# 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.

## Answers and Replies

Orodruin
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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
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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
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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
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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
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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
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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
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## 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).