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What is $$\int_{K} e^{a \cdot x+ b \cdot y} \mu(x,y)$$ where $K$ is the Koch curve and $\mu(x,y)$ is a uniform measure look here.

Attempt: I can evaluate the integral numerically and I have derived a method to integrate $e^x$ over some cantor sets, look here. When I tried using that method to integrate the Koch Curve, I end up unable to express the integral in direct terms of its self. Here's a proof that integration can be done over the Koch Curve...

Information: I'd like a symbolic answer if its available, but infinite series/products for this integral are great too. If there's a reference that actually handles this specific function over fractals and derives a symbolic result, that's good to. Also feel free to change $K$ to any other (non-trivial of course ;) ) variant of the Koch curve if that makes it easier to compute. I warn only that because the goal is to integrate over any fractal rather than just one or two special examples, you shouldn't pick needlessly trivial examples...

Motivation: The derivation of this result allows for integration over a fractal, however the actual reason this is useful, is because of the usefulness of the exponential function. For instance, the concept of average temperature over a fractal is a very interesting concept. $e^x$ type functions allow for rudimentary temperature fields to be constructed and theoretically integrated over fractals. $e^x$ type functions are useful for many kinds of problems, but they seem to be difficult to integrate over fractals. In addition, developing a theory for integrals over fractals, requires a large library of results, and $e^x$ should definitely be included in that list of integrable functions.

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  • $\begingroup$ Interesting question, but what's the purpose? $\endgroup$ – user21820 Apr 28 '15 at 15:01
  • $\begingroup$ @user21820 I added my motivation. $\endgroup$ – Zach466920 Apr 28 '15 at 15:07
  • $\begingroup$ I see. Interesting! $\endgroup$ – user21820 Apr 28 '15 at 15:11
  • $\begingroup$ just for the non mathematicans, uniform measure is $dxdy$ right? $\endgroup$ – tired Apr 28 '15 at 15:14
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    $\begingroup$ @tired The short answer is yes. The long answer is no. The "measure" function just assigns mass to otherwise pathological functions like the above. For instance $\mu(K)$ is 1, which makes sense the measure of the whole object is unity. $\mu(K/3)$ is 1/4 however, since the object can be broken into 4 self-similar parts. You assign mass densities to the object using measure and integrate by adding up all of these densities...Note that regular integration of the above function leads to 0. $\endgroup$ – Zach466920 Apr 28 '15 at 15:18
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not an answer yet, just some thoughts.

Say our Koch curve $K$ starts at $(0,0)$, ends at $(1,0)$ and the midpoint is at $(1/2, 1/(2\sqrt{3}\;))$. Mark seems to have used this, since his computation with $a=b=1$ agrees with mine.

Self-similarity is described by two maps of the plane to itself: $$ L(x,y) = \left(\frac{x}{2}+\frac{y}{2\sqrt{3}},\frac{x}{2\sqrt{3}}-\frac{y}{2}\right), \\ R(x,y) = \left(\frac{x}{2}-\frac{y}{2\sqrt{3}}+\frac{1}{2},-\frac{x}{2\sqrt{3}}-\frac{y}{2}+\frac{1}{2\sqrt{3}}\right), $$ So $L(K)$ is the left half and $R(K)$ is the right half. Set $K$ is the unique nonempty compact set with $K = L(K) \cup R(K)$. Map $L$ shrinks by factor $1/\sqrt{3}$, reflects in the $x$-axis, rotates by $\pi/6$, and fixes the point $(0,1)$. Map $R$ shrinks by factor $1/\sqrt{3}$, reflects in the $x$-axis, rotates by $-\pi/6$ and fixes the point $(1,0)$.

The measure $\mu$ on $K$ is made up of two parts, which are images of $\mu$ under $L, R$, respectively, with half the measure. That is, for integrable $f$ we have $$ \int_K f\,d\mu = \int_{L(K)} f\,d\mu+\int_{R(K)} f\,d\mu = \frac{1}{2}\int_K f\circ L\,d\mu + \frac{1}{2}\int_K f\circ R\,d\mu $$

Now if we write $$ q(a,b) := \int_K e^{ax+by}d\mu(x,y) $$ the self-similarity shows $$ q(a,b) = \frac{1}{2}q\left(\frac{a}{2}+\frac{b}{2\sqrt{3}}, \frac{a}{2\sqrt{3}}-\frac{b}{2}\right)+\frac{1}{2}\exp\left(\frac{a}{2}+\frac{b}{2\sqrt{3}}\right)q\left(\frac{1}{2}-\frac{b}{2\sqrt{3}},-\frac{a}{2\sqrt{3}}-\frac{b}{2}\right) $$

We could use this recursively to evaluate $q(a,b)$ numerically. At each iteration, the point $(a,b)$ where $q$ should be evaluated moves closer to the origin by factor $1/\sqrt{3}$. We stop when we are "close enough" to $(0,0)$, since we know $q(0,0)=1$. But, of course, at each iteration the number of exponentials we have to evaluate doubles, so it is a slow method.

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  • $\begingroup$ You don't answer all my questions, but you make serious headway. I have no problem giving you the bounty. You clearly put a lot of thought into this. Good job! $\endgroup$ – Zach466920 May 13 '15 at 14:55
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This "answer" is in response to your comment that you'd be interested in seeing series/product solutions. As I'm sure you know, it's not difficult (in principle) to compute the integral of $x^p$ or $y^p$ with respect to a self-similar measure. (I have Mathematica code that automates the procedure.) Thus, we can get an approximation by simply writing $$ e^{ax+by}=e^{ax}e^{by}, $$ replacing the exponential expressions with a finite sum approximation, and then integrating. The result is: $$ \left(1+\frac{a}{2}+\frac{19a^2}{120}+\frac{3 a^3}{80}+\frac{92983 a^4}{13023360}+\frac{5935 a^5}{5209344}+\frac{618497323 a^6}{3948161817600}+\cdots\right)\times \\ \left(1+\frac{b}{6 \sqrt{3}}+\frac{b^2}{120}+\frac{b^3}{1008\sqrt{3}}+\frac{83b^4}{2604672}+\frac{601 b^5}{234420480 \sqrt{3}}+\frac{2095657 b^6}{35533456358400}+\cdots\right) $$ Unfortunately, I see no significant simplification beyond this. In particular, I am not able to find closed form expressions for the integrals of the power functions - only exact expressions for specific integers.

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    $\begingroup$ I don't think the integral of the product will be the product of the integrals... Putting $a=b=1$ in your product is $1.8829$ and not the $1.88006$ that you computed for that case. My computation agrees with your $1.88006$. So for a series solution, we will also need to compute integrals of $x^py^q$. $\endgroup$ – GEdgar May 12 '15 at 15:54

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