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

Question

Given $f\in C^{1}({\bf R}^{3})$ , 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 .

Answer 1

$$\left\{\begin{array}{lll} f_{t}+xf_{y}=0\\ f|_{t=0}=f_{0}(x,y) \end{array}\right.$$ 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 : $\quad \frac{dt}{1}=\frac{dy}{x} \quad\implies\quad y-xt=c_1$

A second family of characteristic curves comes from $\quad\frac{dx}{0}=$finite$\quad\implies\quad x=c_2$

A third family of characteristic curves comes from $\quad\frac{df}{0}=$finite$\quad\implies\quad f=c_3$

Thus, the general solution of the PDE, expressed on the form of implicite equation is : $$\Phi\left(x\:,\: (t-xt) \:,\: f \right)=0$$ 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 : $$f(x,y,t)=F\left(x\:,\:(y-xt) \right)$$ where $F$ is any differentiable function of two variables.

CONDITION :

$f(x,y,0)=f_0(x,y)=F\left(x\:,\:(y-0) \right) \quad\implies\quad F\left(x\:,\:y \right)=f_0(x,y)$

Now, the function $F$ is determined. We put it into the above general solution where the second variable is $(y-xt)$. This leads to : $$f(x,y,t)=f_0\left(x\:,\:(y-xt) \right)$$

Answer 2

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}

Answer 3

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 $t=s$, $x_0=x$, $y_0=y-x_0s=y-xt$.