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.

  • $\begingroup$ 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". $\endgroup$ – Chee Han Apr 3 '18 at 17:40
  • $\begingroup$ 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? $\endgroup$ – Taln Apr 3 '18 at 18:04
  • $\begingroup$ 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. $\endgroup$ – Chee Han Apr 3 '18 at 19:43
  • $\begingroup$ Even with @CheeHan's corrections, it doesn't seem like a solution exists. Are there any typos in the original PDE? $\endgroup$ – MasterYoda Apr 3 '18 at 20:48
  • $\begingroup$ @MasterYoda I've added $' for small |y| '$ to my post, but other than that I see no typos. $\endgroup$ – 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. $$u(x,y)=\frac{1}{-y+F(e^y-x)}$$




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.

ADDITION, answering to the comment of MasterYoda :

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).

enter image description here

  • $\begingroup$ Your solution doesn't appear to satisfy the original PDE. $\endgroup$ – MasterYoda Apr 3 '18 at 21:46
  • $\begingroup$ @ MasterYoda : The solution satisfy the PDE. See the addition to my main answer. $\endgroup$ – JJacquelin Apr 4 '18 at 7:26
  • $\begingroup$ Thanks, looks like I did my math wrong. $\endgroup$ – MasterYoda Apr 4 '18 at 23:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.