What is Laplace operator?

In summary: \\f & = \left(\frac{\partial}{\partial x}x^3+\frac{\partial}{\partial y}y^2\right)\\& = \left(\frac{\partial}{\partial x}x^3+\frac{\partial}{\partial y}y^2+\frac{\partial}{\partial z}z^2\right)\\& = \left(\frac{\partial}{\partial x}x^3+\frac{\partial}{\partial y}y^2+\frac{\partial}{\partial z}z^3\right)\\& = \left(\frac{\partial}{\partial x}x^
  • #1
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##\frac {\partial \vec F} {\partial x} ## + ##\frac{\partial \vec F} {\partial y} ## = vector which gives me a direction of the greatest increase of the greatest increase of the function, where ##\vec F ## = gradient of the function. If I multiple the first by ##\hat i## and the second by ##\hat j## then I will get the length of the x-component of vector ##\frac {\partial \vec F} {\partial x} ## plus the length of the y-component of vector ##\frac{\partial \vec F} {\partial y} ##. What does this sum mean?
 
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  • #2
Hi,

Your thread name is 'Laplace operator' ...

If I go from your $$F = \nabla f = {\partial f\over\partial x} {\bf \hat\imath} + {\partial f\over\partial y} {\bf \hat\jmath} $$ in 2 dimensions to your vector $$\frac{\partial \vec F} {\partial x} +\frac{\partial \vec F} {\partial y} = \Biggl({\partial^2 f\over\partial x^2}+{\partial^2 f\over\partial y\partial x} \Biggr) {\bf \hat\imath} +
\Biggl( {\partial^2 f\over\partial y^2} + {\partial^2 f\over\partial x\partial y} \Biggr) {\bf \hat\jmath} $$ Then I do not understand what you mean with
erocored said:
If I multiple the first by ##\hat i## and the second by ##\hat j##​
but let's assume you mean 'take the inner product', then I get
[edit] wrong ! Instead of first inner product then adding up I made the mistake of first adding up !
$$\frac{\partial \vec F} {\partial x} \cdot\hat\imath +\frac{\partial \vec F} {\partial y} \cdot\hat\jmath = {\partial^2 f\over\partial x^2}+{\partial^2 f\over\partial y\partial x} +
{\partial^2 f\over\partial y^2} + {\partial^2 f\over\partial x\partial y} $$And that is NOT the 2 D Laplacian ...

[edit] first doing the inner product and then the addition does yield the Laplacian:
$$\Delta \vec F = \nabla^2 \vec F = \nabla \cdot (\nabla\vec F ) =
{\partial^2 f\over\partial x^2}+{\partial^2 f\over\partial y^2}$$which you probably meant, and which does have a physical meaning .

##\ ##
 
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  • #3
Hello, I am sorry for my mistake. Here is what I meant exactly:
2021-03-25 (2).png
I have vector ##\vec v_1## at point ##\left(x, y \right)## , vector ##\vec v_2## at point ##\left(x+dx, y \right)## and vector ##\vec v_3## at point ##\left(x, y+dy \right)## (##\vec v_1## = ##\vec F\left(x, y \right)##, ##\vec v_2## = ##\vec F \left(x+dx, y \right) ##, ##\vec v_3## = ##\vec F \left(x, y+dy \right) ##).

Then difference between ##\vec v_2## and ##\vec v_1## will be ##d\vec v_{21}##, difference between ##\vec v_3## and ##\vec v_1## will be ##d\vec v_{31}## (##d\vec v_{21}## = ##\vec F \left(x+dx, y \right) ## - ##\vec F\left(x, y \right)##, ##d\vec v_{31}## = ##\vec F \left(x, y+dy \right) ## - ##\vec F\left(x, y \right)##).

Next I will get vector ##\frac {d\vec v_{21}} {dx}## which is collinear with ##d\vec v_{21}##, and so for ##\frac {d\vec v_{31}} {dy}##.

Now if I multiple ##\frac {d\vec v_{21}} {dx}## by ##\hat i## I will get a length of the x-component of that vector and the y-component of vector ##\frac {d\vec v_{31}} {dy}## if multiple by ##\hat j##.

Sum of ##\frac {d\vec v_{21}} {dx} \cdot \hat i## and ##\frac {d\vec v_{31}} {dy} \cdot \hat j## is equal to ##\frac {\partial \vec F} {\partial x} \cdot \hat i## + ##\frac {\partial \vec F} {\partial y}\cdot \hat j##. This is also equal to
##\frac \partial {\partial x} \cdot \left(\frac {\partial f} {\partial x}\cdot \hat i + \frac {\partial f} {\partial y}\cdot \hat j \right) \cdot \hat i## + ##\frac \partial {\partial y} \cdot \left(\frac {\partial f} {\partial x}\cdot \hat i + \frac {\partial f} {\partial y}\cdot \hat j \right) \cdot \hat j## =
##\frac {\partial^2 f} {\partial^2 x} \cdot \hat i \cdot \hat i## + ##\frac {\partial^2 f} {\partial x \cdot \partial y} \cdot \hat i \cdot \hat j## + ##\frac {\partial^2 f} {\partial x \cdot \partial y} \cdot \hat i \cdot \hat j## + ##\frac {\partial^2 f} {\partial^2 y} \cdot \hat j \cdot \hat j## =
##\frac {\partial^2 f} {\partial^2 x}##+##\frac {\partial^2 f} {\partial^2 y}## = sum of the lengths of those dashed lines. I can not understand what does this sum mean.
 
