M3M6: Methods of Mathematical Physics

$$ \def\dashint{{\int\!\!\!\!\!\!-\,}} \def\infdashint{\dashint_{\!\!\!-\infty}^{\,\infty}} \def\D{\,{\rm d}} \def\dx{\D x} \def\dt{\D t} \def\C{{\mathbb C}} \def\CC{{\cal C}} \def\HH{{\cal H}} \def\I{{\rm i}} \def\qqfor{\qquad\hbox{for}\qquad} $$

Dr. Sheehan Olver
s.olver@imperial.ac.uk

Website: https://github.com/dlfivefifty/M3M6LectureNotes

Lecture 12: Ideal fluid flow

Understanding branch cuts and Cauchy transforms allows us to systematically solve equations involving Laplace equation. A classic example is Ideal fluid flow. Consider the case of uniform flow with angle $\theta$ around an infinitesimally small plate on $[-1,1]$. We can model this as $$ \begin{align*} v(x,y) &\sim y \cos \theta - x \sin \theta \\ v_{xx} + v_{yy} &= 0 \\ v(x,0) & = 0 \qqfor -1 < x <1 \end{align*} $$ Using the techniques we developed in the last few lectures, we obtain a nice, simple, closed form expression as the imaginary part of an analytic function:

We divide this task into stages:

1. Rephrasing as a complex-analytical problem: $v(x,y)$ to $\phi(z)$
2. Reduction to inverting a Hilbert transform: $\phi(z)$ to $w(x)$
3. Calculating the inverse Hilbert transform: Finding $w(x)$
4. Calculating its Cauchy transform: $w(x)$ to $\phi(z)$

1. Real and imaginary parts of analytic functions satisfy Laplace's equation

The real and imaginary parts of an analytic function satisfy Laplace's equation: that is if $\phi(z) = \phi(x + \I y) = u(x,y) + \I v(x,y)$ where $u$ and $v$ are the real/imaginary parts, then $$ u_{xx} + u_{yy}= 0 \\ v_{xx} + v_{yy} = 0 $$ To see this, note that the complex-derivative of $\phi(z)$ can be written in terms of two different partial derivatives: $$ \phi'(z) = \lim_{h \rightarrow 0} {\phi(z+h) - \phi(z) \over h} = \lim_{h \rightarrow 0} {u(x+h,y)-u(x,y) + \I (v(x+h,y)-v(x,y)) \over h} = u_x + \I v_x \\ \phi'(z) = \lim_{h \rightarrow 0} {\phi(z+\I h) - \phi(z) \over \I h} = \lim_{h \rightarrow 0} {u(x,y+h)-u(x,y) + \I (v(x,y+h)-v(x,y)) \over \I h} = - \I u_y + v_y $$ Taking a second derivative we get two equations: $$ \phi'(z) = u_{xx} + \I v_{xx} = -u_{xx} -\I v_{yy} $$ which implies $u_{xx} + u_{yy} = 0$ and $v_{xx} + v_{yy} = 0$.

2. Reduce PDE to the Hilbert transform of an unknown function

Therefore we can rewrite the ideal fluid flow equation as a problem of calculating $\phi(z) = u(x,y) + \I v(x,y)$ whose imaginary part is the solution too the ideal fluid flow PDE (we don't use the real part $u$). That is, we want to find analytic $\phi(z)$ that satisfies \begin{align*} \phi(z) &\sim e^{-\I \theta} z \\ \Im \phi(x) &= 0 \qquad\hbox{for}\qquad -1 < x < 1 \end{align*}

Write $$ \phi(z) = e^{-\I \theta} z + c + \CC_{[-1,1]} w(z) $$ for an as-of-yet unknown function $w$ and $C$ an unknown constant, we have that $$ 0 = \Im \phi(x) = -x \sin \theta + \Im c + \Im \CC_{[-1,1]}^+ w(x) = -x \sin \theta + \Im c -{1 \over 2} \HH w(x) $$ In this example, we can take $c = 0$. Therefore, we want to solve $$ \HH w(x) = - 2 x \sin \theta $$ for $w$.

3. Calculating the inverse Hilberrt transform

We now plug the problem into the inverse Hilbert transform formula:

$$ w(x) = {- 1\over \sqrt{1-x^2}} {\cal H}[f \sqrt{1-\diamond^2}](x) + {D \over \sqrt{1-x^2}} $$

where $f(x) = -2 x \sin \theta$. Now that

Similar to last lecture, we find $$ {\cal C}[\diamond \sqrt{1-\diamond^2}](z) = {z \sqrt{z-1} \sqrt{z+1} - z^2 + 1/2 \over 2 \I } $$ and therefore $$ {\cal H}[\diamond \sqrt{1-\diamond^2}](x) = \I({\cal C}^+ + {\cal C}^-)[\diamond \sqrt{1-\diamond^2}](x) = {1 \over 2} - x^2 $$ Thus (relabeling $D$) we have $$ w(x) = 2\sin \theta {D-x^2 \over \sqrt{1-x^2}} $$

Demonstration Here we see that this gives us the right Hilbert transform:

$D$ is arbitrary, but from physical principles we know that we don't want the solution to blow up. If $w$ blows up then so does its Cauchy transform. Therefore, we choose $D = -1$ so that

$$ w(x) = 2 \sin \theta \sqrt{1-x^2} $$

4. Calculating its Cauchy transform

Now recall $$ \CC\left[{\sqrt{1-\diamond^2}}\right](z) = {\sqrt{z-1} \sqrt{z+1} -z \over 2 \I} $$

Therefore, $w(x) = 2 \sin \theta \sqrt{1-x^2}$, implies $$ \CC w(z) = - \I \sin \theta (\sqrt{z-1} \sqrt{z+1} -z) $$ which means $$ \phi(z) = e^{-\I \theta} z - \I \sin \theta (\sqrt{z-1} \sqrt{z+1} -z). $$

Numerical example of two intervals

We note that for obstacles on the real line, represented by a contour $\Gamma$, the problem of ideal fluid flow around $\gamma$ is still reducible to solving the singular integral equation $$ \HH_\Gamma f(x) = - 2 x \sin \theta $$ Even when not solvable exactly, one can solve it numerically: