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I have a query regarding the practical applicability of variable separable method. Usually we decompose the variables w.r.t components corresponding to different dimensions, this method is used to solve a differential equation analytically. When it comes to the practical usage, we are often provided already with the solution in form of model output. Is it possible to decompose them along their dimensions? For example I can write the zonal velocity as

              U(x, y, z, t) = uv(z) * uh(x,  y,  t)

Will this decomposition be unique? I am clueless because it just seems to be a matrix multiplication. I wonder how would be the form of those matrices. Is it possible to obtain uv and uh from the data variable using any software tool?

Edit: An example of such a decomposition is provided here at page number 7 (equation 2.6). I wonder if it is possible to obtain the decomposed RHS values practically from a variable.

Moreover I often see a PDE associated with the decomposition which is absolutely logical. The equation is solved analytically by substituting the decomposed variables and different software tools are able to plot the analytical solution. The situation is little different in the study attached herewith. I wonder how to obtain the decomposed values if I am not getting any way to solve it analytically.

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  • $\begingroup$ Can you further clarify your question? I do not see the matrix multiplication that you talk about in the formula that you write down. $\endgroup$ – Basileios May 10 at 15:57
  • $\begingroup$ @Agni I feel this question is more suited to scicomp rather than ES SE. $\endgroup$ – gansub May 11 at 12:34
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Multiplicative decomposition as defined in the post is often called factorization. It is not always possible to achieve an exact factorization. There would be some residual difference that we could denote as eps(x, y, z, t).

U(x, y, z, t) = uv(z) * uh(x,  y,  t) + eps(x, y, z, t)

Such a decomposition is not unique, however you may choose a pair of functions that minimizes the value of eps(x, y, z, t) according to some metric.

For further reading consider the description of sklearn Non-negative matrix factorization function.

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  • $\begingroup$ I agree! The decomposition depends upon the PDE as well, and the same data variable may have different expressions, so the form of decomposition will change accordingly, which may give slightly different residues each time. $\endgroup$ – Agni Jun 10 at 9:31

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