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I'm trying to prove that the following PDE has no solutions that are non-constant functions $f\in C^1(\mathbb{R}^2)$.

$$ (y+1)u_y + x u_x = 0 $$

I'm not sure what is the theory that would help start formulating a solution for this since the material provided in my class is relatively bad explained and/or I'm missing something.

I tried first solving it with the method of characteristics:

$$ \frac{dy}{dx} = \frac{y+1}{x} \implies \frac{y+1}{x} = K $$

Then I consider the variable change: $$ \xi = \frac{y+1}{x}\\ \eta = y $$ And I get of course: $$ (\eta+1)u_\eta = 0 \implies u=constant $$

But I suspect this doesn not constitute the proof requested. This would just proof there is constants solutions. I've tried also asumming it has such solution $u(x, y)$ non-constant to get to an absurd.

Then $u_x \neq 0$ and $u_y \neq 0$.

And Then it would be

$$(y+1)u_y = -xu_x \quad \text{with} \quad (y+1)u_y\neq 0; xu_x\neq 0; y\neq1; x\neq0$$

But I don't get anything.

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2 Answers 2

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$$ (y+1)u_y + x u_x = 0 $$ $\frac{dy}{y+1}=\frac{dx}{x}=\frac{du}{0}$

First characteristics, from $\quad\frac{dy}{y+1}=\frac{dx}{x}$ : $$\frac{y+1}{x}=c_1$$ Second characteristics, from $\frac{du}{0}=$finite, which implies $u=c_2$ :

General solution on the form of implicit equation : $$\Phi\left(\frac{y+1}{x}\:,\:u\right)=0$$ $\Phi$ is an arbitrary function of two variables, to be determined according to some boundary conditions.

Or equivalently, on explicit form : $$u(x,y)=f\left(\frac{y+1}{x}\right)$$ $f$ is an arbitrary function, to be determined according to some boundary conditions.

If $f$ is a non constant function, $u$ is necessarily a function of $\frac{y+1}{x}$, so is not $C^1$. Thus the PDE has no solutions that are non-constant functions $f\in C^1(\mathbb{R}^2)$.

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  • $\begingroup$ Just a comment, your answer seems to imply that if the problem were, for example, defined for $C^1((0,\infty)\times(0,\infty))$, then the solution would not necessarily be constant, but this is not the case. The solution is constant also. $\endgroup$ Commented May 15, 2018 at 6:44
  • $\begingroup$ @Enredanrestos : So $u(x.y)=\frac{x}{1+y}$ is not a solution of the PDE in the first quadrant? $\endgroup$ Commented May 15, 2018 at 15:47
  • $\begingroup$ Yes it is, I was wrong. The point apparently is that the characteristics dont cross 'inside' the domain.. $\endgroup$ Commented May 15, 2018 at 21:47
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The method of characteristics says that $u$ is constant on each ray through $(x,y) = (0,-1)$. Thus the value at every point is equal to the value at $(0,-1)$, therefore is constant.

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