Separation of variables: Wave equation governing a string with fixed ends

[jonny997]

Homework Statement

Deriving the general solution to wave eq via separation of variables

Relevant Equations

Included in image

I am trying to follow through a derivation of the solution to the wave equation governing a string with fixed ends via separation of variables but I am stuck at the step which concludes ‘m’ must be a natural number 1,2,3,… etc. as opposed to an integer. After analysing each case: m > 0, m = 0 and m < 0, I’m still a bit confused as to why we can rewrite the solution in terms of A_n and B_n, and consider only m > 0 when the coefficients for m>0 and m<0 differ by a negative sign … does this not matter or have i made a mistake somewhere?

 

[kuruman]

I think that you are unnecessarily complicating matters.

If $\varphi''(x)=-m^2\varphi(x)$, then you can immediately assume a solution $$\varphi(x)=C_1~e^{\sqrt{(-m^2)}~x}+C_2~e^{-\sqrt{(-m^2)}~x}$$which you can verify by direct substitution back into the spatial part of the equation.

At $x = 0$, $$0=\varphi(0)=C_1+C_2 \implies \varphi(x)=C(e^{\sqrt{(-m^2)}~x}-e^{-\sqrt{(-m^2)}~x})$$
At $x = \pi$, $$\begin{align} & 0=\varphi(\pi)=C(e^{\sqrt{(-m^2)}~\pi}-e^{-\sqrt{(-m^2)}~\pi}) \nonumber \\ & \implies e^{\sqrt{(-m^2)}~\pi}=e^{-\sqrt{(-m^2)}~\pi}. \end{align}$$What does equation (1) say about the exponent $\sqrt{(-m^2)}~$?

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[jonny997]

I think that you are unnecessarily complicating matters.

If $\varphi''(x)=-m^2\varphi(x)$, then you can immediately assume a solution $$\varphi(x)=C_1~e^{\sqrt{(-m^2)}~x}+C_2~e^{-\sqrt{(-m^2)}~x}$$which you can verify by direct substitution back into the spatial part of the equation.

At $x = 0$, $$0=\varphi(0)=C_1+C_2 \implies \varphi(x)=C(e^{\sqrt{(-m^2)}~x}-e^{-\sqrt{(-m^2)}~x})$$
At $x = \pi$, $$\begin{align} & 0=\varphi(\pi)=C(e^{\sqrt{(-m^2)}~\pi}-e^{-\sqrt{(-m^2)}~\pi}) \nonumber \\ & \implies e^{\sqrt{(-m^2)}~\pi}=e^{-\sqrt{(-m^2)}~\pi}. \end{align}$$What does equation (1) say about the exponent $\sqrt{(-m^2)}~$?

Please note that we prefer that you post your work in $\LaTeX$ which makes the equations more legible and easier to refer to. To learn how, please click on the link "LaTeX Guide", lower left abpve "Attach files." Thank you.

Hey, thanks for the response. I guess eq. (1) implies that m in the exponent has to be an integer?

 

[kuruman]

Hey, thanks for the response. I guess eq. (1) implies that m in the exponent has to be an integer?

Not yet. In general you can write the exponent as a complex number, $\sqrt{(-m^2)}=a+ib$. Put that back in the boundary condition and see what you can say about the real and imaginary parts $a$ and $b$.

 

[jonny997]

Not yet. In general you can write the exponent as a complex number, $\sqrt{(-m^2)}=a+ib$. Put that back in the boundary condition and see what you can say about the real and imaginary parts $a$ and $b$.

Is it necessary to write $\sqrt{(-m^2)}=a+ib$? Can I not just write $e^{imx} = e^{-imx}$

This is what I got writing the exponent out as a + ib:
$$ e^{(a+ib)\pi} = e^{-(a+ib)\pi} \rightarrow e^{2\pi(a+ib)} = 1 \rightarrow e^{2\pi a}(\cos{2\pi b} + i\sin{2\pi b}) = 1 $$
Then I guess a has to be 0 and b has to be an integer?

