iteration to convergence

I am very very new to Matlab and this will become quite apparent, but I am working on a problem where the variable I'm solving for is also in the equation. I was told this could be accomplished in Matlab.

v=(((2*Ke)/deltaZ)*log(Ar))+ sqrt((alpha+v^2)/2)

PS. this is a Hydrogeology equation to determine velocity using heat as a tracer

%Constants

lambdaE=2.8;  %Effecitve Thermal Conductivity

p=2.7; %Density of Fluid and Sediment

c=4.2;  %Heat Capacity of Fluid and Sediment

Ke=[lambdaE/(p*c)];  %Effective Thermal Diffusivity

f=2;  %Frequency

P=(1/f);  %Period

v=.3; %Fluid Velocity (initial)

deltaPhi = 0.004;  %Phase shift between shallow and deep points [Measured in the Lab]

Ar = 0.1; %Amp ratio of shallow and deep points[Measured in the Lab]

deltaZ=.2;  %Distance between shallow and deep points

alpha=sqrt(v^4+(8*pi*Ke/P)^2);

Amp Method

%Ar=exp((deltaZ/(2*Ke))*(v-sqrt((alpha + v^2)/2)))

%VAr

v=(((2*Ke)/deltaZ)*log(Ar))+ sqrt((alpha+v^2)/2)

I really hope someone out there can take pity and give me a brief walk through how this can be solved.

-Ethan

Answer 1

[bym]

I think it converges pretty quick; see below. My tweaks are commented with % *comment *

clc;clear

%%Constants
lambdaE=2.8;  %Effecitve Thermal Conductivity
p=2.7; %Density of Fluid and Sediment
c=4.2;  %Heat Capacity of Fluid and Sediment
Ke=lambdaE/(p*c);  %Effective Thermal Diffusivity
f=2;  %Frequency
P=(1/f);  %Period

deltaPhi = 0.004;  %Phase shift between shallow and deep points [Measured in the Lab]
Ar = 0.1; %Amp ratio of shallow and deep points[Measured in the Lab]
deltaZ=.2;  %Distance between shallow and deep points

Amp Method

%Ar=exp((deltaZ/(2*Ke))*(v-sqrt((alpha + v^2)/2)))
%VAr
tol = 1;
v=zeros(50,1); % *changed v to vector* 
alpha = v;     % *changed alpha to vector*

v(1)=.3;                   %Fluid Velocity (initial)
alpha(1)=sqrt(v(1)^4+(8*pi*Ke/P)^2);
c = 0;      % *counter*
while tol>.00001  % *desired accuracy*
c=c+1;
v(c+1)=(((2*Ke)/deltaZ)*log(Ar))+ ...
       sqrt((alpha(c)+v(c).^2)/2);  % * note changes to alpha & c
alpha(c+1)=sqrt(v(c+1)^4+(8*pi*Ke/P)^2); % *added*
tol = abs(v(c)-v(c+1));
end

[Bittu Patel]

But what if soluion diverges then the while loop will run till infinity how to break that

Answer 2

[Walter Roberson]

The solutions are

(4*Ke*ln(Ar) +/- (8*Ke^2*ln(Ar)^2+alpha*deltaZ^2)^(1/2)) / deltaZ

That is, can be done as a simple quadratic.

[x y]

Wow! Well I take my hat off to you for your mad math skills, but its more of a "how to do iteration" question. Thanks though, that neat to see the equation reworked like that!

[Walter Roberson]

My "mad math skills" just consisted of pasting it in to a symbolic mathematics program ;-)

[Walter Roberson]

while true

new_v=(((2*Ke)/deltaZ)*log(Ar))+ sqrt((alpha+v^2)/2);

if abs(new_v - v) < 0.0001 %or as appropriate for tolerance

break; %we converged

end

v = new_v;

end

[x y]

Sorry, didn't notice you'd updated the comment. I understand what you did and thats a brilliant idea, but it doesn't work? doesn't seem to iterate nor converge.