Looking for feedback and maybe simpler intuition for my proof of the theorem, shown below

The statement of the theorem:


Among all second-order homogeneous PDEs in two dimensions with constant coefficients, show that the only ones that do not change under a rotation of the coordinate system (i.e., are rotationally invariant), have the form $$a\nabla^2u = bu $$


The general PDE of those conditions is written as: $$a_1u_{xx} + 2a_2u_{xy} + a_3 u_{yy} + b_1u_x +b_2u_y +cu = 0$$ A counter-clockwise rotation of a point $x,y$ can be given by the rotation matrix is given by the figure below,

Rotation of an angle

where with some basic geometry, we can derive

$$x' = \|{\mathbf{v}}\|\cos\left(\theta + \tan^{-1}\left(\frac{y}{x}\right)\right) = x\cos\theta - y\sin\theta$$ $$y' = \|{\mathbf{v}}\|\sin\left(\theta + \tan^{-1}\left(\frac{y}{x}\right)\right) = x\sin\theta + y\cos\theta$$ This can be summarized in a matrix transformation $$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}$$

giving the map: \begin{align*} x\mapsto x'=x\cos\theta - y\sin\theta \\ y\mapsto y'=x\sin\theta+y\cos\theta \end{align*}

From here we find derivatives of our new coordinates: $$ \frac{\partial x'}{\partial x} = \cos\theta \quad \frac{\partial y'}{\partial x}=\sin\theta$$ $$\frac{\partial x'}{\partial y} = -\sin\theta \quad \frac{\partial y'}{\partial y} = \cos\theta$$

Now the first derivatives of $u(x',y')$ with respect to $x,y$:

\begin{align*} u_x = \frac{\partial u}{\partial x} = \frac{\partial u}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial u}{\partial y'} \frac{\partial y'}{\partial x} = u_{x'}\cos\theta + u_{y'} \sin\theta \\ u_y = \frac{\partial u}{\partial x} = \frac{\partial u}{\partial x'} \frac{\partial x'}{\partial y} + \frac{\partial u}{\partial y'} \frac{\partial y'}{\partial y} = -u_{x'}\sin\theta + u_{y'} \cos\theta \end{align*}

And then the second derivatives:

\begin{align*} &u_{xx} = u_{x'x'} \cos^2\theta +2u_{x'y'}\sin\theta\cos\theta+ u_{y'y'} \sin^2\theta \\ &u_{xy} = -u_{x'x'} \cos\theta\sin\theta - u_{y'x'} \sin^2\theta + u_{x'y'} \cos^2\theta + u_{y'y'}\sin\theta\cos\theta \\ &u_{yy}= u_{x'x'} \sin^2\theta -2u_{x'y'}\sin\theta\cos\theta+ u_{y'y'} \cos^2\theta \end{align*} Substituting into the general PDE and rearranging factors of partial derivatives, can be written as $$ \color{blue}{\widetilde{a_1}u_{x'x'} + \widetilde{a_2}u_{x'y'} + \widetilde{a_3} u_{y'y'} + \widetilde{b_1}u_{x'} +\widetilde{b_2}u_{y'} +\widetilde{c}u = 0} $$

where: \begin{align*} &\widetilde{a_1} = a_1 \cos^2\theta -2a_2\cos\theta\sin\theta + a_3\sin^2\theta \\ &\widetilde{a_2} = (a_1-a_3)\sin 2\theta + 2a_2 \cos 2\theta \\ &\widetilde{a_3} = a_1 \sin^2\theta + 2a_2\sin\theta\cos\theta +a_3\cos^2\theta \\ &\widetilde{b_1} = b_1\cos\theta - b_2\sin\theta \\ &\widetilde{b_2} = b_1\sin\theta + b_2\cos\theta \\ &\widetilde{c} = c \end{align*}

