pde first order nonlinear - eikonal equation

Question

How to solve this equation?

$$ \left( \dfrac{\partial U}{\partial x } \right)^2 + \left( \dfrac{\partial U}{\partial y } \right)^2 = 1$$ with the Cauchy problem $U|_{y=0} = U(x)$.

I don't know how to start to solve this.

Thank you!

Answer 1

$$ \left( \dfrac{\partial U}{\partial x }\right)^2 + \left( \dfrac{\partial U}{\partial y } \right)^2 = 1 \tag 1$$ $$\dfrac{\partial U}{\partial y } =\pm\sqrt{1-\left( \dfrac{\partial U}{\partial x } \right)^2 }$$

$$\dfrac{\partial^2 U}{\partial x\partial y } =\mp\frac{\dfrac{\partial U}{\partial x }\dfrac{ \partial^2 U}{\partial x^2 } }{\sqrt{1-\left( \dfrac{\partial U}{\partial x } \right)^2 }}$$

$$\dfrac{\partial U}{\partial x }\dfrac{ \partial^2 U}{\partial x^2 } \pm\sqrt{1-\left( \dfrac{\partial U}{\partial x } \right)^2} \dfrac{\partial^2 U}{\partial x\partial y }=0$$

Let $z(x,y)=\dfrac{\partial U}{\partial x }$

$$z\dfrac{\partial z}{\partial x }\pm\sqrt{1-z^2}\dfrac{\partial z}{\partial y }=0 \tag 2$$ This is a quasi linear first order PDE which solving is well known (Use method of characteristics or other classical method) :

$$z=F\left(x\pm\frac{z\:y}{\sqrt{1-z^2}}\right) \tag 3$$ The solution $z(x,y)$ is on the form of implicit equation. $F$ is an arbitrary function, to be determined according to the boundary condition.

Boundary condition :

$U(x,0)=U(x)$ where $U(x)$ is a given function.

In order to avoid confusion between unknown $U(x,y)$ and known $U(x)$ , better change of symbol :

$U(x,0)= f(x)$ where $f(x)$ is a given function.

$\left(\dfrac{\partial u}{\partial x }\right)_{(x,0)}=\frac{df(x)}{dx}=G(x)$

$G(x)$ is a known function since $f(x)$ is given. Thus : $$z(x,0)=G(x)\quad\text{is the boundary condition for equation }(2)$$

Putting it into Eq.$(3)$ with $y=0$ leads to : $$G(x)=F(x)$$ So, now the function $F$ is determined. Putting it into Eq.$(3)$ : $$z=G\left(x\pm\frac{z\:y}{\sqrt{1-z^2}}\right) \tag 4$$ To solve completely the initial PDE $(1)$, the next step is the integration of $z(x,y)=\dfrac{\partial U}{\partial x }$ with $y$ considered as a parameter. $$U(x,y)=\int\dfrac{\partial U}{\partial x }dx=\int z(x,y)dx$$ In general, this cannot be donne without $z(x,y)$ on an explicit form, not on the form of implicit equation $(4)$.

So, further calculus depends on the function $f(x)$, that is the function $U(x)$ appearing in the wording of the problem.

• In the general case, one cannot go further and the solution $U(x,y)$ of the initial PDE cannot be expressed on a closed form. Nevertheless, the problem is loosely considered as implicitly solved.

• For some particular functions $f(x)$, the equation $(4)$ can be solved for explicit $z(x,y)$ and then the solution $U(x,y)$ can be obtained on close form (Of course, if the integral exists on close form, which is far to be the general case).