# How to get sinusodial behaviour from 2nd order ode using function handle?

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Henry B. on 28 May 2021
Answered: Alan Stevens on 28 May 2021
function[dtheta] = FiniteConduct(x,t,alph,L)
n=[1:10]
dtfh= @(t) 1-x./L -2 ./pi.*sum(1 ./n .*exp(-alph.*t.*((n.*pi/L).^2)).*sin(n.*pi.*x./L));
dtheta = dtfh(t(1));
for (k=1:numel(t)-1)
for (n = 1:10)
for (x = [0:.01:.1])
x = x+1;
if(x<L)
n=n+1;
dtheta(k+1) = dtheta(k) + dtfh(t(k+1))
else
x = L;
end
end
end
end
end
There seems to be something wrong with my function handle and the way I have implemented the sigma notation for sum? im trying to use convert this equation below into a function handle. Alan Stevens on 28 May 2021
You coud try implementing the function along these lines (obviously, you will need to use your own values for alpha etc):
alpha = 0.1;
L = 1;
t = 0.1;
x = 0:0.01:1;
for i = 1:numel(x)
theta(i) = dtfh(x(i),t,alpha,L);
end
plot(x,theta)
function theta = dtfh(x,t,alpha,L)
S = 0;
Sold = 100;
n = 0;
while abs(S-Sold)>1e-8
Sold = S;
n = n+1;
S = 1/n*exp(-alpha*t*(n*pi/L)^2)*sin(n*pi*x/L) + S;
end
theta = 1 - x/L - 2/pi*S;
end
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