# Applying method of characteristic equations to $e^y u_x + u_y = u^2$

I have the PDE $e^y u_x + u_y = u^2$, $u(x,0) = x$ for small $|y|$. Also $u = u(x(s), y(s))$.

So $\frac{dx}{ds} = e^{y}$, $\frac{dy}{ds} = 1$, and $\frac{du}{ds} = u^{2}$.

Solving the first ode: $\frac{dx}{ds} = e^{y} \to dx = e^{y}ds$ and integrate both sides to obtain $x = e^{y}s + c_1$.

The second ODE: $\frac{dy}{ds} = 1 \to dy = 1ds$ and integrate both sides to obtain $y = s + c_2$.

The third ODE: $\frac{du}{ds} = u^{2}$, the solution to this ode is $-\frac{1}{u} = s + c_3 \implies u = -\frac{1}{s+c_3}$.

Okay now I want to eliminate the parameter $s$. I solve for $s$ from the solution of the second ode: $s = y -c_2$ then substitute into the solution of the first ode: $x = e^{y}(y -c_2) + c_1 = e^{y}y - e^{y}c_2 + c_1$. Also $u = -\frac{1}{y-c_2 + c_3}$.

This is where I am stuck and am not sure what to do. I was attempting to follow https://en.wikipedia.org/wiki/Method_of_characteristics#Example but I am not doing this correctly I think.

• With the given initial condition, you can show that $c_2 = 0$. Also, you need to substitute $y$ into the characteristic equation $\frac{dx}{ds} = e^y = e^s$ and then integrate; this will give $x = e^s + c_1$ and $c_1$ is found by solving $x(0) = x_0$, where $x_0$ is the parameterisation parameter for the "initial curve". – Chee Han Apr 3 '18 at 17:40
• Why is $e^y = e^{s}$? I know $y = y(s)$. Letting $x(0) = x_0$, from the equation $x(s) = e^s + c_1$, we get $x_0 = e^0 + c_1 = 1 + c_1$. So $c_1 = x_0 - 1$. Is this right? – Taln Apr 3 '18 at 18:04
• Because $y = s + c_2$ and $y(0) = 0$ gives $c_2 = 0$. And yes, $c_1$ is correct. Now you can do the same for $\frac{du}{ds} = u^2$, with $u(0) = x_0$. Then the final step would be to eliminate $s$ and $x_0$ if possible. Depending on the question, you might also want to determine the validity of your solution. Any standard PDE books that discuss first order PDE should have more information about this. – Chee Han Apr 3 '18 at 19:43
• Even with @CheeHan's corrections, it doesn't seem like a solution exists. Are there any typos in the original PDE? – MasterYoda Apr 3 '18 at 20:48
• @MasterYoda I've added $' for small |y| '$ to my post, but other than that I see no typos. – Taln Apr 3 '18 at 20:57

$$e^yu_x+u_y=u^2$$ I agree with your equations which can be written in a summarised form : $$\frac{dx}{e^y}=\frac{dy}{1}=\frac{du}{u^2}=ds$$ A first characteristic equation comes from $\quad\frac{dx}{e^y}=\frac{dy}{1}\quad$ leading to : $$e^y-x=c_1$$ A second characteristic equation comes from $\quad\frac{dy}{1}=\frac{du}{u^2}\quad$ leading to : $$\frac{1}{u}+y=c_2$$ The general solution of the PDE, expressed on the form of implicit equation is : $$\Phi\left((e^y-x) \:,\: (\frac{1}{u}+y) \right)=0$$ $\Phi$is an arbitrary function of two variables.

Or, equivalently the general solution of the PDE on explicit form is : $$\frac{1}{u}+y=F(e^y-x)$$ where $F$ is an arbitrary function. $$u(x,y)=\frac{1}{-y+F(e^y-x)}$$

CONDITION :

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

$F(1-x)=\frac{1}{x}$

Let $\quad X=1-x\quad;\quad x=1-X\quad;\quad F(X)=\frac{1}{1-X}$

So, the function $F$ is determined.

We put it into the above general solution where $\quad X=e^y-x\quad$ then $\quad F(e^y-x)=\frac{1}{1-(e^y-x)}$

$$u(x,y)=\frac{1}{-y+\frac{1}{1-e^y+x}}$$ This is the particular solution of the PDE which fits to the boundary condition.

Note: The above calculus is consistent with your calculus. The difference is that $ds$ is eliminated at the beginning, while in your calculus $ds$ is eliminated later. This doesn't change the final result.

Of course, I checked the solution $u(x,y)=\frac{1}{-y+\frac{1}{1-e^y+x}}$ before publishing my answer. After the MasterYoda's comment I checked it again. Definitively the solution satisfies the PDE (copy below). 