What are the fundamental solutions for the operator $$\mathcal D=a{\partial^2\over\partial x_1^2}+b{\partial^2\over\partial x_2^2}$$ on $\Bbb R^2 $ with standard cordinates $(x_1,x_2)$.

Here $a,b\in \Bbb R$.

Definition: A distribution $E\in D'(R)$is called a fundamental solution of an differential operator $P(D)$ if $$P(D)E= \delta(x).$$

Here $$P(D)=\sum_{|\alpha|\le m} a_\alpha D^\alpha.$$

Differential operator in the above problem looks like a kind of generalization of Laplace operator. Actually, I don't know; is there any generalization of Laplace operator? But if we take $a,b=1$ then it is the Laplace operator and in this case, maybe we can show that $E=\frac1 {2\pi} log|x|, n=2$ is a fundamental solution...

Thank you.

  • $\begingroup$ @enzotib thank you for editing. $\endgroup$ Jun 22, 2013 at 15:54
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    $\begingroup$ When $a=b=1$, by consider radial solution of the problem, i.e. by supposing that $u(r)=u(|(x,y)|)$, you arrive just at the fundamental solution of $\Delta $. I think that in the more general case, the function $u(r)$ with $r=\sqrt{ax^2+by^2}$ that satisfies your equation will be a fundamental solution, i.e. (if I am not making wrong calculations) $E=\frac{1}{2\pi}\log \sqrt{ax^2+by^2}$. "This is just a guess" $\endgroup$
    – Tomás
    Jun 22, 2013 at 16:10
  • $\begingroup$ Thank you for adding this info my friend. Let me check it. $\endgroup$ Jun 22, 2013 at 16:48
  • $\begingroup$ Yes there are generalizations of the Laplace operator, e.g. the Laplace beltrami operator. $\endgroup$ Nov 17, 2013 at 0:23

2 Answers 2


If $a,b$ are positive or negative at the same time, then scaling will do the trick. Let $$x_1' = c x_1, x_2' = d x_2,$$ then $$ \frac{\partial^2 u}{\partial x_1'^2} = \frac{1}{c^2}\frac{\partial^2 u}{\partial x_1^2}, \quad \frac{\partial^2 u}{\partial x_2'^2} = \frac{1}{d^2}\frac{\partial^2 u}{\partial x_2^2}. $$ Comparing with the equation we have $c = 1/\sqrt{a}$ and $d = 1/\sqrt{b}$, the fundamental solution to $$ a\frac{\partial^2 u}{\partial x_1^2}+ b\frac{\partial^2 u}{\partial x_2^2}= \frac{\partial^2 u}{\partial x_1'^2} + \frac{\partial^2 u}{\partial x_2'^2}= \delta_{(0,0)}(x_1,x_2) $$ is $$ E = \frac{1}{4\pi} \ln(x_1'^2 + x_2'^2) = \frac{1}{4\pi} \ln\left(\frac{x_1^2}{a} + \frac{x_2^2}{b}\right). $$ To verify it, notice your operator $\mathcal{D}$ acting $u$ is $$ \mathcal{D}u = \nabla \cdot (A\nabla u) = a\frac{\partial^2 u}{\partial x_1^2}+ b\frac{\partial^2 u}{\partial x_2^2}, \quad \text{ where } A = \begin{pmatrix}a &0\\0&b\end{pmatrix}. $$ $\nabla$ is just $D$, and $\nabla \cdot $ is the divergence operator. You can find that $$ \nabla u = \frac{1}{4\pi}\left(\frac{2 b x_1}{b x_1^2+a x_2^2}, \frac{2 ax_2}{b x_1^2+a x_2^2}\right), $$ and acting $A$ on $\nabla u$ we have: $$ A\nabla u = \frac{1}{4\pi}\left(\frac{2 ab x_1}{b x_1^2+a x_2^2}, \frac{2 a bx_2}{b x_1^2+a x_2^2}\right). $$ When $x_1,x_2\neq 0$, take divergence: $$ \nabla \cdot (A\nabla u) = \frac{\partial }{\partial x_1} \left(\frac{2 ab x_1}{b x_1^2+a x_2^2}\right)+ \frac{\partial }{\partial x_2} \left(\frac{2 a bx_2}{b x_1^2+a x_2^2}\right) = 0. $$ When $x_1,x_2 = 0$, you can refer to my answer in this question: Green's theorem and flux and the divergence is a Dirac delta (for we have a $1/(4\pi)$ factor).

If $a,b$ have different signs, then this is not an elliptic operator anymore but 1D a wave operator, the solutions are plane wave will be rather given in D'Alembert formula, and the fundamental solution is given by the Heaviside step function $H(x_2 -\sqrt{-b/a} x_1 )$.

  • $\begingroup$ Thank you so much dear Shushao Cao. Please elaborate me last two steps of your proof. Thanks $\endgroup$ Jun 22, 2013 at 21:46
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    $\begingroup$ @MatHeMatCian Hi, I added more in my answer, please see if you are okay with it. $\endgroup$
    – Shuhao Cao
    Jun 23, 2013 at 0:21

Let $r=\sqrt{ax^2+by^2}$ and suppose that $u(r)$ satisfies your equation. Note that $$\frac{\partial u(r)}{\partial x}=\frac{au'(r)x}{r}$$

$$\frac{\partial^2 u(r)}{\partial x^2}=au'(r)\left(r-\frac{ax^2}{r}\right)\frac{1}{r^2}+\frac{a^2u''(r)x^2}{r^2}$$

We have analogous formulas for derivative with respect to $y$. We conclude that $$\frac{1}{a}\frac{\partial^2 u(r)}{\partial x^2}+\frac{1}{b}\frac{\partial^2 u(r)}{\partial y^2}=u''(r)+\frac{u'(r)}{r}=0$$

The last equation can be rewriten for $r\neq 0$ as $$(ru'(r))'=0$$

which implies my comment in Op's question. Now it remains to prove that this is a fundamental solution. I dont know if this is true (I guess it is, if not I will erase this answer, but now I dont have time to verify it), but the calculation is the same for the the case $a=b=1$. You can find this calculation in any good book of potential theory.

Remark: The cases $ab=0$ must be considered separately.

Remark 2: By reading @ShubhaoCao interesting answer, I notice that my argument only works for the cases where $ab>0$. In the case where $a>0$ and $b>0$ there is no need of adaptation in my proof. In the case where they are negatives, we multiply the equation by $-1$ and then we are in the positive case. As @ShubhaoCao showed this is the case where $u$ is a fundamental solution.

  • $\begingroup$ Thanks a lot dear Tomas for answering and making valuable remarks. How will we consider the case you mentioned in your first remark. $\endgroup$ Jun 22, 2013 at 21:51

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