Solving equations of the form $y(x) f(x) =0$

Question

When speaking with my advisor recently, we were led in the course of a physics problem to an equation of the form $$y(x) \ f(x) = 0$$ with $f(x)$ known and $y(x)$ unknown. My immediate instinct was to conclude that $y(x) = 0$ for all $x$. Clearly, this must be the right conclusion if there is no $x$ such that $f(x)=0$.

However, my advisor claims that a more general solution for any $f(x)$ is $$y(x) = \sum_{n=1}^m c_n \delta(x-x_n) $$ where the $c_n$ are arbitrary, $\delta()$ is the dirac distribution, and the $x_n$ satisfy $f(x_n) = 0$.

Intuitively, I understand why this is reasonable, but can it be justified rigorously? Is there any reason to prefer this solution over my instinctual one? The case of $f(x) =0$ seems particularly troublesome for this idea.

Follow-up: Thanks to Martin for providing a counter-example for arbitrary $f$. Does the claim hold if we require that $f$ is continuous and vanishes at a finite number of points?

Answer 1

It completely depends on what $f$ looks like. Your advisor's suggestion seems to imply that the thinks that $f=0$ on a finite number of points $x_1,\ldots,x_m$. But unless you specify this as a condition, it is not necessarily the case: consider for instance $$ f(x)=\begin{cases}x^2,&\mbox{ if }x\geq0 \\ 0,&\mbox{ if } x<0 \end{cases} $$ and $y(x)=f(-x)$. Then both $f$ and $y$ are differentiable, $y(x)f(x)=0$ for all $x$, and $y$ is not of the form you mention.

Answer 2

Check this example: $\displaystyle{{\rm y}'\left(x\right) = 0}$ with $\displaystyle{{\rm y}\left(0\right) = y_{0}}$. It has the obvious solution $\displaystyle{{\rm y}\left(x\right) = y_{0},\ \forall\ x}$. Let's assume we insist ($~\small\it\mbox{it's like killing an cockroach with a gun machine}~$) to use a Fourier transform to solve this equation:

\begin{equation} {\rm y}\left(x\right) = \int_{-\infty}^{\infty}{{\rm d}k \over 2\pi}\,\tilde{\rm y}\left(k\right) {\rm e}^{{\rm i}kx}. \quad \mbox{We get}\phantom{A} {\rm i}k\tilde{\rm y}\left(k\right) = 0 \quad\Longrightarrow\quad \tilde{\rm y}\left(k\right) = A\,\delta\left(k\right) \qquad\qquad\left(\Large\tt\mbox{I}\right) \end{equation}

where $A$ is a constant. Then $$ {\rm y}\left(x\right) = \int_{-\infty}^{\infty}{{\rm d}k \over 2\pi}\, A\,\delta\left(k\right){\rm e}^{{\rm i}kx} = {A \over 2\pi} \quad\Longrightarrow\quad {\rm y}\left(0\right) = y_{0} = {A \over 2\pi} \quad\Longrightarrow\quad {\rm y}\left(x\right) = y_{0}\,,\ \forall\ x $$

If you set $\tilde{\rm y}\left(k\right) = 0$ in step $\left(\Large\tt\mbox{I}\right)$, you get the wrong result ${\rm y}\left(x\right) = 0$.