Question

7. Use a suitable Fourier Transform to find the solution of the IVP 2t-r-1 ,2-1 t 〉 0, , u(x, t), uz (x, t) 0asx→00, t〉0, →

Please show all steps to solution.

Answer 1

Apply the Fourier Transform in the variable x to the given PDE: We have $\widehat{D^\alpha u}=(i\xi)^{|\alpha|}\hat{u}$ for a multi-index $\alpha$, and $\hat{\delta_0}(x)\equiv 1$ ; using these, we obtain $\hat{u}_{tt}=c^2(i\xi)^2\hat{u}+1$ , i.e. $\hat{u}_{tt}+c^2\xi^2\hat{u}=1$ for t > 0. The initial conditions transform to $\hat{u}=0,\hat{u}_t=0$ for t = 0.

We readily solve the homogeneous part of this ODE to obtain

úh (t)-c1 cos ct c2 sin ct C Sin
. Moreover,
up(t) =
is a particular solution for the nonhomogeneous part. Thus the general solution of the ODE obtained after the transform is

. Taking the inverse Fourier transform, the solution of our PDE is

u (t) = CI cos

.

- cI(cos ct)c(sin ct)2

Now $(\cos{c\xi t})^{\check{\,}} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{2\pi ix\xi}\cos{c\xi t}\, d\xi = {\frac{1}{2}}(\delta_0(x-ct)+\delta_0(x+ct))$ ,

$(\sin{c\xi t})^{\check{\,}} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{2\pi ix\xi}\sin{c\xi t}\, d\xi = {\frac{i}{2}}(\delta_0(x+ct)-\delta_0(x-ct))$, and

$\left( \frac{1}{c^2\xi^2} \right) ^{\check{\,}} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{2\pi i x \xi} \left( \frac{1}{c^2\xi^2} \right)\, d\xi = -\frac{|x|}{2c^2}$. Combining these, we obtain

$u(x,t)=\frac{1}{2}\left( c_1(\delta_0(x-ct)+\delta_0(x+ct))+ic_2(\delta_0(x+ct)-\delta_0(x-ct)) -\frac{|x|}{c^2}\right)$.

Note: 1. If there are nontrivial initial conditions $u(x,0)=g(x),u_t(x,0)=h(x)$ , then the homogeneous part of the solution becomes similar to D'Alembert's solution; the Dirac delta distributions act on these initial conditions to give the familiar form.

2. I am more comfortable using $\xi$ for denoting the Fourier-transformed variable corresponding to .