Using Separation of Variables to Solve $\frac{\partial{u}}{\partial{t}} = x^2 \frac{\partial^2{u}}{\partial{x}^2}$ on $1 < x < e$, $0 < t < \infty$

Question

I have the following PDE problem.

Proceeding as follows, use the method of separation of variables to solve

$\dfrac{\partial{u}}{\partial{t}} = x^2 \dfrac{\partial^2{u}}{\partial{x}^2}$ on $1 < x < e$, $0 < t < \infty$

subject to

$u(1, t) = 0$ and $u(e, t) = 0$ and $u(x, 0) = f(x)$

I am then instructed as follows. I have included my solution below each instruction.

A) Try $u(x,t) = X(x)T(t)$ to obtain a pair of ODEs

Let $u(x, t) = X(x)T(t)$.

$\therefore T'X = x^2X''T$

$\implies x^2 \dfrac{X''}{X} = \dfrac{T'}{T}$

$\therefore \dfrac{T'}{T} = - \lambda = x^2 \dfrac{X''}{X}$

$X'' + \dfrac{1}{x^2} \lambda X = 0$ and $T' + \lambda T = 0$

B) Solve the differential equation for $X(x)$ by trying $X(x) = x^p$.

$X(x) = x^p$: $p(p - 1)x^{p - 2} + \dfrac{1}{x^2} \lambda x^p = 0$

$\therefore$ We require that $p^2 - p + \lambda = 0$.

$\therefore p = \dfrac{1}{2} \pm \dfrac{\sqrt{1 - 4\lambda}}{2}$

C) What condition must $\lambda$ satisfy in order to obtain oscillatory solutions?

For oscillatory solutions, we require $1 - 4\lambda < 0$

$\implies 1 < 4\lambda \implies \lambda > \dfrac{1}{4}$

D) Determine these oscillatory solutions. Note that $x^a = e^{a \ln x}$.

This is the part that I'm confused with.

My solution is different from the instructor's solution.

My Solution

Let $1 - 4\lambda = - \beta^2 \implies \lambda = \dfrac{\beta^2 + 1}{4}$.

$\therefore p = \dfrac{1 \pm \sqrt{-\beta^2}}{2}$

$\implies p = \dfrac{1}{2} \pm \dfrac{i \beta}{2} $

$\therefore X(x) = x^{1/2} \pm \frac{i \beta}{2} = x^{1/2}x^{\pm i \beta/2} = \sqrt{x}e^{\pm (i \beta \ln(x))/2 } = \sqrt{x} \left[ A\cos \left( \dfrac{\beta \ln(x)}{2} \right) + B\sin \left( \dfrac{\beta \ln(x)}{2} \right) \right]$

Instructor's Solution

$1 - 4\lambda = -4\beta^2$

$\implies \lambda = \dfrac{1 + 4\beta^2}{4}$

$\implies p = \dfrac{1}{2} \pm i \beta$

$\implies X(x) = x^{1/2 \pm i \beta} = \sqrt{x}e^{\pm i \beta}$

$\implies X(x) = \sqrt{x}[A \cos(\beta \ln x) + B \sin(\beta \ln x)]$

Notice that the instructor has $1 - 4\lambda = -4\beta^2$, whereas I have $1 - 4\lambda = - \beta^2$ Since we found in C) that we need $1 - 4\lambda < 0$, it seems to me that $1 - 4\lambda = - \beta^2$ would be correct? I cannot understand why the instructor's solution of $1 - 4\lambda = -4\beta^2$ would be correct?

I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

Both the solutions are correct.

After when you have decided to declare $1 - 4 \lambda < 0$ to get oscillatory solutions, now it's more a matter of choice, what do you put $1-4\lambda$ equal to.

You can even choose $1-4\lambda = -\beta $. It must be mentioned that $\beta >0$. But that will give you a $\sqrt{\beta}$ in solutions.

Similarly, $-\beta^2$ will give $\beta/2$ at some point and $-4 \beta^2$ will give $\beta$.

The reasons to avoid $\sqrt{\beta}$ or for that matter even $\beta /2$ (as your instructor did) could only be aesthetics or if you ever program this solution into computer, square-roots and divisions might take more time to calculate.

Sometimes you might want the solution to look like a particular expression, you can choose accordingly.