<s>[INST] For the partial differential equation $u_{xx}-u_{yy}=-1$ find $u(1/2,1/2)$ 
Let , $u(x,y)$ be a solution of $u_{xx}-u_{yy}=-1$ , $x\in \mathbb R$ , $y>0$ with $u(x,0)=0$ , $u_y(x,0)=0$ , $x\in \mathbb R$. Then , find $u\left(\frac{1}{2},\frac{1}{2}\right)$.

Attempt :
Solution is $u(x,y)=f_1(y+x)+f_2(y-x)-\frac{x^2}{2}$. With the help of initial conditions I can't eliminate the arbitrary functions $f_1$ and $f_2$. How I can eliminate these two functions ? 
 [/INST] $$
u(x, y) = f_1(y + x) + f_2(y-x) - \frac {x^2}2 \implies u_y = f_1' + f_2'
$$
Now, substitute ICs
\begin{align}
u(x, 0) &= f_1(x) + f_2(-x) - \frac {x^2}2= 0 \tag 1 \\
u_y(x, 0) &= f_1'(x) + f_2'(-x) = 0  \tag 2
\end{align}
Differentiate $(1)$
$$
f_1'(x) - f_2'(-x) -x = 0 \tag 3
$$
And solve it in conjunction with the $(2)$
$$
2f_1'(x) = x \implies f_1 = \frac {x^2}4 + C \tag 4
$$
and
$$
f_2(-x) = \frac {x^2}2 - \frac {x^2}4 - C = \frac {x^2}4 - C
$$
or
$$
f_2(x) = \frac {x^2}4 - C \tag 5
$$
So, finally you can write
$$
u(x, y) = \frac {(y+x)^2}4 + \frac {(y-x)^2}4 - \frac {x^2}2
$$
</s>