# The Hiemenz boundary layer

This is a particular case of a Falkner-Skan boundary layer corresponding to \beta=1, that is the boundary layer arround a stagnation point. In this case, there is a easy solution ofthe Pohlhausen profile that we use as an initial guess for Newton and as a comparison for the final result.

See blasius.m and falkner_skan.m for other details. We use the polynomial closure at order 4, so that \displaystyle \frac{u(\eta)}{\bar u_e}=(2 \eta - 2 \eta^3 +\eta^4) + \frac{1}{6}\Lambda (\eta- 3 \eta^2 + 3 \eta^3 -\eta^4) with \Lambda= \delta^2 d\bar u_e/d\bar x in the Hiemenz case \bar u_e=\bar x then \Lambda= \delta^2, The Von K'arm'an equation \displaystyle \frac{d}{d \bar x} (\bar u_e^2\tilde \delta_{2}) + {\tilde \delta_{1}} {\bar u_e} \frac{d \bar u_{e}}{d \bar x} = \frac{u'(0)}{\tilde \delta} \bar u_{e}, \;\;\mbox{ reads }\;\; 2 \tilde \delta_{2} + {\tilde \delta_{1}} = \frac{u'(0)}{\tilde \delta }, which is an equation were \delta_1/\delta=(36 -\Lambda)/120 and \delta_2/\delta= 37/315 - \Lambda/945 - (\Lambda^2)/9072, and u'(0) = (2 + \Lambda/6) and remember that \Lambda= \delta^2, we then substitute in VK: \displaystyle \delta \left(\frac{3}{10}-\frac{\delta^2}{120}\right)-\frac{\frac{\delta^2}{6}+2}{\delta}+2 \delta \left(-\frac{\delta^4}{9072}-\frac{\delta^2}{945}+\frac{37}{315}\right)=0 we solve and find numericaly \delta=2.65562 this gives \Lambda=7.05 and \delta_1=0.640617, and H=2.30809 and \tau= \frac{u'(0)}{\tilde \delta }=1.1957

Dependency

clear all; clf;
%setenv("GNUTERM","X11")

% parameters
L=10; % box height
N=100; % number of gridpoints
beta=1; % pressure gradient parameter

For the differentiation matrices, since we as well need the third derivative, we build directly the matrices from chebdif.m as in diffmat.m instead of using the function dif1D.m.

% differentiation matrices
scale=-2/L;
[y,DM] = chebdif(N,3);
dy=DM(:,:,1)*scale;
dyy=DM(:,:,2)*scale^2;
dyyy=DM(:,:,3)*scale^3;
y=(y-1)/scale;
I=eye(N); Z=zeros(N,N);

% initial guess from order 4 Pohlhausen profile
d99=2.65562;
lambda=d99^2;
eta=y/d99; deta=dy*d99; % rescaled vertical coordinate
u0=1-(1-eta).^3.*(1+(1-lambda/6)*eta);
u0(eta>1)=1;
A=deta; A(1,:)=I(1,:); u0(1)=0; % set up integration
g0=A\u0; % compute integral

sol=g0;

% Newton iterations
quit=0;count=0;
while ~quit

% the present solution
g=sol; gy=dy*g; gyy=dyy*g; gyyy=dyyy*g;

# The FS equation

Here is the nonlinear equation that we wish to solve \displaystyle 0=g_{yyy}+gg_{yy} + \beta (1 - g_y^2) and the analytical Jacobian is \displaystyle A=\partial_{yyy}+g\partial_{yy}+g_{yy}I -2 \beta g_y \partial_{y}

% nonlinear function
f=gyyy+g.*gyy+beta*(1-gy.^2);

% analytical jacobian
A=dyyy+diag(g)*dyy+diag(gyy)-2*beta*diag(gy)*dy;

# Boundary conditions

The boundary conditions are homogeneous Dirichlet and Neuman at the wall and Neuman 1 at the top. The constraint matrix for the boundary conditions is thus \displaystyle C=\left(\begin{array}{c} I|_0\\ \partial_y|_0\\ \partial_y|_L \end{array}\right) so that the homogeneous and nonhomogeneous boundary conditions are expressed \displaystyle Cg=\left(\begin{array}{c} 0\\ 0\\ 1 \end{array}\right)

% Boundary conditions
loc=[1,2,N];
C=[I(1,:); dy(1,:); dy(N,:)];
f(loc)=C*g-[0;0;1];
A(loc,:)=C;

% convergence test
res=norm(f);
disp([num2str(count) '  ' num2str(res)]);
if count>50|res>1e5; disp('no convergence'); break; end
if res<1e-5; quit=1; disp('converged'); continue; end

% Newton step
sol=sol-A\f;
count=count+1;
end

To recover the velocity profile from g is \displaystyle u=g_y

% the solution
u=dy*g;

% Compute the displacement thickness and normalize it to unity
INT=([diff(y)',0]+[0,diff(y)'])/2; % integration weights
mt=INT*(1-u);

% Showing the result
plot(u,y,'b-',u0,y,'r--'); xlabel('u'); ylabel('y'); ylim([0,y(end)]);
title('Falkner-skan boundary layer');
legend('Falkner-Skan','Polhausen');
grid on

set(gcf,'paperpositionmode','auto');
print('-dpng','-r80','hiemenz.png');

% Compute the displacement thickness
w=([diff(y)',0]+[0,diff(y)'])/2; % integration weights

d1num=w*(1-u)
d1pol=(36-lambda)/120*d99

shearnum=gyy(1)
shearpol=(2+lambda/6)/d99

# Result and Plot

The Hiemenz flow f''' + ff'' + (1 - f'^2)=0 as solution f''(0)=1.2325 (compare to 1.1957 for Pohlhausen4), the displacement thickness is \int (1-f')d \eta=0.6479 (compare to 0.640617 for Pohlhausen4).

• Please check the convergent case \beta \rightarrow \infty so f'''+1-f'^2=0. f''_0=1.15, and $_0^(1-f’)d=0.779$