1
$\begingroup$

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...

$\endgroup$

1 Answer 1

1
$\begingroup$

$$(u_t )^2 + (u_x )^2 = u^2 $$ $$u(x, 0) = Ae^{−\sqrt{1+x^2}}$$ 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) :

https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf , Eq.(2.208).

https://web.stanford.edu/class/math220a/handouts/firstorder.pdf , example 11.

https://people.cam.cornell.edu/~zc227/extras/math6200_pres.pdf , §.1.

Of course, the boundary conditions are diferent, but with the same method you will find the solution : $$v(x,t)=-\sqrt{(t+1)^2+x^2}$$ $$u(x,t)=A\:e^{-\sqrt{(t+1)^2+x^2}}$$

$\endgroup$
2
  • $\begingroup$ So u is a continuous function of x for all t, can we say anything else about the behavior of u? $\endgroup$
    – dxdydz
    Dec 11, 2018 at 0:34
  • $\begingroup$ $u(x,t)$ decreases when $x$ and $t>0$ increase. One can give the maximum and the minimum. $\endgroup$
    – JJacquelin
    Dec 11, 2018 at 13:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .