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Let's say $\boldsymbol{\vec F}$ is a field vector with a liner relationship

$$\boldsymbol{\vec F}(\boldsymbol{\vec x})=\boldsymbol A \boldsymbol{\vec x}$$

where $\boldsymbol{\vec x}$ is a vector of size $n$ and $\boldsymbol A$ is a constant $n\times n$ square matrix.

For a given point $\boldsymbol{\vec x}_0$, how can I find the hyper-surface $S(\boldsymbol{\vec x})=0$ which crosses $\boldsymbol{\vec x}_0$ and it is perpendicular to the vector field $\boldsymbol{\vec F}$?

I believe this there is an explicit form of the surface which depends on $\boldsymbol{\vec x}_0$ and $\boldsymbol A$.


On hyper-surface $S$, there will be $n-1$ degree of freedom.

I am looking for a solution for an $n$ dimension vector field. This image is just for illustration.

contour

PS. This question is a special case (linear form) of my previous question.

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    $\begingroup$ In 2D its the integral curve of the perpendicular field $F^\perp $ passing through $x_0$. I don't know enough to generalise this. $\endgroup$ – Calvin Khor Dec 31 '17 at 14:04
  • $\begingroup$ @CalvinKhor, do you suggest any tag to invite related people in matrix algebra to this question? $\endgroup$ – ar2015 Dec 31 '17 at 14:07
  • $\begingroup$ I do not feel like it is a metrix algebra question. Maybe tag with surfaces and/or submanifolds, a hypersurface is orientable iff it admits a normal vector field. $\endgroup$ – Calvin Khor Dec 31 '17 at 14:19
  • $\begingroup$ It seems your previous more general question has an answer in preparation $\endgroup$ – Calvin Khor Dec 31 '17 at 14:20
  • $\begingroup$ @CalvinKhor Nothing yet. $\endgroup$ – ar2015 Dec 31 '17 at 14:25
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Such orthogonal hypersurfaces need not exist if $n\geq3$. Relevant here is Frobenius' theorem that gives the necessary and sufficient local integrability condition. The following source

http://staff.ustc.edu.cn/~wangzuoq/Courses/16F-Manifolds/Notes/Lec11.pdf

contains a readable introduction. Note that your vector field ${\bf F}$ defines at each point ${\bf x}$ an $(n-1)$-dimensional orthogonal hyperplane, called a distribution in the quoted notes.

Frobenius' theorem expands to a condition on the matrix $A$ in your question. The case $n=3$ should be intuitively comprehensible. About this case I can say the following: If the vector field ${\bf F}$ has orthogonal surfaces these can be regarded as level surfaces of some scalar function $g$. This $g$ has an associated gradient field $\nabla g$ which then is orthogonal to these same surfaces, hence parallel to ${\bf F}$. It follows that there is a scalar function $\lambda$ such that $${\bf F}=\lambda\>\nabla g\ .$$ We now compute $${\rm curl}({\bf F})=\nabla\lambda\times\nabla g\ .$$ It follows that ${\rm curl}({\bf F})$ is orthogonal to $\nabla g$, hence to the given ${\bf F}$. In this way we have derived the integrability condition $${\rm curl}({\bf F})\cdot{\bf F}\equiv0\ .\tag{1}$$ This is a necessary condition for the existence of such a family of orthogonal surfaces. Frobenius' theorem says that this condition is also (locally) sufficient. (Unfortunately I could not find a proof of this simple case of Frobenius' theorem on the web.)

In the case of your ${\bf F}:=A{\bf x}$ this means the following: Being a linear function of ${\bf x}$ the field ${\bf F}$ has constant curl ${\bf c}=(A_{32}-A_{23},A_{13}-A_{31}, A_{21}-A_{12})$. The condition $(1)$ then expands to $$A^{\top}{\bf c}={\bf 0}\ .\tag{2}$$ If $A$ is nonsingular then $(2)$ enforces ${\bf c}={\bf 0}$, which means that $A$ has to be a symmetric matrix. Rodrigo de Azevedo's solution shows that in this case we indeed can find orthogonal surfaces.

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  • $\begingroup$ Many thanks Christan. I have difficulties understanding this theorem. Would you please explain it in a plain language? For me it is unclear what distribution, involutive, Lie bracket, image of $d_{ι_N}$ and sub-bundle mean. $\endgroup$ – ar2015 Jan 9 '18 at 1:43
  • $\begingroup$ Many thanks for your comprehensive answer. Would you please also clarify the meaning of (locally)? Do concepts of curl and Helmholtz decomposition have meaning in higher dimensions too? $\endgroup$ – ar2015 Jan 11 '18 at 5:48
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Given a linear vector field $\mathrm v : \mathbb R^n \to \mathbb R^n$ defined by $\rm v (x) := A x$ and a point $\mathrm x_0 \in \mathbb R^n$, we would like to find a function $f : \mathbb R^n \to \mathbb R$ such that vector field $\rm v$ is orthogonal to the following level set

$$\mathcal S := \left\{ \mathrm x \in \mathbb R^n \mid f (\mathrm x) = 0 \right\}$$

and $\mathrm x_0 \in \mathcal S$. Let us demand that $\rm v$ be orthogonal to all level sets of $f$. Hence,

$$\nabla f (\mathrm x) = \mathrm A \mathrm x$$

Assuming that matrix $\rm A$ is symmetric and integrating the set of linear PDEs above, we obtain

$$f (\mathrm x) = \frac 12 \mathrm x^\top \mathrm A \, \mathrm x + f_0$$

Since $\mathrm x_0 \in \mathcal S$, the value of the integration constant is

$$f_0 = - \frac 12 \mathrm x_0^\top \mathrm A \, \mathrm x_0$$

and, thus,

$$f (\mathrm x) = \frac 12 \mathrm x^\top \mathrm A \, \mathrm x - \frac 12 \mathrm x_0^\top \mathrm A \, \mathrm x_0$$

If matrix $\rm A$ is not symmetric, then I do not know how to integrate the linear PDEs.


Example

Let $n=2$. Given $\mathrm A = \begin{bmatrix} 2 & 1\\ 1 & 2\end{bmatrix}$ and $\mathrm x_0 = \begin{bmatrix} 2\\ 0\end{bmatrix}$, we obtain the function

$$f (x_1, x_2) = x_1^2 + x_1 x_2 + x_2^2 - 4$$

The following plot depicts the level set (an ellipse) and the vector field. The length of the arrows is not proportional to $\| \rm A x \|_2$. The purpose is to show that the vector field is orthogonal to the ellipse.

ellipse and vector field


Addendum

Let $n = 2$. The constraint $\nabla f (\mathrm x) = \mathrm A \mathrm x$ produces the following linear PDEs

$$\partial_1 f = a_{11} \, x_1 + a_{12} \, x_2 \tag{1}$$

$$\partial_2 f = a_{21} \, x_1 + a_{22} \, x_2 \tag{2}$$

Integrating equation (1) with respect to $x_1$, we obtain

$$f (x_1,x_2) = \frac 12 \, a_{11} \, x_1^2 + a_{12} \, x_1 \, x_2 + g (x_2)$$

Differentiating $f$ with respect to $x_2$ and using equation (2), we obtain the ODE

$$g' (x_2) = a_{22} \, x_2 + \underbrace{(a_{21} - a_{12})}_{= 0} \, x_1 = a_{22} \, x_2$$

which allows us to conclude that $a_{21} = a_{12}$, i.e., matrix $\rm A$ is symmetric. Integrating the ODE,

$$g (x_2) = \frac 12 \, a_{22} \, x_2^2 + c$$

where $c$ is an integration constant. Thus,

$$f (x_1,x_2) = \frac 12 \, a_{11} \, x_1^2 + a_{12} \, x_1 \, x_2 + \frac 12 \, a_{22} \, x_2^2 + c$$

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In order to satisfy the perpendicularity condition, the vector field ${\bf F}(x)$ should be collinear to the normal vector of the hypersurface $S(x)=0$, i.e. $\forall x\in\mathbb R^n\; \exists k(x)\in\mathbb R:$ $$\tag{1} \nabla S(x)=k(x){\bf F}(x). $$ Let $F(x)=(f_1(x),\ldots,f_n(x))$, then (1) can be rewritten in the coordinate form as $$ \frac{\partial S}{\partial x_1}=k(x)f_1(x)\;,\ldots,\;\frac{\partial S}{\partial x_n}=k(x)f_n(x); $$ or $$ k(x)=\frac{\frac{\partial S}{\partial x_1}}{f_1(x)}=\;\ldots\;= \frac{\frac{\partial S}{\partial x_n}}{f_n(x)}. $$ We can equate the adjacent fractions and obtain $$ \left\{\begin{array}{rcl} f_2(x)\dfrac{\partial S}{\partial x_1}&=&f_1(x)\dfrac{\partial S}{\partial x_2}\\ &\vdots&\\ f_n(x)\dfrac{\partial S}{\partial x_{n-1}}&=&f_{n-1}(x)\dfrac{\partial S}{\partial x_n}\\ \end{array}\right. $$ or finally $$\tag{2} \left\{\begin{array}{lll} f_2(x)\dfrac{\partial S}{\partial x_1}-f_1(x)\dfrac{\partial S}{\partial x_2}&=&0\\ &\vdots&\\ f_n(x)\dfrac{\partial S}{\partial x_{n-1}}-f_{n-1}(x)\dfrac{\partial S}{\partial x_n}&=&0.\\ \end{array}\right. $$ Hence, the question is: does the system (2) have nontrivial solutions?

1) n=2

Consider at first the simple case when $n=2$. In this case the system (2) consists of one equation $$\tag{2a} f_2(x)\dfrac{\partial S}{\partial x_1}-f_1(x)\dfrac{\partial S}{\partial x_2}=0. $$ This equation can be solved using the method of characteristics: $$\tag{3} \frac{dx_1}{f_2(x)}=\frac{dx_2}{-f_1(x)}. $$ The ode (3) always have a solution for any initial point (except for the case $f_1(x)=f_2(x)=0$), thus, the curve $S(x)=0$ exists.

