@Jjjones98 The diagonal decomposition of a symmetric matrix A is expressed as A = P Λ P⁻¹, where Λ is a diagonal matrix with eigenvalues of A on the diagonal, and P is a matrix whose columns are a linearly independent set of eigenvectors corresponding to those eigenvalues.

The eigenvectors of a matrix are certainly not uniquely defined—any nonzero multiple of an eigenvector is also an eigenvector and, in the case of a repeated eigenvalue, any linear combination of the corresponding eigenvectors is also an eigenvector. It follows that the matrix P is not unique; it may be modified in very many ways, but the identity A = P Λ P⁻¹ holds regardless of those choices.

In a linear algebra course one shows that among the various choices of the matrix P there exists an orthogonal matrix V, that is, a matrix so that V^T V = I. It follows that V⁻¹ = V^T, and therefore A = V Λ V^T.  I hope that this answers your question.

Nota bene: Throughout this discourse I am assuming that A is real and symmetric.

@Jjjones98 In my calculations I showed that A = U Λ U ⁻¹. Note that this involves U inverse, not U transpose. This addresses your original question where you had asked for A=UDU^{-1}.

If U were an orthogonal matrix, then you could have replaced the inverse by the transpose, but the U we have calculated is not orthogonal, therefore you can't.

If you really want to express the diagonal decomposition in terms of the transpose, you need to rescale the columns of U so that each column is a vector of unit length in the Euclidean norm.  In this particular case it just happens that all columns of U are of length sqrt(2), therefore if you define V = U / sqrt(2), then you will have A = V Λ V ^T, where V ^T is the transpose of V.

@Kitonum Very clever!  It took me some thinking to see how that works.

When we think of a "perturbation", we are thinking of how an expression deviates from a reference/base value.

Your question asks, among other things, for perturbing sin(x/eps) for small esp.  What does that mean?  What is the base value from which the perturbation of  sin(x/eps) is measured?

@student_md

@Markiyan Hirnyk Let me point out that the paragraph that yourself quoted from Wiki refers to the Dirac delta function.  In my post I used function in that same customary sense.  I trust that people who are knowledgeable about the subject are quite comfortable with that usage.

@Jjjones98 Here is how one derives the expression for the Green's function by hand.

Consider the fourth order ODE y''''(x) = Dirac(x-u). The Dirac function is zero to the left and to the right of the point x=u. Therefore we may solve the ODE y''''(x)=0 on the interval 0 < x < u and on u < x < 1 separately. Let's call those two solutions y_L(x) and y_R(x).

On the first interval we have:
y_L''''(x)=0, y_L(0)=0, y_L'(0)=0, y_L(u)=a, y_L'(u)=b.

On the second interval we have:
y_R''''(x)=0, y_R(u)=c, y_R'(u)=d, y_R(1)=0, y_R'(1)=0.

Here a, b, c, d are to be determined. Toward that end, you calculate y_L(x) and y_R(x) by hand. Then you determine the four constants a, b, c, d by solving the following system of four linear equations in the four unknowns a,b c, d:
  y_L(u)=y_R(u),
  y_L'(u)=y_R'(u),
  y_L''(u)=y_R''(u),
  y_L'''(u) - y_R'''(u) = 1.

Doing this is quite straightforward by hand, if you have the patience for it. Alternatively, just do it in Maple with dsolve() as I have shown. It does exactly the same thing.

@das1404 In my worksheet I refer to some equations through their labels.  That's not available in older Maples.  Here I am posting an edited version which avoids such references.  I saved the result as a *.mws file from within Maple 2017.  See if that works for you.

As to what you refer to as "Tom", I can't tell what it is intended to be since I am unable to view the result of your work.  When I execute your worksheet in Maple 2017, it stops with an error message before it hits the animation part.

Perhaps you can edit my worksheet and add the "Tom" part yourself.

Download cone-map.mws

@Jjjones98 One does not "prove" a definition.  One just "gives" a definition.

For instance, in your problem you should begin by defining what you mean be the Green's function.  Then investigate what that definition implies.  If you do things right, you will arrive the piecewise form that you gave in your original post.

You have said that you have proved various continuity and differentiability properties of your problem's Green's function.  What did you take as the definition of the Green's function?

Actually all the information you need is in Markiyan's post.  He gives a link to a wikipedia article where the Green's function is defined.  Then he uses Maple to derive the consequences of that definition.  His final result is not exactly what you are looking for but it takes just one small extra step to get there as shown here:

Download mw.mw

@tomleslie That's very nice!

@Preben Alsholm Thanks for spotting and pointing out the typo in my worksheet.  I could see that the result was incorrect but was unable to locate the source of the error.

Additionally, thanks to torabi, who pointed out the typo in the definition of the function G.

After fixing those errors, the worksheet produces the figure shown in the cited article.

Here is the figure:

and here is the corrected worksheet: frac-dsolve-ver2.mw

@Daniel Skoog That's very nice.  I didn't know about that application.

@denkasyan Your expression is different from what I have given.  Don't retype; just copy and paste my code into your worksheet.

@denkasyan Perhaps this is what you are asking:

restart;
with(PolyhedralSets):
y := -20;
p := PolyhedralSet([-x >= 0, -y >= 0, -z >= 0, -(1/2)*x >= 0, (-x-y)*(1/2) >= 0, (-x-z)*(1/2) >= 0]);
Plot(p);

If that's not what you want, then you will have to explain more clearly what it is that you are after.

@denkasyan  Saying "y = -20 is a constant" does not tell me what it is meant to do. How is it related to the polyhedron?