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
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 '18 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 '18 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.