Consider the subcase when ${\bf F}(x)$ is linear: $$ f_1(x)= a_{11}x_1+a_{12}x_2,\quad f_2(x)= a_{21}x_1+a_{22}x_2. $$ In this case the ode (3) has the form $$ \frac{dx_1}{a_{21}x_1+a_{22}x_2}=\frac{dx_2}{-a_{11}x_1-a_{12}x_2}. $$ It is a homogeneous differential equation, thus, a closed form solution can be found.

2) n>2

In the case when $n\ge 3$, there is a possibility that (2) has no nontrivial solutions. This is because the distribution spanned by the vector fields corresponding to the equations of (2) is not necessarily involutive.

The system (2) can be expressed in terms of the Lie derivative: $$\tag{2b} \mathcal L_{{\bf X}_1}S=0,\;\ldots,\;\mathcal L_{{\bf X}_{n-1}}S=0, $$ where $$ {\bf X}_1= \left( f_2(x),-f_1(x),0,\ldots,0 \right)^T,\;\ldots\; {\bf X}_{n-1}= \left( 0,\ldots,0,f_n(x),-f_{n-1}(x) \right)^T $$ The key fact about the system (2b) is that it implies $$ \mathcal L_{[{\bf X}_i,{\bf X}_j]}S= \mathcal L_{{\bf X}_i}(\mathcal L_{{\bf X}_j}S)-\mathcal L_{{\bf X}_j}(\mathcal L_{{\bf X}_i}S)=0 $$ for any $i,j=1,..,n-1$, where $[{\bf X}_i,{\bf X}_j]$ is the Lie bracket.

The distribution $D(x)=span\{{\bf X}_1,\;\ldots\;{\bf X}_{n-1}\}$ is involutive iff the Lie bracket $[{\bf X}_i,{\bf X}_j]$ of any pair of the vector fields ${\bf X}_1,\;\ldots\;{\bf X}_{n-1}$ belongs to $D(x)$, i.e. $[{\bf X}_i,{\bf X}_j]$ is a linear combination of the original vector fields. If the distribution is not involutive (let, for example, $[{\bf X}_i,{\bf X}_j]\notin D(x)$), then we can add the equation $\mathcal L_{[{\bf X}_i,{\bf X}_j]}S=0$ (which is not a linear combination of other equations) to (2b) without changing the solution set.

Now consider the counterexample. Let $n=3$, ${\bf F}(x)=(x_3,x_1,x_2)$. The system (2) becomes $$\tag{2c} \left\{\begin{array}{lll} x_1\dfrac{\partial S}{\partial x_1}-x_3\dfrac{\partial S}{\partial x_2}&=&0\\ x_2\dfrac{\partial S}{\partial x_2}-x_1\dfrac{\partial S}{\partial x_3}&=&0.\\ \end{array}\right. $$ The Lie bracket of the vector fields $$ {\bf X}_1=(x_1,-x_3,0)^T,\quad {\bf X}_2=(0,x_2,-x_1)^T $$ can be calculated as $$ [{\bf X}_1,{\bf X}_2]=\frac{\partial {\bf X}_2}{\partial x} {\bf X}_1- \frac{\partial {\bf X}_1}{\partial x} {\bf X}_2 $$ $$ =\left(\begin{array}{ccc} 0&0&0\\0&1&0\\-1&0&0 \end{array}\right) \left(\begin{array}{c} x_1\\-x_3\\0 \end{array}\right)- \left(\begin{array}{ccc} 1&0&0\\0&0&-1\\0&0&0 \end{array}\right) \left(\begin{array}{c} 0\\x_2\\-x_1 \end{array}\right)= \left(\begin{array}{c} 0\\-x_1-x_3\\-x_1 \end{array}\right). $$ Hence, the system (2c) is equivalent to the system $$\tag{2d} \left\{\begin{array}{rrrll} x_1\dfrac{\partial S}{\partial x_1}&-x_3\dfrac{\partial S}{\partial x_2}&&=&0\\ &x_2\dfrac{\partial S}{\partial x_2}&-x_1\dfrac{\partial S}{\partial x_3}&=&0\\ &(-x_1-x_3)\dfrac{\partial S}{\partial x_2}&-x_1\dfrac{\partial S}{\partial x_3}&=&0\\ \end{array}\right., $$ but the system (2d) has no nontrivial solutions. From the last two equations $$ (-x_1-x_3-x_2)\dfrac{\partial S}{\partial x_2}=0; $$ we are not interested in the solutions on the plane $-x_1-x_3-x_2=0$, so we can assume that $\dfrac{\partial S}{\partial x_2}=0$. Then it is easy to see that $$ \dfrac{\partial S}{\partial x_2}=\dfrac{\partial S}{\partial x_1}=\dfrac{\partial S}{\partial x_3}=0, $$ so the only possible solution is trivial.

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