# Boas 4.12.18, 2nd Derivatives of Imp. Multivariable Integral

## Homework Statement

Show that u(x, y) = y/π ∫-∞ f(t) dt / ((x - t)2+y2) satisfies uxx + uyy = 0.

Leibniz' Rule

## The Attempt at a Solution

I'm not even sure Leibniz' Rule can be applied here since there seems to be a discontinuity in the integrand when x=t and y=0. When I use it and take the second derivatives, I get terms that have no way of cancelling (for example the -2(x-t) and -2y terms). I also tried integration by parts first but it did not simplify things.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
2021 Award
You are correct. The function does not satisfy the Laplace equation in two dimensions at ##y = 0##. The Green's function of the Poisson equation in 2D is proportional to ##\ln(r^2/r_0^2)##. It is a matter of rather straight-forward computations to show that your function corresponds to a source term of the form ##f(x) \delta'(y)##, i.e.,
$$u_{xx} + u_{yy} \propto f(x) \delta'(y).$$

So my goal is to show the Laplace equation is satisfied for all other values of y?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
2021 Award
For other values of ##y##, you should indeed get zero.

Are my first derivatives correct?

With (x-t)2+y2=g for compactness:

ux=-2xy/π∫f(t)dt/g2 + 2y/π∫t*f(t)dt/g2

uy=1/π∫f(t)dt/g - 2y2/π∫f(t)dt/g2

Orodruin
Staff Emeritus
Homework Helper
Gold Member
2021 Award
I suggest you remove the integral, ##\pi##, and ##f(t)## for readability. They will always be there. I also strongly recommend against splitting the ##(x-t)## into two terms in the first derivative wrt ##x##, but yes, they look correct.

$$u_x = \frac {-2y(x-t)}{g^2}$$

$$u_y = \frac {1}{g} - \frac {2y^2}{g^2}$$

Orodruin
Staff Emeritus
Homework Helper
Gold Member
2021 Award
Yes, that looks fine. Now the second derivatives.

$$u_{xx} = \frac {1}{g^2} - \frac {4(x-t)^2}{g^3}$$

$$u_{yy} = \frac {8y^3}{g^3} - \frac{6y}{g^2}$$

edit: oops I forgot the -2y in ##u_{xx}##

Orodruin
Staff Emeritus
Homework Helper
Gold Member
2021 Award
You are missing the inner derivative of the first term in ##u_{xx}## and a factor 2 in the second term (one from the ##g^2## in the denominator and one from the inner derivative). You also have some sign errors.

Edit: Ok, what is missing is your factor ##-2y##. So what do you get when you do the sum?

$$u_{xx} + u_{yy} = 8y \left( \frac {y^2-(x-t)^2}{g^3} - \frac {1}{g^2}\right)$$

Alright, I see it now. Thank you!

Auxillary question, why is the factor of π included in u(x,y)? Is u a function that comes up in the sciences?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
2021 Award
The factor 1/pi is obviuosly irrelevant for whether or not you get zero. It is however an important overall factor in the definition of the Green’s function.

• mishima
Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Show that u(x, y) = y/π ∫-∞ f(t) dt / ((x - t)2+y2) satisfies uxx + uyy = 0.

Leibniz' Rule

## The Attempt at a Solution

I'm not even sure Leibniz' Rule can be applied here since there seems to be a discontinuity in the integrand when x=t and y=0. When I use it and take the second derivatives, I get terms that have no way of cancelling (for example the -2(x-t) and -2y terms). I also tried integration by parts first but it did not simplify things.
Your expression is ambiguous. I genuinely cannot tell if you mean
$$u(x,t) = \frac{y}{\pi} \int_{-\infty}^{\infty} \frac{f(t)}{(x-t)^2+y^2}\hspace{4ex}(1)$$
or
$$u(x,t) = \frac{y}{\pi \int_{-\infty}^{\infty} \frac{f(t)}{(x-t)^2 + y^2} \, dt} \hspace{5ex}(2)$$
or, possibly, something else. Please use parentheses to make everything clear and unabmiguous (or, better still, use LaTeX to typeset your expressions, just as I have done in the above).