Solution to Laplace equations in 1D

[Mondo]

In general I know how to solve it, here I would just like to close some doubts I have:
Laplace equation -> [imath]\frac{d^2V(x)}{dx^2}[/imath] first doubt here in regards to the notation: why do we mark second derviatives differently in nominator and denominator? In other words why [imath]dx^2[/imath] and not [imath]d^2x[/imath] in denominator.
Next moving forward to the solution [imath]\int{\frac{d^2V(x)}{dx^2}}dx = \int{0}dx[/imath] -> [imath]\int{\frac{dV(x)}{dx}}dx + A = \int{0}[/imath] -> [imath]V(x) + Ax + B = 0[/imath] -> [imath]V(x) = -(Ax + B)[/imath]

My solution differs from the generic on by a sign on the right hand side. It appears to me the generic solution was obtained by expanding constants from [imath]\int{0}dx[/imath] instead. but then what happened to the constants from the left hand side? In my case I abandoned RHS constants as it always evaluates to 0 -> [imath]0 * dx = 0[/imath]. Is it just a preferred solution to get a positive sign on the RHS or mine is incorrect per se?

Thank you.

 

[mario99]

In general I know how to solve it, here I would just like to close some doubts I have:
Laplace equation -> [imath]\frac{d^2V(x)}{dx^2}[/imath] first doubt here in regards to the notation: why do we mark second derviatives differently in nominator and denominator? In other words why [imath]dx^2[/imath] and not [imath]d^2x[/imath] in denominator.

Because [imath]dx[/imath] is one thing! [imath]dx \neq d \times x.[/imath]

So [imath]dx^2 \ \text{means} \ dx \times dx \ \text{or} \ (dx)^2[/imath], but mathematicians are fancy, so they just write [imath]dx^2[/imath].

[imath]dx[/imath] means something that is very very very small. It is called a differential element. You can also say [imath]dx[/imath] is an infinitely small change in the variable [imath]x[/imath].

Next moving forward to the solution [imath]\int{\frac{d^2V(x)}{dx^2}}dx = \int{0}dx[/imath] -> [imath]\int{\frac{dV(x)}{dx}}dx + A = \int{0}[/imath] -> [imath]V(x) + Ax + B = 0[/imath] -> [imath]V(x) = -(Ax + B)[/imath]

My solution differs from the generic on by a sign on the right hand side. It appears to me the generic solution was obtained by expanding constants from [imath]\int{0}dx[/imath] instead. but then what happened to the constants from the left hand side? In my case I abandoned RHS constants as it always evaluates to 0 -> [imath]0 * dx = 0[/imath]. Is it just a preferred solution to get a positive sign on the RHS or mine is incorrect per se?

Thank you.

I did not understand your second question, but this is called:

[imath]\displaystyle \frac{d^2V(x)}{dx^2} = 0[/imath]

a second order differential equation.

You need to find a solution [imath]V(x)[/imath], when you differentiate it two times you get zero, so [imath]V(x) = Ax + B[/imath] is the general guess.

Or you can find the solution in the same way you did, but you have to correct this:

[imath]\displaystyle \int 0 \ dx = A[/imath] because [imath]\displaystyle \frac{dA}{dx} = 0.[/imath]

Also you have to understand this: we do not care about the negative signs in the general solution. Therefore, you can write [imath]V(x) = -(Ax + B)[/imath] as [imath]V(x) = Ax + B[/imath]. Thanks to the arbitrary constants [imath]A[/imath] and [imath]B[/imath], they can be anything.

A fancy student will do this:

[imath]V(x) = -(c_1x + c_2) = -c_1x - c_2[/imath]

Let [imath]A = -c_1[/imath] and [imath]B = -c_2[/imath]

Then

[imath]V(x) = Ax + B[/imath]

I told you the arbitrary constants can be anything!

 

[Mondo]

Thanks for an answer @mario99

Because dxdxdx is one thing! dx≠d×x.dx \neq d \times x.dx=d×x.

True but so is [imath]d^2V(x)[/imath], a second derivative if function [imath]V(x)[/imath], so it could be written as [imath]\frac{dV(x)^2}{dx^2}[/imath]. Although the later would probably introduce a confusion with the order - is V(x) squared? So I think I answered my own question.

As for the second part of my question: I was mostly interested in integration constants. They should happen on both sides of the equation and they do, as you explained. The point is at the end of calculation, we just arrange them into two constants as needed.

Thank you.

 

[mario99]

Thanks for an answer @mario99

True but so is [imath]d^2V(x)[/imath], a second derivative if function [imath]V(x)[/imath], so it could be written as [imath]\frac{dV(x)^2}{dx^2}[/imath]. Although the later would probably introduce a confusion with the order - is V(x) squared? So I think I answered my own question.

As for the second part of my question: I was mostly interested in integration constants. They should happen on both sides of the equation and they do, as you explained. The point is at the end of calculation, we just arrange them into two constants as needed.

Thank you.

[imath]\displaystyle \frac{d}{dx}[/imath] is a differential operator. So if you want to take the second derivative it must be written [imath]\displaystyle \frac{d^2}{dx^2}[/imath]. It is wrong to write [imath]\displaystyle \frac{d}{dx^2}[/imath]. It does not make sense to write [imath]\displaystyle \frac{d V(x)^2}{dx^2}.[/imath] This just means the function is squared but the differential operator is invalid.

You can write this:

[imath]\displaystyle \frac{d}{dx}\left(\frac{dV(x)^2}{dx}\right) = \frac{d^2V(x)^2}{dx^2}[/imath]

But this does mean something else. You are taking the second derivative of the function [imath]\displaystyle V(x) \ \text{squared} \ [/imath] [imath]\displaystyle \rightarrow [V(x)]^2[/imath].