Fundamental solution of a differential operator

I'm trying to solve this question, but I don't know how to deal with it:

If we have $b=(b_1,,\dots,b_n)\in\mathbb{R}^n$ and $\beta\in\mathbb{R}$, prove that the differential operator $b_1\frac{\partial}{\partial x_1}+\dots b_n\frac{\partial}{\partial x_n}-\beta$ has the following distribution $T$ as fundamental solution: $$\left<T,\varphi\right>=\int_{0}^{\infty}\varphi(tb)e^{\beta t}\;dt$$

Thanks in advance for any help.

• Can you manage simple cases like $(b_1,...,b_n)=(1,0,...,0)$? And conclude from this to the general case?
– Vobo
Commented Jan 23, 2014 at 12:48

Let us call

$$L = -\beta + \sum_{i=1}^n b_i\frac{\partial}{\partial x_i};\qquad L^\ast = -\beta - \sum_{i=1}^n b_i\frac{\partial}{\partial x_i}.$$

That a distribution $T$ is a fundamental solution of $L$ means $LT = \delta$, where $\delta$ is the Dirac distribution. With the dual $L^\ast$ of $L$, since by definition

$$\langle LT,\varphi\rangle = \langle T, L^\ast \varphi\rangle,$$

that means $\langle T, L^\ast\varphi\rangle = \varphi(0).$

So you have to compute

$$-\int_0^\infty \left(\beta\varphi(tb) + \sum_{i=1}^n b_i\frac{\partial\varphi}{\partial x_i}(tb)\right)e^{\beta t}\,dt$$

and see that the result is $\varphi(0)$. Integration by parts after seeing that

$$\sum_{i=1}^n b_i\frac{\partial\varphi}{\partial x_i}(tb) = \frac{d}{dt} \varphi(tb)$$

looks like a very promising approach.

With regard to a point brought up by mkl314, the formula

$$\langle T,\varphi\rangle = \int_0^\infty \varphi(tb)e^{\beta t}\,dt$$

does not define a tempered distribution for $\beta > 0$, the integral generally does not converge for $\varphi \in \mathscr{S}(\mathbb{R}^n)$ then. When treating tempered distributions, the cases $\beta < 0$ and $\beta > 0$ must be distinguished ($\beta = 0$ fits both ways), and lead to different fundamental solutions of $L$. However, in $\mathscr{D}'(\mathbb{R}^n)$, no such distinction is necessary; since $\varphi \in \mathscr{D}(\mathbb{R}^n)$ has compact support, the integral converges for all $\beta \in \mathbb{R}$, and defines a distribution. And

$$\langle T_1, \varphi\rangle = -\int_{-\infty}^0 \varphi(tb) e^{\beta t}\,dt$$

gives another fundamental solution of $L$ in $\mathscr{D}'(\mathbb{R}^n)$.

• Thanks for the answer, Daniel. I understand the general argument, but I have two (maybe trivial) doubts: How do you get the expression of $L^{\ast}$ (I'm not very used to duals). Commented Jan 23, 2014 at 19:07
• And I tried integrating by parts (and using that $\varphi$ has compact support) but I only get this... $-\displaystyle\int_{0}^{\infty}\beta\varphi(tb) dt-\int_{0}^{\infty}\frac{d}{dt}\varphi(tb)e^{\beta t}dt=-\displaystyle\int_{0}^{\infty}\beta\varphi(tb)dt-\left(\left[e^{\beta t}\varphi(tb)\right]_{0}^{\infty}-\int_{0}^{\infty}\beta e^{\beta t}\varphi(tb)dt\right)=-\displaystyle\int_{0}^{\infty}\beta\varphi(tb)dt+\varphi(0)+\int_{0}^{\infty}\beta e^{\beta t}\varphi(tb)dt$ Commented Jan 23, 2014 at 19:08
• It's the definiton of the derivative of distributions, basically. $\langle \frac{\partial}{\partial x_i} T,\varphi\rangle := \langle T, -\frac{\partial}{\partial x_i}\varphi\rangle$. Regarding your second comment, you have forgotten a factor $e^{\beta t}$ in the first integral, so the two integrals you get after integration by parts cancel. Commented Jan 23, 2014 at 19:11
• Now it's completely clear, Daniel. Thanks a lot!! Commented Jan 23, 2014 at 19:17

By the Lojasiewicz-Hörmander theorem, there is a fundamental solution $T\in S'(\mathbb{R}^n)$ of the differential operator $L=(b,\nabla)-\beta$ with constant coefficients. In case $\beta\leqslant 0$, this fundamental solution is indeed defined by the identity $$\langle T,\varphi\rangle = \int\limits_0^{\infty}\varphi(bt)e^{\beta t}dt\quad\forall\,\varphi\in S(\mathbb{R}^n).$$ But in case $\beta >0$, it is another identity: $$\langle T,\varphi\rangle = -\int\limits_{-\infty}^{0}\varphi(bt)e^{\beta t}dt\quad\forall\,\varphi\in S(\mathbb{R}^n).$$ For further details, read the post by Daniel Fischer.