Greens Function and Linear First Order Differential Equations
Given a linear differential operator $D$ and a function $\rho$, consider the differential equation
\begin{align}
D f(x) =\rho(x)
\end{align}
The Greens function $G(x, x_0)$ for this differential equation is defined by another differential equation, namely
\begin{align}
D G(x, x_0) = \delta(x-x_0)
\end{align}
where $\delta(x-x_0)$ is the Dirac delta function. The solution to the original differential equation is
\begin{align}
f(x)=\chi(x)+\int_{\mathbb{R}} G(x,x_0) \rho(x_0) dx_0
\end{align}
where $\chi(x)$ is a solution to the homogenous differential equation $D\chi(x)=0$.
We can prove that $f$ as given above is actually a solution to original differential equation simply by applying the differential operator $D$ to $f$: \begin{align} D f(x) & = D\left[\chi(x)+\int_{\mathbb{R}} G(x,x_0) \rho(x_0)\, \mathrm{d}x_0\right]\\ & = 0 + \int_{\mathbb{R}} D G(x,x_0) \rho(x_0)\, \mathrm{d}x_0\\ & = \int_{\mathbb{R}} \delta(x-x_0) \rho(x_0)\, \mathrm{d}x_0\\ & = \rho(x)\,. \end{align} Consider now the general linear first-order differential equation \begin{align} y' + a(x) y = b(x)\,. \end{align} Here we have the differential operator $D = \left[\frac{d}{dx} + a(x)\right]$, and the source term is $\rho(x)=b(x)$. To obtain a solution using the method above we must first determine the Greens function $G(x, x_0)$ by solving the equation $DG(x,x_0)=\delta(x-x_0)$: \begin{align} \left[\frac{d}{dx} + a(x)\right]G(x, x_0) &= \delta(x - x_0)\,. \end{align} We try out the following solution: \begin{align} G(x, x_0) = e^{A(x_0) - A(x)}\cdot\theta(x-x_0)\,, \end{align} in which $A'(x)=a(x)$, and $\theta(x-x_0)$ is the Heaviside step function. We show that this is indeed a correct Greens function by applying $D$: \begin{align} D G(x, x_0) &= \left[\frac{d}{dx} + a(x)\right]\left(e^{A(x_0) - A(x)}\cdot\theta(x - x_0)\right)\\ &= -a(x)\cdot e^{A(x_0) - A(x)}\cdot \theta(x-x_0) + e^{A(x_0) - A(x)}\cdot \delta(x-x_0)\\ & \qquad + a(x) e^{A(x_0) - A(x)}\theta(x-x_0)\\ & = e^{A(x_0) - A(x)}\delta(x-x_0)\\ &= \delta(x-x_0)\,, \end{align} where we used that $\theta'(x-x_0)=\delta(x-x_0)$, and the fact that $\delta(x-x_0)=0$ for $x\neq x_0$.
The function $\chi(x)$ is determined by solving the homogenous differential equation \begin{align} \left[\frac{d}{dx} + a(x)\right]\chi(x) &= 0 \end{align} which we recast as \begin{align} \chi'(x) &= -a(x)\chi(x)\,, \end{align} the solution to which is \begin{align} \chi(x) &= c_1\cdot e^{-A(x)}\,. \end{align} We are now ready to plug $\chi(x)$ and $G(x, x_0)$ into the formula for $f(x)$: \begin{align} f(x) &= \chi(x) + \int_{\mathbb{R}} G(x, x_0) \rho(x_0)\,\mathrm{d}x_0\\ &= c_1\cdot e^{-A(x)} + \int_{\mathbb{R}} e^{A(x_0) - A(x)}\cdot\theta(x-x_0)\cdot b(x_0)\,\mathrm{d}x_0\\ &= e^{-A(x)}\left(c_1 + \int\limits_{-\infty}^{x} e^{A(x_0)}\cdot b(x_0)\,\mathrm{d}x_0\right)\\ &= e^{-A(x)}\left(c_1 + \left[\int e^{A(x)} \cdot b(x)\,\mathrm{d}x + c_2\right] \right)\\ &= e^{-A(x)} \int e^{A(x)} \cdot b(x)\,\mathrm{d}x + c\cdot e^{-A(x)}\,. \end{align} This is the general formula for the solution to a linear first-order differential equation, derived using the Greens function method.
