# Non Linear PDE Using Charpit's Method .

Using Charpit's method . How Can I solve this : $$u^{2}(p^{2}+q^{2})=x^{2}+y^{2}$$ where$$du=p dx+q dy \ , p =u_{x} ,q=u_{y}$$ So we have $$f(x,y,p,q,u)=u^{2}(p^{2}+q^{2})-x^{2}+y^{2}=0$$ and $$\frac{dx}{-2pU}=\frac{dy}{-2qU}=\frac{dU}{-2U(p^{2}+q^{2})}=\frac{dp}{-2x+p(p^2+q^2)}=\frac{dq}{-2y+q(p^2+q^2)}$$ the solution is $$u^2=b+x\sqrt{(x^2+a^2)}+ a\ ln(x+\sqrt{(x^2+a^2)})+y\sqrt{(y^2-a^2)}+ a\ ln(y+\sqrt{(y^2-a^2)})$$ I don't how he come up with that solution !

Thank you .

## 1 Answer

Use the change of variable $$X=x^2,\quad Y=y^2,\quad U=u^2\\ P=\frac{\partial U}{\partial X},\quad Q=\frac{\partial U}{\partial Y}$$ Note that $$P=\frac{\partial U}{\partial u}\frac{\partial u}{\partial x}\frac{\partial x}{\partial X}=\frac{up}{x}\\ Q=\frac{\partial U}{\partial u}\frac{\partial u}{\partial y}\frac{\partial y}{\partial Y}=\frac{uq}{y}$$ Our pde in terms of the new variables is \begin{align} & P^2X+Q^2Y=X+Y\\\implies& X(P^2-1)+Y(Q^2-1)=0 \end{align}

General method for solving $$f(x,y,u,p,q)=f_1(x,p)+f_2(y,q)=0$$

In this case Charpit's auxiliary equations become $$\frac{dp}{\frac{\partial f_1}{\partial x}}=\frac{dq}{\frac{\partial f_2}{\partial x}}=\frac{du}{-p\frac{\partial f_1}{\partial p}-q\frac{\partial f_2}{\partial q}}=\frac{dx}{-\frac{\partial f_1}{\partial p}}=\frac{dy}{-\frac{\partial f_2}{\partial q}}$$ Cross-multiplying 1st and 4th ratios we get \begin{align} &\frac{\partial f_1}{\partial x}dx+\frac{\partial f_1}{\partial p}dp=0 \implies df_1=0\\ \therefore\;&f_1(x,p)=a \end{align} where $a$ is a constant. And, $$f_2(y,q)=-f_1(x,p)=-a$$ We can write $p,q$ as $$p=F_1(x,a),\quad q=F_2(y,a)$$ Finally, integrating we get the required solution $$u=\int F_1(x,a)\,dx+\int F_2(y,a)\,dy$$

Current problem

\begin{align} F_1(X,a)=&\sqrt{1+\frac{a}{X}},\quad F_2(Y,a)=\sqrt{1-\frac{a}{Y}}\\ \implies U=&\int F_1(X,a)\,dX+\int F_2(Y,a)\,dY\\ \implies u^2=& 2\int\sqrt{x^2+a}\,dx+2\int\sqrt{y^2-a}\,dy\\ =& b+x\sqrt{x^2+a}+\sqrt a\ln|x+\sqrt{x^2+a}|+y\sqrt{y^2-a}-\sqrt a\ln|y+\sqrt{y^2-a}| \end{align} Here we have assumed $a>0$ and $b$ is an arbitrary constant.