# Solve the given first order Partial differential equation.

• chwala
chwala
Gold Member
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.

Last edited:
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.

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