# Finding the Fundamental Solution of the Operator $\frac{\partial}{\partial x}+c$

I am trying to find the Fundamental Solution for the Operator $\frac{\partial}{\partial x}+c$. I approach it by considering a radial function $v(r)$, and considering the ODE $v'(r)+c v=0$. This ODE has the solution $v=\frac{A}{e^{cr}}$, where $A\in\mathbb R$. Then,take an interval centered at $x$ with width $2\epsilon$. Then: $$1="\int_{x-\epsilon}^{x+\epsilon}\delta(r)"=\int_{x-\epsilon}^{x+\epsilon}u'(r) dr+\int_{x-\epsilon}^{x+\epsilon}cu(r)dr$$ I am unsure as to how this leads to a Fundamental Solution. If anyone could guide me in the right direction it would be appreciated.

• Your question isn't clear. Saying you are searching for the fundamental solution of the operator $\partial_{x} + c$ implies you are looking at 1D problems, in which case I don't know why you start integrating in $\mathbb{R^{3}}$. If you are looking at 3D problems, the operator would be $\partial_{x_{1}} + \partial_{x_{2}} + \partial_{x_{3}} + c$ no? – Mattos Dec 20 '17 at 4:36
• @Mattos yes sorry, will fix – Felicio Grande Dec 20 '17 at 4:37
• @Mattos, fixed but i'm not sure if im thinking about this question right regardless – Felicio Grande Dec 20 '17 at 4:56
• I'm not sure why you take a radial variable now when the problem is 1D i.e $r = x$. The fundamental solution is given by $L f = \delta (x)$ where $L$ is a linear operator. In your problem, $L = \partial_{x} + c$, so you need to solve $$\partial_{x} f + cf = \delta(x) \implies f' + cf = \delta (x) \implies (e^{cx} f)' = \delta(x) e^{cx} \implies f = e^{-cx} \int \delta(x) e^{cx} dx$$ Also, as far as I'm aware, the integral of the delta function only equals 1 when integration is done over $\mathbb{R}$, not $2 \epsilon$. – Mattos Dec 20 '17 at 10:55
• @Mattos so if the delta function was concentrated at $0$, the function would just be $f=e^{-cx}$? – Felicio Grande Dec 20 '17 at 19:14