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how can i solve integral equation ,symbolically methode by using a matlab

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aissa ghaddab
aissa ghaddab on 16 Jul 2021
Commented: Walter Roberson on 17 Jul 2021
y(x)+ integral(k(x,t)*y(t) , t , a, b) =f(x)

Answers (2)

Sulaymon Eshkabilov
Sulaymon Eshkabilov on 16 Jul 2021
You can start by initiating the variables via syms, e.g.:
syms t x
y = expression1;
f = expression2;
Sol = solve(y+integral(@(t)expression3)-expression2==0)

Walter Roberson
Walter Roberson on 16 Jul 2021
syms a b f(x) k(x,t) y(x)
eqn = y(x) + int(k(x,t)*y(t), t, a, b) == f(x)
eqn = 
But what are you trying to solve for? For example do you have known f(x) and y(x) and a and b, and the task is to find k(x,t) ? Or do you have known functions and known a and the task is to find b?
  2 Comments
Walter Roberson
Walter Roberson on 17 Jul 2021
That appears to be an inhomogeneous Fredholm equation of the second kind; it looks like it would typically be examined using Liouville–Neumann series https://en.wikipedia.org/wiki/Liouville%E2%80%93Neumann_series . For the usual approach to hold,
syms x t a b real
Kcond = int(abs(exp(x*t)),t,a,b)
Kcond = 
which is (exp(b*x)-exp(a*x))/x .
This can be true for particular ranges of x, but clearly cannot be true for arbitrary real a, b, and x ranges. Picking a and b arbitrarily,
Kcond13 = subs(Kcond, [a,b], [1, 3])
Kcond13 = 
vpasolve(Kcond13 == 1)
ans = 
limit(Kcond13, x, -inf)
ans = 
0
so it could be true from -inf to about -0.357 but not for greater x.
What this tells us is that the Liouville Neumann approach is not suitable for this equation -- that there is no general contraction mapping.
No matter which non-zero a and b you choose, there is some x for which (exp(b*x)-exp(a*x) is arbitrarily large, and that increases non-linearly, so at some point that is going to be greater than x. So it looks to me as if that particular set of equations does not have a contraction mapping, and so probably there is no solution.

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