# Solve the First Order PDE equation

Problem: For equation: $$u=xu_x+yu_y+\frac{1}{2}(u_x^2+u_y^2)$$ find the solution with $$u(x,0)=\frac{1}{2}(1-x^2)$$

Here is what I have so far:

Let $$F(x,y,z,p,q)=z-xp-yq-\frac{1}{2}(p^2+q^2)$$ with $$p=u_x,\\ q=u_y$$

Characteristic Equations:

$$\begin{eqnarray} \frac{dx}{dt}&=& -x-p \\ \frac{dy}{dt}&=&-y-q \\ \frac{dz}{dt}&=&-px-p^2-qy-q^2 \\ \frac{dp}{dt}&=&0 \\ \frac{dq}{dt}&=&0 \end{eqnarray}$$ I know that I need to get Characteristic strips( Solutions of Characteristic Equations) but I am stuck. Please help. Thanks

By inspection one find two solutions : $$u(x,y)=\frac12(1-x^2)+y$$ $$u(x,y)=\frac12(1-x^2)-y$$ Both satisfy the PDE and the condition. Of course this oversimplified approach doesn't say if they are more solutions or not.

$$\underline {\text{HINT to solve with the characteristics method}}$$ .

A preliminary transformation in order to simplify the job :

Change of function : $$\quad u(x,y)=\frac12 (v^2-x^2-y^2)\quad$$ with $$v=v(x,y)$$ .

$$u_x=vv_x-x\quad$$ and $$\quad u_y=vv_y-y$$

$$x(vv_x-x)+y(vv_y-y)+\frac12(vv_x-x)^2+\frac12(vv_y-y)^2=\frac12(v^2-x^2-y^2)$$

After simplification : $$\boxed{(v_x)^2+(v_y)^2=1}$$ With condition $$v(x,0)=2u(x,0)+x^2+0^2=(1-x^2)+x^2=1$$ $$\boxed{v(x,0)=1}$$ I suppose that you can take it from here.

From last two equations: $$p=a, ~~q=b$$ where $a$, $b$ are arbitrary constants. First three equations can be written as $$dz=pdx+q dy=adx+bdy.$$ Integration gives$$z=ax+by+c$$ You can find $c={a^2+b^2\over 2}$

This is a special case in Charpit's method and the equation is called as Clairaut's equation.

• $u(x,y)=ax+by+c$ doesn't agree with the condition $u(x,0)=\frac12(1-x^2)$ . – JJacquelin Apr 29 '19 at 15:16