Since we require rotational invariance, the original equation and the transformed PDE must have the same value,namely $0$, on all of $u$. This can be written as

$$\small{a_1u_{xx} + 2a_2u_{xy} + a_3 u_{yy} + b_1u_x +b_2u_y +cu = \widetilde{a_1}u_{x'x'} + \widetilde{a_2}u_{x'y'} + \widetilde{a_3} u_{y'y'} + \widetilde{b_1}u_{x'} +\widetilde{b_2}u_{y'} +\widetilde{c}u} $$

whence: \begin{align} \tag{1} a_1& = a_1 \cos^2\theta -2a_2\cos\theta\sin\theta + a_3\sin^2\theta&\\ \tag{2} 2a_2& = (a_1-a_3)\sin 2\theta + 2a_2 \cos 2\theta& \\ \tag{3} a_3& = a_1 \sin^2\theta + 2a_2\sin\theta\cos\theta +a_3\cos^2\theta& \\ \tag{4} b_1& = b_1\cos\theta - b_2\sin\theta&\\ \tag{5} b_2& = b_1\sin\theta + b_2\cos\theta&\\ \tag{6} c &= c& \end{align}

Excluding the trivial case where $\{a_i\},\{b_i\},c = 0$, we can make several conclusions. Note that the deductions below are made with the understanding that any arbitrary angle $\theta$ must be valid, hence it is erroneous to apply $\theta =0$ in order to reach equality.

  1. can only be true when $a_1=a_3$ and $a_2=0$,
  2. implies $a_2=0$ and $a_1=a_3$,
  3. like (1) is only true when $a_1=a_3$ and $a_2=0$,
  4. is true when $b_1=b_2=0$
  5. like (4) is true when $b_1=b_2=0$,
  6. implies $c\in \mathbb{R}$ is valid.

All together we then know $a1=a3$, $a_2=b_1=b_3=0$ and $c=c$. Returning to the rotated PDE we now know: $$\widetilde{a_1} = a_1, \widetilde{a_2} = 0, \widetilde{a_3} = a_1 , \widetilde{b_1} = 0, \widetilde{b_2} = 0, \widetilde{c} = c $$ So the PDE under a rotation, $u(x',y')$, becomes \begin{align*} &a_1u_{x'x'} + a_1u_{y'y'} +cu = 0\\ \Rightarrow& a_1(u_{x'x'}+u_{y'y'}) = -cu \\ \Rightarrow& a\nabla^2{u} = bu \end{align*} where we have chosen $a_1=a, -c = b$ for all $a,b\in\mathbb{R}$. This is in terms of the new rotation $u(x',y')$, and so it remains to show that $\nabla^2{u(x,y)} = \nabla^2{u(x',y')}$, consider from the derivatives before: \begin{align*} &u_{xx} = u_{x'x'} \cos^2\theta +2u_{x'y'}\sin\theta\cos\theta+ u_{y'y'} \sin^2\theta \\ &u_{yy}= u_{x'x'} \sin^2\theta -2u_{x'y'}\sin\theta\cos\theta+ u_{y'y'} \cos^2\theta \end{align*} Sum them together $$u_{xx}+u_{yy}= u_{x'x'} (\sin^2\theta+\cos^2\theta) -2u_{x'y'}\sin\theta\cos\theta+ 2u_{x'y'}\sin\theta\cos\theta+ u_{y'y'} (\cos^2\theta+\sin^2\theta)=u_{x'x'}+u_{y'y'}$$ hence, $\nabla^2{u(x,y)} = \nabla^2{u(x',y')}$ as required.

Therefore a rotation applied to any second order homogeneous 2D PDE with constant coefficents will transform to a PDE of the form $a\nabla^2{u} = bu$ under the rotated coordinate $x',y'$, which we have shown to be equivalent under the regular coordinates $x,y$. This is the only PDE that is invariant under rotation. $$\tag*{$\blacksquare$}$$

Additional Remark

Also was just curious about rotational invariant functions and operators. Anything that solves laplaces equation ($\nabla^2=0$) is called a harmonic function, and satisfies properties such as the mean value property and the maximal principle. I assumed at first harmonic functions meant they were radial, but I think it is more along the lines of being symmetric?

The laplacian is rotationally invariant but the laplace equation has some solutions which are radial (rotationally invariant) and some which are not. I also noticed that the converse is not true, i.e, a radial function does not imply $\nabla^2 = 0$, like $f(x,y)=x^2+y^2$

See bounty remark below

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    To understand why you just end up with $1$ and $\nabla^2$ as possible operators it might be useful to consider the PDE in Fourier space. If you rotate in real space you rotate in Fourier space and the only vector quantity that is invariant under rotations is the vector norm $\|\vec{k}\|^2 \sim -\nabla^2$. This is very rough, but I’m sure one can make a precise but simple argument along these lines that avoids lengthy calculations. For general order it seems the only rotationally invariant PDEs will be on the form $f(\nabla^2)u = 0$ where $f$ is some polynomial (so radial in Fourier space). – Winther Mar 15 '19 at 00:46
  • @Winther Thank you for the insight. I will look further into making this concrete – Hushus46 Mar 17 '19 at 03:32

2 Answers2


We say that a linear operator $L$ is rotationally invariant if and only if $L$ commutes with the orthogonal group, i.e. $[L, O] = 0$ for every $O \in \text{O}(n)$.

Hence what you are proving is that if $L$ is a second order linear operator then \begin{align} LO[f](x) = L[f(O x)] = [Lf](O x) = OL[f](x) \end{align} if and only if $L = a\Delta-bI$. Moreover, this is equivalent to showing \begin{align} L[f](x, y) = O^{-1}LO[f](x, y) \end{align} for every function $f$, that is, $L$ remains fixed under the conjugation action of orthogonal transformations.

Example: Let us look at an example. Consider $f(x, y) = x e^y$ and $L=\Delta$. Observe \begin{align} O[f]=&\ f(\cos\theta x - \sin\theta y, \sin\theta x+\cos\theta y) \\ =&\ (\cos\theta x-\sin\theta y)e^{\sin\theta x+\cos\theta y} \end{align} where \begin{align} O = \begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}. \end{align} Then we see that \begin{align} g(x, y):=LO[f](x, y)= e^{\sin\theta x+\cos\theta y}(x\cos\theta-y\sin\theta) \end{align} and finally \begin{align} O^{-1}[g](x, y) =&\ g(\cos\theta x+\sin\theta y, -\sin\theta x+\cos\theta y)\\ =&\ e^{\sin\theta\cos\theta x+\sin^2\theta y-\sin\theta\cos\theta x+\cos^2\theta y}(\cos^2\theta x+\sin\theta \cos\theta y+\sin^2\theta x-\sin\theta\cos\theta y)\\ =&\ xe^y. \end{align} Hence \begin{align} O^{-1}LO [f](x, y) = xe^y. \end{align} Also, note that $\Delta f =x e^y$. Thus, $L[f](x, y) = O^{-1}LO[f](x, y)$.

Radial Function: In fact the only radial harmonic solution defined on the entire $xy$-plane are constants. This is a simple consequence of the mean-value identity and the maximum principle for harmonic function. Hence $L$ being rotationally invariant doesn't mean \begin{align} f(Ox) = f(x) \text{ for all } O \in \text{O}(2)\ \ \implies \ \ \Delta f =0. \end{align}

Last Remark: Unfortunately, I don't think there are much easier ways to show the only rotationally invariant second order differential operators are given by $L=a\Delta-bI$ other than direct computation.

Jacky Chong
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Perhaps writing the real variables $x$ and $y$ as complex variables $z$ and $\bar{z}$ could provide some information as expected.

Define \begin{align} z&=x+iy,\\ \bar{z}&=x-iy, \end{align} which yields \begin{align} \frac{\partial}{\partial x}&=\frac{\partial}{\partial z}+\frac{\partial}{\partial\bar{z}},\\ \frac{\partial}{\partial y}&=i\left(\frac{\partial}{\partial z}-\frac{\partial}{\partial\bar{z}}\right). \end{align}

