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Given the partial differential equation \begin{equation} (1-x)\left[- f(x,y) + \frac{\partial f(x,y)}{\partial x} \right] + (1-y)\left[- f(x,y) + \frac{\partial f(x,y)}{\partial y} \right] = 0 \end{equation} in the interval $0 \leq x < 1$, $0 \leq y < 1$, the following solution can be obtained by separation of variables: \begin{equation} f(x,y) = C \, \exp(x+y) \end{equation}

Is this the general solution?

Thank you.

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  • $\begingroup$ One integration constant is definitely not enough to parametrize the general solution. What you probably want here is probably the Method of Characteristics. $\endgroup$ – Igor Khavkine Jun 4 '15 at 16:33
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With the method of characteristics, I obtain the following solution. The characteristic equations are \begin{eqnarray} \frac{d x(s)}{ds} & = & 1-x(s),\\ \frac{d y(s)}{ds} & = & 1-y(s),\\ \frac{d F(x(s),y(s))}{ds} & = & [2-x(s)-y(s)]F(x(s),y(s)). \end{eqnarray} The solution of the third characteristic equation is \begin{equation} F(x(s),y(s)) = F(x(0),y(0)) \, \exp[x(s)-x(0)+y(s)-y(0)]. \end{equation} The solution of the first two characteristic equations is \begin{eqnarray} \frac{1-x(s)}{1-x(0)} &=& \exp(-s),\\ \frac{1-y(s)}{1-y(0)} &=& \exp(-s). \end{eqnarray} Setting $y(0)=0$, the solution of the first two characteristic equations implies that $x(0)$ is a function of $\frac{1-x(s)}{1-y(s)}$: substituting this into the expression above for $F(x(s),y(s))$, we have \begin{equation} F(x,y) = G\left( \frac{1-x}{1-y} \right) \exp(x+y). \end{equation}

This is the 'more general' solution proposed by Robert Israel. My question is: Is this the general solution?

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  • $\begingroup$ The answer may depend on what you mean by "the general solution". Given the order and the type of the equation (scalar, first order) a generic subset of the solution is expected to depend on one function of one variable, $G(z)$ here. So a rough answer is Yes, but it's more difficult to rule out the possibility that there are no other solutions, that is, those that escape the above parametrization. $\endgroup$ – Igor Khavkine Jun 5 '15 at 12:12
  • $\begingroup$ Thank you for your reply. May you please suggest a reference where I could find a justification of your statement "Given the order and the type of the equation (scalar, first order) a generic subset of the solution is expected to depend on one function of one variable"? $\endgroup$ – James Jun 5 '15 at 13:52
  • $\begingroup$ That was not a rigorous comment and trying to make it rigorous introduces lots of complications that are likely irrelevant to you, especially since you haven't specified what notion of "general solution" is important to you. For the purposes of this question, it is sufficient to notice that the method of characteristics allows you to parametrized solutions within a neighborhood of a codim-1 "initial-value" submanifold (one that transversely intersects the characteristics) by the restriction of the solution to this submanifold. $\endgroup$ – Igor Khavkine Jun 5 '15 at 14:22
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A more general solution is

$$ f(x,y) = G\left( \dfrac{1-y}{1-x}\right) \exp(x+y) $$

for arbitrary differentiable function $G$.

EDIT: To see that this is the general solution, write $f(x,y) = g(x,y) \exp(x+y)$. The equation becomes

$$ (1-x) \dfrac{\partial g}{\partial x} + (1-y) \dfrac{\partial g}{\partial y} = 0$$

With a change of variables $s = (1-y)/(1-x), t = x$ (i.e. $y = st-s+1, x=t$), the equation becomes $$ (1-t) \dfrac{\partial g}{\partial t} = 0 $$ Thus for $x < 1$, $g$ must be a function of $s = (1-y)/(1-x)$ alone.

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  • $\begingroup$ This solves the problem, thank you. $\endgroup$ – James Jun 6 '15 at 11:39

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