Partial Differential Equations fourier transform with a sin(t) term Using fourier transform and properties of fourier transform solve the given problem
$$\frac{\partial u}{\partial t}+\sin(t)\frac{\partial u}{\partial x}=0$$
$$u(x,0)=\sin(x)$$
What I've gotten so far is:
$$F(u(z,t))=A(z)e^{iz\cos(t)}$$
where
$$F(f(x))(z)$$ is the fourier transform of $f(x)$
I'm stumped after that and not sure how to get to the final conclusion. Any advice is appreciated!
 A: Applying Fourier transform with respect to space variable $x$, you have
$$
\frac{d\hat{y}}{dt}+{\rm i}k\sin t\,\hat{y}=0
$$
with the initial condition
$$
\hat{y}(0)={\rm i\sqrt{\frac{\pi}{2}}}(\delta(k-1)-\delta(k+1)),
$$
which has the solution
$$
\hat{y}={\rm i\sqrt{\frac{\pi}{2}}}(\delta(k-1)-\delta(k+1)) \exp({\rm i}k-{\rm i}k\cos t).
$$
Taking the inverse Fourier transform you find
$$
\mathscr{F^{-1}}(\hat{y})=\sin(x+\cos t-1).
$$
A: I don't think this PDE can really be solved by using fourier transform.
Instead, this PDE can be solved by using method of characteristics:
$\dfrac{\partial u}{\partial t}+\sin(t)\dfrac{\partial u}{\partial x}=0$
$\csc(t)\dfrac{\partial u}{\partial t}+\dfrac{\partial u}{\partial x}=0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=\csc(t)$ , letting $t(0)=\dfrac{\pi}{2}$ , we have $-\cos(t)=s$
$\dfrac{dx}{ds}=1$ , letting $x(0)=x_0$ , we have $x=s+x_0=-\cos(t)+x_0$
$\dfrac{du}{ds}=0$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)=f(x+\cos(t))$
$u(x,0)=\sin(x)$ :
$f(x+1)=\sin(x)$
$f(x)=\sin(x-1)$
$\therefore u(x,t)=\sin(x+\cos(t)-1)$
