00:01
We are going to determine whether each of the following functions is a solution of laplace equation uxx plus uyy equals 0.
00:13
That is the sum of the second order partial derivative respect to x and with respect to y is equal to zero.
00:24
And here we have some functions parts a, b, c, d, e and f.
00:33
So first we are going to simplify a little bit notation because when we write the laplace equation this here, it is clear that more formally we should write something like this.
00:58
And that's because the partial derivatives are a function of x and wiles.
01:06
So both partial derivatives must be evaluated at any point xy.
01:12
But we are going to write this way in order to simplify notation that is why in all the examples have written that you depends on xy for all this function but when we are going to calculate the derivatives the partial derivative we are not going to use evaluation and xy to simplify but it is clear that any of the derivatives or every derivatives get to be evaluated at xy that said we're going to start with part a that is using the simplification i just mentioned u equals x square plus y square so to arrive to these partial derivatives i need the first order of partial derivatives so we start with partial derivative of u respect to is to x and the partial derivative of u with respect to y is to y and then the second order partial derivative respect to x is derivative of this respect to x is 2 and the second order partial derivative respect to y is the derivative of this expression respect to y is 2 so the sum of the second order partial derivatives of u is 4 and so is not equal 0 so and we then conclude that the function u equals x squared plus y square is not a solution to la plus equation we go now in part b use u equal x square minus y square so i'm going to do these partial derivatives respect to x of and second order directly that is partial derivative of u respect to x is 2x and then the second order respect to x is 2.
04:13
Now the second the first order partial derivative respect to y is negative to y and the second order with respect to y is negative 2.
04:29
So the sum of the second order partial derivative the respect to x and y is 2 minus 2 equals 0 then the function u equals x square minus y square is a solution to la plus equation part c the function is x cubed plus 3x y square so we start with the partial derivative or respect to x is 3x square.
05:27
Now this term, x is a variable, 3 way square is constant, and the derivative is 3y square.
05:39
And the second order with respect to x is 6x and that's it, because this turn doesn't depend on x.
05:51
Now the first order partial derivative, respect to y, this term doesn't depend on y, and this one does depend on y, so it's 6xy.
06:05
Because 3x is a constant and derivative of y square is 2y.
06:10
So we get 6xy and the second order partial derivative with respect to y is 6x.
06:24
So the sum of this second order partial derivatives is 6x plus 6x equals 2.
06:34
12 x which is not this null function so this function is not a solution of the plus equation.
07:04
When we go for d, natural logarithm of score root of x squared plus y square.
07:20
Let's see the first order partial derivative respect to x is partial derivative respect to x of the argument.
07:38
Over the argument.
07:45
That is, the partial derivative respect to x of the natural logarithm is 1 over the argument, that is 1 over squares of square plus square square times the partial derivative respect to x of the argument.
08:01
So here we have applied not only the definition of the derivative of the natural logarithm, but also the chain rule.
08:11
And so this is the numerative we have the partial derivative respect to x of square root of x squared plus y square that is one -half times x squared plus y -square to the negative one -half that is because square root of x -quare plus y -square is written as a square plus y -square to the one -half so applying properties of the derivative we get this times the partial derivative with respect to x of the argument x square or the base square plus y square over this square root here.
08:57
And so this partial derivative is, okay, let's see that we get one -half times x square plus y -square to the negative one -half times 2x.
09:29
Over the square root here.
09:34
Now we can see that this negative power in the numerator goes to the denominator as x squared plus y square to the one -half which is the same as the square root of x square plus y -square and because it's going to be multiplied by the same expression we get simply x squared plus y -square.
09:54
These two cancelled out and so we get x over x squared plus y square.
10:07
Now we do the second order partial derivative respect to x, that is this, it is then the derivative respect to x of this expression, that is a quotient respect to x.
10:26
So we get the derivative of the numerator with respect to x is 1, so we get 1 times x squared plus y square minus minus x times the partial derivative respect to x of the denominator is 2x over x square plus y square square and we simplify this as y square minus x square because it's s square minus 2x square it's negative x square over x square plus y square squared squared so the second order partial derivative and up being y squared minus x square over x squared plus y squared square square.
11:28
And now to calculate the second order partial derivative respect to y, we notice that this function is symmetric respect to x and y.
11:41
That is if we replace x by y and vice versa, we get the same function.
11:48
It means that we can find the partial derivatives here respect to y by simply interchanging x and y.
12:04
Then by symmetry we have that the second order partial derivative of u respect to y will be x square that is we're changing here y by x and here vice versa x.
12:25
So we get minus y square over.
12:29
And if we interchange here, we get the same.
12:31
So it's x squared plus y square square square...