# One dimensional wave equation

Could someone help me out on the following questions?

Q. Consider the free vibrations of a string of length L clamped at x = 0 and constrained at x = L such that $$u_x \left( {L,t} \right) = - ku\left( {L,t} \right),k > 0$$.

(a) Show that the eigenvalues are given by the positive roots of: $$k\tan \left( {\omega L} \right) + \omega = 0$$. How many positive solutions for omega does this equation have?

(b) The string is released from rest so that $$u_t \left( {x,0} \right) = 0$$, and the string is given an initial deformation given by u(x,0) = (hx)/L, where h is constant. Use the principle of linear superposition to find the solution u(x,t).

Wave equation: $$\frac{{\partial ^2 u}}{{\partial t^2 }} = c^2 \frac{{\partial ^2 u}}{{\partial x^2 }}$$

For part (a) I just tried a separable solution $$u\left( {x,t} \right) = T\left( t \right)\phi \left( x \right)$$. Where lambda is a constant I obtain the two ODEs:

$$T''\left( t \right) + \lambda c^2 T\left( t \right) = 0$$

$$\phi ''\left( x \right) + \lambda \phi \left( x \right) = 0$$

The boundary conditions (the string being clamped at x = 0 and constrained at x = L in the specified way) results in the following ODE problem.

$$\frac{{d^2 \phi }}{{dt^2 }} + \lambda \phi = 0,\phi \left( 0 \right) = 0,\phi '\left( L \right) + k\phi \left( L \right) = 0$$

I found that if lambda is not positive then the ODE has trivial solutions so I take lambda to be positive.

$$\phi \left( x \right) = A\cos \left( {\omega x} \right) + b\sin \left( {\omega x} \right),\lambda = \omega ^2 > 0$$

Plugging in the boundary conditions of the ODE above I get a A = 0 and $$F\left( {\sin \left( {\omega L} \right) + k\omega \cos \left( {\omega L} \right)} \right) = 0$$. I take F not equal to zero to avoid a trivial solution to the ODE. So I get

$${\sin \left( {\omega L} \right) + k\omega \cos \left( {\omega L} \right)}$$ = 0

which does not correspond to the given result. I also can't figure out how many positive solutions for omega the equation that I needed to derive in part (a) has.

I don't think I can do part (b) without having completed part (a) because I need to sum over a sequence of functions whose arguments contain omega to find the solution to the PDE.

Can someone give me some help? Thanks.

Last edited:

HallsofIvy
Homework Helper
Benny said:
Could someone help me out on the following questions?

Q. Consider the free vibrations of a string of length L clamped at x = 0 and constrained at x = L such that $$u_x \left( {L,t} \right) = - ku\left( {L,t} \right),k > 0$$.

(a) Show that the eigenvalues are given by the positive roots of: $$k\tan \left( {\omega L} \right) + \omega = 0$$. How many positive solutions for omega does this equation have?

(b) The string is released from rest so that $$u_t \left( {x,0} \right) = 0$$, and the string is given an initial deformation given by u(x,0) = (hx)/L, where h is constant. Use the principle of linear superposition to find the solution u(x,t).

Wave equation: $$\frac{{\partial ^2 u}}{{\partial t^2 }} = c^2 \frac{{\partial ^2 u}}{{\partial x^2 }}$$

For part (a) I just tried a separable solution $$u\left( {x,t} \right) = T\left( t \right)\phi \left( x \right)$$. Where lambda is a constant I obtain the two ODEs:

$$T''\left( t \right) + \lambda c^2 T\left( t \right) = 0$$

$$\phi ''\left( x \right) + \lambda \phi \left( x \right) = 0$$

The boundary conditions (the string being clamped at x = 0 and constrained at x = L in the specified way) results in the following ODE problem.

$$\frac{{d^2 \phi }}{{dt^2 }} + \lambda \phi = 0,\phi \left( 0 \right) = 0,\phi '\left( L \right) + k\phi \left( L \right) = 0$$

I found that if lambda is not positive then the ODE has trivial solutions so I take lambda to be positive.

$$\phi \left( x \right) = A\cos \left( {\omega x} \right) + b\sin \left( {\omega x} \right),\lambda = \omega ^2 > 0$$

Plugging in the boundary conditions of the ODE above I get a A = 0 and $$F\left( {\sin \left( {\omega L} \right) + k\omega \cos \left( {\omega L} \right)} \right) = 0$$. I take F not equal to zero to avoid a trivial solution to the ODE. So I get

$${\sin \left( {\omega L} \right) + k\omega \cos \left( {\omega L} \right)}$$ = 0
You were good up to this point but I think perhaps you got $\phi$ and $\phi'$ confused. The general solution to your equation is, as you say
$$\phi \left( x \right) = A\cos \left( {\omega x} \right) + b\sin \left( {\omega x} \right),\lambda = \omega ^2 > 0$$
(Any reason for using b instead of B here?)
Certainly, since
$$\phi(0)= A= 0$$
we have
$$\phi \left( x \right) = b\sin \left( {\omega x} \right)$$
so that
$$\phi' \left( x \right) = b\cos \left( {\omega x} \right)$$
$$\phi'(L)+ k\phi(L)= b\omega cos(\omega L)+ kbsin(\omega L)$$
$$= b(\omega cos(\omega L)+ k sin(\omega L))= 0$$
In order that b not be 0, so that $\phi$ be non-trivial, you must have
$$\omega cos(\omega L)+ k sin(\omega L)= 0$$
or, dividing by $cos(\omega L)$,
$$\omega+ k tan(\omega L)= 0$$
exactly the condition given.

It looks I just pulled an "F" from out of nowhere but I think I just misread my working while I was transferring it from my work book and onto PF. So there was no particular reason for choosing b, I just wasn't thinking straight, not enough sleep.

Anyway thanks for pointing out my mistake. I see how to get to that equation now.

$$k\tan \left( {\omega L} \right) + \omega = 0$$

$$\tan \left( {\omega L} \right) = - \frac{\omega }{k}$$

I'm not sure how to determine how many positive solutions there are for omega. I think I need to actually find an explicit solution for omega to solve the PDE, is it possible or ncessary to solve for omega to solve the PDE?

One more thing, for a PDE what's the difference, if any, between a boundary condition and an initial condition? Any help would be great thanks.

HallsofIvy
tangent goes to negative infinity at all odd multiples of $\pi$ so the graph of $y= tan(\omega L)$ will cross the straightline graph of $y= -\frac{\omega}{k}$ infinitely often.