Solve the given first order Partial differential equation.

[chwala]

Homework Statement

This is my own question - set by me

Relevant Equations

Pde

Solve the given PDE for $u(x,t)$;

$\dfrac{∂u}{∂t} +8 \dfrac{∂u}{∂x} = 0$

$u(x,0)= \sin x$

$-∞ <x<∞ , t>0$


In my working (using the method of characteristics) i have,

$x_t =8$
$x(t) = 8t + a$

$a = x(t) - 8t$ being the first characteristic.

For the second characteristic,

$u(x(t),t) = f(a) = \sin a = \sin (x(t)-8t)$

thus the solution is,

$u(x,t) = \sin (x-8t)$

Insight welcome. Cheers.

 

[chwala]

I also read that we may use another approach i.e thinking of $z=u(x,t)$ as a surface in $\mathbb{R^3}$ and have the following lines,

Denote by $\Gamma$ the curve on the surface,

$\Gamma_a = \begin{cases} x=x_0(a) & \\ t=t_0(a) \end{cases}$


$\Gamma_a = \begin{cases} x=x_0(a) & \\ t=t_0(a) & \\ z=z_0(a) =f(x_0(a),y_0(a)). \end{cases}$

...

$r(s) =(x(s),t(s),z(s))$

$\dfrac{dx}{ds} = 8$
$x_0 =a$


$\dfrac{dt}{ds} = 1$
$t_0 =0$

$\dfrac{dz}{ds} = 0$
$z_0 =f(a)=\sin a$

For 1st two characteristic equation,
$x=8s+a$
$t=s$

For third one, $z= \sin a$

Inverting the transformation, we get
$s=S(x,t) = t$
$a=M(x,t)= x-8s=x-8t$
$u(x,t) = Z(S(x,t),M(x,t)=\sin(x-8t)$.

Insight welcome guys.