Thanks to these relations, we have \begin{align} u_x&=u_z+u_{\bar{z}},\\ u_y&=i\left(u_z-u_{\bar{z}}\right),\\ u_{xx}&=u_{zz}+2u_{z\bar{z}}+u_{\bar{z}\bar{z}},\\ u_{xy}&=i\left(u_{zz}-u_{\bar{z}\bar{z}}\right),\\ u_{yy}&=-\left(u_{zz}-2u_{z\bar{z}}+u_{\bar{z}\bar{z}}\right). \end{align} As a consequence, $$ a_1u_{xx}+2a_2u_{xy}+a_3u_{yy}+b_1u_x+b_2u_y+cu=0 $$ is equivalent to \begin{equation} \left(a_1+2ia_2-a_3\right)u_{zz}+2\left(a_1+a_3\right)u_{z\bar{z}}+\left(a_1-2ia_2-a_3\right)u_{\bar{z}\bar{z}}+\left(b_1+ib_2\right)u_z+\left(b_1-ib_2\right)u_{\bar{z}}+cu=0.\tag{1} \end{equation}

Now, for rotational transformation, we have $$ z\to e^{i\theta}z $$ for some $\theta\in\left[0,2\pi\right)$. Under this transformation, it is straightforward that Eq. $(1)$ becomes \begin{equation} e^{-2i\theta}\left(a_1+2ia_2-a_3\right)u_{zz}+2\left(a_1+a_3\right)u_{z\bar{z}}+e^{2i\theta}\left(a_1-2ia_2-a_3\right)u_{\bar{z}\bar{z}}+e^{-i\theta}\left(b_1+ib_2\right)u_z+e^{i\theta}\left(b_1-ib_2\right)u_{\bar{z}}+cu=0.\tag{2} \end{equation}

Finally, note that the rotational invariance is equivalent to the arbitrariness of $\theta$. Hence compare Eqs. $(1)$ and $(2)$, and the invariance implies the following cases.

  • If $c\ne 0$, the invariance forces \begin{align} a_1+2ia_2-a_3&=0,\\ a_1-2ia_2-a_3&=0,\\ b_1+ib_2&=0,\\ b_1-ib_2&=0. \end{align} These results indicate that $a_1=a_3$ and $a_2=b_1=b_2=0$, as is expected.
  • If $c=0$ and $a_1+a_3\ne 0$, obviously the same result apply, and we still have the expected conclusion.
  • If $c=0$ and $a_1+a_3=0$ and $a_1+2ia_2-a_3\ne 0$ (i.e., $a_1+ia_2\ne 0$), we have \begin{align} a_1-2ia_2-a_3&=0,\\ b_1+ib_2&=0,\\ b_1-ib_2&=0, \end{align} which, however, do not admit any real solution: the four equalities yield $a_1=a_2=0$ and violate the condition $a_1+ia_2\ne 0$.
  • If $c=0$ and $a_1+a_3=0$ and $a_1+2ia_2-a_3=0$, these conditions lead to $a_1=a_2=a_3=0$, making the equation no longer second order.

To sum up, the desired conclusion is completely proven.

