I have been given the following definition of the fundamental solution

Let $L$ be a differential operator with $C^{\infty}$ coefficients. Then the solution to the equation $L(U_{\xi})=\delta_{\xi}$ is called a fundamental solution.

I wanted to try this on some example. I would like to find the fundamental solution of

$$L(u) = u''-c^{2} u \quad c \geq 0$$

In my lecture notes it is writen that If one searches fundamental solutions, which are represented through functions, then these must satisfy

$$\int_{\mathbb{R}} U(x,\xi)L^*(\phi) dx=\phi(\xi) \quad \forall \phi \in \mathcal{D}(\mathbb{R})$$

My working

First I calculated the adjoint differential operator and found that $\langle L(u),\psi \rangle= \langle u,L(\psi) \rangle$, thus $L$ is self adjoint.

How I understood it, I need to solve


which means

$$\frac{d^2}{dx^2} U(x,\xi)- c^2 U(x,\xi)=\delta_{\xi}$$

Let $H(x,\xi)$ denote the Heaviside Function, where the 'jump' exists at $\xi \in \mathbb{R}$. I know that $\frac{d}{dx}H(x,\xi)=\delta_{\xi}$, though I am not really sure how I can use this to solve the equation (or if I can use this). I don't know how to tackle this equation. I would be thankful if someone could help me.


I tried the following




I think the solution should be split in 3 parts, $x<\xi, x=\xi$ and $x>\xi$. If I solve the equation (1) for $x \neq \xi$, I get something like

$$U(x,\xi)= \begin{array}{cc} \{ & \begin{array}{cc} c_1e^{c(x-\xi)}+c_2 e^{-c(x-\xi)} & x < \xi \\ c_3e^{c(x-\xi)}+c_4 e^{-c(x-\xi)} & x > \xi \\ \end{array} \end{array} $$ Further I know that if $U(x,\xi )$ is a fundamental solution it must satisfy: Let $\phi$ be a test function, then


$$\int_{-\infty}^{\infty}U(x,\xi)(\phi''(x)-c^2 \phi(x))dx=\phi(x)$$

I assume there will be a 'jump' at $x=\xi$ but I am not really sure how to get there. Also I am not really sure how to calculate the coefficients $c_1,c_2,c_3,c_4$

I tried to split the integral (2) up in two parts. For $\epsilon>0$


$$\int_{-\infty}^{\xi-\epsilon }U(x,\xi)(\phi''(x)-c^2 \phi(x))dx+\int_{\xi+\epsilon}^{\infty}U(x,\xi)(\phi''(x)-c^2 \phi(x))dx$$

I tried to integrate by parts such that I get something like

$$(\text{something})+\int{\phi'(x)(\text{something else})}dx \tag a$$

for both integrals in (3), in the hope that, since (a) must equal $\phi(x)$, I could use this somehow to get the solution.

My actual result of integration by parts of (3) is:

\begin{align} &c_1e^{-c \epsilon}+c_2e^{c \epsilon} \phi(\xi - \epsilon)+(-c_1 c e^{-c \epsilon}+c_2 c e^{c \epsilon})\phi(\xi - \epsilon)\\ &+(-c_3e^{\epsilon c}-c_4^{-c \epsilon}) \phi(\xi+\epsilon)+(c_3 c e^{\epsilon c}-c_4 ce^{-\epsilon c})\phi(\xi + \epsilon)\\ &+\int_{\infty}^{\xi- \epsilon}\phi'(x)(c_1ce^{c(x- \xi)}(\xi-x+1)+c_2c^{-c \ (x - \xi)}(x- \xi-1))dx\\ &+\int_{\xi+\epsilon}^{\infty}\phi'(x)(c_3ce^{c(x-\xi)}(\xi - x +1)+c_4c^{e^{-c(x-\xi)}}(x-\xi-1)) \end{align}

  • 1
    $\begingroup$ You should probably read the wiki page on Greens functions, specifically the example given. $\endgroup$ Nov 17, 2021 at 15:57
  • $\begingroup$ @mattos Thanks, I tried to solve it this way. My solution would be $U(x,\xi)=k_1 e^{cx}+k_2e^{-cx}$, but when I try to check if $U''-c^2U=\delta_{\xi}$, I get as a result $U''-c^2U=0$ and not $\delta_{\xi}$ $\endgroup$
    – John.W
    Nov 17, 2021 at 16:15
  • $\begingroup$ Consider the even-simpler case of $c=0$ first. $\endgroup$ Nov 17, 2021 at 17:14
  • $\begingroup$ @paulgarrett I did that, for c=0 we have the equation $\frac{d^2}{dx^2}U(x,\xi)=\delta_{\xi}$, I can write $\frac{d^2}{dx^2}U(x,\xi)=\frac{d}{dx}H(x,\xi)$, thus by integrating I get $\frac{d}{dx}U(x,\xi)=H(x,\xi)$. If I now Integrate again I get $U(x,\xi)=xH(x,\xi)$. But how do I use this for the general case? $\endgroup$
    – John.W
    Nov 17, 2021 at 17:20
  • 1
    $\begingroup$ @John.W, I'll write something out a little later, if no one else does... $\endgroup$ Nov 17, 2021 at 18:08

1 Answer 1


Yes, your edit is one way to approach $u''-c^2u=\delta$ (since constant coefficient differential operators are translation invariant, without loss of generality we can consider $\delta$ at $0$).

That is, quite rigorously, away_from$0$_, $\delta$ is locally $0$. The idea that distributions are "local", and do/can have pointwise values (etc.) away from "singular support" is standard, perhaps unglamorous, but does rigorize the otherwise possibly-too-cavalier talk about values of $\delta$. Note that we cannot talk about any "value" of $\delta$ at $0$, because that point is in its singular support...

So, yes, as you realized, you want to patch together a solution to the left, and a solution to the right, to get the $\delta$. As in general with linear equations, there is great ambiguity in the constants, since we can always adjust a solution to the inhomogeneous equation by solutions to the homogeneous one.

(An analogous problem of construction of Green's functions for Sturm-Liouville problems on finite intervals uses the additional boundary-value conditions to eliminate that ambiguity.)

In your example, and with $c>0$, we could attempt to use the best-behaved solutions in $x<0$ and $x>0$, namely, the exponential functions that do not blow up as $x\to -\infty$ and $x\to+\infty$: $e^{cx}$ and $e^{-cx}$, respectively. The symmetry of the problem under $x\to -x$ implies that we can take a linear combination $C\cdot e^{c|x|}$ for some constant.

Indeed, to avoid having the first derivative already have a jump, introducing a $\delta$, we want the limit at $0$ from the left to match that from the right. The jump in first derivatives will produce the $\delta$.

Another way to think about this is to write $D$ for $d/dx$, and the equation as $(D-c)(D+c)u=\delta$. Thus, we can first solve $(D-c)v=\delta$, and then solve $(D+c)u=v$. The equation $(D-c)v=\delta$ can be simplified by letting $v=e^{cx}w$, turning it into $e^{cx}w'=\delta$. We can multiply through by $e^{-cx}$ (compactly supported distributions can rigorously be multiplied by smooth functions), to get $w'=\delta$. So $w$ could be the Heaviside function $H$, and $v=e^{cx}H$. And so on.


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