Revision as of 10:37, 24 April 2008 (view source) Yoder2 (Talk) (→‎Partial Differential Equations (Continued): Fix transcription errors for this section.) ← Older edit		Revision as of 10:45, 24 April 2008 (view source) Yoder2 (Talk) (→‎Partial Differential Equations (Continued): more typos, link to new page.) Newer edit →
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−	We may discritise this using a finite difference equation.	+	We may discretize this using a finite difference equation.
	<center><math>		<center><math>
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	e.g. in <math>\mathbb{R}^{2}</math>, consider the curve<center><math>		e.g. in <math>\mathbb{R}^{2}</math>, consider the curve<center><math>
−	\overrightarrow(\tau)=(x(\tau),y(\tau))</math></center>	+	\overrightarrow{\tau}(\tau)=(x(\tau),y(\tau))</math></center>
−	We shall comonly drop the vector notation in the rest of this discussion.	+	We shall commonly drop the vector notation in the rest of this discussion.
	[Figure: Arbitrary curve , looping from <math>\tau=0</math> to <math>\tau=10,000</math>.]		[Figure: Arbitrary curve , looping from <math>\tau=0</math> to <math>\tau=10,000</math>.]
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	<center><math>		<center><math>
	\frac{dc}{dt}=\frac{d^{2}c}{d\tau^{2}}</math></center>		\frac{dc}{dt}=\frac{d^{2}c}{d\tau^{2}}</math></center>
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	This gives us two 1-D heat equations		This gives us two 1-D heat equations
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−	If we cannot solve these analytically, we can again discritize the	+	If we cannot solve these analytically, we can again discretize the
	solution.		solution.
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	[Figure showing the notation]		[Figure showing the notation]
−	[Review of Iterative solutions to partial differential equations.	+	[[Solving differential equations iteratively_OldKiwi]]
−	A "Button" on the kiwi:]	+
−		+
−	<center><math>	+
−	\frac{dx}{dt}=2</math></center>	+
−	<center><math>	+
−	\frac{x(t+\Delta t)-x(t)}{\Delta t}=2</math></center>	+
−	<center><math>	+
−	x(t+\Delta t)=x(t)+\Delta t2</math></center>	+
−	Now pick an initial t, say <math>t=0</math>. Assume a boundary condition, <math>x(0)=7</math>.	+
−		+
−	Then <math>x(0)=x_{0}</math>, so <math>x_{0}=7</math>.	+
−		+
−	Then <math>x(\Delta t)=x_{1}</math>, so <math>x_{1}=7+\Delta t2=7.2</math> (We pick <math>\Delta t=0.2</math>)	+
−		+
−	Then <math>x(2\Delta t)=x_{2}</math>, so <math>x_{2}=7.2+\Delta t2=7.4</math>	+
−		+
−	[Plot of solution]	+
	=How does this apply to valley-finding?=		=How does this apply to valley-finding?=

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Revision as of 10:45, 24 April 2008

The Last Class

Welcome to the last class of the semester! The kiwi is going to be a great resource.

Partial Differential Equations (Continued)

[Note: I do not think the difference between $ \partial $ and $ d $ is very important -- feel free to fix it if you disagree. I'm also a little sloppy with the vector symbols]

$ \frac{d}{dt}=u_{xx} $

We may discretize this using a finite difference equation.

$ \frac{u(t+\Delta t,x)-u(t,x)}{\Delta t}=\frac{u(t,x+\Delta x)-2u(t,x)+u(t,x-\Delta x)}{\Delta x^{2}} $ $ u(t+\Delta t,x)=u(t,x)+\frac{\Delta t}{\Delta x^{2}}(u(t,x+\Delta x)-2u(t,x)+u(t,x-\Delta x)) $

where $ u(t,x) $ is the initial guess. But we can also look at this

a different way.$ =\frac{\Delta t}{\Delta x^{2}}u(t,x+\Delta x)+(1-2\frac{\Delta t}{\Delta x^{2}})u(t,x)+\frac{\Delta t}{\Delta x^{2}}u(t,x-\Delta x) $

What does this mean? Let's consider an example.

We fix t, and let $ \frac{\Delta t}{\Delta x^{2}}=\frac{1}{3} $. Then$ u(t+\Delta t,x)=\frac{1}{3}u(t,x+\Delta x)+\frac{1}{3}u(t,x)+\frac{1}{3}u(t,x-\Delta x) $

This is the same as convolution:

[Figure: [1/3,1/3,1/3]*[(x-delx),*(x),(x+delx)]u(t,.) ]

At each successive time step, the convolution iterates on the results of the previous convolution until the solution stabilizes. (That is, until we reach the steady state solution, where $ \frac{du}{dt}=0\Longrightarrow u_{xx}=0 $)

We can use heat flow to smooth out parametric curves.

e.g. in $ \mathbb{R}^{2} $, consider the curve$ \overrightarrow{\tau}(\tau)=(x(\tau),y(\tau)) $

We shall commonly drop the vector notation in the rest of this discussion.

[Figure: Arbitrary curve , looping from $ \tau=0 $ to $ \tau=10,000 $.]

