3
$\begingroup$

Theorem 1 in page 4 of the book Numerical Solution of Partial Differential Equations in Science and Engineering by L. Lapidus:

The general solution of the quasilinear PDE $$a(x,y,u)u_x + b(x, y,u)u_y = c(x, y,u),$$ is given by $$G(v,w)=0,$$ where $G$ is an arbitrary function and where $v(x, y, u) = c_1$, and $w(x,y,u)=c_2$ form a solution of the equations $$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$$

My question is how can I prove this theorem?

In that book, it explains how can interpret quasilinear PDEs geometrically, but I can't understand how to obtain $\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}$.

Thanks in advance for any explanation or clarification about this theorem.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Assume that $u(x,y)$ is a solution then $$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy$$ $$\Rightarrow \frac{\partial u}{\partial y}=\frac{du-\frac{\partial u}{\partial x}dx}{dy}$$ The equation becomes $$a\frac{\partial u}{\partial x}+b\frac{du-\frac{\partial u}{\partial x}dx}{dy}=c$$ $$a\frac{\partial u}{\partial x}dy+bdu-b\frac{\partial u}{\partial x}dx=cdy$$ $$\frac{\partial u}{\partial x}\bigg(ady-bdx\bigg)=cdy-bdu$$ The equation holds for $$(1) ady-bdx=0\Rightarrow \frac{dy}{b}=\frac{dx}{a}$$ $$(2) cdy-bdu=0\Rightarrow \frac{du}{c}=\frac{dy}{b}$$

$\endgroup$
3
  • $\begingroup$ Thanks @Occupy for your question, can i ask one another question? How G is answer of the pdes? i mean a construction proof. $\endgroup$
    – niksirat
    May 21, 2016 at 2:12
  • $\begingroup$ @user626 You can check the Cauchy problem to get more insight. $\endgroup$
    – AnilB
    May 21, 2016 at 10:49
  • $\begingroup$ @ccupy-gezi I read Cauchy problem but i can't find relation between my question and it. $\endgroup$
    – niksirat
    May 28, 2016 at 9:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .