How to find solve for second order pde with initial conditions using Wolfram Mathematica? I have next task:
$$
\frac{\partial^2 u}{\partial x \partial y} = 0,~ u(x,x^2) = 0,~ \frac{\partial u}{\partial x}(x, x^2) = \sqrt{|x|},~|x| < 1
$$
I write this, but it don't work:
weqn = D[u[x,y],x,y] == 0 
ic = {u[x,x^2] == 0, Derivative[0,1][u][x,x^2] == Sqrt[Abs[x]], Abs[x] < 1} 
sol = DSolve[{weqn, ic},u,{x,y}]

And I got next error:

DSolve: Equation or list of equations expected instead of Abs[x]<1 in the first argument {$\mathrm u^{(1,1)}[x,y] == 0$, {$\mathrm{u[x,x^2] == 0, u^{(0,1)}[x,x^2]==\sqrt{Abs[x]},~ Abs[x]<1}$}}.

When I try to find a solution to a similar problem, when in conditionals the second argument like $ \mathrm x ^ 2 $, I can't do it.
Where did I go wrong?
I want to use Wolfram Mathematica to solve the problem.

$$\frac{\partial^2 u}{\partial x \partial y} =0$$
integrating with respect to $y$ we get $$\frac{\partial u}{\partial x} = f(x) $$ and similaty integrating with respect to $x$ we get $$u(x,y) =F(x) +G(y)$$
We have $$f(x)=\frac{\partial u}{\partial x} (x, x^2 ) =\sqrt{|x|}$$
therefore $$F(x) = \int \sqrt{|x|} dx $$ and hence $$F(x) =\frac{2}{3}\text{sign} (x)|x|^{\frac{3}{2}}$$
and since $$0=u(x,x^2 ) =\frac{2}{3}\text{sign}(x)|x|^{\frac{3}{2}} +G(x^2 )$$
we get $$G(y) =-\frac{2}{3}\text{sign}(y)|y|^{\frac{3}{4}}$$
Hence the solution is $$ u(x,y) =\frac{2}{3}\text{sign}(x)|x|^{\frac{3}{2}} -\frac{2}{3}\text{sign}(y)|y|^{\frac{3}{4}}$$
