Laplace in Cyl. form: one step I'm not sure about

• kostoglotov
In summary: I will leave the derivative to you).This is only true when the variables are independent of each other, which they are not in this case. When you have several variables, you need to use the chain rule to take derivatives. So in this case, you should have$$\frac{\partial r}{\partial x} = \frac{x}{r} \;\;\;\; \frac{\partial r}{\partial y} = \frac{y}{r}$$Also, the term "homogeneous" refers to a specific type of PDE, not to the ability to separate variables. A homogeneous PDE is one
kostoglotov
I have completed the exercise, but I did something weird in one step to make it work, and I'd like to know more about what I did...or if what I did was at all valid.

1. Homework Statement

Show that Laplace's equation

$$\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}=0$$

when converted to cylindrical coordinates gives

$$\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2}+\frac{\partial^2u}{\partial z^2}=0$$

Homework Equations

obviously

$$x = r\cos{\theta}$$
$$y = r\sin{\theta}$$
$$z = z$$

The Attempt at a Solution

It was a long exercise, so I'll skip ahead to just before the step I'm asking about

$$r^2\frac{\partial^2u}{\partial r^2}+\frac{\partial^2u}{\partial \theta^2}+\frac{\partial^2u}{\partial z^2} = \frac{\partial^2u}{\partial x^2}(r^2\cos^2{\theta}+r^2\sin^2{\theta})+\frac{\partial^2u}{\partial y^2}(r^2\cos^2{\theta}+r^2\sin^2{\theta})+\frac{\partial u}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial y}(-r\sin{\theta})+\frac{\partial^2u}{\partial z^2}$$

$$r^2\frac{\partial^2u}{\partial r^2}+\frac{\partial^2u}{\partial \theta^2} = r^2\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\right)+\frac{\partial u}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial y}(-r\sin{\theta})$$

Now, I got rid of the second partial of u to z since it just takes up space, and is not interesting right now. What I am interested in is the last two terms of the RHS. By the chain rule,

$$\frac{\partial u}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial y}(-r\sin{\theta}) = \frac{\partial u}{\partial r}\frac{\partial r}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial r}\frac{\partial r}{\partial y}(-r\sin{\theta})$$

$$\frac{\partial r}{\partial x} = \frac{1}{\cos{\theta}} \ and \ \frac{\partial r}{\partial y} = \frac{1}{\sin{\theta}}$$

So

$$\frac{\partial u}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial y}(-r\sin{\theta}) = -r\left(\frac{\partial u}{\partial r}+\frac{\partial u}{\partial r}\right)$$

Now...here's where I need more insight, I have an inkling that the following is NOT right

$$\frac{\partial u}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial y}(-r\sin{\theta}) = -2r\frac{\partial u}{\partial r}$$

My intuition that this was wrong came from a few places. First off, it did not look like it would give the solution I was being directed to find...but that isn't sufficient for real understanding.

My understanding is that when using partial derivatives, especially when using them with the chain rule, one is finding rates of change in given directions, thus it's kind of like a vector (or basis vector - I haven't studied linear algebra yet) situation...and one cannot simply add basis vectors to each other...so I opted for this

$$\frac{\partial u}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial y}(-r\sin{\theta}) = -r\frac{\partial u}{\partial r}$$

and of course, by reinserting the second partial for u on z and rearranging one can show the very thing the problem asked us to show.

However, it bothers me that I have

$$-r\left(\frac{\partial u}{\partial r}+\frac{\partial u}{\partial r}\right) = -r\frac{\partial u}{\partial r}$$

What is going on here?

Like I said, I've completed the exercise, but I feel like my insight is lacking.

Your partial derivatives ##\partial r / \partial x## and ##\partial r/\partial y## are wrong. You cannot simply solve for ##r## from ##x = r\cos\theta## and differentiate wrt x, ##\theta## depends on x and y! You need to differentiate the full expression which defines r in terms of x and y, i.e., ##r^2 =x^2+y^2##.

kostoglotov
Orodruin said:
Your partial derivatives ##\partial r / \partial x## and ##\partial r/\partial y## are wrong. You cannot simply solve for ##r## from ##x = r\cos\theta## and differentiate wrt x, ##\theta## depends on x and y! You need to differentiate the full expression which defines r in terms of x and y, i.e., ##r^2 =x^2+y^2##.

So even though we can consider theta as being constant when we partial differentiate $\frac{\partial x}{\partial r}$ or $\frac{\partial y}{\partial r}$, I need to understand that $\frac{\partial r}{\partial x}$ is not simply the reciprocal of $\frac{\partial x}{\partial r}$. So is this what makes PDE's so difficult? Because this then suggests to me that only ordinary differential fractions are separable...I haven't done anything beyond ODE's yet. Is the ability to flip and separate the differential fraction what they mean by homogenous?

kostoglotov said:
Because this then suggests to me that only ordinary differential fractions are separable...I haven't done anything beyond ODE's yet.

If by this you mean that ##dy/dx = 1## implies ##dy = dx##, then yes, this is not generally true when you have several variables. If you have several ##y## and several ##x##, you will have
$$dy^i = \frac{\partial y^i}{\partial x^j} dx^j$$
(here using Einstein summation convention, the ##j## is a summation index). When we talk about a separable PDE, we talk about a PDE which can be broken down into a set of ODEs by using separation of variables, e.g., ##u(x,t) = X(x)T(t)##.

kostoglotov
Basically the chain rule with partial derivatives is abit different, it is ##\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}## and similar for theta. Using this expression you can easily see that ##-r\frac{\partial u}{\partial r}## gives the desired result.

This is from a theorem in calculus regarding the partial derivation of a composition of function of 2 variables with 2 other functions of 2 variables each (just as the chain rule with ordinary derivatives is from the composition of a function of 1 variable with another function of 1 variable).

Here our function of two variables is ##u(x,y)## and we take the composition of this function with the functions ##x=x(r,\theta)=rcos\theta## and ##y=y(r,\theta)=rsin\theta##

So basically your mistake is when you say that ##\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}##, this is not correct, the correct is ##\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}## with ##r^2=x^2+y^2## and ##\theta=arctan(y/x)##.

Last edited:
kostoglotov

1. What is Laplace in cylindrical form?

Laplace in cylindrical form is a mathematical representation of the Laplace equation in cylindrical coordinates. It is used to solve problems in physics and engineering involving cylindrical systems, such as cylinders, pipes, and rotating bodies.

2. How is Laplace in cylindrical form different from Laplace in Cartesian form?

In Laplace in cylindrical form, the equation is expressed in terms of cylindrical coordinates (r, θ, z) instead of Cartesian coordinates (x, y, z). This allows for a more efficient and accurate solution to problems involving cylindrical systems.

3. What is the one step to solving Laplace in cylindrical form?

The one step to solving Laplace in cylindrical form is to convert the Laplace equation from Cartesian form to cylindrical form by substituting the appropriate expressions for r, θ, and z. This will result in a simpler equation that can then be solved using standard techniques.

4. What are some applications of Laplace in cylindrical form?

Laplace in cylindrical form has various applications in physics and engineering, such as in fluid dynamics, electromagnetism, and heat transfer. It is commonly used to solve problems involving cylindrical geometries, such as flow in pipes, heat transfer in cylindrical objects, and electromagnetic fields around cylindrical conductors.

5. Are there any limitations to using Laplace in cylindrical form?

While Laplace in cylindrical form is a powerful tool for solving many problems, it does have its limitations. It is most effective for problems with cylindrical symmetry and may not be suitable for more complex geometries. Additionally, it may not be the best approach for problems involving non-linear or time-dependent equations.

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