  1. Solutions to the Laplace equation and harmonic functions are exactly the same. As you mentioned, one way to define the harmonic functions is to take them as the solutions to the Laplace equation.
  2. Harmonic functions are not necessarily radial. Radial harmonic functions are called the fundamental solutions to the Laplace equation. In 2-D, it is $\log r$; in 3-D, it is $1/r$. These solutions are essential, and can be made use of to construct Green functions to help solve Poisson equations.
  3. Let $f$ be a harmonic function, and suppose it yields a separation of variables as $f(r,\theta)=F(r)\Theta(\theta)$. Then $F$ complies with a radial equation, and $\Theta$ is called a spherical harmonic function. These functions are essential in, say, quantum mechanics.
  4. In general, $f$ can be expressed as $$ f=\sum_nF_n\Theta_n, $$ where each $F_n$ complies with a radial equation, and each $\Theta_n$ is a spherical harmonic function. This expression can be obtained by solving the Laplace equation by separation of variables.
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  • When you mention $\Theta$, I assume you are referring to functions that come from [this](https://en.wikipedia.org/wiki/Spherical_harmonics). Also, what sort of books/references would you recommend that I can use to learn more about this field. I guess it could be classified as (Harmonic Function Theory?) I really would like to learn more about this class of solutions – Hushus46 Mar 21 '19 at 05:58
  • Also just to make sure, in 2-D the equation is $f(r) = \log r = \log \sqrt{x^2+y^2}=f(x,y)$ and in 3-D $f(r)=\frac{1}{r}=\frac{1}{\sqrt{x^2+y^2+z^2}}=f(x,y,z)$ ? – Hushus46 Mar 21 '19 at 06:15
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    @Hushus46: You are absolutely right. For spherical harmonic functions, you may refer to [this page](http://mathworld.wolfram.com/SphericalHarmonic.html), together with its references, for more information. Usually, this field is covered in Laplace Equations, Fourier Transforms, and Harmonic Analysis. If you are interested in their images and visualizations, I suggest that you refer to any textbook for Quantum Mechanics, in the chapter "Schrodinger equations for hydrogen atoms". Textbooks for mathematics could be abstract. – hypernova Mar 21 '19 at 09:39
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    @Hushus46: Besides, I also suggest that you try this at first. For a 2-D Laplace equation, in polar coordinates, we always have $f(r,\theta)=\sum_{n\in\mathbb{Z}}\hat{f}_n(r)e^{in\theta}$. This is the Fourier expansion of $f$ in $\theta$, since, as per the physical meaning of $\theta$, $f$ periodic in $\theta$. Substitute this form of $f$ into the 2-D Laplace equation in polar coordinates. Note that, due to the orthogonality of $e^{in\theta}$'s, the $r$-dependent coefficient in front of each $e^{in\theta}$ must vanish, and you will have the ODE for each $\hat{f}_n$. – hypernova Mar 21 '19 at 09:51
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    @Hushus46: In the 2-D case, the spherical harmonic functions are quite simple; they are $e^{in\theta}$'s, i.e., $\sin n\theta$'s and $\cos n\theta$'s. In the general m-D case, however, it becomes rather complicated. In this case, with general spherical coordinates, we have $f(r,\theta_1,\theta_2,...,\theta_{m-1})=\sum_{n_1,n_2,...,n_{m-1}\in\mathbb{Z}}\hat{f}_{n_1,n_2,...,n_{m-1}}(r)e^{2in_1\theta_1+2in_2\theta_2+...+2in_{m-2}\theta_{m-2}+in_{m-1}\theta_{m-1}}$. You may still substitute this form into the Laplace equation in general spherical coordinates, yet the calculation becomes tedious. – hypernova Mar 21 '19 at 09:57
  • Thank you for the detailed responses!! Im on my phone at university, so I haven’t had a chance to digest everything you’ve written yet, but I just want to know what is $\hat{f(r)}$ ? – Hushus46 Mar 21 '19 at 10:01
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    @Hushus46: Lastly, you are absolutely right for the $\log r$ and $1/r$ arguments. – hypernova Mar 21 '19 at 10:01
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    @Hushus46: Don't worry. That's definitely fine. Just take your time. $\hat{f}$ is just a notation. In [Fourier transform](https://en.wikipedia.org/wiki/Fourier_transform), if the original function is denoted as $f$, its Fourier transform is usually denoted by $\hat{f}$. From what I stated above, each $\hat{f}_n(r)$ is an unknown function to be determined. We just know its existence as per Fourier transform, but we need to determine it by substituting the expression into Laplace equation to figure out its governing equation. – hypernova Mar 21 '19 at 10:06
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    @Hushus46: For Fourier series and Fourier transforms, you may refer to [this](http://people.bu.edu/andasari/courses/Fall2015/LectureNotes/Lecture13_20Oct2015.pdf), from page 22 on, for a quick view. – hypernova Mar 21 '19 at 10:12
  • ah okay, yes I have detailed with fourier transforms and series in my PDE class, particularly for the heat and wave equation. The professor used a different notation but thats okay, it seems the hat notation is the general consensus! – Hushus46 Mar 21 '19 at 10:14
  • i will have alot of things to learn after exam season over the summer thanks to you! Appreciate it – Hushus46 Mar 21 '19 at 10:15
  • @Hushus46: My pleasure to be able to help. Take your time and good luck to your exams! – hypernova Mar 21 '19 at 16:13