# Thread: wave partial differential eqn

1. ## wave partial differential eqn

Can someone help me solve this: the vibrating string partial differential equation if an external force per unit mass proportional to displacement acts on the string.

So I have uxx - ku = (1/a^2)utt

How do I solve this? I've only seen it without external forces (here Pauls Online Notes : Differential Equations - Vibrating String). I've just started studying PDE's.

2. Originally Posted by PvtBillPilgrim
Can someone help me solve this: the vibrating string partial differential equation if an external force per unit mass proportional to displacement acts on the string.

So I have uxx - ku = (1/a^2)utt

How do I solve this? I've only seen it without external forces (here Pauls Online Notes : Differential Equations - Vibrating String). I've just started studying PDE's.

Do you have boundary and initial conditions to go with this PDE?

3. Originally Posted by danny arrigo
Do you have boundary and initial conditions to go with this PDE?
That's what I was wondering.

I assume that the vibrating string has fixed endpoints.

$\displaystyle u(0,t) = u(L,t) = 0$

But you need some initial conditions as well to determine all arbitrary constants.

4. Originally Posted by Rincewind
That's what I was wondering.

I assume that the vibrating string has fixed endpoints.

$\displaystyle u(0,t) = u(L,t) = 0$

But you need some initial conditions as well to determine all arbitrary constants.
It could also have a free endpoint where $\displaystyle u_x(0,t) =0,\;\; \text{or}\;\;\;u_x(L,t) = 0$.

5. Originally Posted by danny arrigo
It could also have a free endpoint where $\displaystyle u_x(0,t) =0,\;\; \text{or}\;\;\;u_x(L,t) = 0$.
Yup. Anything is possible. I wonder if PvtBillPilgrim will clarify.

6. Originally Posted by Rincewind
That's what I was wondering.

I assume that the vibrating string has fixed endpoints.

$\displaystyle u(0,t) = u(L,t) = 0$

But you need some initial conditions as well to determine all arbitrary constants.
I want these conditions.
Basically, u(x,0)=f(x)
du/dt(x,0)=g(x) where this is a partial derivative
u(0,t)=0=u(L,t)
How does the -ku play into the final solution? It's solved in the website I gave in the original post, but with no external forces.

7. Originally Posted by PvtBillPilgrim
I want these conditions.
Basically, u(x,0)=f(x)
du/dt(x,0)=g(x) where this is a partial derivative
u(0,t)=0=u(L,t)
How does the -ku play into the final solution? It's solved in the website I gave in the original post, but with no external forces.
Use the usual separation of variables $\displaystyle u = T(t) X(x)$ so your equation separates

$\displaystyle \frac{T''}{a^2 T} = \frac{X'' - k X}{X} = c$

From the second

$\displaystyle X'' - k X = c X$ or $\displaystyle X'' + \omega^2 X = 0$ so $\displaystyle c = - k - \omega^2$. I'm thinking you can take it from here.