# How to find a fundamental solution of $L(f)=f''-c^2f$

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

$$L(U(x,\xi))=\delta_{\xi}$$

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.

Edit

I tried the following

$$L(U(x,\xi))=\delta_{\xi}$$

1.

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

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

2.

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

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 $$\epsilon>0$$

3.

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

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}

• You should probably read the wiki page on Greens functions, specifically the example given. Nov 17, 2021 at 15:57
• @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 $\delta_{\xi}$ Nov 17, 2021 at 16:15
• Consider the even-simpler case of $c=0$ first. Nov 17, 2021 at 17:14
• @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? Nov 17, 2021 at 17:20
• @John.W, I'll write something out a little later, if no one else does... Nov 17, 2021 at 18:08

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.