# Solving the transport equation with an initial condition

We have $$\begin{cases} u_t + x^2 u_x = 0\\ u(x,0) = x \end{cases}$$ I found that the general solution is $$u(x,t) = f\left(-\frac{1}{x} - t\right)$$ Applying the initial condition $$u(x,0) = f\left(-\frac{1}{x}\right) = x$$ Then I thought we would have $$u(x,t) = -\frac{1}{x} - t$$ which checks out but it does not satisfy the initial condition. So I am wondering where I went wrong.

• Well, if $f\left(-\frac{1}{y}\right) = y,$ as you have shown, then $f\left(-\left(\frac{1}{x} + t\right)\right) = f\left(-\frac{(1+tx)}{x}\right) = \frac{x}{1 + tx}$. Essentially, $f$ acts on the entirety of its input, it doesn't magically know that some part comes from $x$ and behave differently on just that part. – stochasticboy321 Jun 30 '17 at 17:25
• @stochasticboy321 Can you explain this further I still do not understand – justanewb Jun 30 '17 at 19:47
• $f$ is a function of a single variable. You want to determine $f(u),$ where $u = -(x^{-1} + t)$ is your canonical variable. You found that for any input, say $y$, $f(-1/y) = y$. Therefore, $f(u) = -1/u$. Now plug in the value of $u$ in terms of $x,t$. The idea is that $f$ is a map from $\mathbb{R}$ to $\mathbb{R}$. It cannot sniff out that the argument given to it is composed of two different variables. Analogously, suppose $g(x) = x^2.$ What is $g(2x)$? $g(xy)$? – stochasticboy321 Jul 1 '17 at 2:39

You correctly found that the general solution is $$u(x,t) = f\left(-\frac{1}{x} - t\right) \tag 1$$ Applying the initial condition $$u(x,0) = f\left(-\frac{1}{x}\right) = x$$ The goal is to determine the function $f$.
Let $\quad X=-\frac{1}{x}\quad\to\quad x=-\frac{1}{X}\quad$ , thus : $$f(X)=-\frac{1}{X}$$ Now the function $f$ is determined. We have to put it into equation (1) where the variable is $$\quad X=-\frac{1}{x} - t$$ This no longer $X=-\frac{1}{x}$ as above, since $t$ is no longer equal to $0$. $$f\left(-\frac{1}{x}-t\right)= -\frac{1}{-\frac{1}{x} - t}$$
$$u(x,t) = -\frac{1}{-\frac{1}{x} - t} = \frac{x}{1+xt}$$