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Solve $$(y-u)u_x + (u-x)u_y = x-y, \qquad u=0 \text{ when } xy=1.$$

I tried to solve the equation above using characteristic method $$\begin{cases} x'=y-u \\ y' = u - x \\ u' = x-y \end{cases}. \tag{1}$$

Adding together the first and the second equation in $(1)$ we get that $x'+y'=y-x$. Thus (using third equation in $(1)$) $$u = x+ y + C \implies C = u - x- y.$$

Solving the first and the second equation yields to $$\frac{1}{2} y^2 - uy = ux - \frac{1}{2}x^2 + \tilde{C} \implies \tilde{C} = \frac{1}{2}(y^2 +x^2)-u(x+y).$$

That gives us $$F\big(u-x-y, \frac{1}{2}(y^2 +x^2)-u(x+y) \big) = 0.$$

How can I finish this example?

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  • $\begingroup$ I don't agree with your intermediate results. For example I think that $C=u+x+y$ instead of $C=u-x-y$. Since they are not enough steps edited I can't check your calculus for the second characteristic equation. HINT : See my answer below. $\endgroup$
    – JJacquelin
    Commented Apr 28, 2020 at 14:36

1 Answer 1

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$$(y-u)u_x+(u-x)u_y=x-y$$

System of characteristic ODEs (Charpit-Lagrange) : $$\frac{dx}{y-u}=\frac{dy}{u-x}=\frac{du}{x-y}$$ A first characteristic equation comes from :

$\frac{dx}{y-u}=\frac{dy}{u-x}=\frac{du}{x-y}=\frac{dx+dy+du}{(y-u)+(u-x)+(x-y)}=\frac{dx+dy+du}{0} \quad\implies\quad dx+dy+du=0$

$$u+x+y=c_1$$

A second characteristic equation comes from :

$\frac{dx}{y-u}=\frac{dy}{u-x}=\frac{du}{x-y}=\frac{xdx+ydy+udu}{x(y-u)+y(u-x)+u(x-y)}=\frac{xdx+ydy+udu}{0} \quad\implies\quad xdx+ydy+udu=0$

$$x^2+y^2+u^2=c_2$$

General solution of the PDE expressed on the form of implicit equation $\Phi(c_1,c_2)=0$ or equivalently $c_2=F(c_1)$ :

$$x^2+y^2+u^2=F(x+y+u)$$ $\Phi$ or $F$ are undetermined functions until no condition is specified.

With condition $u=0$ when $xy=1$ :

Then $x^2+y^2+0=F(x+y+0)=x^2+\frac{1}{x^2}=F(x+\frac{1}{x})$

Let $X=x+\frac{1}{x} \quad\implies\quad x=\frac12\left(X\pm\sqrt{X^2-4}\right)$

$F(X)=\frac14\left(X\pm\sqrt{X^2-4}\right)^2+\frac{4}{\left(X\pm\sqrt{X^2-4}\right)^2}$ and after simplification : $$F(X)=X^2-2$$ Now $F(X)$ is determined. We put it into the above general solution where $X=x+y+u$ Then $F(x+y+u)=(x+y+u)^2-2$ .

$$x^2+y^2+u^2=(x+y+u)^2-2$$ Solving for $u$ leads to : $$\boxed{u(x,y)=\frac{1-xy}{x+y}}$$

NOTE:

Another method consists in the change of variables $\zeta=x+y$ ; $\eta=xy$ which leads to a separable PDE.

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  • $\begingroup$ In which region is the solution valid? I suppose there are some constraints imposed by the choice $X = x + 1/x$, namely $|X| \ge 2$. $\endgroup$ Commented Sep 19, 2022 at 6:19
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    $\begingroup$ @esoteric-elliptic : This is strictly the answer to the OP's question. No more. The above solution satisfies both the PDE and satisfies the specified condition. One hasn't to be more papist than the Pope. $\endgroup$
    – JJacquelin
    Commented Sep 19, 2022 at 6:52
  • $\begingroup$ That's fair - I was just asking a follow-up question, if you may be kind enough to answer. Besides, solving a PDE does entail mentioning the region in which one has solved it, especially if it is not the whole of $\Bbb R^2$. Thanks! Lastly, there isn't a unique choice for the substitution, you could have picked $X = -x -1/x$ as well. Perhaps that leads to a different solution? $\endgroup$ Commented Sep 19, 2022 at 8:51
  • $\begingroup$ Of course since $F$ is an arbitrary function one could pick $X=-(x+1/x)$ or $X=(x+1/x)^2$ or many others. The function $F$ changes wrt the "picked" argument function without changing the general solution. But the calculus would be more complicated. Then why not chosing the simplest ? I don't understand your question " solving a PDE does entail mentioning the region in which one has solved it" since the solution involves an arbitrary function. I suppose that you mean "solving and satisfying a given boundary condition". OK. One have to study the region where the solution exists. $\endgroup$
    – JJacquelin
    Commented Sep 19, 2022 at 10:15

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