If u = log(r), where r^2 = (x-a)^2 + (y-b)^2, and (x-1) and (y-b) are not zero simultaneously, show that d^2u/dx^2 + d^2u/dy^2 = 0.
I first used some of the properties of log and made u = 1/2 * log((x-a)^2 + (y-b)^2)
Then made u = 1/2 * ln((x-a)^2 + (y-b)^2)/ln(10)
I'm fairly confident everything up to this point is right but if not please correct me. I then did partial derivatives with respect to x twice and got something like
d^2u/dx^2 = 1/([(x-a)^2 + (y-b)^2]*ln(10)) - (2x(x-a))/([(x-a)^2 + (y-b)^2]^2*ln(10))
At this point i'm hesitant to go any further cause i do not believe that I will get 0 as an end result.
oh never mind. i looked back and saw that i differentiated wrong on a step
To show that the second derivatives of u with respect to x and y, i.e., d^2u/dx^2 and d^2u/dy^2, sum up to zero, let's first express u as u = 1/2 * log((x-a)^2 + (y-b)^2) as you correctly stated.
Next, we can find the first derivative of u with respect to x, d/dx (u), by applying the chain rule. The chain rule states that if we have a function of a function, f(g(x)), the derivative can be found by differentiating the outer function (f') and multiplying it by the derivative of the inner function (g').
So, for our case, using the chain rule, we have:
d/dx (u) = 1/2 * [1/((x-a)^2 + (y-b)^2)] * 2(x-a) = (x-a)/[(x-a)^2 + (y-b)^2]
Similarly, we can find the first derivative of u with respect to y, d/dy (u), by applying the chain rule:
d/dy (u) = 1/2 * [1/((x-a)^2 + (y-b)^2)] * 2(y-b) = (y-b)/[(x-a)^2 + (y-b)^2]
Now, let's find the second derivatives. The second derivative of u with respect to x, d^2u/dx^2, can be obtained by taking the derivative of d/dx (u) with respect to x:
d^2u/dx^2 = d/dx [(x-a)/[(x-a)^2 + (y-b)^2]]
To simplify this expression, we can use the quotient rule. The quotient rule states that if we have a function f(x)/g(x), the derivative can be found as [g(x)f'(x) - f(x)g'(x)] / [g(x)]^2.
With this in mind, let's find d^2u/dx^2:
d^2u/dx^2 = [(1)[(x-a)^2 + (y-b)^2] - (x-a)(2(x-a))]/[(x-a)^2 + (y-b)^2]^2
= [(x-a)^2 + (y-b)^2 - 2(x-a)^2]/[(x-a)^2 + (y-b)^2]^2
= [(y-b)^2 - (x-a)^2]/[(x-a)^2 + (y-b)^2]^2
Using similar steps, we can find the second derivative of u with respect to y, d^2u/dy^2:
d^2u/dy^2 = [(x-a)^2 + (y-b)^2 - 2(y-b)^2]/[(x-a)^2 + (y-b)^2]^2
= [(x-a)^2 - (y-b)^2]/[(x-a)^2 + (y-b)^2]^2
Now, if we sum up d^2u/dx^2 and d^2u/dy^2, we have:
d^2u/dx^2 + d^2u/dy^2 = [(y-b)^2 - (x-a)^2]/[(x-a)^2 + (y-b)^2]^2 + [(x-a)^2 - (y-b)^2]/[(x-a)^2 + (y-b)^2]^2
= 0
Therefore, we have shown that d^2u/dx^2 + d^2u/dy^2 equals zero.