Method of characteristics non-linear PDE Consider the following initial-value problem:
$$xu_x-uu_t=t$$
$$u(1,t)=t$$
I've come to the follow characteristic equations:
$$\frac{\mathrm{d}x}{\mathrm{d}\tau }=x,\,\,\,\frac{\mathrm{d}t}{\mathrm{d}\tau }=-u,\,\,\,\frac{\mathrm{d}u}{\mathrm{d}\tau }=t$$
I can solve the first easily;
$$\int \frac{1}{x}\mathrm{d}x=\int \mathrm{d}\tau $$
$$\therefore x=e^\tau $$
The second and third applying the chain rule;
$$\frac{\mathrm{d}u}{\mathrm{d}t}=-\frac{t}{u}$$
$$\int u \mathrm{d}u=-\int t\mathrm{d}t$$
$$\therefore u^2=F(\xi )-t^2$$
Putting back in the initial conditions, I wind up with the following;
$$F(\xi )=2\xi ^2$$
$$\therefore u=\sqrt{2\xi ^2-t^2}$$
Now, from here I cannot find a way to evaluate what $\xi$ is in terms of the natural variables $x$ and $t$.
According to Maple, the solution is as follows:
$$u(x,t)=\frac{t}{\tan{\left(\ln{\left(\frac{e^{\frac{\pi}{4}}}{x}\right)}\right)}}$$
How do I find this solution algebraically?
 A: $$xu_x-uu_t=t $$
Your system of ODEs is correct. Written equivalently as :
$$\frac{dx}{x}=\frac{dt}{-u}=\frac{du}{t}=d\tau$$
A first characteristic equation comes from $\frac{dt}{-u}=\frac{du}{t}$ :
$$u^2+t^2=c_1 \tag 1$$
A second characteristic equation comes from $\frac{dx}{x}=\frac{dt}{-u}=\frac{dt}{-\sqrt{c_1-t^2}}$
$\ln|x|=-\int\frac{dt}{\sqrt{c_1-t^2}}=-\tan^{-1}\left(\frac{t}{\sqrt{c_1-t^2}}\right)+$constant.
$$\ln|x|+\tan^{-1}\left(\frac{t}{\sqrt{c_1-t^2}}\right)=c_2$$
$$\ln|x|+\tan^{-1}\left(\frac{t}{\sqrt{(u^2+t^2)-t^2}}\right)=c_2$$
$$\ln|x|+\tan^{-1}\left(\frac{t}{u}\right)=c_2 \tag 2$$
The general solution of the PDE expressed on the form of implicit equation $c_2=\Phi(c_1)$ is :
$$\ln|x|+\tan^{-1}\left(\frac{t}{u}\right)=\Phi(u^2+t^2) \tag 3$$
where $\Phi$ is an arbitrary function, to be determined according to the boundary condition.
CONDITION : $u(1,t)=t$
$$\ln|1|+\tan^{-1}\left(\frac{t}{t}\right)=\Phi(t^2+t^2)$$
$$\Phi(2t^2)=\frac{\pi}{4}$$
This implies that the function $\Phi$ is a constant function $=\frac{\pi}{4}$. 
Such case can be encountered when the boundary condition is specified on a characteristic curve, which is the present case.
Nevertheless the function $\Phi$ is determined. We put it into the general solution $(3)$. 
$$\ln|x|+\tan^{-1}\left(\frac{t}{u}\right)=\frac{\pi}{4}$$ 
$$\tan^{-1}\left(\frac{t}{u}\right)=-\ln|x|+ \frac{\pi}{4}$$
$$u(x,t)=\frac{t}{\tan\left(-\ln|x|+\frac{\pi}{4}\right)}$$
This is the same result than Maple since $\ln(\frac{e^{\pi/4}}{x})=\frac{\pi}{4}-\ln(x)$ .
