Separation of Variables, but not equal to constant

[Foracle]

TL;DR

Is the following PDE separable
$$\frac{1}{T} \frac{\partial T}{\partial t} = k \frac{1}{Q} \frac{\partial Q}{\partial q} - \frac{\partial q}{\partial t} \frac{1}{Q} \frac{\partial Q}{\partial q}$$
where $\frac{\partial q}{\partial t}$ can depend on $q$ and $t$?

Suppose I have 2 variables q and t (time), where q is some reparameterization of x (position) : $x \to q = x f(t)$.

Suppose I have a partial differential equation :
$$\frac{\partial u(q,t)}{\partial t} = k \frac{\partial u(q,t)}{\partial q}$$
where k = constant
Then I do a separation of variables $u(q,t) = Q(q)T(t)$
The differential equation becomes (after some manipulation):
$$\frac{1}{T} \frac{\partial T}{\partial t} = k \frac{1}{Q} \frac{\partial Q}{\partial q} - \frac{\partial q}{\partial t} \frac{1}{Q} \frac{\partial Q}{\partial q}$$
where I have used the fact that $\frac{\partial Q}{\partial t} = \frac{\partial q}{\partial t} \frac{\partial Q}{\partial q}$

Now the left hand side is only dependent on $t$, while the right hand side depends on both $q$ and $t$. Since both sides still depend on $t$, can I say that
$$(LHS) = (RHS) = g(t)$$
(LHS = Left hand side, RHS = Right hand side, g(t) is some function of time)

Additional question :
I have seen on a research paper where the author says that for the above equation to be separable, $\frac{\partial q}{\partial t}$ has to be constant so that RHS only depends on $q$, hence $(LHS) = (RHS) = constant$.
But since $q$ still depends on time ($q = x f(t)$), doesn't this mean RHS still depends on time and it should be $(LHS) = (RHS) = g(t)$ instead?

 

[pasmith]

Summary:: Is the following PDE separable
$$\frac{1}{T} \frac{\partial T}{\partial t} = k \frac{1}{Q} \frac{\partial Q}{\partial q} - \frac{\partial q}{\partial t} \frac{1}{Q} \frac{\partial Q}{\partial q}$$
where $\frac{\partial q}{\partial t}$ can depend on $q$ and $t$?

Suppose I have 2 variables q and t (time), where q is some reparameterization of x (position) : $x \to q = x f(t)$.

Suppose I have a partial differential equation :
$$\frac{\partial u(q,t)}{\partial t} = k \frac{\partial u(q,t)}{\partial q}$$
where k = constant
Then I do a separation of variables $u(q,t) = Q(q)T(t)$
The differential equation becomes (after some manipulation):
$$\frac{1}{T} \frac{\partial T}{\partial t} = k \frac{1}{Q} \frac{\partial Q}{\partial q} - \frac{\partial q}{\partial t} \frac{1}{Q} \frac{\partial Q}{\partial q}$$
where I have used the fact that $\frac{\partial Q}{\partial t} = \frac{\partial q}{\partial t} \frac{\partial Q}{\partial q}$

If by \dfrac{\partial}{\partial t} you mean differentiation with respect to t with q held constant, then by definition <br /> \frac{\partial q}{\partial t} = 0 and Q depends on q alone.

Now the left hand side is only dependent on $t$, while the right hand side depends on both $q$ and $t$. Since both sides still depend on $t$, can I say that
$$(LHS) = (RHS) = g(t)$$
(LHS = Left hand side, RHS = Right hand side, g(t) is some function of time)

Additional question :
I have seen on a research paper where the author says that for the above equation to be separable, $\frac{\partial q}{\partial t}$ has to be constant so that RHS only depends on $q$, hence $(LHS) = (RHS) = constant$.
But since $q$ still depends on time ($q = x f(t)$), doesn't this mean RHS still depends on time and it should be $(LHS) = (RHS) = g(t)$ instead?

You need to provide more context. Are you changing variables from (x,t) to (q = xf(t),t)? If so, differentiation with respect to t with x held constant is not the same thing as differentiation with respect to t with q held constant, and you need to be clear which of the two you mean when you write \dfrac{\partial}{\partial t}.

 

[Foracle]

Are you changing variables from (x,t) to (q=xf(t),t)?

Yes, I am doing that.

you need to be clear which of the two you mean when you write $\frac{∂}{∂t}$

When I do $\frac{\partial u(q,t)}{\partial t}$, I am not holding anything constant. So
$$\frac{\partial }{\partial t} = \frac{\partial q}{\partial t} \frac{\partial}{\partial q} + \frac{\partial}{\partial t}$$
The $\frac{\partial}{\partial t}$ on the 2nd term of the RHS holds q constant.

 

[Foracle]

I have another question that is kind of related.

I am currently studying this research paper by O. Hamidi. What bothers me is in the equation (5), why did he choose to write $\tau (x,t) = G(x) + K(t)$ instead of the usual separation of variables $\tau (x,t) = G(x)K(t)$?
Both substitutions give different solution, and both methods seem good to me. Why do people use one over the other?