Boas 4.12.18, 2nd Derivatives of Imp. Multivariable Integral

[mishima]

Homework Statement

Show that u(x, y) = y/π ∫_-∞^∞ f(t) dt / ((x - t)²+y²) satisfies u_xx + u_yy = 0.

Homework Equations

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]

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.,
$$
u_{xx} + u_{yy} \propto f(x) \delta'(y).
$$

 

[mishima]

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

 

[Orodruin]

For other values of $y$, you should indeed get zero.

 

[mishima]

Are my first derivatives correct?

With (x-t)²+y²=g for compactness:

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

u_y=1/π∫f(t)dt/g - 2y²/π∫f(t)dt/g²

 

[Orodruin]

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.

 

[mishima]

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

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

 

[Orodruin]

Yes, that looks fine. Now the second derivatives.

 

[mishima]

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

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

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

 

[Orodruin]

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?

 

[mishima]

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

Alright, I see it now. Thank you!

 

[mishima]

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

 

[Orodruin]

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.

 

[Ray Vickson]

Homework Statement

Show that u(x, y) = y/π ∫_-∞^∞ f(t) dt / ((x - t)²+y²) satisfies u_xx + u_yy = 0.

Homework Equations

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).