I am asked to solve this IVP
$$ \begin{cases} u_x+2xu_t=0\\ u(x,0)=0\\ u(0,t)=t, \qquad t\geq 0 \end{cases} $$
I already have found the general solution, that is:
$$\frac{1}{dx}=\frac{2x}{dt}=\frac{0}{du} $$
skipping some computations, yields
$$u(x,t)=\Phi(x^2-t)$$
At this point, I started to use separately the conditions:
$$ \begin{cases} u(x,0)=\Phi(x^2)=0 \quad\rightarrow\quad \zeta_1:=x^2 \quad\rightarrow\quad \zeta_1 - x^2=0\\ u(0,t)=\Phi(-t)=t \quad\rightarrow\quad \zeta_1:=-t \quad\rightarrow\quad t=-\zeta_1 \end{cases} $$
How could I end in order to solve it? Can anyone help me to combine the two conditions?
Thanks in advance.
I agree with $\quad u(x,t)=\Phi(x^2-t)$
Let $\quad X=x^2-t$
Conditions :
$$\begin{cases} u(x,0)=0= \Phi(x^2) \quad\implies\quad \Phi(X)=0\\ u(0,t)=t= \Phi(-t),\quad t\geq 0 \quad\implies\quad \Phi(X)=-X,\quad X\leq 0 \end{cases}$$
Graphical representation of the function $\Phi(X)$ :
Obviously, the analytical form of the function $\Phi(X)$ is : $$\Phi(X)=-X\:\text{H}(-X)$$ H is the Heaviside step function.
Putting this function into the general solution $\quad u(x,t)=\Phi(x^2-t)\quad$ leads to :
$$ u(x,t)=-(x^2-t)\:\text{H}\big(-(-x^2-t)\big)$$ $$ u(x,t)=(t-x^2)\:\text{H}\big(t-x^2\big)$$
Final checking :
For $t<x^2$ the solution of the PDE is trivial : $u=0$.
For $t>x^2$ the solution $\quad u=-x^2+t\quad$ satisfies the PDE because $u_x=-2x\Phi'$ and $u_t=\Phi'$ hence $u_x+2xu_t=0$.
At $t=0 \quad\to\quad H(-x^2)=0 \quad\to\quad u(x,0)=0\:\:$ which satisfies the boundary condition.
At $x=0 \quad\to\quad H(t)=1 \quad\to\quad u(0,t)=-0+t=t\:\:$ which satisfies the boundary condition.