To start with, let me apologize for my ignorance as I know next to nothing about partial differential equations. My question is about the tensor product of Banach spaces but actually I do not understand those either, so let me rephrase it as a Hilbert space question. Alright! Let us consider the Lebesgue Hilbert space $L^2(X)$ and why not also $L^2(Y)$ where $X$ and $Y$ are the domains of our respective functions.

Now the tensor product of $L^2(X) \otimes L^2(Y)$ is readily seen to be the closure of the space of functions that have the form $fg$ where $f$ is in $L^2(X)$ and $g$ is in $L^2(Y)$. Perfect. This is of course (Banach spaces so think isomorphic) also equal to $L^2(X \times Y)$. Nothing interesting here. (For an interesting related discussion check out my question Sequence of measurable functions and the excellent answer by Nate which gives a couple of very interesting tools.)

So let me start with a nice PDE. The Poisson equation. $\Delta u = -f$ where - if you have a inclination to thinking in practical terms - you could see $f$ as a normalized charge density function and then $u$ would be the potential this charge density gives us.

So, where does our solution live? As $u$ would be the convolution with the so called fundamental solution of the Laplace equation this would mean that the solution $u$ is at least as smooth as $f$ itself (that is why I took this equation...). Also, if we would look on the whole of $\mathbf R^3$ then we might have a problem with integrability: the maximum principle would state that if the function would be bounded (like if it would have compact support and $f$ continuous) then it would be a constant. That messes up the integrability. So as I know nothing about these things let us just consider the domain to be the unit cube $Q$. Integrable for sure!

So the solution would live in $L^2([0,1])$ tensored with itself three times. Actually, all its derivatives should be in this space as well, so we have the Sobolev space $W^{2, 2}(Q)$ if we take $f$ to be at least $C^2$. Also, as I have kinda shown here (need to restrict) here (for $L^2$ you can also use Plancherel) this is equal to the $L^2$ space intersected with the $L^2$ functions that have Laplacian as well in $L^2$.

Now the question...

As I have asked before in Non-separable linear PDE and about the excellent answer of Willy there I was wondering about solutions in these spaces.

If we solve for instance the heat equation by separation of variables in a 1D-heat bar we find two ODEs for both the position variable and the time variable. These belong to a set of simultaneous eigenvalues $\lambda_n$, and have corresponding solutions $s_n$ and $t_n$. Then we construct the solution as $$u = \sum_n \lambda u_n t_n.$$ Question: This is a solution in the product space. What makes it so that one cannot write a posteriori a solution of a non-separable equation in this way? The simultaneous $\lambda_n$ would not correspond to such an equation?

Any omissions and mistakes are solely due to my misunderstanding and/or misconception. I appreciate any constructive comment.

Oh blimey... For now, let me make it more abstract to find my source of misunderstanding. Let $H_1$ and $H_2$ be Hilbert spaces. A simple tensor (tensor of rank one) $x_1 \otimes x_2$ is the identification of $x_1$ with its dual element $x_1^*$. Kind of like $x_1 \times x_2$ where $x_1$ is in $H_1$ and $x_2$ is in $H_2$. Then the inner product defined on $H_1 \otimes H_2$ is given by $$\langle u_1 \otimes u_2, v_1 \otimes v_2 \rangle = \langle u_1, v_1 \rangle_{H_1} \langle u_2, v_2 \rangle_{H_2}.$$ Next, extend by linearity and complete under this inner product.

More abstractly, to each simple tensor $x_1 \otimes x_2$ associate the rank one operator $$ \begin{align} T:H_1^* &\to H_2\\ x^* &\mapsto x^*(x_1) x_2. \end{align} $$ We can use this to construct a linear mapping between $H_1 \otimes H_2$ and the space of finite rank operators from $H_1^*$ to $H_2$. These are a subspace of the Hilbert-Schmidt operators which has scalar product for an orthonormal basis $(e_n)$ of $H_1$, $$\langle \Lambda_1, \Lambda_2 \rangle = \sum_n \langle \Lambda_1 e_n^*, \Lambda_2 e_n^*\rangle.$$ Hence, we can identify the tensor product $H_1 \otimes H_2$ as the Hilbert Schmidt operators from $H_1^*$ to $H_2$.

Also to my knowledge, whenever the spaces are separable: $$L^2(X) \otimes L^2(Y) \cong L^2(X \times Y) \cong L^2(X; L^2(Y)).$$

So, if I have $f$ in $L^2$ and $g$ in $L^2$, then the set of $fg$ is in $L^2 \otimes L^2$. Furthermore, it is a dense set in $L^2(X \times Y)$.

