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Consider the initial value problem for the wave equation $\frac{\partial^2}{\partial t^2}u(x,t) = c^2 \frac{\partial^2}{\partial x^2}u(x,t)$ , $-\infty < x < \infty$, t > 0 with initial conditions u(x,0) = f(x) and $\frac{\partial}{\partial t} u(x,0) = g(x)$ for -$\infty<x<\infty$.

Using a Fourier transform in x, show that the solution is $u(x,t) = \frac{1}{2}(f(x-ct)+f(x+ct)) + \frac{1}{2c}\int_{x - ct}^{x+ct}g(x')dx'$. Hint, you will need to consider the result of $\int_{x - ct}^{x+ct}e^{-iku}du$.

I tried to do this question by first taking the fourier transform w.r.t. x and then arrived at solution for pde (let F(u) be fourier transform of solution u)$F(u) = Acos(ckt)+Bsin(ckt)$. By applying the initial conditions, my solution then becomes $F(u) = F(f)cos(ckt)+\frac{F(g)}{ck}sin(ckt)$. I'm not sure what to do next to arrive to the answer above. Thanks

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The function $u(x,t)$ is derived from solving the two conditions separately.

$$ u(x,t) = u_{1}(x,t) +u_{2}(x,t) \tag{1} $$

if you take the pde as follows

$$ \begin{align}\begin{cases} \frac{\partial^{2} }{\partial t^{2}}u(x,t) = c^{2} \frac{\partial^{2} }{\partial x^{2}}u(x,t) \\ - \infty < x < \infty , t > 0 \\ u(x,0) = f(x) \\ \frac{\partial}{\partial x}u(x,0) = g(x) \end{cases} \end{align} \tag{2}$$

First part

$$ \begin{align}\begin{cases} \frac{\partial^{2} }{\partial t^{2}}u_{1}(x,t) = c^{2} \frac{\partial^{2} }{\partial x^{2}}u_{1}(x,t) \\ - \infty < x < \infty , t > 0 \\ u_{1}(x,0) = f(x) \\ \frac{\partial}{\partial x}u_{1}(x,0) = 0 \end{cases} \end{align} \tag{3}$$

and we solve the first part

$$\hat{U}_{1}(\omega,t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} u_{1}(x,t) e^{i \omega x} dx \tag{4} $$

$$ u_{1}(x,t) = \int_{-\infty}^{\infty} \hat{U}_{1}(\omega, t) e^{-i \omega x} d \omega \tag{5} $$

If you take the fourier transform you get

$$ \frac{\partial \hat{U}_{1}}{\partial t^{2}} =-c^{2} \omega^{2} \hat{U}_{1} \tag{6} $$

when you apply the initial conditions you get $$ \hat{U}_{1}(\omega, 0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x) e^{i \omega x} \tag{7}$$ $$ \frac{\partial }{\partial t}\hat{U}_{1}(\omega, 0) = 0 \tag{8}$$

$$ \hat{U}_{1}(\omega,t) = A(\omega) \cos( c\omega t) + B(\omega)\sin(c \omega t) \tag{9} $$ $$B(\omega) = 0 \\ A(\omega) = \hat{U}_{1}(\omega, 0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x) e^{i \omega x} dx \tag{10} $$ now if you use the inverse fourier transform you get $$ u_{1}(x,t) = \int_{-\infty}^{\infty} \hat{U}_{1}(\omega, 0) \cos( c\omega t) e^{- i \omega x} d \omega \tag{11} $$

$$ u_{1}(x,t) = \frac{1}{2} \int_{-\infty}^{\infty} \hat{U}_{1}(\omega, 0) \bigg[ e^{- i \omega (x-ct)} + e^{-i \omega (x+ct)} \bigg]d \omega \tag{12}$$

$$ u_{1}(x,t) = \frac{1}{2}\big[f(x-ct) +f(x+ct) \big] \tag{13} $$

Second part

$$ \begin{align}\begin{cases} \frac{\partial^{2} }{\partial t^{2}}u_{2}(x,t) = c^{2} \frac{\partial^{2} }{\partial x^{2}}u_{2}(x,t) \\ - \infty < x < \infty , t > 0 \\ u_{2}(x,0) = 0 \\ \frac{\partial}{\partial x}u_{2}(x,0) = g(x) \end{cases} \end{align} \tag{14}$$

you get

$$\hat{U}_{2}(\omega, t) = A(\omega) \cos( c\omega t) + B(\omega) \sin(c\omega t) \tag{15} $$

$$ \hat{U}_{2}(\omega,0) = 0 \\ \frac{\partial }{\partial t}\hat{U}_{2}(\omega, 0) = G(\omega) \tag{16}$$

$$ \hat{U}_{2}(\omega,t) = G(\omega) \frac{\sin(c \omega t}{c\omega } = \frac{\pi}{c}G(\omega)F(\omega) \tag{17}$$

then you get

$$ f(x) = \begin{align}\begin{cases} 0 & |x| > ct \\ 1 & |x| <ct \end{cases} \end{align} \tag{18}$$

there's an exercise earlier in the book. This is Haberman. It is actually the $f(x)$ function above nearly

$$ f(x) = \begin{align}\begin{cases} 0 & |x| > a \\ 1 & |x| <a \end{cases} \end{align} \tag{19}$$

determine the Fourier transform

$$ F(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x)e^{i \omega x} dx = \frac{1}{2\pi} \int_{-a}^{a} e^{i \omega x} dx = \frac{e^{i\omega x}}{2\pi i\omega}\Big|_{-a}^{a} \tag{20}$$

this gives

$$ F(\omega) = \frac{1}{2\pi i\omega} (e^{i \omega a} - e^{-i \omega a}) = \frac{1}{\pi \omega}\sin(\omega a) \tag{21} $$

$$u_{2}(x,t) = \frac{\pi}{c}\big[ \frac{1}{2\pi} \int_{-\infty}^{\infty} g(\hat{x}) f(x-\hat{x}) d\hat{x} \big] \tag{22}$$ $$ f(x-\hat{x}) = \begin{align}\begin{cases} 0 & |x-\hat{x}| > ct \\ 1 & |x-\hat{x}| <ct \end{cases} \end{align} \tag{23}$$

$$ u_{2}(x,t) = \frac{1}{2c} \int_{x-ct}^{x+ct} g(\hat{x}) d\hat{x} \tag{24} $$

$$ u(x,t) = \frac{1}{2}\big[f(x-ct) +f(x+ct) \big] +\frac{1}{2c} \int_{x-ct}^{x+ct} g(\hat{x}) d\hat{x} \tag{25} $$

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  • $\begingroup$ Thanks for the detailed solution. Can you explain how you got line 18. @Ryan Howe $\endgroup$ – stutsy Oct 20 '18 at 6:06
  • $\begingroup$ there is an edit at line $19$ $\endgroup$ – user3417 Oct 20 '18 at 6:17

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