# Separation of Variables, but not equal to constant

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• Foracle

#### Foracle

TL;DR 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}##

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)

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?

Last edited:

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 $$\frac{\partial q}{\partial t} = 0$$ 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)

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 $\dfrac{\partial}{\partial t}$.

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.

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?