Given the PDE $u_t + uu_x = 0$, where $u(x, 0) = f(x)$, I need to verify that $u=f(x-tu)$ is a solution to the PDE. How does one do this?
Of course, intuitively one understands that substitution would be the first method to try. So, $u_t = -f_t(x-ut)(u+tu_t)$, $u_x = f_x(x-ut)(1-tu_x)$. Thus $u_t+uu_x=-f_t(x-ut)(u+tu_t)+uf_x(x-ut)(1-tu_x)=-f_t(x-ut)(f(x-ut)+tu_t)+f(x-ut)f_x(x-ut)(1-u_x)$.
Now, if $t=0$, we get: $-f_t(x)f(x)+f(x)f_x(x)(1-u_x)$.
So, how does one verify that the above reduces to zero? I'm lost.
Also, I'm wondering that if $u=f(x-ut)$, does this mean that $u$ is defined in terms of itself? How can that be?
Best Answer
You forcing function $f$ is a function of $1$ variable, so there should be no $f_t$ or $f_x$; just $f'$. The chain rule gives $$\frac{\partial}{\partial t} f(x-tu) = f'(x-tu) \frac{\partial}{\partial t} [x-tu] = f'(x-tu)\cdot[-u - tu_t].$$ Likewise $$\frac{\partial}{\partial x} f(x-tu) = f'(x-tu) \cdot [1 - tu_x]$$ Then for $u$ satisfying $u = f(x-tu)$, we have \begin{align*} u_t + uu_x &= f'(x-tu)\cdot[-u-tu_t] + f'(x-tu) \cdot [u - tu u_x] \\ &= -t f'(x-tu)\cdot[u_t + u u_x]. \end{align*} So either $u_t + uu_x = 0$ or $f'(x-tu) = -1/t$, but the latter isn't possible because $f'(y)$ needs to actually needs to actually depend on $y$; i.e., any definition of $f'(x-tu)$ should include the quantity $x-tu$.