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A string is at rest and in a straight line. At t = 0 it is subjected to a constant force distribution perpendicularly from above and along the entire string. This force distribution remains constant for all times $t >> 0$. Determine the wave-function of the string y = u(x, t) if it is bounded (u = 0 at the ends) and has length L. (The motion is planar: no point on the string ever leaves the xy plane.)

My suggestion:

Set the force equal to some constant , F. Then the wave-equation becomes inhomogeneous:

$$\alpha u_{xx}-u_{tt}=F\\ u_x(x,0)=0\\ u(0,t)=u(L,t)=0$$

Homogenize, by setting $u_{tt}=0$, solve the homogeneous eqn $u_{xx}=0$ and get $u(x)=Fx+C_1$

Set $v(x,t)=u(x,t)-F$ obtain the New PDE:

$$\alpha v_{xx}-v_{tt}=0\\ v_x(x,0)=F\\ v(0,t)=v(L,t)=0$$

Ansatz: $$u(x,t)=\sin\frac{n\pi}{L}xu(t)$$

Insert in PDE and get:

$$v(x,t)=\sum_{n=1}^\infty \sin\frac{n\pi}{L}x\bigg[C_2\cos\alpha\frac{n\pi}{L}t+C_3\sin\alpha\frac{n\pi}{L}t\bigg]$$

Use IC $v_x(x,0)=F$:

$$v_x(x,t)=\sum_{n=1}^\infty \frac{n\pi}{L}\cos\frac{n\pi}{L}x\bigg[-C_2\alpha\frac{n\pi}{L}\sin\alpha\frac{n\pi}{L}t+\alpha\frac{n\pi}{L}C_3\cos\alpha\frac{n\pi}{L}t\bigg]$$

$$v_x(x,t)=\sum_{n=1}^\infty \frac{n\pi}{L}\cos\frac{n\pi}{L}x\bigg[-C_2\alpha\frac{n\pi}{L}\sin0+\alpha\frac{n\pi}{L}C_3\cos0\bigg]$$

$$F=v_x(x,0)=\sum_{n=1}^\infty \bigg[\alpha\frac{n\pi}{L}\bigg]^2C_3\cos\frac{n\pi}{L}x$$

Relevant Fourier integrals:

$$\alpha_k=\frac{2}{L}\int_0^Lg(x)cos\frac{n\pi}{L}xdx$$

$$\alpha_0=\frac{1}{L}\int_0^Lf(x)dx$$

Clearly, here we see that $$g(x)=f(x)=\bigg[\alpha\frac{n\pi}{L}\bigg]^2C_3/F$$

But this does not seem correct, since I now have 2 unknown coefficients, $C_2,C_3$ and only one IC on $t$.

Using D'Alembert formula we could get:

$$u(x,t)=\frac{1}{2}\bigg(\sin\frac{n\pi}{L}(x+Ct)+\sin\frac{n\pi}{L}(x-Ct)\bigg)+Ft$$

But this seems odd.

Thanks

Luthier415Hz
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1 Answers1

2

Too long for a comment.

There are two "big" mistakes in your solution:

  1. If the string is initially at rest and in a straight line, the initial condition is not simply $u_x(x,0)=0$, but$^{(*)}$ $$ u(x,0)=0\quad \text{and}\quad u_t(x,0)=0; \tag{1} $$

  2. The solution to $\alpha u_{xx}=F$ is not $u(x)=Fx+C_1$, but $u(x)=\frac{Fx^2}{2\alpha}+C_1x+C_2$. Applying the boundary conditions $u(0)=u(L)=0$ to this solution one can determine $C_1$ and $C_2$; the final result is $$ u(x)=\frac{F}{2\alpha}x(x-L). \tag{2} $$

Therefore, rewriting the original PDE in terms of $v(x,t):=u(x,t)-\frac{F}{2\alpha}x(x-L)$, one obtains the homogeneous wave equation $$ \alpha v_{xx}-v_{tt}=0, \tag{3} $$ with initial conditions $$ v(x,0)=-\frac{F}{2\alpha}x(x-L), \qquad v_t(x,0)=0, \tag{4} $$ and boundary conditions $$ v(0,t) = v(L,t) = 0, \tag{5} $$ which can be solved with the method of separation of variables or Fourier series.

$^{(*)}$ Notice that $u_x(x,0)=0$ plus the boundary conditions $u(0,t)=u(L,t)=0$ is equivalent to $u(x,0)=0$, but the latter is a simpler way of expressing the same condition.

Gonçalo
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