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I have a first order PDE:


With the initial condition:


I have calculated result in Mathematica: $u(x,y)=\dfrac{y}{x}$ , but I am trying to solve the equation myself, but I had no luck so far. I tried with method of characteristics, but I could not get the correct results. I would appreciate any help or maybe even whole procedure.


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The function $u$ which you call the final result does not solve the PDE. – Did Apr 19 '13 at 10:57
up vote 0 down vote accepted



Follow the method in

$\dfrac{dx}{dt}=1$ , letting $x(1)=1$ , we have $x=t$

$\dfrac{dy}{dt}=\dfrac{y}{x}+1=\dfrac{y}{t}+1$ , letting $y(1)=y_0$ , we have $y=y_0t+t\ln t=y_0x+x\ln x$

$\dfrac{du}{dt}=\dfrac{1}{x}=\dfrac{1}{t}$ , letting $u(1)=f(y_0)$ , we have $u(x,y)=\ln t+f(y_0)=\ln x+f\left(\dfrac{y}{x}-\ln x\right)$

$u(1,y)=y$ :


$\therefore u(x,y)=\ln x+\dfrac{y}{x}-\ln x=\dfrac{y}{x}$

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There must be (at least) an error here, but I fail to see it:

General first order two variables PDE:

$$a(x,y)u_{x}(x,y)+b(x,y)u_{y}(x,y)+c(x,y)u(x,y)=f(x,y) $$

In this case

$a(x,y)=x$, $b(x,y)=(x+y)$, $c(x,y)=0$ and $f(x,y)=1$

Now we will find some $$\xi(x,y)=c $$

$c \in \mathbb{R}$ constant solves the differential equation

$$\frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}=\frac{x+y}{x}=1+\frac{y}{x} $$

In order to solve this define a new function $$u(x)=\frac{y(x)}{x}$$ so that

$$\frac{du}{dx}=\frac{dy}{dx}\frac{1}{x}-\frac{y(x)}{x^2}=\left(1+u\right)\frac{1}{x}-\frac{u}{x}=1 $$

Hence $u(x)=x+c$, $\frac{y}{x}=x+c \rightarrow c=\frac{y}{x}-x$

$$\boxed{\xi(x,y)=\frac{y}{x}-x} \Rightarrow (\xi +x)x=y$$

Now let $u(x,y)=v(x,\xi)$ and take derivatives

$$u_{x}=v_{x}+v_{\xi}\left( -\frac{y}{x^2}-1\right) =v_{x}+\left( -\frac{(\xi+x)}{x}-1\right)v_{\xi}$$

$$ u_{y}=v_{\xi}\frac{1}{x}$$

Substitution into the PDE yields

$$ x\left( v_{x}+\left( -\frac{(\xi+x)}{x}-1\right)v_{\xi}\right)+(x+(\xi+x)x)v_{\xi}\frac{1}{x}=1$$

$$xv_{x}-(\xi+x)v_{\xi} -xv_{\xi}+v_{\xi}+\xi v_{\xi}+xv_{\xi}=1$$


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