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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.

Chill2Macht
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niksirat
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1 Answers1

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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 $$(1) ady-bdx=0\Rightarrow \frac{dy}{b}=\frac{dx}{a}$$ $$(2) cdy-bdu=0\Rightarrow \frac{du}{c}=\frac{dy}{b}$$

AnilB
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