 

[kuruman]

Is it necessary to write $\sqrt{(-m^2)}=a+ib$? Can I not just write $e^{imx} = e^{-imx}$

Yes, it is necessary because you don't know that $m$ is purely imaginary until you show it to be so. Also, $e^{imx} \neq e^{-imx}$. This equation comes from the boundary condition and you cannot replace $\pi$ with $x$.

Then I guess a has to be 0 and b has to be an integer?

It's not a guess, it's a conclusion and it follows from the premises. You did it, and thanks for using $\LaTeX.$

 

[jonny997]

Yes, it is necessary because you don't know that $m$ is purely imaginary until you show it to be so.

It's not a guess, it's a conclusion and it follows from the premises. You did it!

Okay thank you. I’m still a bit confused though… In the book I’m following m is restricted further and only m = 1,2,3,… is considered. If m = 0 the solution vanishes. I’m not understanding why the case in which m<0 resolves to that of m>0 when the coefficients for cosine differs by a -ve sign. Hopefully, that makes sense lol.

 

[kuruman]

. . . when the coefficients for cosine differs by a -ve sign

What coefficients for cosine? It looks like you tried to do this in your head without following through with the algebra. We have found that the exponent is purely imaginary, i.e. $\sqrt{(-m^2)}=ib$. This makes the boundary condition $e^{i\pi b}=e^{-i\pi b}$ which implies, as you said, that $b=n$ where $n$ is an integer. Note that whether $n$ is positive or negative does not affect the boundary condition in any way.

Now let's put this back into the spatial wave function and write $$\varphi(x)=C(e^{i~n~x}-e^{-i~n~x})=C~2i\sin(nx)=A\sin(nx)~~~~(A\equiv 2iC).$$This last form is a solution of $$\frac{\varphi''(x)}{\varphi(x)}=-n^2~~~~~(n \text{ integer})$$and satisfies the boundary conditions.

 

[jonny997]

What coefficients for cosine? It looks like you tried to do this in your head without following through with the algebra. We have found that the exponent is purely imaginary, i.e. $\sqrt{(-m^2)}=ib$. This makes the boundary condition $e^{i\pi b}=e^{-i\pi b}$ which implies, as you said, that $b=n$ where $n$ is an integer. Note that whether $n$ is positive or negative does not affect the boundary condition in any way.

Now let's put this back into the spatial wave function and write $$\varphi(x)=C(e^{i~n~x}-e^{-i~n~x})=C~2i\sin(nx)=A\sin(nx)~~~~(A\equiv 2iC).$$This last form is a solution of $$\frac{\varphi''(x)}{\varphi(x)}=-n^2~~~~~(n \text{ integer})$$and satisfies the boundary conditions.

Sorry I’ll try to explain it better. We can write the spatial part of the solution as $\varphi (x) = A\sin{nx}$ and the temporal part as $\psi (t) = B\cos{nt} + C\sin{nt}$, with $n \in \mathbb{Z}$. Part of the full solution then is $$ u_{n}(x,t) = \varphi (x)\psi (t) = (AB\cos{nt} + AC\sin{nt})\sin{nx}$$ Then comparing what happens for the following 3 cases:
$n = 0: u_{n}(x,t) = 0$
$n > 0: u_{n}(x,t) = (AB\cos{|n|t} + AC\sin{|n|t})\sin{|n|x}$
$n < 0: u_{n}(x,t) = (-AB\cos{|n|t} + AC\sin{|n|t})\sin{|n|x}$

I should be able to write $u_{n}(x,t) = (a_{n}\cos{nt} + b_{n}\sin{nt})\sin{nx}$ and take a linear superposition over all n > 0, i.e., n = 1,2,3 etc. to obtain the full solution u(x,t). The coefficient of the cosine term for n < 0 and n > 0 differs by -1 though, does this not matter?