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I have a problem with partial differential equation

$$u_xu_y=xy$$ where $$u(0,y)=-y.$$

Therefore $x_0(s)=0$,$y_0(s)=s$,$u_0(s)=-s.$ I wrote my equation as $F(x,y,u,p,q)=pq-xy=0$ and defined the characteristic equations: $$\frac{dx}{dt}=q$$ $$\frac{dy}{dt}=p$$ $$\frac{du}{dt}=2pq$$ $$\frac{dp}{dt}=y$$ $$\frac{dq}{dt}=x.$$

I know that $F(x_0(s),y_0(s),u_0(s),p_0(s),q_0(s))=0$ and $p_0(s)x_0'(s)+q_0(s)y_0'(s)=u_0'(s)$, then $p_0(s)q_0(s)=0$ and $q_0(s)=-1$, therefore $p_0(s)=0.$ Then $p(t,s)=0$ and $q(t,s)=-1$ and then I solved characteristic equations: $x(t,s)=-t, y(t,s)=s, u(t,s)=-s.$ But u(x,y)=-y is not a solution, and I don't know what I did wrong.

Mmath
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2 Answers2

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That's a good start, but the point where you go wrong is when you claim that $p(t,s)$ and $q(t,s)$ are constant functions; that's not correct, since their derivatives are $dp/dt=y$ and $dq/dt=x$, not $dp/dt=0$ and $dq/dt=0$.

To solve the characteristic equations correctly, note that the system decouples into two subsystems that you can solve separately, $$ \frac{dx}{dt} = q ,\qquad \frac{dq}{dt} = x $$ and $$ \frac{dy}{dt} = p ,\qquad \frac{dp}{dt} = y , $$ together with the last equation for $u(t,s)$, which you can integrate rather easily once you have computed $q(t,s)$ and $p(t,s)$: $$ \frac{du}{dt} = 2pq . $$ And of course you also need to use the right initial conditions at $t=0$ for all these equations, but you have already determined those conditions correctly ($x(0,s)=x_0(s)=0$, etc.), so I'm not writing them here.

There are many ways to solve the subsystems. For example, note that $d^2 x/dt^2=x$. Or use $x+q$ and $x-q$ as new variables.

Do you think you can take it from here?

Hans Lundmark
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I solved it and now I have: $$x(t,s)=-\frac{1}{2}(e^t-e^{-t})$$ $$q(t,s)=-\frac{1}{2}(e^t+e^{-t})$$ $$y(t,s)=\frac{s}{2}(e^t+e^{-t})$$ $$p(t,s)=\frac{s}{2}(e^t-e^{-t})$$ and $u(t,s)=-\frac{s}{4}(e^{2t}+e^{-2t})-\frac{s}{2}.$

I checked the initial conditions and I think that it is a good solution, but I saw that

$$u(x,y)=xy-\frac{s}{2}$$ and I have a problem with $s$, because the function should depend only on $x$ and $y$.

Mmath
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    Excellent! Maybe it's easier to go back to $x$ and $y$ if you write the expressions as $x=-\sinh t$, $y=s \cosh t$ and $u=-s \cosh^2 t = -s \cosh t \cdot \cosh t$? (Then you can use $\cosh t = \sqrt{1 + \sinh^2 t}$.) – Hans Lundmark Mar 24 '21 at 10:44
  • It's great idea! I have solution, much thanks! – Mmath Mar 24 '21 at 11:14