# Behavior of the solution of the eikonal equation

Consider the nonlinear ﬁrst-order initial-value problem: $$(u_t )^2 + (u_x )^2 = 1$$ with initial condition $$u(x, 0) = {−\sqrt{1+x^2}}$$. Find its solution for all $$t>0$$ using the method of characteristics.

Let $$x(s)=\langle x(s),t(s) \rangle, p= \langle p_1(s),p_2(s) \rangle = \langle u_x,u_t \rangle , z = u(x,t)$$

$$F(p,z,x)=p_1^2+p_2^2-1, \frac{dF}{dp}=\langle 2p_1,2p_2 \rangle, \frac{dF}{dz}=0, \frac{dF}{dx}=\langle 0,0 \rangle$$

Characteristics:

\begin{align} p_s &= \frac{dF}{dx}-\frac{dF}{dz}p=-\langle 0,0 \rangle -0\langle p_1,p_2 \rangle=\langle 0,0 \rangle \\ z_s &= \frac{dF}{dp}p= \langle 2p_1,2p_2\rangle \cdot \langle p_1,p_2 \rangle =2p_1^2+2p_2^2=2\\ x_s &= \frac{dF}{dp}= \langle 2p_1,2p_2 \rangle \end{align}

IVP:

$$x_0 =r \quad t_0 =0, \quad u_0=z_0={−\sqrt{1+r^2}}$$

$$u_x(x,0) = -\frac{x}{{\sqrt{1+x^2}}}\implies p_{1_0}= -\frac{r}{{\sqrt{1+r^2}}}=p_1$$

$$p_{1_0}^2+p_{2_0}^2-1 = 0 \implies p_{2_0}=\pm \sqrt{1-\frac{r^2}{1+r^2}}=\pm \frac{1}{{\sqrt{1+r^2}}}=p_2$$

Suppose $$p_{2_0}=\frac{1}{{\sqrt{1+r^2}}}$$, then $$t_s=2p_2=\frac{2}{{\sqrt{1+r^2}}}$$ with $$t_0 =0 \implies t = \frac{2}{{\sqrt{1+r^2}}}s \implies s=\frac{t{\sqrt{1+r^2}}}{2}$$

$$x_s = 2p_1 = -\frac{2r}{{\sqrt{1+r^2}}}$$ with $$x_0 =r$$ implies $$x=-\frac{2r}{{\sqrt{1+r^2}}}s+r=-tr+r=r(1-t) \implies r =\frac{x}{1-t}.$$

$$z_s=2$$ with $$z_0={−\sqrt{1+r^2}}$$ implies $$z=u(x,t)=2s{−\sqrt{1+r^2}}=t\sqrt{1+r^2}-\sqrt{1+r^2}=\sqrt{1+r^2}(t-1)=\sqrt{1+\frac{x^2}{(1-t)^2}}(t-1)=\sqrt{(1-t)^2+x^2},$$ but this solution doesn't satisfy the IVP. This seems bizarre to me, since I parametrize the variables to fit the initial conditions. How can this be?

You found $$p_{2_0}=\pm \frac{1}{{\sqrt{1+r^2}}}$$

Why do you suppose $$p_{2_0}=\frac{1}{{\sqrt{1+r^2}}}$$ instead of $$p_{2_0}=-\frac{1}{{\sqrt{1+r^2}}}$$ ?

One have to examine the both cases and chose which one agrees to the boundary condition.

The supposition with sign $$+$$ leads to $$z=\sqrt{(1-t)^2+x^2}$$ which doesn't agrees with $$z_0=-\sqrt{1+x^2}$$.

The supposition with sign $$-$$ leads to $$z=-\sqrt{(1-t)^2+x^2}$$ which agrees with $$z_0=-\sqrt{1+x^2}$$.

Thus the solution is : $$u(x,t)=-\sqrt{(1-t)^2+x^2}$$ $$u_x=-\frac{x}{\sqrt{(1-t)^2+x^2}}$$

$$u_t=-\frac{-(1-t)}{\sqrt{(1-t)^2+x^2}}$$

$$(u_x)^2+(u_t)^2=\frac{x^2}{(1-t)^2+x^2}+\frac{(1-t)^2}{(1-t)^2+x^2}=\frac{x^2+(1-t)^2}{(1-t)^2+x^2}$$ $$(u_x)^2+(u_t)^2=1$$