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On Eigen(patch(ix))values - (RWLA,MCT,GT,AM Part VIII)

March 16th, 2011 No comments

So in the continuation of this series, I have been thinking long and hard about the curious property of the existence of eigen(patch(ix))values that I have talked about in a previous post.  I began to question whether such eigen(patch(ix))values are limited to a finite set (much as in finite matrices) or whether there was some other fundamental insight, like, if 1 is an eigen(patch(ix))value, then all elements of  \mathbb{R} are too (or all of  \mathbb{R} minus a finite set).  In my latest attempt to understand this, the question comes down to, using the "star" operator, whether

 a(x) \star p(x,y) = \lambda a(x)

has discrete values of  \lambda or, "what values can lambda take for the equation to be true," in direct analogy with eigenvalues when we're dealing with discrete matrices.  I am not using yet "integral transform notation" because this development seemed more intuitive to me, and thus I'm also limiting the treatment to "surfaces" that are smooth and defined on  [0,1] \times [0,1] , like I first thought of them. Thus, the above equation translates to:

 \int_0^1 a(1-y) p(x,y) dy = \lambda a(x)

and, if we recall our construction of the patch (or patchix if we relax the assumption that integrating with respect to x is 1)  p(x,y) = f_1(x) g_1(y) + f_2(x) g_2(y) :

 \begin{array}{ccc} \lambda a(x) & = &\int_0^1 a(1-y) \left(f_1(x) g_1(y) + f_2(x) g_2(y) \right) dy \\ & = & f_1(x) \int_0^1 a(1-y) g_1(y) dy + f_2(x) \int_0^1 a(1-y) g_2(y) dy \\ & = & B_1 f_1(x) + B_2 f_2(x) \end{array}

where  B_1, B_2 are constants.  It is very tempting to divide  \lambda as

 a(x) = \frac{B_1}{\lambda} f_1(x) + \frac{B_2}{\lambda} f_2(x)

must hold provided  \lambda \neq 0 .  So we have excluded an eigen(patch(ix))value right from the start, which is interesting.

We can systematically write the derivatives of  a(x) , as we're going to need them if we follow the algorithm I delineated in one of my previous posts (NB: we assume a finite number of derivatives or periodic ones, or infinite derivatives such that the subsequent sums we'll write are convergent):

 \begin{array}{ccc} a(x) & = & \frac{B_1}{\lambda} f_1(x) + \frac{B_2}{\lambda} f_2(x) \\ a'(x) & = & \frac{B_1}{\lambda} f'_1(x) + \frac{B_2}{\lambda} f'_2(x) \\ a''(x) & = & \frac{B_1}{\lambda} f''_1(x) + \frac{B_2}{\lambda} f''_2(x) \\ \vdots & \vdots & \vdots \\ a^k(x) & = & \frac{B_1}{\lambda} f^k_1(x) + \frac{B_2}{\lambda} f^k_2(x) \\ \vdots & \vdots & \vdots \end{array}

provided, as before,  \lambda \neq 0 .  We want to calculate the constants  B_1, B_2 , to see if they are restricted in some way by a formula, and we do this by integrating by parts as we did in a previous post to obtain the cool "pasquali series." Thus, we have that if  B_1 = \int_0^1 a(1-y) g_1(y) dy , the tabular method gives:

 \begin{array}{ccccc} \vert & Derivatives & \vert & Integrals & \vert \\ \vert & a(1-y) & \vert & g_1(y) & \vert \\ \vert & -a'(1-y) & \vert & G_1^1(y) & \vert \\ \vert & a''(1-y) & \vert & G_1^2(y) & \vert \\ \vert & \vdots & \vert & \vdots & \vert \end{array}

and so,

 \begin{array}{ccc} B_1 & = & \int_0^1 a(1-y) g_1(y) dy \\ & = & a(1-y) G_1^1(y) \vert_0^1 + a'(1-y) G_1^2(y) \vert_0^1 + \ldots \\ & = & \sum_{i = 0}^\infty a^i(1-y) G_1^{i + 1} \vert_0^1 \end{array}

if we remember the alternating sign of the multiplications, and we are allowed some leeway in notation.  Ultimately, this last bit means:  \sum_{i=0}^\infty a^i(0) G_1^{i+1}(1) - \sum_{i=0}^\infty a^i(1) G_1^{i+1}(0) .

Since we have already explicitly written the derivatives of  a(x) , the  a^i(0), a^i(1) derivatives can be written as  \frac{B_1}{\lambda} f_1^i(0) + \frac{B_2}{\lambda} f_2^i(0) and  \frac{B_1}{\lambda} f_1^i(1) + \frac{B_2}{\lambda} f_2^i(1) respectively.

We have then:

 B_1 = \sum_{i=0}^\infty \left( \frac{B_1}{\lambda} f_1^i(0) + \frac{B_2}{\lambda} f_2^i(0) \right) G_1^{i+1}(1) - \sum_{i=0}^\infty \left( \frac{B_1}{\lambda} f_1^i(1) + \frac{B_2}{\lambda} f_2^i(1) \right) G_1^{i+1}(0)

Since we aim to solve for  B_1 , multiplying by  \lambda makes things easier, and also we must rearrange all elements with  B_1 in them, so we get:

 \lambda B_1 = B_1 \sum_{i=0}^\infty \left( f_1^i(0) G_1^{i+1}(1) - f_1^i(1) G_1^{i+1}(0) \right) + B_2 \sum_{i=0}^\infty \left( f_2^i(0) G_1^{i+1}(1) - f_2^i(1) G_1^{i+1}(0) \right)

Subtracting both sides the common term and factoring the constant we endeavor to solve for, we get:

 \left( \lambda - \sum_{i=0}^\infty \left( f_1^i(0) G_1^{i+1}(1) - f_1(1) G_1^{i+1}(0) \right) \right) B_1 = B_2 \sum_{i=0}^\infty \left(f_2^i(0) G_1^{i+1}(1) - f_2^i(1) G_1^{i+1}(0) \right)

or

 B_1 = \frac{B_2 \sum_{i=0}^\infty f_2^i(1-y) G_1^{i+1}(y) \vert_0^1}{\lambda - \sum_{i=0}^\infty f_1^i(1-y) G_1^{i+1}(y) \vert_0^1} = \frac{B_2 D}{\lambda - C}

A similar argument for  B_2 suggests

 B_2 = \frac{B_1 \sum_{i=0}^\infty f_1^i(1-y) G_2^{i+1}(y) \vert_0^1}{\lambda - \sum_{i=0}^\infty f_2^i(1-y) G_2^{i+1}(y) \vert_0^1} = \frac{B_1 E}{\lambda - F}

where the new constants introduced emphasizes the expectation that the sums converge.  Plugging in the one into the other we get:

 B_1 = \frac{\left( \frac{B_1 E}{\lambda - F} \right) D}{\lambda - C} = \frac{B_1 E D}{(\lambda - F) (\lambda - C)}

and now we seem to have additional restrictions on lambda:  \lambda \neq F and  \lambda \neq C .  Furthermore, the constant  B_1 drops out of the equation, suggesting these constants can be anything we can imagine (all of  \mathbb{R} without restriction), but then we have the constraint:

 (\lambda - F)(\lambda - C) = ED

which is extraordinarily similar to its analogue in finite matrix or linear algebra contexts.  Expanding suggests:

 \lambda^2 - (F + C) \lambda + (CF - ED) = 0

which we can solve by the quadratic equation of course, as:

 \lambda_{1,2} = \frac{(F + C) \pm \sqrt{(F-C)^2 + 4ED} }{2}

So not only is  \lambda not equal to a few values, it is incredibly restricted to two of them.

So here's a sort of conjecture, and a plan for the proof.  The allowable values of  \lambda is equal to the number of x terms  a(x) (or  p(x,y) ) carries.  We have already shown the base case, we need only show the induction step, that it works for  k and  k+1 terms.