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I have searched a bit on the forum and I don't think this question was already answered. I don't really have high mathematical skills so I wouldn't know how to properly solve that but I am really interested in knowing the solutions of this partial differential equation :

$$\frac{\partial f}{\partial x}+a\frac{\partial f}{\partial y}f=0\tag{1}$$

Where $a>0$. I have already asked somewhere else online and I know there are an infinite number of solutions but I wanted to have a rigorous approach and to know if it is possible to find a solution that satisfies to the following conditions:

$$\begin{align}x=0&\Rightarrow f(x,y)=b\\ y=0&\Rightarrow f(x,y)=0\\ x=y=0&\Rightarrow f(x,y)=0\\ x\to\infty&\Rightarrow f(x,y)→0\\ y\to\infty&\Rightarrow f(x,y)→c\end{align}$$

Where $b,c\in\mathbb{R}^+$

Thank you for your help !

(If the question is not adapted to this forum I will remove it)

Thomas Andrews
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Antoine Curé
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    But $f$ is not partially differentiable on $x=0$ because it is discontinuous there. – Jan Safronov May 25 '21 at 17:48
  • Use the method of characteristics. I get the solution to be in implicit form $$f = g(y - axf)$$ which you can check by differentiation solves the problem \begin{align} f_{x} &= -(af + axf_{x}) g' \ \implies f_{x} &= \frac{-afg'}{(1 + axg')} \ f_{y} &= (1 - axf_{y}) g' \ \implies f_{y} &= \frac{g'}{(1 + axg')} \ \implies f_{x} + aff_{y} &= 0 \end{align} Now use your conditions to determine the form of $g$ and hence $f$. – Matthew Cassell May 25 '21 at 23:54

2 Answers2

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Except for notation, this is the well known inviscid Burger's equation. To make it look like most references, replace $x,y$ by $t,x$ and set $u(x,t) = af(t,x)$; the problem becomes $$ u_t+uu_x = 0, \qquad u(x,0) = ab, \qquad \lim_{t\to\infty}u(x,t) = 0, $$ with some other conditions.

There is no solution. The short reason is that, applying the intial value $u=ab$ only, there is the constant solution $u(x,t) = ab$ (or $f(x,y) = b$). Then applying the limit condition gives a contradiction. Applying more conditions doesn't remove the difficulty.

For more information about this equation, one standard reference is Hyperbolic Systems of Conservation Laws and the Mathematical Theory of Shock Waves, by Peter Lax. In particular the statement about the contradicion holds because solutions are unique, in other words the constant solution $u=ab$ causes trouble but is there some better one? No, but this is not obvious so it requires some work.

Bob Terrell
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That looks like a candidate for "separation of variables". Look for a solution if the form f(x,y)= X(x)Y(y) where X is a function of x only and Y is a function of y only. The equation becomes $X'Y+ aX^2YY'= 0$. Dividing both sides by $X^2Y$ gives $\frac{X'}{X^2}+ aY'= 0$ or $\frac{X'}{X^2}= -aY'$. Since the left side is a function of x only while the right side is a function of y only, the only way they can be equal for all x and y is if they are constants.

Call that constant C. The equations become $\frac{X'}{X^2}= C$ and $-aY'= C$

Those are both easy to solve. We can write the first one as $\frac{dX}{X^2}= Cdx$ and, integrating both sides $-X^{-2}= Cx+ D$, $X^2= -\frac{1}{Cx+ D}$. $X= \frac{1}{\sqrt{-Cx- D}}$

$dY= -aCdx$ so $Y= -aCx+ E$.

So $f(x,y)= X(x)Y(y)= -\frac{aCx+ E}{\sqrt{-Cx+ D}}$.

Now you need to determine the constants C, D and E so that this function, or linear combinations of such functions (for different C), satisfy the boundary conditions.

user247327
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