  • $\begingroup$ Does the lack of answers imply that my question is too stupid or that there is nobody that can answer it here? If so, would it more appropriate to ask on MO? $\endgroup$ – Jonas Teuwen Nov 11 '12 at 17:22
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    $\begingroup$ No, it mostly meant that your motivation was too distracting. :p I glanced at this question when it was first posted and was too lazy to read through the whole thing. I probably would've posted an answer the first time around if paragraphs 3 - 5 were much more condensed and to the point. $\endgroup$ – Willie Wong Jan 8 '13 at 9:36
  • $\begingroup$ @WillieWong I will rewrite it if you still think it is a 'logical fallacy' after my comment! $\endgroup$ – Jonas Teuwen Jan 8 '13 at 11:08
  • $\begingroup$ In any case, I have rewritten it slightly. I do not want people to assume I am that stupid that I think tensor spaces are that trivial. $\endgroup$ – Jonas Teuwen Jan 8 '13 at 11:17

That brings me to the following: if an equation is not separable what does this mean about the solution the spaces lives in? If the solution would be in a kind of $L^2$ space, then would it be impossible that $L^2([0,1]^3)$ would be the space the solution can live in? As this would be the same as our tensor product it would imply the solution is actually possible to be written in such a way!

The answer is simply that you made a logical fallacy. And nothing whatsoever can be said about the solution space.

You tried to take the converse of the statement $$ \text{solution writable as } f(x)\cdot g(y) \implies \text{solution is in a tensor product space}$$ which is false for any nontrivial vector spaces!

Let $V,W$ be vector spaces, an element of $V\otimes W$ is not always rank 1 (that is to say, expressible as $v\otimes w$ for $v\in V$ and $w\in W$). Given $v,v'\in V$ and $w,w'\in W$, in general $v\otimes w + v'\otimes w' \in V\otimes W$ cannot be "factored" as $v''\otimes w''$. (Just think about $V = W = \mathbb{R}$ and the space of two-by-two matrices....)

So just because a solution lives in $L^2(Q)$ does not imply that said solution can be written as $f(x)g(y)h(z)$ for $f,g,h\in L^2([0,1])$.

For the revised question:

This is a solution in the product space. What makes it so that one cannot write a posteriori a solution of a non-separable equation in this way? The simultaneous $λ_n$ would not correspond to such an equation?

Nothing prevents you from writing a solution that way. You've basically just described the Fourier series. (Any $L^2$ function $f$ on $Q$ can be written as the Fourier series $$ f(x,y,z) \approx \sum_{j,k,l = -\infty}^\infty a_{jkl} \exp[ 2\pi i(jx + ky + l z)]$$ Chain this with some bijection of $\mathbb{Z}^3 \to \mathbb{Z}$ you end up with a single sum like you described.)

What fails is for each $\exp[\ldots]$ or $u_nt_n$ to solve the equation. Because, let's face it: if every solution can be written as a linear combination of solutions which are pure tensors, then the linear equation is separable...

  • $\begingroup$ Oops - logical fallacy is for me more like a error in verbal reasoning -. Anyway, I meant to say that it is a dense set in the tensor product. That is, a Schauder basis. Is that not true? When we separate we can find sequences of basis functions and then we construct the basis for the solution space as a Schauder basis. $\endgroup$ – Jonas Teuwen Jan 8 '13 at 11:07
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    $\begingroup$ Still wrong. Think back to the two-by-two matrices which can be thought of as $\mathbb{R}^1\otimes \mathbb{R}^1$. The rank-1 objects are not invertible, hence have determinant 0. It is very, very far from dense. I think you are confusing the linear span of a Schauder basis $\operatorname{span}(\{v_1, v_2, \ldots\})$ with the union of the one dimensional subspaces $\cup_j \operatorname{span}(\{v_j\})$. The latter are the "separable functions" and they are not dense in $L^2$. $\endgroup$ – Willie Wong Jan 8 '13 at 11:24
  • $\begingroup$ Alright, I seem to be totally misunderstanding things, so I have written how I think it is (but first: lunch). (I'll look at your edit later). $\endgroup$ – Jonas Teuwen Jan 8 '13 at 12:04
  • $\begingroup$ Alright - thanks. Indeed, this is what I figured since I asked the question. But then I was wondering 'how bad' this could be - as in a splitting in a good and a bad part. Different question so I'll accept this. $\endgroup$ – Jonas Teuwen Jan 8 '13 at 14:16

Another way of looking at this is as a question about bases of various Hilbert-Sobolev spaces. Your representation of the solution of Poisson's equation is valid in an $L^2$ sense on a product region (0,1)x(0,1) because the tensor product of bases of $L^2$ spaces is a basis of the $L^2$ space on the product space.

When $L^2$ is replaced by a Sobolev space $H^1$ or $H^1_0$ then you need to choose appropriate bases. This can be done. See Auchmuty and Rivas, Laplacian Eigenproblems ... and Tensor Products of Hilbert Spaces, JMAA 435, (2016), 842-859 (and an addendum on their web-sites)


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