Solving Wave Equation in 1+1D via Fourier Transforms with Dirac Delta function initial condition

Question

I'm trying to use the Fourier transform method to solve the following PDE:

This is a an infinite string with a pulse for it's initial condition. (At $t=0$, the string is stricken sharply so that the impulse $\Delta J$ is transferred instantaneously at position $x=0$).

I wrote the PDE in terms of the Fourier Transform:

And then did algebra to get it here:

My issue comes with doing the inverse Fourier transform, as I am running into 2 integrals I don't know how to solve.

Piecing together my professors notes, somehow this is the solution to the first integral?

But even then, I'm left with a second integral that has no solution that makes sense:

If I run this through Mathematica, I get a Sign(c) Sign(t) function, which has no $x$ dependence. Not only is this not a function of $U(x, t)$, neither second derivative is also not equal to the delta function, so I don't believe this is the correct answer.

If anyone knows how to approach these integrals, I would greatly appreciate it.

Edit: I think something was up with my assumptions in Mathematica, because now I am getting the following:

Which looks more like a wave equation? Issue is, I'm not sure what the second* derivative for $\text{Sign}(c \cdot t \pm x)$ looks like with respect to either $t$ or $x$, so I am not able to confirm if this is correct.

Answer 1

Hints:

1. Instead of Fourier transformation, it is easier to introduce light-cone coordinates $$x^{\pm}~=~x\pm ct,\tag{A}$$ which factorize OP's PDE: $$ \frac{\partial^2u}{\partial x^+\partial x^-}~=~-\frac{\Delta J}{2c\rho} \delta(x^+)\delta(x^-). \tag{B}$$

2. The complete solution can now be easily obtained: $$ u~=~-\frac{\Delta J}{8c\rho}{\rm sgn}(x^+){\rm sgn}(x^-)+f_L(x^+) + f_R(x^-), \tag{C}$$ where $f_L,f_R$ are arbitrary functions.

3. Next invoke various boundary conditions. Presumably OP is interested in the causal/retarded solution $$ u~=~\frac{\Delta J}{2c\rho}\theta(x^+)\theta(-x^-), \tag{D}$$ which vanishes for $t<0$.

Answer 2

Note that any function that is a linear combination of terms of the form $f(x \pm ct)$ satisfies the homogeneous wave equation (with no source term). Your proposed solution meets this criterion, so it cannot be a solution to the inhomogeneous equation with the impulse source term.

The solution Mathematica proposes is almost correct, we just have to "turn it on" at $t = 0$, so that $U = 0$ for $t < 0$. The correct solution can be written as $$ U = \frac1{4} \frac{\Delta J}{\rho c} u(t)\left[\mathrm{sgn}(ct - x) + \mathrm{sgn}(ct + x)\right]$$ where $u(t) = \frac12[1+\mathrm{sgn}(t)]$ is the unit (or Heaviside) step function.

To verify this solution, plugging it into the left-hand side of the wave equation gives $$(\partial_t^2-c^2\partial_x^2)U=\frac{\Delta J}{\rho}\left\{2\delta(t)\delta(x)+\frac1{4c}\delta'(t)[\mathrm{sgn}(ct-x)+\mathrm{sgn}(ct+x)\right\}.$$ Using the identity $\delta'(y)f(y) = f(0)\delta'(y) - f'(0)\delta(y)$, the second term becomes $-\delta(t)\delta(x)$, so we end up with the correct source term $\frac{\Delta J}{\rho}\delta(t)\delta(x).$