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  • #4
BvU said:
[edit] first doing the inner product and then the addition does yield the Laplacian:
$$\Delta \vec F = \nabla^2 \vec F = \nabla \cdot (\nabla\vec F ) =
{\partial^2 f\over\partial x^2}+{\partial^2 f\over\partial y^2}$$which you probably meant, and which does have a physical meaning .

##\ ##
Thank you! But is it correct that the Laplace operator is equal to the sum of these dashed lines, and if so, what does this sum mean?
 
  • #5
erocored said:
I am sorry for my mistake
I don't think you made one. In contrast I screwed up grossly o:) o:) and edited my reply to correct it.

I am afraid I let you make you drawing without that helping to shed light on your original question: it's imply too complicated (at least for me). In my case, what helps here to see and fix the mistake is a 'numerical' example (below). But it still doesn't clearly show what the result signifies :frown:$$
\begin{align*}
f & = x^3y^2 + x^2y^3 \\ \ &\ \\

\vec F =\vec \nabla f & =
\Bigl (3x^2y^2 + 2xy^3 \Bigr )\; {\bf \hat\imath} + \Bigl (2x^3y + 3x^2y^2 \Bigr )\;\hat\jmath \\ \ &\ \\
{\partial \vec F \over\partial x } & =
\Bigl (6xy^2 + 2y^3 \Bigr )\; {\bf \hat\imath} + \Bigl (6x^2y + 6xy^2 \Bigr )\;\hat\jmath \\ \ &\ \\
{\partial \vec F \over\partial y } & =
\Bigl (6x^2y + 6xy^2 \Bigr )\; {\bf \hat\imath} + \Bigl (2x^3 + 6x^2y \Bigr )\;\hat\jmath \\ \ &\ \\

\frac{\partial \vec F} {\partial x} \cdot\hat\imath +\frac{\partial \vec F} {\partial y} \cdot\hat\jmath
& =\Bigl (6xy^2 + 2y^3 \Bigr ) + \Bigl (2x^3 + 6x^2y \Bigr ) \\ \ \\

& ={\partial^2 f\over\partial x^2}+{\partial^2 f\over\partial y^2} = \Delta f

\end{align*}
$$

So after this error of mine we can finally come to the Laplacian
BvU said:
which does have a physical meaning .

The Laplacian is the divergence of a gradient: $$\Delta f = \vec \nabla\cdot(\vec\nabla f) $$ and it plays an important role in physics in the Laplace equation and in the (more general) Poisson equation.

It can be shown that a conservative force field is the gradient of some potential. And the divergence indicates the presence (or absence) of sources.

I wouldn't know how to come to that insight directly from the definitions. Only in steps in vector calculus.

Google things like Conservative vector field , Laplacian vector field , Divergence theorem ,
physical interpretation of ... (any of: gradient, curl, divergence, laplacian :smile: ),

##\ ##
 
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  • #6
Thank you!
 
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  • #7
The last smiley was a realization that my reply is rather circular ... :wink:
 
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What is the Laplace operator?

The Laplace operator, also known as the Laplacian, is a mathematical operator that is used to describe the second-order spatial variation of a function. It is represented by the symbol ∇² and is commonly used in fields such as physics, engineering, and mathematics.

What does the Laplace operator do?

The Laplace operator takes a function as input and outputs the sum of the second-order partial derivatives of that function. In other words, it measures how quickly a function changes at a specific point in space. It is often used to describe physical phenomena such as heat flow, fluid dynamics, and electrical potential.

How is the Laplace operator used in physics?

In physics, the Laplace operator is used to describe the behavior of physical systems. For example, in fluid dynamics, it is used to describe the velocity and pressure of a fluid at a specific point in space. In electromagnetism, it is used to describe the electric and magnetic fields at a given point. It also plays a crucial role in the Schrödinger equation, which is used to describe the behavior of quantum particles.

What is the difference between the Laplace operator and the gradient operator?

The gradient operator, represented by the symbol ∇, is a vector operator that describes the first-order spatial variation of a function. It takes a function as input and outputs a vector whose components are the partial derivatives of the function. The Laplace operator, on the other hand, is a scalar operator that describes the second-order spatial variation of a function. It takes a function as input and outputs a scalar value representing the sum of the second-order partial derivatives of the function.

What are some applications of the Laplace operator?

The Laplace operator has a wide range of applications in various fields, including physics, engineering, and mathematics. Some common applications include solving differential equations, image processing, signal processing, and optimization problems. It is also used in computer graphics and computer vision for tasks such as edge detection and image smoothing.

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