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If I have a function $f(x,y)$, is it always true that I can write it as a product of single-variable functions, $u(x)v(y)$?


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I am asking this because for the time-independent Schroedinger equation, it seems to assume that this is true. – H Taylor Aug 26 '11 at 18:04
Are you sure that is assumed? Or (for a linear equation) you first find all solutions of that kind, then take linear combinations? (That would be a conventional thing to do for certain PDEs.) – GEdgar Aug 26 '11 at 18:26
@GEdgar: As well as show the separable solutions are orthogonal and form a full basis. – anon Aug 26 '11 at 18:34
Thanks. The thing is, they just seem to multiply the solutions of the $x$-dependent equation by $e^{-iEt/\hbar}$ -- the solution of the $t$-dependent equation... – H Taylor Aug 26 '11 at 18:49
some linear pdes (heat/wave/laplace) are solved by assuming separation of variables for a solution and taking "linear combinations" to get series solutions. the assumption of separation of variables is convenient because it turns the pdes into systems of odes, which are easier to solve. i believe this is how fourier obtained his series. – yoyo Aug 26 '11 at 20:55
up vote 5 down vote accepted

No, it is not. It is unusual that you can do so. For example, $f(x,y)=x+y$ cannot be.

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Unfortunately, the answer is no. For example, let $f(x,y)=x-y$.

Note that $f(1,1)=0$. If $f(x,y)=u(x)v(y)$ for all $x$, $y$, then $0=f(1,1)= u(1)v(1)$. But if $u(1)v(1)=0$, then $u(1)=0$ or $v(1)=0$.

Suppose for example that $u(1)=0$. Then $u(1)v(y)=0$ for all $y$. If $f(x,y)=u(x)v(y)$ for all $x$, $y$, then $1-y=0$ for all $y$. This is clearly not true.

The example $f(x,y)=x-y$ is not really special. "Most" functions $f(x,y)$ are not expressible as $u(x)v(y)$.

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It is true that linear combinations of such functions are dense. Depending on exactly what space of functions you're working with, this should follow from the locally compact form of the Stone-Weierstrass theorem. Thus it is not unreasonable to expect that we can find solutions to a linear PDE by taking the closure of the space of linear combinations of separable solutions.

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If $f(x,y)=x^2+y^2$ could be written as $u(x)v(y)$, then since $f(0,0)=0$, this means either $u(0)=0$ or $v(0)=0$. If this were the case, $f(x,y)$ would be equal to zero on either the entire $y$-axis or the entire $x$-axis. This clearly is not the case, so $f(x,y)$ can not be written as $u(x)v(y)$.

You might try graphing a few functions that are of the form $u(x)v(y)$ and see what the graph looks like. This might give some insight to why the statement is not true.

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If a multivariable function $f(x_1,\dots,x_n)$ is separable then the Hessian $H\ln f$ is diagonal, which isn't usually the case. Moreover, if $f$ decomposes into $g_1(x_1)\cdots g_n(x_n)$, then

$$\int \frac{\partial\ln f}{\partial x_i} dx_i=g_i(x_i)+C$$

where the integration is done in a single-variable sense.

Can someone prove a $C^2$ function $f$ is always separable when $H\ln f$ is diagonal?

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The question is not as simple as it seems to be! The answer is YES. For example, let f(x,y)=x−y. Then we can find z1=x-y, z2=x+y, and f(x,y)=f(z1,z2)=z1. This problem is actually related to diagonalization of a function in a specific basis. A separable multivariable function f(x1,…,xn) means that if we can find the basis or the tansformation.

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This is not correct. The answer to the original question is NO. – azimut May 24 at 10:11
Yes, your are right, but I say the other aspect of this question. – Lin Zhang May 24 at 14:28

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