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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, $u(x)v(y)$?

Thanks.

H Taylor
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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
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    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
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    @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
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    @H: But then what do they do with them? Add many of them together to match the boundary/initial conditions? – GEdgar Aug 26 '11 at 20:58
  • @yoyo: Thanks. I guess it is the thing applied mathematicians seem to like to do: first make an assumption, carry out calculations, then verify that the assumption is valid...? I am guessing that one way of verifying that is to check that this way gives the correct number of linear independent eigenfunctions for a given eigenvalue? – H Taylor Aug 26 '11 at 21:17
  • cont'... The thing I don't know is the number of such l.i. eigenfns to expect when given a PDE. Assuming that this method is right, which it surely is, the number of l.i. eigenfns is 2 for each eigenvalue? How could I tell the number of l.i. eigenfns given a PDE? Is it true that for any 2nd order PDEs there are 2 l.i. efns even if it involves more than one variable? – H Taylor Aug 26 '11 at 21:23
  • @GEdgar: Indeed. But all are of the form $C_n u_n(x)e^{-iE_nt/\hbar}$ where $C_n$ is a constant. – H Taylor Aug 26 '11 at 21:24
  • Actually the general solution is the sum of the series containing terms in my comment above and an integral $\int dE C(E) u_E(x) e^{-iEt/\hbar}$ – H Taylor Aug 26 '11 at 21:29

5 Answers5

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

Ross Millikan
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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 $u(x)v(y)$.

André Nicolas
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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.

Qiaochu Yuan
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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.

ShawnD
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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?

anon
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