The problem is the following:
I have the bidimensional eikonal equation with non-constant propagation:
$u_x^2+u_y^2=u^2$
The goal is:
i) To find the characteristic strips for the parametrization $x=f(s)$, $y=g(s)$ and $z=u(f(s),g(s))=h(s)$;
ii)To find the integral surfaces passing through the circle $x=cos(s)$, $y=sin(s)$, $z=1$;
iii)To find the integral surfaces through the line $x=s$, $y=0$, $z=1$.
Writing the equation in general form,$F(x(s),y(s),u,p(s),q(s))=0$, i.e., $p^2+q^2-u^2=0$, where $p=u_x, q=u_y$,we have, for the item i), the following system of EDO's:
$\frac{dx}{ds}=F_p=2p$
$\frac{dy}{ds}=F_p=2q$
$\frac{dz}{ds}=pF_p+qF_q=2(p^2+q^2)$
$\frac{dp}{ds}=-F_x-pF_u=2(u_xu+pu)$
$\frac{dq}{ds}=-F_y=2(u_yu+qu)$
For item ii), I know the answer is $z=\exp[\pm (1-\sqrt{x^2+y^2})]$, but I don't know how to find it;
For item iii), I know the answer is $u=\exp(\pm y)$, but I don't know how to find it.
Suggestions to solve the problem?
Best Answer
Write the PDE as $F(x,y,u,u_{x},u_{y})=0$, where $F(x,y,z,p,q)=p^{2}+q^{2}-z^{2}$. The characteristic equations are $$\begin{cases}\dot{x}(s)=F_{p}(x(s),y(s),z(s),p(s),q(s))=2p(s)\\ \dot{y}(s)=F_{q}(x(s),y(s),z(s),p(s),q(s))=2q(s)\\ \dot{z}(s)=p(s)F_{p}(x(s),y(s),z(s),p(s),q(s))+q(s)F_{q}(x(s),y(s),z(s),p(s),q(s))=2(p(s)^{2}+q(s)^{2})=2z(s)^{2}\\ \dot{p}(s)=-F_{x}(x(s),y(s),z(s),p(s),q(s))-p(s)F_{z}(x(s),y(s),z(s),p(s),q(s))=2p(s)z(s)\\ \dot{q}(s)=-F_{y}(x(s),y(s),z(s),p(s),q(s))-q(s)F_{z}(x(s),y(s),z(s),p(s),q(s))=2q(s)z(s)\end{cases}$$
Suppose that we have initial data $(x^{0},y^{0},z^{0},p^{0},q^{0})$. We first solve the ODE for $z(s)$, obtaining $$z(s)=\dfrac{z^{0}}{1-2z^{0}s}$$
We substitute this result in to the ODEs for $p(s),q(s)$ to obtain $$\begin{cases}\dot{p}(s)=2p(s)\dfrac{z^{0}}{1-2z^{0}s}, & p(s)=\dfrac{p^{0}}{1-2z^{0}s}\\ \dot{q}(s)=2q(s)\dfrac{z^{0}}{1-2z^{0}s}, & q(s)=\dfrac{q^{0}}{1-2z^{0}s} \end{cases}$$
Lastly, we substitute these results into the ODEs for $x(s),y(s)$ to obtain $$\begin{cases}\dot{x}(s)=2\dfrac{p^{0}}{1-2z^{0}s}, & x(s)=-\dfrac{p^{0}}{z^{0}}\ln\left|1-2z^{0}s\right|+x^{0}\\ \dot{y}(s)=2\dfrac{q^{0}}{1-2z^{0}s}, & y(s)=-\dfrac{q^{0}}{z^{0}}\ln\left|1-2z^{0}s\right|+y^{0}\end{cases}$$ for $s$ in the appropriate domain.
Suppose our boundary condition is $u=1$ on $\Gamma=S^{1}$. Fix a point $(x,y)\in\mathbb{R}^{2}$, and let $s$ and $(x^{0},y^{0})=(\cos\theta^{0},\sin\theta^{0})$, for some $\theta^{0}\in[0,2\pi]$ be such that $(x,y)=(x(s),y(s))$. Clearly, $z^{0}=1$, so it remains to solve for $x^{0},y^{0},p^{0},q^{0},s$. Since $\frac{d}{d\theta}u(\cos\theta,\sin\theta)_{\theta^{0}}=Du(x^{0},y^{0})\cdot (-y^{0},x^{0})=0$, we see that $(p^{0},q^{0})=\lambda(x^{0},y^{0})$, where $\lambda$ is some nonzero scalar. Since ${p^{0}}^{2}+{q^{0}}^{2}={z^{0}}^{2}=1$, we see that $\left|\lambda\right|=1$.
$$x^{2}+y^{2}=\left({x^{0}}^{2}+{y^{0}}^{2}\right)\left[\lambda^{-1}(1-\lambda\ln\left|1-2s\right|)\right]^{2}\Rightarrow s=\dfrac{1-\exp[\lambda^{-1}(1-\sqrt{x^{2}+y^{2}})]}{2}$$
We conclude that $$u(x,y)=u(x(s),y(s))=z(s)=\exp\left[-\lambda^{-1}(1-\sqrt{x^{2}+y^{2}})\right]=\exp\left[\pm(1-\sqrt{x^{2}+y^{2}})\right]$$
Suppose now our boundary condition is $u=1$ on $\Gamma=\mathbb{R}\times\left\{0\right\}$. Proceeding as above, we fix a point $(x,y)\in\mathbb{R}^{2}$, and let $s$ and $(x^{0},0)$ be such that $(x,y)=(x(s),y(s))$. As before, $z^{0}=1$. Since $u=1$ on $\Gamma$, we obtain that $p^{0}=u_{x}(x^{0},0)=0$. From the identity ${p^{0}}^{2}+{q^{0}}^{2}={z^{0}}^{2}$, we obtain that $\left|q^{0}\right|=1$. Substituting this information into the equations above, we see that $$x=x^{0}, y=-q^{0}\ln\left|1-2s\right|\Rightarrow s=\dfrac{1-e^{\pm y}}{2}$$ Substituting this result into the equation for $z(s)$, we conclude that $$u(x,y)=u(x(s),y(s))=z(s)=e^{\pm y}$$