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?
Best Answer
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 : $$u(x,t)=-\sqrt{(1-t)^2+x^2}$$ $u_x=-\frac{x}{\sqrt{(1-t)^2+x^2}}$
$u_t=-\frac{-(1-t)}{\sqrt{(1-t)^2+x^2}}$
$(u_x)^2+(u_t)^2=\frac{x^2}{(1-t)^2+x^2}+\frac{(1-t)^2}{(1-t)^2+x^2}=\frac{x^2+(1-t)^2}{(1-t)^2+x^2}$ $$(u_x)^2+(u_t)^2=1$$