# Wave equation 1D inhomogeneous Laplace/Fourier Transforms vs Green's Function

I am trying to solve the following 1D inhomogeneous wave equation. Forgive me if I a miss any rigorous mathematical concept.

$$\frac{\partial^2 u}{\partial x^2} - \frac{1}{c^2}\frac{\partial^2 u}{\partial t^2} = \delta(x)f(t)$$

where $c$ is the wave velocity, $f(t)$ any force function and $\delta (x)$ is the dirac delta function. I don't know if I am doing right (and not how to apply also it) but I want that the energy vanishes at infinity $x^+_-$. The $\delta(x)$ is used to simulate a point source at $x=0$.

I used the Fourier Transform $\mathcal{F}$ first to get to:

$$\frac{\partial^2 U_{\omega}(\omega,x)}{\partial x^2} + \frac{\omega^2}{c^2} U_{\omega}(\omega,x) = \delta(x)F(\omega)$$

$$\frac{\partial^2 U_{\omega}(\omega,x)}{\partial x^2} + k^2 U_{\omega}(\omega,x) = \delta(x)F(\omega)$$ (A)

with $k^2 = \frac{\omega^2}{c^2}$ and the subscript $U_{\omega}(\omega,x) = \mathcal{F_t}(u(t,x))$ meaning Fourier Transformed from $t$.

That's the Helmholtz equation but let's continue applying now the Laplace $\mathcal{L}$ transform and applying the conditions $U_{\omega}(\omega,0) = 0$ and $\frac{dU_{\omega}(\omega,0)}{dx} = 0$

$$s^2U_{\omega s}(\omega,s) - sU_{\omega}(\omega,0) - \frac{dU_{\omega}(\omega,0)}{dx}+ k^2 U_{\omega s}(\omega,s) = \mathcal{L}(\delta(x))F_{\omega}$$ $$s^2U_{\omega s}(\omega,s) + k^2 U_{\omega s}(\omega,s) = F_{\omega}$$ $$U_{\omega s}(\omega,s)\left(s^2 + k^2 \right) = F_{\omega}$$ $$U_{\omega s}(\omega,s) = F_{\omega}\frac{1}{s^2 + k^2}$$ $$U_{\omega}(\omega,x) = F_{\omega} \frac{\sin(kx)}{k} u(x)$$

with $u(x)$ as the Heaviside step function and $U_{\omega s}(\omega,s) = \mathcal{L_x}(U_{\omega}(\omega,x))$

the solution would than be

$$u(t,x) = \mathcal{F}^{-1}\{F_{\omega} \frac{\sin(kx)}{k} u(x)\}$$ (B)

If I use the Helmholtz approach from (A) with green's function I would get to :

$$u(t,x) = \mathcal{F}^{-1}\{ G(x) \ast \left[ F_{\omega} \delta(x) \right] \}$$

$$u(t,x) = \mathcal{F}^{-1}\{F_{\omega} \frac{j e^{jkx}}{2k}\}$$ (C)

Where $j = \sqrt{-1}$ imaginary unit and $G(x) = \frac{j e^{jkx}}{2k}$ is the Green's function for the inhomogeneous Helmholtz equation.

Why the difference between (B) and (C), what am I missing?

Edit:

I used The boundary conditions are: $$U_{\omega}(\omega,0) = \frac{dU_{\omega}(\omega,0)}{dx} = 0$$

I need the energy vanishes at infinity $x^+_-$ or $\lim_{|x| \to \infty} u(t,|x|) = 0$ Don't know how to apply this though above.

• I am completely confused by your original problem statement. Is your source function really a delta in space times a function of time, or are you trying to inject a green's function approach before you even begin? Commented Sep 30, 2013 at 16:44
• @RonGordon my source function yes is really a delta in space times a function of time. I am applying it to simulate a punctual source.
– imbr
Commented Sep 30, 2013 at 20:51
• @eusoubrasileiro one of solutions includes the initial data $u(x)$, while the other does not, so you are missing some details. I'm travelling right now, but if no one answers this in a week I'll look into it. Cheers Commented Sep 30, 2013 at 22:01
• I edited the question to better describe what I need.
– imbr
Commented Oct 3, 2013 at 14:33

I think you think this question too complicated.

Similar to how to solve this PDE:

Let $\begin{cases}p=x+ct\\q=x-ct\end{cases}$ ,

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}$

$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$

$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial t}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial t}=c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}$

$\dfrac{\partial^2u}{\partial t^2}=\dfrac{\partial}{\partial t}\left(c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial t}+\dfrac{\partial}{\partial q}\left(c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial t}=c^2\dfrac{\partial^2u}{\partial p^2}-c^2\dfrac{\partial^2u}{\partial pq}-c^2\dfrac{\partial^2u}{\partial pq}+c^2\dfrac{\partial^2u}{\partial q^2}=c^2\dfrac{\partial^2u}{\partial p^2}-2c^2\dfrac{\partial^2u}{\partial pq}+c^2\dfrac{\partial^2u}{\partial q^2}$

$\therefore\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}-\dfrac{1}{c^2}\left(c^2\dfrac{\partial^2u}{\partial p^2}-2c^2\dfrac{\partial^2u}{\partial pq}+c^2\dfrac{\partial^2u}{\partial q^2}\right)=\delta\left(\dfrac{p+q}{2}\right)f\left(\dfrac{p-q}{2c}\right)$

$4\dfrac{\partial^2u}{\partial pq}=\delta\left(\dfrac{p+q}{2}\right)f\left(\dfrac{p-q}{2c}\right)$

$\dfrac{\partial^2u}{\partial pq}=\dfrac{1}{4}\delta\left(\dfrac{p+q}{2}\right)f\left(\dfrac{p-q}{2c}\right)$

$u(p,q)=F(p)+G(q)+\dfrac{1}{4}\int_b^q\int_a^p\delta\left(\dfrac{r+s}{2}\right)f\left(\dfrac{r-s}{2c}\right)dr~ds$

$u(x,t)=F(x+ct)+G(x-ct)+\dfrac{1}{4}\int_b^{x-ct}\int_a^{x+ct}\delta\left(\dfrac{r+s}{2}\right)f\left(\dfrac{r-s}{2c}\right)dr~ds$