms involving trig functions and exponentials. (I "cheated" and used Wolfram Mathematica!). Once you've derived the explicit expression for ##x(t,t_{0})##, you can directly verify that it does indeed solve the last two equations of the theorem:
$$\frac{d^{2}x\left(t,t_{0}\right)}{dt^{2}}-x\left(t,t_{0}\right)=g\left(t\right)=e^{t}\sin\left(t\right)\tag{5}$$and:
$$x\left(t_{0},t_{0}\right)=\left.\frac{dx\left(t,t_{0}\right)}{dt}\right|_{t=t_{0}}=0\tag{6}$$You next note:
psie said:
I'm a little unsure whether or not ##(4)## is the general solution...
Well, the general solution ##x## to a ##2^{nd}##-order ODE must involve two arbitrary constants of integration, which can be used to set arbitrary initial conditions on ##x,x'## at a particular point. So solution ##(4)## cannot be general since by eq.##(6)## it has the very specific initial values ##x=x'=0## at ##t=t_{0}##. But this is easily remedied: simply add to ##x## your arbitrary solution ##x_{h}(t)## of the homogeneous equation ##x_{h}''-x_{h}=0## to get the general solution ##x_{g}##:$$x_{g}\left(t,t_{0}\right)\equiv x\left(t,t_{0}\right)+x_{h}\left(t\right)=x\left(t,t_{0}\right)+c_{1}e^{t}+c_{2}e^{-t}\qquad(c_{1},c_{2}\text{ arbitrary constants})$$##x_{g}## still satisfies the inhomogeneous equation ##(5)##, but the initial conditions ##(6)## generalize to:$$x_{g}\left(t_{0},t_{0}\right)=c_{1}e^{t_{0}}+c_{2}e^{-t_{0}},\quad\left.\frac{dx_{g}\left(t,t_{0}\right)}{dt}\right|_{t=t_{0}}=c_{1}e^{t_{0}}-c_{2}e^{-t_{0}}$$Evidently, by appropriately choosing ##c_{1},c_{2}## you have the freedom to set any desired initial conditions on ##x_G## and ##x_G'## at ##t=t_{0}##, as is required for the general solution. Hence, you can express the general solution of your problem in terms of the Green's function as:$$x_{g}\left(t,t_{0}\right)=\intop_{t_{0}}^{t}\sinh\left(t-\tau\right)e^{\tau}\sin\left(\tau\right)d\tau+c_{1}e^{t}+c_{2}e^{-t}$$

Last edited:
DeBangis21, psie and topsquark

## What is a Green's function?

A Green's function is a type of solution used to solve inhomogeneous differential equations subject to specific boundary conditions. It acts as an impulse response for the system described by the differential equation, allowing the construction of the general solution from the inhomogeneous term and the boundary conditions.

## How do you construct a Green's function for a given differential operator?

To construct a Green's function for a given differential operator, you need to solve the differential equation $$\mathcal{L}G(x, x') = \delta(x - x')$$, where $$\mathcal{L}$$ is the differential operator and $$\delta$$ is the Dirac delta function. The solution $$G(x, x')$$ must also satisfy the boundary conditions of the original problem.

## How is the Green's function used to find the general solution of an inhomogeneous differential equation?

Once the Green's function $$G(x, x')$$ is known, the general solution to the inhomogeneous differential equation $$\mathcal{L}u(x) = f(x)$$ can be found using the integral $$u(x) = \int G(x, x') f(x') \, dx'$$. This integral represents the convolution of the Green's function with the inhomogeneous term $$f(x)$$.

## What are the boundary conditions, and how do they affect the Green's function?

Boundary conditions are constraints that the solution to a differential equation must satisfy at the boundaries of the domain. These conditions affect the form of the Green's function because $$G(x, x')$$ must also satisfy these boundary conditions. Different types of boundary conditions (Dirichlet, Neumann, or mixed) will lead to different forms of the Green's function.

## Can Green's functions be used for non-linear differential equations?

Green's functions are primarily used for linear differential equations. For non-linear differential equations, the superposition principle does not hold, making the concept of a Green's function inapplicable in the straightforward manner used for linear equations. However, in some cases, linearization techniques or perturbative methods can be used to approximate the solutions of non-linear problems using Green's functions.

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