# What Are The Necessary Conditions For a Function to be Separable?

A common step in solving many PDEs is to write a multivariable solution function as a product of two more or single variable functions. For example, if given,

$$\alpha\,\frac{\partial U(x,t)}{\partial x} = \frac{\partial U(x,t)}{\partial t},$$

one may begin be writing,

$$U(x,t) = X(x) T(t).$$

My question is, what are the necessary conditions for writing a multi variable function as a product of several single variable functions? Furthermore, are there any cases where this cannot be done?

• Pretty sure you could say that if your pde looks like this (just taking two independent variables - the generalization would work as well): $$f_1(x,U,U_x,U_{xx},\cdots)+f_2(x,U,U_x,U_{xx},\cdots)+\cdots+f_n(x,U,U_x,U_{xx},\cdots)+g_1(y,U,U_y,U_{yy},\cdots)+g_2(y,U,U_y,U_{yy},\cdots) +\cdots+g_m(y,U,U_y,U_{yy},\cdots)=0,$$ those would be sufficient conditions. I'm less sure about necessary conditions. You might also need the pde to be linear. Sep 25 '19 at 19:10
• There are certainly loads of non-separable pde's out there. Sep 25 '19 at 19:12

Of course most two-variable functions $$U(x,t)$$ cannot be written as $$X(x) T(t)$$. Those that can are, well, those that happen to be of that form...

The point in the context of PDEs like the heat equation is that you want to find a whole sequence of such separable solutions $$U_n(x,t)=X_n(x) T_n(t)$$ that you can use in order to write the general solution as a linear combination of these simple solutions: $$U(x,t)=\sum_{n=0}^\infty c_n U_n(x,t)$$.

• Yes, but it is important to note that the linearity of the differential operators are important for this to work. Sep 26 '19 at 13:07
• @mathreadler: Certainly! Sep 26 '19 at 13:29

A sufficient and necessary condition is the following.

A function $$F:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}:(x,y)\mapsto F(x,y)$$ can be written as $$F(x,y)=f(x)g(y)$$ for some functions $$f,g:\mathbb{R}\rightarrow\mathbb{R}$$ if and only if for every $$x,y,w,z\in\mathbb{R}$$ it holds that $$F(x,y)F(z,w)=F(x,w)F(z,y).$$

The implication "$$\implies$$" follows directly from the commutativity of multiplication and the other implication is obtained by first noticing that the zero function $$F=0$$ can obviously be written as a product of functions of a single variable, and if there exists some $$(a,b)\in\mathbb{R}^2$$ such that $$F(a,b)\neq 0$$, then we get that $$\forall x,y\in\mathbb{R}$$ it holds $$F(x,y)F(a,b)=F(x,b)F(a,y),$$ which implies that

$$F(x,y)=F(x,b)\frac{F(a,y)}{F(a,b)}=f(x)g(y).$$