Does this equation imply separation of variables?

Asked Mar 08 '18 at 08:08

Active Mar 08 '18 at 08:08

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I have the following equation:

\begin{equation} \dfrac{q(t,r)}{s(t,r)}=B(r), \qquad \qquad (1) \end{equation}

with $q>0$ and $s>0$.

Apart from the trivial case $q(t,r)=\mathrm{const.} \times s(t,r)$ (i.e., $B(r)= \mathrm{const.}$) does this equation imply necessarily the separation of variables for $q$ and $s$

$$ q(t,r)=\alpha(t)Q(r), \qquad s(t,r)=\alpha(t)S(r)? $$

Taking the derivative of $(1)$ with respect to time leads to

\begin{equation} \dfrac{\partial_t\, s}{s}-\dfrac{\partial_t\, q}{q}=- \dfrac{s}{q}\, \partial_t\left(\dfrac{q}{s}\right)=0, \end{equation} implying that $q/s$ is a constant in time, i.e., it does not give any further information.

Thank you in advance for your help!

asked Mar 08 '18 at 08:08

Frank

you forget the case where $s(t,r)=\phi(r)q(t,r)$ not only the trivial case of constant .. – user577215664 Mar 08 '18 at 08:19
1

@Isham but then $\phi (r)=1/ B(r)$. It is a different version of the same equation. – Frank Mar 08 '18 at 08:21
yes of course ....but B is not a constant then – user577215664 Mar 08 '18 at 08:21
But that would mean that $s/q$ does not depend on time. Does this imply necessarily separation of variables for $s$ and $q$? – Frank Mar 08 '18 at 08:24

Does this equation imply separation of variables?

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