# Using Fourier transform to solve for pde

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.

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

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|

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|

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}|

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

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

• Thanks for the detailed solution. Can you explain how you got line 18. @Ryan Howe Oct 20, 2018 at 6:06
• there is an edit at line $19$
– user3417
Oct 20, 2018 at 6:17