Why aren't these the characteristics to my PDE?

Question

I am solving this Cauchy problem: (x+y^2)u_x+yu_y+(x/y-y)u=1 where the initial condition is $u(x,1)=0$. The solution is $(x(s,t),y(s,t),z(s,t))=(e^{2t}+(s-1)e^t,e^t,1/(s-1)+e^{t(1-s)}/(1-s))$. I thought that this meant that my characteristics where $(e^{2t}+(s-1)e^t,e^t)$. However, if $g(t)=u(e^{2t}+(s-1)e^t,e^t)$ then $$g'(t)=u_x(2e^{2t}+(s-1)e^t)+u_ye^t=u_ x(x+y^2)+u_yy\neq 0 \quad \text{(according to my equation)}$$ Why aren't these the caracteristics? How can I find them?

Answer 1

The solution is a function of $x$ and $y$ only. Why $s$ and $t$ are still appearing in your proposed solution ?

Since the steps of your calculus are not edited one cannot says where exactly is the hitch.

$$(x+y^2)u_x+y\,u_y=1-(\frac{x}{y}-y)u$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{x+y^2}=\frac{dy}{y}=\frac{du}{1-(\frac{x}{y}-y)u}$$ A first characteristic equation comes from solving $\frac{dx}{x+y^2}=\frac{dy}{y}$ : $$y-\frac{x}{y}=c_1$$ A second characteristic equation comes from solving $\frac{dy}{y}=\frac{du}{1+c_1u}$ : $$(1+c_1u)y^{-c_1}=c_2$$ The general solution of the PDE on the form of implicit equation $c_2=F(c_1)$ is : $$(1+(y-\frac{x}{y})u)y^{-(y-\frac{x}{y})}=F(y-\frac{x}{y})$$ $F$ is an arbitraty function until no condition is taken into account.

$$\boxed{u(x,y)=\frac{y^{(y-\frac{x}{y})}F(y-\frac{x}{y})-1}{y-\frac{x}{y}}}$$

CONDITION $u(x,1)=0$ : $$u(x,1)=\frac{1^{(1-\frac{x}{1})}F(1-\frac{x}{1})-1}{(1-\frac{x}{1})}=0$$

$$F(1-x)=1$$ Thus $F$ is the constant function $F=1$.

The particular solution which satisfies both the PDE and the condition is :

$$\boxed{u(x,y)=\frac{y^{(y-\frac{x}{y})}-1}{y-\frac{x}{y}}}$$