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$u(x, y)$ is said to be harmonic if it is a solution to the partial differential equation $u_{x x}+u_{y y}=0,$ where $c$ is a constant.Show the given function is harmonic.$$u(x, y)=2 x y$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 2

Partial Derivatives

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Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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02:01

$u(x, y)$ is said to be ha…

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01:36

A function of two variable…

02:36

So in order for us to show that this functions harmonic, we need to find the second full partial derivatives with respect to both X and Y and plug it into that equation up there and see if it actually spits out zero or not. So I'll first go ahead, start by finding the second full partial with prospective X, so that would give us use of X. And I remember we treat this. Why here is a constant. So we just take the derivative of X, which would just be one that gives us two why. And now if we do the second full partial of this with respect text, well, why is a constant? So if I multiply that by a constant, that still constant. So the derivative of that is just going to be zero. So our second full partial is zero. Now, let me come over here and write this out again. We're going to go ahead and take the partial. This with respect. Why? So it's gonna be the same idea. Um, two x is a constant. So take the river. Just why that gives one. So we get the first partial with respect to why is just two X and then just like before when we came over here? To take this to X is a constant with respect to why so that will just be zero. So we have our second partial is zero. Then we just come down here and then do you xx plus you? Why? Why and will you xx zero You y y zero Add those up and we get zero. So this checks out for it being harmonic.

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