I’m tackling to solve a 1st order quasi PDE as below $$\frac{\partial u}{\partial t} + \frac{\sqrt {u}}{x} \frac{\partial u}{\partial x} = 0$$
Here, the I.C & B.C are $$u(0,x)=u_0 $$ and $$u(0<t<t_0 ,x_0 ) = u_0 (1-t/t_0 )^2 $$ $$u(t > t_0 ,x_0 )=0 $$
I think I could get the general solution using method of characteristics but I have no idea how to get unique solution from the I.C & B.Cs.
Here is my trials..
The characteristics can be written,
$$\frac{dt}{1}=\frac{xdx}{\sqrt{u}}=\frac{du}{0}$$
So, from the last term I can get $${du} = {0} \rightarrow u=k_1$$
And with other relation,
$$k_2 = x^2 - 2t \sqrt{u}$$
The general solution would be $$u(t,x)=f\left( x^2 - 2t \sqrt{u} \right)$$
Applying I.C & B.C I got
$$u(0, x) = u_0 = f(x^2)$$ $$u(t,x_0)|_{0<t<t_0}= u_0\left(1-t/t_0 \right)^2 = f\left({x_0}^2-2t(1-t/t_0)\sqrt{u_0}\right)$$ $$u(t, x_0)|_{t>t_0} = 0 = f\left({x_0}^2 - 2t\sqrt{u} \right)=f\left({x_0}^2 \right)$$
How can I find the unique solution from these?
$$u(t,x)=f\left( x^2 - 2t \sqrt{u} \right) \qquad\text{OK.}$$ If $u(0,x)=u_0(x)$ is a given function $$u(0, x) = u_0(x) = f(x^2)\quad\implies\quad f(X)=u_0(\sqrt{X})$$ The solution on implicit form is : $$u(t,x)=u_0\left( \sqrt{x^2 - 2t \sqrt{u}} \right)$$ In case of $u_0(x)=$constant, the solution is $\quad u(t,x)=u_0\quad$ which obviously satisfies the PDE.