… It then showed that this is just an iterative sequence, where $v_{n+1} =v_{n} + a(u_{n}-v_{n})$ in every iteration. Now he wants to show that this sequence converges to the function I stated earlier. We can see that $a\|u_{n}\|^{2} + b\|v_{n}\|^{2} \to \infty$ as $n\to\infty$ I know from looking at examples in other notes that $a$ and $b$ are positive, but how can we really find the value of $a$ and $b$? A: This is in fact an example of a linear elliptic equation: $$a\|u\|^2+b\|v\|^2=c$$ As pointed out in the comments, this is a graph of a function $u(x)=\pm \sqrt{ax^2+bx+c}$ which is always of one of two forms: the graph of a convex function, or a plane. In this case, it’s a graph of a convex function (a plane in this case). In fact, the first term in the functional determines when the graph is a parabola, as the second one is negative, and the third term $c$ determines its vertex at which the function is infinite. Also, you can see this as a factoring of polynomials, as $\frac{a}{2}(x-\sqrt{x^2+2c})(x+\sqrt{x^2+2c})+bx+c=0$. In the special case of $c=0$, we get a parabola. In this case, the iterates behave as follows: $$u_{n+1}=\pm \sqrt{a(u_n-v_n)^2+b(u_n-v_n)+c}$$ and, assuming $u_0,v_0 eq0$, we have:  u_{n+1}=\pm \sqrt{a(u_n-v_n)^2+b(u_n-v_n)+c}=\pm\sqrt