A short extract from a book of mine states that:

If $$\color{red}{A(x,y)\frac{\partial p}{\partial x}+B(x,y)\frac{\partial p}{\partial y}=0\tag{A}}$$ where $p=p(x,y)$ and $A$ and $B$ are also functions of $x$ and $y$.

Also $$\color{blue}{\rm d p=\frac{\partial p}{\partial x}\rm d x+\frac{\partial p}{\partial y}\rm d y=0\tag{B}}$$ The forms of $\color{red}{(\rm A)}$ and $\color{blue}{(\rm B)}$ are very alike and become the same if we require that $$\color{#180}{\frac{\rm d x}{A(x,y)}=\frac{\rm d y}{B(x,y)}\tag{C}}$$ By integrating this expression the form of $p$ can be found.

I need to apply the above method to solve $$x\frac{\partial u}{\partial x}-2y\frac{\partial u}{\partial y}=0$$

The book solution tells me that:

If we seek a solution of the form $u(x,y)=f(p)$, we deduce from $\color{#180}{(\rm C)}$ that $u(x,y)$ will be constant along lines of $(x,y)$ that satisfy $$\frac{\rm d x}{x}=\frac{\rm d y}{-2y}$$ which on integrating gives $x=cy^{-1/2}$. $\color{purple}{\text{Identifying the constant of integration}}$ $\color{purple}{c}$ $\color{purple}{\text{with}}$ $\color{purple}{p^{1/2}}$ (to avoid fractional powers), $\color{purple}{\text{we conclude that}}$ $\color{purple}{p=x^2y}$. Thus the general solution of the PDE is $$u(x,y)=f(x^2y)$$ where $f$ is an arbitrary function.

I have understood every step of this proof with the exception of the part marked $\color{purple}{\rm purple}$. Specifically, I don't understand why the constant of integration is $p^{1/2}$.

So basically; Could someone please explain to me why on earth $c=p^{1/2}$?

Or put in another way, Can I make $c=p$ such that $p=x\sqrt{y}$. So the general solution of the PDE is $$u(x,y)=f(x\sqrt{y})$$ where $f$ is an arbitrary function?

Many thanks.


1 Answer 1


Using $p^{1/2}$ is simply a label to make the result look cleaner. Specifically, the choice eliminates a fractional power of $y$ in the result.

Consider the curves $x=cy^{-1/2}$, which you have already understood are curves upon which the function $u(x,y)$ is constant. Thus, you can see that for some function $g$:

$$u(x,y) = g(c)$$

That is, $u$ is simply dependent upon the constant $c$, because the value of $u$ only depends upon which curve you are on. Now, let's just decide to define another constant $p$ such that $c=p^{1/2}$. Since $u$ is a function of $c$ then clearly $u$ is also a function of $p$ so we can say that for some function $h$:

$$u(x,y) = h(p)$$

But, we know that:

$$p = c^2 = (xy^{1/2})^2 = x^2y$$

Thus, relabeling the function $h$ by $f$, we have the result:

$$u(x,y) = f(x^2y)$$

Note that it would have also been correct to not worry about the business of redefining a new constant $p$ and to simply say that $u(x,y)=g(c)=g(xy^{1/2})$. It's just that whoever wrote this solution wanted to eliminate the fractional exponent.

  • $\begingroup$ If $p^{1/2}$ is just a label; Can I make $c=p$ such that $p=x\sqrt{y}$. So the general solution of the PDE is $u(x,y)=f(x\sqrt{y})$ where $f$ is an arbitrary function? $\endgroup$
    – BLAZE
    Commented Jan 7, 2016 at 4:56
  • 1
    $\begingroup$ Yes, as I noted in the last paragraph of my answer, you don't need to make the substitution of $c=p^{1/2}$. It is simply to make the result look cleaner. The fact is that u is dependent on a constant value and whether you call this constant $c$ or $p^{1/2}$ or something as random as $s^5$ doesn't matter. In any case it is still just a constant. $\endgroup$
    – wgrenard
    Commented Jan 7, 2016 at 6:22
  • 1
    $\begingroup$ And to convince yourself that it doesn't matter you can always plug both $f(x^2y)$ and $f(xy^{1/2})$ back into the PDE and see that both work. $\endgroup$
    – wgrenard
    Commented Jan 7, 2016 at 6:26
  • $\begingroup$ Perfect answer, thanks for your time. $\endgroup$
    – BLAZE
    Commented Jan 7, 2016 at 22:39

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