How do I integrate the following?

$$\frac{1}{\Psi}\frac{\partial \Psi}{\partial x} = Cx$$

where $C$ is a constant.

I'm supposed to get a Gaussian function out of the above by integrating but don't know how to proceed.

  • $\begingroup$ I'm particularly confused by the the partial derivative. I'm not sure how to integrate it with that in the expression. $\endgroup$
    – Joebevo
    Oct 18 '12 at 8:19
  • $\begingroup$ That's a math question. And the solution to $\Phi'(x)=A(x)\Phi(x)$ is $\Phi(x)\propto\text{exp}(\int^xA(y)\text d y)$. $\endgroup$
    – Nikolaj-K
    Oct 18 '12 at 8:27
  • $\begingroup$ There is no problem with the partial derivative here. It is just telling us that $\Psi(x, \mathrm{and\,possibly\,others\,but\,just\,differentiate\,with\,respect\,to}\,x)$. Here, the two variables can be treated as 'independent'. $\endgroup$
    – MoonKnight
    Oct 18 '12 at 9:26

You can do the following. From

$$\frac{1}{\Psi}\frac{\partial \Psi}{\partial x} = Cx,$$

We can write the following interal equation

$$\int \frac{1}{\Psi} \mathrm{d}\Psi = C \int x \mathrm{d}x,$$

$$\ln \Psi + \kappa = \frac{1}{2} C x^{2}.$$

where $\kappa$ is our constant of integration. The above can then be simplified to get your Gaussian form

$$\Psi = e^{\ln \Psi} = e^{\frac{1}{2}Cx^{2} + \kappa} = \kappa' e^{\frac{1}{2}Cx^{2}},$$

where $\kappa' = e^{\kappa}$. Note. As a check you can now differentiate both sides of

$$\ln \Psi + \kappa = \frac{1}{2} C x^{2},$$

with respect to $x$ to get you original equation.

Edit. Based on the comment below. In this case, the function $\Psi$ I have assumed is a function of $x$ only as there is nothing to suggest otherwise. In this case, a partial derivative is not required and the derivative can be treated as odinary. However, if we have $\Psi = \Psi(x, \xi)$ then we would need to incorporate the variable $\xi$ in the constant of integration. The answer would become

$$\Psi(x, \xi) = e^{\ln \Psi(x, \xi)} = e^{\frac{1}{2}Cx^{2} + \kappa(\xi)} = \kappa'(\xi) e^{\frac{1}{2}Cx^{2}},$$

where this can be checked by taking the partial derivative $\partial \Psi(x, \xi) /\partial x$.

I hope this helps.

  • $\begingroup$ Interesting.. you converted the partial into a normal derivative. Is that just standard procedure? $\endgroup$
    – Joebevo
    Oct 18 '12 at 9:37
  • 2
    $\begingroup$ No. As I state in the comment above. $\Psi$ must be a function of $x$. It could be a function off someother variables too; hence the partial derivative. However, these extra variabes are of no concern to us here - as the equation only envolves $x$ - this is why we can treat this as an ordinary derivative in this case. Scince you did not explicity state the dependent variables for $\Psi$ we can treat it in this way. If you did include the dependent variables in the expression for $\Psi$ we may then have to include a constant of intergration that is a function of these extra variables. $\endgroup$
    – MoonKnight
    Oct 18 '12 at 9:44

You have a derivative of $\log \Psi$ on the left-hand side.


Sorry, I can't comment in the right place due to low rep.

@Killercam, you never need to ''treat this [the partial derivative] as an ordinary derivative''. Doing so ignores the possibility of other variables, and doesn't find the most general solution.

The only change in Killercam's derivation, is that $\kappa$ should be a function of any variables which are held constant during the partial derivative, $\partial_x$.

For example, consider the function on $\mathbb{R}^3$

$$ \Psi(x,y,z) = f(y,z)\,e^{\frac{1}{2}Cx^2}. $$

If the partial derivative holds $y$ and $z$ constant, we find that $\partial_x\Psi = Cx\Psi$. Exactly as required.

  • $\begingroup$ I understand this. Of course the function $\Psi$ is not defined explicityly enough to treat other variables it maybe dependent on. However, here we can only assume that it is dependent on $x$, which allows us to treat it as ordinary. Of course, if it is indeed a function of $x$ and $y$ (and perhaps $z$ as you have) also, then your answer is indeed correct, and yes we cannot treat it as ordinary, it must reamin partial and the constants treated accordingly. That said, what I have done above is right - and treating it as ordinary is fine as there is no expicit reason not to in this case... $\endgroup$
    – MoonKnight
    Oct 22 '12 at 7:38
  • $\begingroup$ It's clear that you know how to do this problem; I wasn't trying to put that in doubt. However, I disagree with the statement I quoted, and again with "treating it as ordinary is fine as there is no explicit reason not to". The explicit reason is that it is not the same operator. Making incorrect statements only leaves the possibility of confusion for those less familiar with this kind of material. $\endgroup$ Oct 22 '12 at 16:41
  • $\begingroup$ They are exactly the same for a function of a single variable. A partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant (as you well know). So for a function of a single variable $\partial \sim \mathrm{d}$. For a single variable, again, a partial derivative is exactly equivelent to an ordinary derivative. $\endgroup$
    – MoonKnight
    Oct 22 '12 at 17:01
  • $\begingroup$ With ordinary derivatives, we said that $\mathrm{d}f /\mathrm{d} x$ was the rate of change of $f$ with respect to $x$. The same is true with partial derivatives; that is $(\partial f/\partial x)_{y}$ is still the rate of change of $f$ with respect to $x$. The '$\partial$' symbol is used to indicate that $f$ does not depend only on $x$, i.e. there will be other rates of change for $f$ with respect to the other variables. The subscript $y$ tells us that $y$ is held constant, meaning that we don't have $f(x)$; we have $f(x, y)$. For $f(x)$, $\partial f/\partial x = \mathrm{d} f / \mathrm{d} x$. $\endgroup$
    – MoonKnight
    Oct 22 '12 at 17:05
  • $\begingroup$ Look, I'm not sure this is going to be helpful for anyone to read. Neither of us disagree about the definitions of partial derivatives, nor about the solution to the problem. I only posted as I felt that your first post may have said something unclear. The edit you made to your answer is sufficient to point out the assumption in your first proof, and to avoid any misinterpretation. I'd like to leave it at that. $\endgroup$ Oct 22 '12 at 18:20

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