As an example, a problem from the book Numerical solutions of Partial Differential Equations: Finite difference methods by G.D. Smith.

The PDE: $$ \frac{\partial{U}}{\partial{x}} + \frac{x}{\sqrt{U}}\frac{\partial{U}}{\partial{y}} = 2x $$

With boundary condition $U(x,0) = 0$

I have some basic understanding of how to solve this: $$ dx = \frac{\sqrt{U}}{x}dy = \frac{dU}{2x}$$

This yields $U = x^2+ Const_X$. For the characteristic going through $(x_R,0)$, we get $Const_X = -x_r^2$ since at $y=0$, $U=0$

We can also get: $$ \frac{dy}{dU} = \frac{1}{2\sqrt{U}} \to y= \sqrt{U}$$

(The additional constant can be shown to be 0, since for $y=0$, $U=0$)

Giving us the characteristic curve $x^2-y^2 = x_R^2$

But when you plug $U = x^2 -x_r^2$ back to the PDE, you get that along this characteristic, $\frac{x}{\sqrt{U}}\frac{\partial{U}}{\partial{y}} = 0$ ... Which contradicts $\sqrt{U} = y$ along the very same curve.

Given how little explanation was given in class for this method, I would've assumed I made a wrong assumption somewhere, except this is the exact solution given in the book (3rd edition, so if it was wrong, I'd assume someone would've figured it out by then)


1 Answer 1


You obtained correctly a first family of characteristic curves : $$x^2-U=c_1$$ A second family of characteristic curves is : $$y-\sqrt{U}=c_2$$ The general solution is obtained on the form of implicit equation : $$y-\sqrt{U}=F(x^2-U) \tag 1$$ where $F$ is an arbitrary function.

Condition :

$U(x,0)=0=0-\sqrt{0}=F(x^2-0)\quad\implies\quad F=$constant function $=0$.

Putting this function into equation $(1)$ leads to the solution : $\quad y-\sqrt{U}=0\quad\implies\quad U(x,y)=y^2$

So, the solution satisfying the PDE and the boundary condition is a function of $y$ only : $$U(x,y)=y^2$$

Final checking :

$U_x=0$ and $U_y=2y\quad;\quad U_x+\frac{x}{\sqrt{U}}U_y=0+\frac{x}{y}2y=2x\quad$ is OK.

and the condition $U(x,0)=0^2=0\quad$ is satisfied.


Another approach to answer to the question : $$ \frac{\partial{U}}{\partial{x}} + \frac{x}{\sqrt{U}}\frac{\partial{U}}{\partial{y}} = 2x $$ The change of function and variable : $\quad\begin{cases}X=x^2\\ \sqrt{U(x,y)}=V(X,y) \end{cases}\quad$ leads to : $$ V\frac{\partial{V}}{\partial{X}} + \frac{\partial{V}}{\partial{y}} = 1 $$ This a well-known Burger's equation with inhomogeneous RHS. The literature is extensive about these kind of equations, as well as a lot of questions in StackExchange. For example : Confused about Burger's Equation with an inhomogeneous RHS I hope that studying this topic of Burger's equations with examples will clarify and make you aware that there is no contradiction.

  • $\begingroup$ Is there a simple explanation for why these two are separate characteristic curve? from what I understand, the Lagrange-Charpit equations should hold for any solution. I derived an equation that shows $y$ as a function of $x$ and vice-versa, but it seems that if my derivation is correct, I get a contradiction. How do I argue (mathematically) that these are two different characteristic curves? (There's also a boundary condition for $x=0, y\geq 0$ but it didn't seem relevant at the time). $\endgroup$ Commented Jan 22, 2018 at 20:42
  • $\begingroup$ See the addition to my first answer. $\endgroup$
    – JJacquelin
    Commented Jan 24, 2018 at 8:07
  • $\begingroup$ thank you. At first I didn't understand your answer, but now I understand these are 2 different characteristics and why it's so. This whole topic wasn't explained well in class (since the focus of the course is actually numerical methods for PDEs), which is why I was confused. While the topic is very interesting, so is my thesis subject, on which I will focus for now :) $\endgroup$ Commented Jan 24, 2018 at 16:22
  • $\begingroup$ Yes, the topic of Burger's PDEs is very interesting, but far to be the simplest. It's a valuable attempt to study the theory in addition to the numerical methods. Successful continuation. $\endgroup$
    – JJacquelin
    Commented Jan 24, 2018 at 17:20

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