How can I solve a piecewise ODE
이전 댓글 표시

I am trying to solve the second order ODE as seen above, which corresponds to the conductive heat transfer and the Joule heating of the profile seen below. As seen in the figure below, the profile has varying thickness causing the parameters with the subscript 'i' to change for each of the intervals.

The temperature in the beams with length L1, L2 and L3 will now be denoted as
,
and
, respectively. (The subscripts stand for hot-beam, cold-beam and flexure)
, The six boundary conditions that must be taken into account are stated below and are indicated in the figure above:
(1) 

(2) 

(3) 

(4) 

(5)
(6) 
The solution must be a continuous function and should look something like the graph seen below.

So far I have tried to solve this problem using the function dsolve and using each interval as a seperate ODE. This did however not result in the desired solution, as the boundary conditions are not met and the resulting graph is not a continuous function. The lines of code that I used have been added below.
I would like to know what I should do different or which function to use to solve this problem. Thanks in advance!
Dion
clear all; close all; clc;
%% Constants
% Dimensions
h=10e-6; % height
b=[20e-6 60e-6 20e-6]; % width
L=[200e-6 140e-6 40e-6]; % length
P=2*h+2*b; % perimeter
A=h*b; % area
% Thermal properties
U=15000; % conduction heat transfer coefficient
k_p=50; % thermal concuctivity
T_inf=293.15; % ambient temperature
V=[5 5 5]; % applied voltage
r=22e-6; % resistivity
R=r*L./A; % electrical resistance
%% Differential Equations
syms T_h(x) T_c(x) T_f(x)
DT_h = diff(T_h);
DT_c = diff(T_c);
DT_f = diff(T_f);
ode1 = diff(T_h,x,2) == ((U*P(1))/(A(1)k_p))(T_h-T_inf)-(V(1)^2/(A(1)*k_p*R(1)*L(1)));
ode2 = diff(T_c,x,2) == ((U*P(2))/(A(2)k_p))(T_c-T_inf)-(V(2)^2/(A(2)*k_p*R(2)*L(2)));
ode3 = diff(T_f,x,2) == ((U*P(3))/(A(3)k_p))(T_f-T_inf)-(V(3)^2/(A(3)*k_p*R(3)*L(3)));
odes=[ode1 ode2 ode3];
%% Conditions
cond1 = T_h(0) == 293.15;
cond2 = T_f(L(3)) == 293.15;
cond3 = T_h(L(1)) == T_c(0);
cond4 = T_c(L(2)) == T_f(0);
cond5 = A(1)*DT_h(L(1)) == A(2)*DT_c(0);
cond6 = A(2)*DT_c(L(2)) == A(3)*DT_f(0);
conds = [cond1 cond2 cond3 cond4 cond5 cond6];
%% Solve
[T_h(x), T_c(x), T_f(x)]=dsolve(odes,conds);
%% Plotting
x1=linspace(0,L(1),1000);
x2=linspace(0,L(2),1000);
x3=linspace(0,L(3),1000);
xtot=[x1 L(1)+x2 L(2)+L(1)+x3];
T1=subs(T_h,x,x1);
T2=subs(T_c,x,x2);
T3=subs(T_f,x,x3);
Ttot=[T1 T2 T3];
figure()
plot(xtot, Ttot)
grid on
댓글 수: 9
darova
2021년 3월 18일
Can you show points (
) and temperatures (boundary conditions) on the picture?
Dion Stam
2021년 3월 23일
darova
2021년 3월 23일
WHere are points
?
Dion Stam
2021년 3월 24일
darova
2021년 3월 24일
Can you explain (5) and (6) conditions. Are those related to the whole region or some points?
Dion Stam
2021년 3월 24일
darova
2021년 3월 24일
I don't quite understand these conditions

They mean that form of curves are the same. But
, 
Dion Stam
2021년 3월 24일
darova
2021년 3월 25일
Sorry im lost. I don't know how to proceed
답변 (1개)
Emilien Flayac
2021년 6월 24일
0 개 추천
If this can still be useful, you could try to put your problem into the following form:
where
would be evaluated on grids and Fwould contain your dynamical constraint as well as the boundary conditions (1)-(6).
Then, you could use any method to find the roots. Since you already used the Symbolic toolbox, have a look at vpasolve.
카테고리
도움말 센터 및 File Exchange에서 Ordinary Differential Equations에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!