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I’m solving the PDE for function $u(x,y)$ with parameter $\lambda$ \begin{align} \lambda uu_x+u_y=0. \end{align} Using the method of characteristics, \begin{align} \frac{\mathrm du}{\mathrm ds}=0&\rightarrow u=\Phi(C),\\ \frac{\mathrm dx}{\mathrm ds}=\lambda u&\rightarrow x=\lambda su+C,\\ \frac{\mathrm dy}{\mathrm ds}=1&\rightarrow y=s. \end{align} Combining we find that \begin{align} u=\Phi\left(x-\lambda yu\right), \end{align} which agrees with what Wolfram Alpha tells me. The problem is this doesn’t work for all solutions, specifically the case $u=x/y$, which upon substitution into the PDE you’ll see works for $\lambda=1$, but fails to yield an identity when plugged into the general solution.

I’ve tried taking the different initial conditions $(x_0,y_0)=(0,C)$, but that solution has the same issue. Could someone enlighten me as to how I’m losing this solution? Is there an assumption I am taking at some point to do so?

Eli Bartlett
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    How do the characteristic curves ($u=$ constant) look like for $u=x/y$, do you see a problem? Useful to draw them for $u=x/y$ and compare to those of, say, $u=x/(y+1)$. (For example for the case where we imagine the IC of the PDE were specified along the $x$-axis.) – Winther Jul 30 '22 at 00:16
  • @Eli Bartlett: Contrary to what you claim $u=\Phi\left(x-\lambda yu\right)$ is correct in the particular case $u=x/y$ and $\lambda=1$. Proof : in this case $u=\Phi\big(x-(1)y(x/y)\big)=\Phi(0)=$constant which obviously is a solution of the PDE $\lambda uu_x+u_y=0$. – JJacquelin Jul 30 '22 at 06:14
  • @JJacquelin using your method you must evaluate the whole equation, no? So, taking $u=x/y$ one gets that $u=\Phi(x-\lambda yu)\rightarrow x/y=\Phi(0)$, which is not true for all $x$. – Eli Bartlett Aug 03 '22 at 01:49
  • @Winther I take it you're pointing out that $u=x/y$ has a singularity at $y=0$, meaning we lose that solution? If that is so: I was under the impression that if one didn't do something that restricts the domain of the solution (divide by 0 unwittingly, discard an absolute value,...) that you wouldn't lose solutions. That is what I see in solving ODEs, for example losing a particular solution (like $y\equiv0$ because you must divide by $y$ to get the general solution. – Eli Bartlett Aug 03 '22 at 01:55

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The short answer: the way you have solved it assumes the initial conditions can be specified along $y=0$. But this is not possible for $u=x/y$ as the function is not defined here.

Longer answer: Lets review the whole idea behind the method of characteristics: we specify the initial conditions of the PDE along a curve (for example the $x$-axis) and then find characteristic lines where the function stays constant (or vary in a way we can compute). Then to find $u(x,y)$ we simply follow the characteristic lines from $(x,y)$ till it hits the curve at $(x_0,y_0)$ and then we can just read off the function value $u(x,y)$ from the initial conditions $u(x_0,y_0)$. Notice that there are in general two parameters, $x_0$ and $y_0$, that specify the point on the initial curve. Often we make the simplification that the initial conditions are set on a simple line, say the $x$-axis, and then it reduces to a single parameter (your $C$).

The way you have solved the equation, by taking $\frac{dy}{ds}=1\to y=s$ instead of $y=s+y_0$, you have implicitly assumed that the initial conditions of the PDE can be specified along $y=0$ (for $s=0$ with $x_0 = C$). This is not possible for $u=x/y$. For $u = x/y$ the function is divergent here and as a consequence the characteristic curves you find all pass through $x=y=0$ so if $u(0,0)$ was set to be finite this would mean $u$ has to be constant for this to be consistent (and this is a solution).

If we go back and do it more carefully, we find

$$u(s) = u_0 = f(x_0,y_0)$$ $$x(s) = su_0+x_0$$ $$y(s) = s+y_0$$ where $f(x_0,y_0)$ is our initial condition.

We have seen that taking $y_0=0$ is not allowed if we want a solution like $u=x/y$. If we instead imagine the initial conditions are specified along the curve $y=1$ (i.e. $y_0=1$) then $x(s) = (y-1)u + x_0$ so $u = F(x-(y-1)u)$ for some function $F(t) = f(t,1)$.

Now plugging in $u=x/y$ (i.e. $u(x_0,1) = x_0$ so the initial conditions is $F(x)=x$) we find $x/y = F(x/y)$ which is exactly what we expect.

Winther
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