Separation of Variables in PDEs, regarding the separation constant

Question

I know the mechanics behind the separation of variables and how it is used to solve PDEs; however, I am confused by the assumption behind the separation constant.

Taking the wave equation as an example: $\frac{d^2(u)}{dt^2} =c^2\frac{d^2(u)}{dx^2}$ with the separation of variables, where $u(t,x)=X(x)T(t)$ we derive the expression: $\frac{T''}{T}=c^2\frac{X''}{X}=\omega$ where $\omega$ is the separation constant.

So here, many texts and videos state that since the left hand side is a function of only time and the right hand side is a function of only space (x); therefore, the $\omega$ has to be a constant.

I am confused about $c^2$, from the derivation of the wave equation we know that $c$ represents the velocity and is not dimensionless. So why is it not taken into consideration? Have I overlooked something? Thanks in advance :)

Answer 1

It is not necessary to repeat the argument so that $$\frac{T''}{T}=c^2\frac{X''}{X}=\omega=\text{constant}$$ You have well set out the argument.

More generally, consider the next PDE : $$\frac{d^2(u)}{dt^2} =\left(c_1(x)c_2(y)\right)^2\frac{d^2(u)}{dx^2}$$ where $c$ is no longer a constant but the product of a function of $x$ and a function of $y$. The separation of variables leads to : $$\frac{1}{\left(c_2(y)\right)^2}\frac{T''}{T}=\left(c_1(x)\right)^2\frac{X''}{X}=\omega=\text{constant}$$ $\omega$ is any constant. There is no reason to say that a constant $\omega$ be related to a function $c_1(x)$ or to another function $c_2(y)$. So, it would be senseless to ask " Why $c_1$ or $c_2$ are not taken into consideration? ".

That is the same for the simpler case where $c_1(x)=c=$constant and $c_2(x)=1$.

In fact $c$ is effectively taken into consideration, but not at this step of calculus. It is taken into consideration at next step when the equation $c^2\frac{X''}{X}=\omega$ is solved.

For each value of $\omega$ the solutions involve $c$. They are as many solutions with $c$ into them than they are different $\omega$ and also the linear combinations of those solutions. So, they are an infinity of solutions until no boundary conditions is specified.

Answer 2

$c$ isn't taken into consideration because to get to this linear form of the wave equation from the derivation, $c$ is assumed to be constant. The dimensions of $c$ just ensure that the equation is dimensionally consistent. When authors say the left-hand side is only a function of $t$ and the right-hand side is only a function of $x$, they mean to say $$ \underbrace{\frac{\partial}{\partial x}\left(\frac{T''(t)}{T(t)}\right)}_{0} = \frac{\partial}{\partial x}\left(c^2\frac{X''(x)}{X(x)}\right) $$ and also $$ \frac{\partial}{\partial t}\left(\frac{T''(t)}{T(t)}\right) = \frac{\partial}{\partial t}\left(c^2\frac{X''(x)}{X(x)}\right) = c^2\underbrace{\frac{\partial}{\partial t}\left(\frac{X''(x)}{X(x)}\right)}_{0}.. $$ Thus the equation is constant in both space and time.

Answer 3

$\frac{T''}{T}=c^2\frac{X''}{X}$ is a constant that does not depend on $x$ or $t$ iff $\frac{1}{c^2}\frac{T''}{T}=\frac{X''}{X}$ is a constant that does not depend on $x$ or $t$. If the first constant is $C$ and the second is $D$, then $c^2D=C$. The constants $C$ and $D$ are not arbitrary; they are determined by endpoint conditions, and no matter where you put the constant $c^2$, the solutions will be the same, and the $c^2$ will be properly incorporated into the problem.