Ytukyg

Given $fin C^{1}({bf R}^{3})$ , then find the solution of the following pde :

$$left{begin{array}{lll}
f_{t}+xf_{y}=0\
f|_{t=0}=f_{0}(x,y)
end{array}right.$$

I tried to use the characteristic method to solve this question , but I failed . Any comment or advice I will be grateful .

$$left{begin{array}{lll} f_{t}+xf_{y}=0\ f|_{t=0}=f_{0}(x,y) end{array}right.$$
The characteristic ODEs are : $quad frac{dt}{1}=frac{dy}{x}=frac{dx}{0}=frac{df}{0}$

A first family of characteristic curves comes from : $quad frac{dt}{1}=frac{dy}{x} quadimpliesquad y-xt=c_1$

A second family of characteristic curves comes from $quadfrac{dx}{0}=$finite$quadimpliesquad x=c_2$

A third family of characteristic curves comes from $quadfrac{df}{0}=$finite$quadimpliesquad f=c_3$

Thus, the general solution of the PDE, expressed on the form of implicite equation is :
$$Phileft(x:,: (t-xt) :,: f right)=0$$
where $Phi$ is any differentiable function of three variables.
( Doesn't matter the order of the variables since the function $Phi$ is arbitrary ).

Or alternatively, on explicit form :
$$f(x,y,t)=Fleft(x:,:(y-xt) right)$$
where $F$ is any differentiable function of two variables.

CONDITION :

$f(x,y,0)=f_0(x,y)=Fleft(x:,:(y-0) right) quadimpliesquad Fleft(x:,:y right)=f_0(x,y)$

Now, the function $F$ is determined. We put it into the above general solution where the second variable is $(y-xt)$. This leads to :
$$f(x,y,t)=f_0left(x:,:(y-xt) right)$$

The method of characteristics should tell you that the characteristic curves
$$
t(s),, x(s),,y(s),,z(s)=f(t(s),x(s),y(s))
$$
follow the ODE (up to re-parametrization)
$$
frac{dt}{ds}=1,, frac{dx}{dt}=0,, frac{dy}{dt}=x,, frac{dz}{ds}=0
$$
which has solutions
$$t=s+t_0,, x=x_0,, y=y_0+x_0s,, z=z_0$$ so that for all $s$ (discounting shocks etc., i.e., in a neighborhood of $s=0$)
begin{align}
f(t,x,y)&=z(s)=f(t_0+s,x_0,y_0+x_0s)\
&=z(0)=f(0,x_0,y_0)=f(0,x,y-xt)\
&=f_0(x,y-xt).
end{align}
where obviously the initial values where chosen to conform to the given boundary condition, that is, $t_0=0$ so that $t=s$, $x_0=x$, $y_0=y-x_0s=y-xt$.

If the PDE was just $f_{t} + f_{y} = 0$, then you would make a travelling wave ansatz $f(x,y,t) = f(y-t)$. As the derivative in $y$ has a non-constant coefficient $x$, it stands to reason that when differentiating our solution with respect to $t$, we should get an $x$ out the front. Hence, we can guess that $$f(x,y,t) = f_{0}(x, y - xt)$$ solves the problem, which you can check by differentiation it does.

Alternatively, make the guess $f(x,y,t) = g(alpha y + gamma t)$ which, after differentiation, yields

begin{align}
f_{t} &= gamma g' \
f_{y} &= alpha g' \
therefore f_{t} + x f_{y} &= 0 \
implies gamma g' + x alpha g' &= 0 \
implies gamma &= -x, quad alpha = 1 \
implies f(x,y,t) &= g(y - xt)
end{align}

Applying the initial condition yields

begin{align}
f(x,y,0) &= g(y) \
&= f_{0}(x,y) \
implies f(x,y,t) &= f_{0}(x, y - xt)
end{align}