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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 $D=\partial^2_t-\partial^2_x$).

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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 $\frac{d}{dx}f(x)=0$.

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  • $\begingroup$ 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
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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 $\mathrm{ker}(A)\cup\mathrm{ker}(B)\subsetneq\mathrm{ker}(AB)$.

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  • $\begingroup$ 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

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