# Solving $u_x^2+u_y^2=n_0^2$ Using Method of characteristics

$$\begin{cases} u_x^2+u_y^2=n_0^2\\ u(x,2x)=1\\ \end{cases}$$

For constant $$n_0$$

This is the Eikonal equation which is a non-linear PDE, to use the method of characteristics we "bring it down" to an ode by substituting the partial derivatives.

$$P_1=u_x$$ $$P_2=u_y$$ $$P_1^2+P_2^2-n_0^2=0$$ And the characteristics strips are:

$$\begin{cases} \frac{dx_i}{dt}=\frac{\partial F}{\partial P_i}\\ \frac{du}{dt}=\sum_{i=1}^{n}P_i\frac{\partial F}{\partial P_i}\\ \frac{dP_i}{dt}=-\frac{\partial F}{\partial x_i}-P_i\frac{\partial F}{\partial u}\\ \end{cases}$$

In our case it is:

$$\begin{cases} \frac{dx}{dt}=\frac{\partial F}{\partial P_1}\iff x_t=2P_1\\ \frac{dy}{dt}=\frac{\partial F}{\partial P_2}\iff y_t=2P_2\\ \frac{du}{dt}=P_1\frac{\partial F}{\partial P_1}+P_2\frac{\partial F}{\partial P_2}\iff u_t=2P_1^2+2P_2^2=2(P_1^2+P_2^2)=2n_0^2\\ \frac{dP_1}{dt}=-\frac{\partial F}{\partial x}-P_1\frac{\partial F}{\partial u}\iff P_{1_t}=-0-P_1\cdot 0=0\\ \frac{dP_2}{dt}=-\frac{\partial F}{\partial y}-P_2\frac{\partial F}{\partial u}\iff P_{2_t}=-0-P_2\cdot 0=0\\ \end{cases}$$

We have $$P_{1_t},P_{2_t}=0$$ as $$x,y,u$$ does not appear in $$F$$?

• You have $dP_{i}/dt = 0 \implies P_{i} = c_{i}$ for $i = 1, 2$ and where $c_{i}$ are constants. Mar 11, 2020 at 14:58

$$u_x^2+u_y^2-n_0^2=0=p^2+q^2-n_0^2=F(x,y,p,q,u)$$ $$F_x=0\quad;\quad F_y=0\quad;\quad F_p=2p\quad;\quad F_q=2q\quad;\quad F_u=0$$ Charpit-Lagrange system of characteristic ODEs :
$$\dfrac{dp}{F_x+pF_u}=\dfrac{dq}{F_y+qF_u}=\dfrac{du}{-pF_p-qF_q}=\dfrac{dx}{-F_p}=\dfrac{dy}{-F_q}$$ $$\dfrac{dp}{0}=\dfrac{dq}{0}=\dfrac{du}{-2p^2-2q^2}=\dfrac{dx}{-2p}=\dfrac{dy}{-2q}=dt$$ Note that this is the same as : $$\begin{cases} \frac{dp}{dt}=0\\ \frac{dq}{dt}=0\\ \frac{du}{dt}=-2p^2-2q^2\\ \frac{dx}{dt}=-2p\\ \frac{dy}{dt}=-2q \end{cases}$$ A first characteristic equation comes from $$dp=0$$ $$p=c_1$$ A second characteristic equation comes from $$dq=0$$ $$q=c_2$$ A third characteristic equation comes from $$\dfrac{du}{-2c_1^2-2c_2^2}=\dfrac{dx}{-2c_1}=\dfrac{dy}{-2c_2}=\frac{du-c_1dx-c_2dy}{-2c_1^2-2c_2^2-c_1(-2c_1)-c_2(-c_2)}=\frac{du-c_1dx-c_2dy}{0}$$
$$du-c_1dx-c_2dy=0\quad\implies\quad u-c_1x-c_2y=c_3$$ $$u=c_1x+c_2y+c_3$$ Since we are not looking for the general solution but only for a particular solution which satisfies the boundary condition $$u(x,2x)=1$$ , there is no need for further calculus about more characteristic equations.
Condition : $$\quad u(x,2x)=c_1x+c_2(2x)+c_3=(c_1+2c_2)x+c_3=1\implies\begin{cases}c_1=-2c_2\\c_3=1\end{cases}$$
$$u=-2c_2x+c_2y+1$$
$$p^2+q^2=(-2c_2)^2+c_2^2=n_0^2\quad\implies c_2=\pm\frac{n_0}{\sqrt{5}}$$ $$\boxed{u(x,y)=\frac{\pm n_0}{\sqrt{5}}(-2x+y)+1}$$