Solve the Cauchy problem $$(x − y − 1)u_x + (y − x − u + 1)u_y = u,$$ if $u=1$ on $x^2+(y+1)^2=1.$

Attempt. $$\frac{dx}{x-y-1}=\frac{dy}{y-x-z+1}=\frac{dz}{z}$$

so $$\frac{dx+dy}{(x-y-1)+(y-x-z+1)}=\frac{dz}{z}\iff d(x+y+z)=0$$ so $g_1(x,y,z)=x+y+z=c_1.$ I have not managed to find the second relation $g_2(x,y,z)=0$, needed in order to get $F(g_1,g_2)=0$ for some $F$. (As far as I am concerned, there are no standard procedures in these cases, one has to work on trial-and -error to find the exact expressions).

Thanks in advance.

  • 1
    $\begingroup$ $$\frac{dx-dy}{2(x-y)-2-z}=\frac{dz}{z}$$ $\endgroup$ – Nosrati Sep 25 '18 at 8:22

$$(x − y − 1)u_x + (y − x − u + 1)u_y = u \tag 1$$ $$\frac{dx}{x-y-1}=\frac{dy}{y-x-u+1}=\frac{du}{u}\quad\text{is correct}$$ You rightly found a first characteristic equation : $$x+y+u=c_1$$ A second characteristic equation comes from $$\frac{dx-dy}{(x-y-1)-(y-x-u+1)}=\frac{du}{u}=\frac{d(x-y-1)}{2(x-y-1)+u}$$ With $v=x-y-1$ $$\frac{du}{u}=\frac{dv}{2v+u}$$ is a separable ODE which solution is $v=-u+c_2u^2$ . Thus $x-y-1=-u+c_2u^2$ . The second characteristic equation is : $$\frac{x-y-1+u}{u^2}=c_2$$ The general solution of the PDE can be expressed on the form of implicit equation : $$\frac{x-y-1+u}{u^2}=F(x+y+u) \tag 2$$ where $F$ is an arbitrary function.

Boundary condition in order to determine the function $F$ :

$u=1$ on $x^2+(y+1)^2=1$ , so $\frac{x-y-1+1}{1^2}=F(x+y+1)$ $$F(x+y+1)=x-y$$ Let $X=x+y+1=\pm\sqrt{1-(y+1)^2}+y+1$

$(X-y-1)^2=1-(y+1)^2.\quad$ To be solved for $y$ which leads to

$y=\frac{X-2\pm\sqrt{2-X^2}}{2}\quad;\quad x=\frac{X\mp\sqrt{2-X^2}}{2}\quad;\quad x-y=\mp\sqrt{2-X^2}+1$ $$F(X)=1\mp\sqrt{2-X^2}$$ So, $F(X)$ is determined. We put it into the above general solution $(2)$ where $X=x+y+u$. $$\frac{x-y-1+u}{u^2}=1\mp\sqrt{2-(x+y+u)^2}$$ The solution which complies to the PDE and the boundary condition is expressed on the form of an implicit equation : $$x-y-1+u-u^2\pm u^2\sqrt{2-(x+y+u)^2}=0 \tag 3$$ To express $u(x,y)$ on explicit form we have to solve a polynomial equation of sixth degree. Thus there is no closed form for $u(x,y)$. The final answer is the above implicit form $(3)$.

  • $\begingroup$ Very nice. A small remark: shouldn't you have $\ldots \sqrt{2-(x+y+1)^2}=0$ in equation (3) (and the above)? $\endgroup$ – Nikolaos Skout Sep 26 '18 at 23:16
  • $\begingroup$ In equation (3) we have $\sqrt{2-(x+y+1)^2}$ only on the boundary. Everywhere we have $\sqrt{2-(x+y+u)^2}$. $\endgroup$ – JJacquelin Sep 27 '18 at 6:19

For the second constant of integration $$\frac{dx}{x-y-1}=\frac{dy}{y-x-z+1}=\frac{dz}{z}$$ $$\frac{d(x+y)}{x+y-k}=\frac{dy}{2y-k+1}$$ $$z^2=C({2y + {1-k}})$$ Where $k=x+y+z$ $$z^2=C({y-x-z + 1})$$

  • $\begingroup$ Thank you. But since $z=u(1,0)=1$, we get $1^2=C\cdot 0,$ a contradiction. Why does this happen? $\endgroup$ – Nikolaos Skout Sep 25 '18 at 11:56
  • $\begingroup$ hi @NikolaosSkout you need to write $$h(k)=C$$ Where h is any arbitrary function. Then use initial condition to determine this function $\endgroup$ – Isham Sep 25 '18 at 14:25

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