Consider the wave equation in one dimension:

$$\frac{\partial^2 u}{\partial t^2}-\frac{\partial^2 u}{\partial x^2}=0.$$

The most general solution of this can be written as $F(x-t)+G(x+t)$ for arbitrary functions $F, G$. It is commonly said that this is a consequence of the factorization

$$\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}=\left( \frac{\partial }{\partial t}-\frac{\partial}{\partial x}\right)\left( \frac{\partial }{\partial t}+\frac{\partial}{\partial x}\right).$$

Is this a general fact?

More precisely, assume that the differential operator $D$ factors into the product of two (commuting) operators $A, B$, that is $$D=AB=BA.$$ Is it true that $$ \{u\in C^{n}\ :\ Du=0\}=\{F+G\ :\ AF=0\ \text{and}\ BG=0\}?$$ Here $n$ is the order of $D$ (which is $2$ for $D=\partial^2_t-\partial^2_x$).


If I understand the question correctly, the answer is no, because it is not even true for the square of an operator: solutions to $\frac{d^2}{dx^2}f(x)=0$ are not just sums of two of the constant solutions of $\frac{d}{dx}f(x)=0$.

  • $\begingroup$ I'm sorry, I forgot to thank you. You clearly show that the problem is subtler than I thought. Of course we can say that every solution of $Au=0$ is also solution of $ABu=0$ but not the other way round. $\endgroup$ – Giuseppe Negro Mar 2 '12 at 18:39

Clearly, $\mathrm{ker}(A)\cup\mathrm{ker}(B)\subset\mathrm{ker}(AB)$, and if the operators are linear, then $\mathrm{ker}(A)\oplus\mathrm{ker}(B)\subset \mathrm{ker}(AB)$. But take $f(x,t)=x$. Then, given $A=\partial_{t}-\partial_{x}$ and $B=\partial_{t}+\partial_{x}$, $f$ is in neither $\mathrm{ker}(A)$ nor in $\mathrm{ker}(B)$. But clearly, $f\in \mathrm{ker}(AB)$, so in general $\mathrm{ker}(A)\cup\mathrm{ker}(B)\subsetneq\mathrm{ker}(AB)$.

  • $\begingroup$ The clear inclusion is that the sum $\text{ker}(A) + \text{ker}(B)$ lies in $\text{ker}(AB)$, not just the union, and the question is whether this inclusion is strict or not. $\endgroup$ – Qiaochu Yuan Mar 1 '12 at 2:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.