# Solve IVP of $(u_t )^2 + (u_x )^2 − u^2 = 0$ using method of characteristics

Consider the nonlinear ﬁrst-order initial-value problem: $$(u_t )^2 + (u_x )^2 − u^2 = 0$$ with initial condition $$u(x, 0) = Ae^{−\sqrt{1+x^2}}$$.

(a) Find its solution for all $$t > 0$$ using the method of characteristics.

(b) Describe its behavior for all $$t > 0$$. Does it remain a continuous function of $$x$$ for all $$t$$?

(c) Find $$\lim u(x, t)$$.

My attempt:

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

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

Characteristics:

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

IVP:

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

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

$$p_{1_0}^2+p_{2_0}^2-u_0^2 = 0 \implies p_{2_0}^2=A^2e^{-2\sqrt{1+r^2}}-\frac{A^2e^{−2\sqrt{1+r^2}}r^2}{1+r^2}=\frac{Ae^{−\sqrt{1+r^2}}}{{\sqrt{1+r^2}}}$$

What's the next step? Solving $$p_s,z_s,x_s$$ seems a little bit painful...

$$(u_t )^2 + (u_x )^2 = u^2$$ $$u(x, 0) = Ae^{−\sqrt{1+x^2}}$$ HINT:
$$u(x,t)=Ae^{v(x,t)}\quad;\quad u_x=uv_x\quad;\quad u_t=uv_t$$ $$(v_t )^2 + (v_x )^2 =1$$ $$v(x,0)=-\sqrt{1+x^2}$$ This kind of PDE is well known (Eikonal PDE) :
Of course, the boundary conditions are diferent, but with the same method you will find the solution : $$v(x,t)=-\sqrt{(t+1)^2+x^2}$$ $$u(x,t)=A\:e^{-\sqrt{(t+1)^2+x^2}}$$
• $u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum. – JJacquelin Dec 11 '18 at 13:54