I feel like asking a question that show how long I have to go. Clarification. In differential equations, I start with a rate of change and find the indefinite integral to find the function. In integral equations I start with an integral of some ambiguous function and try to find the derivative???
Asked
Active
Viewed 41 times
1 Answers
0
No, you still want to find a continuous function solving it. An example for an integral equation is
$$x(t)=x_0(t)+\int_a^b K(t,s)x(s)\,ds$$
There is in general no transformation to a differential equation. Depending on the kernel $K$, a solution might not be differentiable.
Differential equation are relatively easy because they can be transformed into a nice integral equation. The initial value problem $x'(t)=f(t,x(t))$, $x(a)=x_a$ has the equivalent integral equation
$$x(t)=x_a+\int_a^t f(s,x(s))\,ds$$
Where the differential equation requires the consideration of differentiable functions, the integral equation allows all continuous functions as input, the differentiability of the solutions is then a consequence, not a condition for solvability.
Lutz Lehmann
- 126,666
-
Then diff eqs are a subset of integral eq's? – Chris Jan 03 '14 at 23:19
-
In this sense, yes. But they have such a special structure that this is rarely used outside the existence theorems. – Lutz Lehmann Jan 04 '14 at 07:10