We can prove that $f$ as given above is actually a solution to original differential equation simply by applying the differential operator $D$ to $f$: \begin{align} D f(x) & = D\left[\chi(x)+\int_{\mathbb{R}} G(x,x_0) \rho(x_0)\, \mathrm{d}x_0\right]\\ & = 0 + \int_{\mathbb{R}} D G(x,x_0) \rho(x_0)\, \mathrm{d}x_0\\ & = \int_{\mathbb{R}} \delta(x-x_0) \rho(x_0)\, \mathrm{d}x_0\\ & = \rho(x)\,. \end{align} Consider now the general linear first-order differential equation \begin{align} y' + a(x) y = b(x)\,. \end{align} Here we have the differential operator $D = \left[\frac{d}{dx} + a(x)\right]$, and the source term is $\rho(x)=b(x)$. To obtain a solution using the method above we must first determine the Greens function $G(x, x_0)$ by solving the equation $DG(x,x_0)=\delta(x-x_0)$: \begin{align} \left[\frac{d}{dx} + a(x)\right]G(x, x_0) &= \delta(x - x_0)\,. \end{align} We try out the following solution: \begin{align} G(x, x_0) = e^{A(x_0) - A(x)}\cdot\theta(x-x_0)\,, \end{align} in which $A'(x)=a(x)$, and $\theta(x-x_0)$ is the Heaviside step function. We show that this is indeed a correct Greens function by applying $D$: \begin{align} D G(x, x_0) &= \left[\frac{d}{dx} + a(x)\right]\left(e^{A(x_0) - A(x)}\cdot\theta(x - x_0)\right)\\ &= -a(x)\cdot e^{A(x_0) - A(x)}\cdot \theta(x-x_0) + e^{A(x_0) - A(x)}\cdot \delta(x-x_0)\\ & \qquad + a(x) e^{A(x_0) - A(x)}\theta(x-x_0)\\ & = e^{A(x_0) - A(x)}\delta(x-x_0)\\ &= \delta(x-x_0)\,, \end{align} where we used that $\theta'(x-x_0)=\delta(x-x_0)$, and the fact that $\delta(x-x_0)=0$ for $x\neq x_0$.
The function $\chi(x)$ is determined by solving the homogenous differential equation \begin{align} \left[\frac{d}{dx} + a(x)\right]\chi(x) &= 0 \end{align} which we recast as \begin{align} \chi'(x) &= -a(x)\chi(x)\,, \end{align} the solution to which is \begin{align} \chi(x) &= c_1\cdot e^{-A(x)}\,. \end{align} We are now ready to plug $\chi(x)$ and $G(x, x_0)$ into the formula for $f(x)$: \begin{align} f(x) &= \chi(x) + \int_{\mathbb{R}} G(x, x_0) \rho(x_0)\,\mathrm{d}x_0\\ &= c_1\cdot e^{-A(x)} + \int_{\mathbb{R}} e^{A(x_0) - A(x)}\cdot\theta(x-x_0)\cdot b(x_0)\,\mathrm{d}x_0\\ &= e^{-A(x)}\left(c_1 + \int\limits_{-\infty}^{x} e^{A(x_0)}\cdot b(x_0)\,\mathrm{d}x_0\right)\\ &= e^{-A(x)}\left(c_1 + \left[\int e^{A(x)} \cdot b(x)\,\mathrm{d}x + c_2\right] \right)\\ &= e^{-A(x)} \int e^{A(x)} \cdot b(x)\,\mathrm{d}x + c\cdot e^{-A(x)}\,. \end{align} This is the general formula for the solution to a linear first-order differential equation, derived using the Greens function method.
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