And consider the family

$ c(t,\tau)=(x(t,\tau),y(t,\tau)) $

satisfying the 2D heat equation

$ \frac{dc}{dt}=\frac{d^{2}c}{d\tau^{2}} $

This gives us two 1-D heat equations

$ \frac{dx}{dt}=\frac{d^{2}x}{d\tau^{2}} $

$ \frac{dy}{dt}=\frac{d^{2}y}{d\tau^{2}} $

The solution to these equations is

$ x(t,\tau)=x(0,\tau)*G(t,\tau) $ $ y(t,\tau)=y(0,\tau)*G(t,\tau) $

If we cannot solve these analytically, we can again discretize the solution.

$ x_{i,j+1}=x_{j}+\frac{\Delta t}{(\Delta\tau)^{2}}(x_{i+1,j}-2x_{i,j}+x_{i-1,j}) $ $ y_{i,j+1}=y_{j}+\frac{\Delta t}{(\Delta\tau)^{2}}(y_{i+1,j}-2y_{i,j}+y_{i-1,j}) $

Note that $ \Delta\tau\neq\Delta x $. $ \Delta\tau $ is the spacing between the parameters, and is a constant. However, $ \Delta x $ is the spacing of the points in the x direction, and it is certainly not constant around the entire curve! And a little note on notation:

$ x(0,0)=x_{0,0} $ $ x(\Delta t,\Delta\tau)=x_{1,1} $ $ x(2\Delta t,\Delta\tau)=x_{2,1} $

[Figure showing the notation]

Solving differential equations iteratively_OldKiwi

How does this apply to valley-finding?

Use estimated density $ p(x) $ to define the energy of the curve in

the feature space. The simplest formulation is $ E=\int_{curve}p(c(\tau))d\tau $

It's funny that rivers solve PDEs. They find the minimal energy route through the valleys. This is exactly what we want, because the valleys separate the mountains, which are our clusters.

Look for a curve that minimizes E. For example, in 2D feature space, look for a curve in $ \mathrm{R}^{2} $, say $ c(\tau)=(x(\tau),y(\tau)) $,

such that $ E=\int_{curve}p(x(\tau),y(\tau))d\tau $

Gradient descent from $ \frac{dc}{dt}=E' $.$ \frac{dc}{dt}=-\nabla p(x,y) $

$ \frac{dx}{dt}(t,\tau)=-\frac{d}{dx}p(x,y) $ $ \frac{dy}{dt}(t,\tau)=-\frac{d}{dx}p(x,y) $

Numerically, we need to discritize the left hand side, but not the

right hand side$ \frac{x(t+\Delta t,\tau)-x(t,\tau)}{\Delta\tau}=-p_{x}(x(t,\tau),y(t,\tau)) $ which we can rewrite as$ x(t+\Delta t,\tau)=x(t,\tau)-\Delta\tau p_{x}(x(t,\tau),y(t,\tau)) $

{[}Figure of density points and approximated density landscape.]

{[}Start with curve around both mountains, this is the initial guess for the boundaries of the feature space]

{[}Note that the curve is discritized, each point is assigned to a different value of $ \tau $]

We can represent the points on the curve in a matrix

$ \left(\begin{matrix} x(1), & y(1)\\ x(2), & y(2)\\ \vdots\\ x(20) & y(20)\end{matrix}\right) $

Then we can iterate using the equation above to find the boundaries which minimizes the energy, and is thus the best seperation of the clusters.

Issues with this approach

• We must fix the number of clusters
• We must assume the curve has some degree of smoothness.\\ {[}Figure of this phenomenon]
• The points tend to collapse on each other or move away from each other.\\ {[}Figure of this phenomenon]

How can we overcome these difficulties? With (as the spanish developers might call it) {}"De Le-vel set."

The Level Set Approach

This method was developed by Osher and Sethian {[}spelling?].

Assume we are evolving a curve in $ \mathbb{R}^{2} $. Instead of doing this, we evolve a surface in $ \mathbb{R}^{3} $ for which the {}``zero level set, that is the intersection with the z=0 plane is the curve we wish to evolve.

{[}Figure of an evolving curve]

{[}Figure of evolving surface instead of curve]

Although evolving such a surface can be challenging, it is definitely worth it, because our curve may undergo topology changes. How can this happen? Even though the surface evolves without topology changes, the zero-set (our curve) can change topologies and have singularities.

From the Dromadaire (Dromedary?) to the Chameau (Camel)

{[}Figure t=0: of the one-hump animal, yielding a single curve]

{[}Figure t=1: The top flattens, the curve turns into a peanut]

{[}Figure t=2: The valley dips so we get the Chameau]

{[}Figure t=3: Below the surface and the two curves appear]

We evolve the surface until it converges, and then we can look at the resulting zero-set and count the clusters. We do not need ot know them a-priori.

What PDE describes this surface evolution?

Suppose $ \frac{d}{dt}(x(t,\tau),y(t,\tau))=f(t,\tau)\overrightarrow{N}(t,\tau) $

where $ \overrightarrow{N}(t,\tau) $ is normal to the curve. This is a very general equation because any curve evolution can be expressed in this fashion. Then we can consider a surface evolving$ \phi(x,y,t):=\mathbb{R}^{2}\times\mathbb{R}\rightarrow $a zero-level set at t, $ \left\{ x,y|\phi(x,y,t)=0\right\} $ is the

curve we are evolving. The Euler form of the level set is $ \frac{d\phi}{dt}=f(t,\tau)||\nabla\phi|| $

Conclusion

And that's it for the semester! Have fun!