Solve for $u$ the PDE $(x − y − 1)u_x + (y − x − u + 1)u_y = u$ if $u=1$ on $x^2+(y+1)^2=1.$

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).

• $$\frac{dx-dy}{2(x-y)-2-z}=\frac{dz}{z}$$ – 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 $$(3)$$.

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

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