Question

Method of Separation of Variables

Answer 1

I have no idea where to start with. If you don't mind, please teach me how to solve this kind of question. I have been struggling with these questions lately. Thank you so much.

Answer 2

Here's a start: Like a lot of differential equations, the solution starts with a good guess:

\[\theta(x,t)=f(t)g(x)\]

Plug this into the differential equation, and then divide both sides by \[f(t)g(x)\].
This moves everything that depends on to one side of the equation, and everyrthing
that depends on x to the other side. (Separating the variables - hence the name)

The only way the two sides can be equal is if both sides are equal to a constant, call
it \[ -\omega^2\].

As a result you get two regular differential equations that are easy to solve in place of the one wave equation you started with.

The general solution will then be a linear combination of the solutions with different values of
\]\omega\].

Answer 3

So i did what you said, then i got:
\[f''(t)=9g''(x)\]
rearrange it, and it becomes:
\[F''(t)/G''(x) = 9\]

Answer 4

here is the starting step:
we have to factorize the unknown function, like this:
\[\theta \left( {x,t} \right) = X\left( x \right) \cdot T\left( t \right)\]
wherein \(X\) depends on \(x\) only, and \(T\) depends on \(t\) only
then, after a substitution into the PDE, we get:
\[\frac{{\ddot T}}{T} = 9\frac{{X''}}{X}\]
wherein the dot notation stands for time derivative, and the superscript notation stands for \(x-\) derivative.
Such formula means that both right and left sides are constant, so we set:
\[\frac{{X''}}{X} = k\quad \; \Rightarrow \;\frac{{\ddot T}}{T} = 9k\]
the solution of the first equation, is:
\[X\left( x \right) = {c_1}{e^{\sqrt k x}} + {c_2}{e^{ - \sqrt k x}}\]
from which we get:
\[X'\left( x \right) = \sqrt k \left( {{c_1}{e^{\sqrt k x}} - {c_2}{e^{ - \sqrt k x}}} \right)\]

Answer 5

next, I apply the initial cndition on \(X'(x)\), and I can write this:
\[\begin{gathered}
X'\left( 0 \right) = \sqrt k \left( {{c_1} - {c_2}} \right) = 0\quad \Rightarrow {c_2} = {c_1} = c \hfill \\
\hfill \\
X'\left( {4\pi } \right) = c\sqrt k \left( {{e^{4\pi \sqrt k }} - {e^{ - 4\pi \sqrt k }}} \right) = 0\quad \Rightarrow {e^{4\pi \sqrt k }} - {e^{ - 4\pi \sqrt k }} = 0 \hfill \\
\end{gathered} \]

Answer 6

then we have this:
\[\begin{gathered}
{e^{4\pi \sqrt k }} - {e^{ - 4\pi \sqrt k }} = 0 \Rightarrow {e^{4\pi \sqrt k }} - \frac{1}{{{e^{4\pi \sqrt k }}}} = 0 \Rightarrow \frac{{{e^{8\pi \sqrt k }} - 1}}{{{e^{4\pi \sqrt k }}}} = 0 \Rightarrow \hfill \\
\hfill \\
\Rightarrow {e^{8\pi \sqrt k }} - 1 = 0 \Rightarrow {e^{8\pi \sqrt k }} = 1 = {e^{i2\pi n}},\quad n \in \mathbb{N} \hfill \\
\end{gathered} \]
namely:
\[8\pi \sqrt k = i2\pi n\quad \Rightarrow {k_n} = \frac{{ - {n^2}}}{{16}},\quad n \in \mathbb{N}\;\]
Now, please use such value of \(k_n\) and then find the functions \(X_n(x)\) and \(T(t)\)

Answer 7

@Michele_Laino okay. So, i did it according to your step, everything seems to make sense until the part where:
\[e^{8 \pi \sqrt{k}} = 1 = e^{i2 \pi n} \]
where do you get the \[e^{i \pi 2n}\]

Answer 8

It is simple:
we have to solve this equation:
\[\huge {e^{ix}} = 1\]
in order toget all solution, we have to use complex numbers. Now we can note this:
\[\Large {e^{i2\pi n}} = \cos \left( {2\pi n} \right) + i\sin \left( {2\pi n} \right) \equiv 1,\quad \forall n \in \mathbb{Z}\]
so we can write:
\[\huge {e^{ix}} = {e^{i2\pi n}}\]
and finally:
\[\huge {x_n} = 2\pi n,\quad \forall n \in \mathbb{Z}\]

Answer 9

solutions*

Answer 10

I have applied the formula of Eulero:
\[\Large {e^{i\theta }} \equiv \cos \theta + i\sin \theta \]

Answer 11

oh... i see. now that makes sense. i think you just do it differently with my lecturer, that's why i got confused. But after I subs. the k value into X and T. I just want to make sure I am doing this right. So I substitute the k value into T''/T = 9k and the same for X? is that right? so what do I do after that? Lol sorry, still not really good at this.

Answer 12

after the substitution, I can write this:
\[\huge {X_n} = {A_n}\cos \left( {\frac{{nx}}{4}} \right)\]

Answer 13

furthermore, we have:
\[\large \begin{gathered}
\frac{{\ddot T}}{T} = \frac{{ - 9{n^2}}}{{16}},\quad \Rightarrow Tt = {a_1}{e^{i\Omega t}} + {a_2}{e^{ - i\Omega t}}, \hfill \\
\hfill \\
\Omega = \frac{{3n}}{4} \hfill \\
\end{gathered} \]

Answer 14

oops..
\[\huge T\left( t \right) = {a_1}{e^{i\Omega t}} + {a_2}{e^{ - i\Omega t}}\]

Answer 15

so the requested eigenfunctions, are:
\[\large \begin{gathered}
\Theta \left( {x,t} \right) = {X_n}\left( x \right)T\left( t \right) = \hfill \\
\hfill \\
= {A_n}\cos \left( {\frac{{nx}}{4}} \right)\;\left\{ {{a_1}{e^{i\Omega t}} + {a_2}{e^{ - i\Omega t}}} \right\}, \hfill \\
\hfill \\
n \in \mathbb{N},\quad \Omega = \frac{{3n}}{4},\quad {a_1},{a_2} \in \mathbb{R} \hfill \\
\end{gathered} \]

Answer 16

more correctly, I have to write this:
\[{\Omega _n} = \frac{{3n}}{4}\]
so the space of solutions, is generated by these eigenfunctions:
\[\begin{gathered}
{f_n}\left( {x,t} \right) = \cos \left( {\frac{{nx}}{4}} \right) \cdot {e^{i{\Omega _n}t}} \hfill \\
\hfill \\
{g_n}\left( {x,t} \right) = \cos \left( {\frac{{nx}}{4}} \right) \cdot {e^{ - i{\Omega _n}t}} \hfill \\
\end{gathered} \]

Answer 17

I have applied the method which I learnt at the university

Answer 18

Your Xn does match to the given Xn, but the only difference is that it does not have An as you applied. just wonder does that mean th T(t) would be different as well?

Answer 19

Yeah, I know. I have come across a lot of different methods to do this while I am trying to search this on Internet. And I still couldn't understand the basic steps or methods to solve this kind of question.

Answer 20

please I got the same eigenfunctions \(X_n(x)\). There the quantities \(A_n\) are the corresponding amplitudes

Answer 21

we can write the functions \(T(t)\) as below:
\[{T_n}\left( t \right) = {B_{1n}}\cos \left( {{\Omega _n}t} \right) + {B_{2n}}\sin \left( {{\Omega _n}t} \right),\quad n \in \mathbb{N}\]

Answer 22

wherein \(B_{1n},B_{2n}\) are real numbers

Answer 23

Thanks. I will try to solve it myself for the sake of my exam but if by any chance i find difficulties with this type of question again, can I just tag you to ask for your help?

Answer 24

yes! :)