Airat Posted July 18, 2016 Posted July 18, 2016 Any idea how to solve equation? [math] \int_0^1 C(yW(x))W^3(x)dx=F(y)[/math] [math]W= 0.5(\cosh(kx)-cos(kx)-\frac{\cosh(k)+\cos(k)}{\sinh(k)+sin(k)}(\sinh(kx)-sin(kx)))[/math] Need to find C F(y) is known function. k- is known constant
studiot Posted July 18, 2016 Posted July 18, 2016 (edited) I don't understand. How can F(y) be a function? The integral (if it exists, which I haven't checked) is a definite integral and therefore a pure number and therefore a constant, not a function at all. and what is y anyway? Edited July 18, 2016 by studiot
Raider5678 Posted July 19, 2016 Posted July 19, 2016 Step one: simplify. I'm not entirely sure what that would look like, I'm thinking something like this: C (yW*4(x*3))d W*2 would be W squared, but I don't know how to make the number smaller on my tablet. Also, this is probably wrong considering I don't do equations like this ever lol
Airat Posted July 19, 2016 Author Posted July 19, 2016 (edited) [math]C( \xi )[/math] - is function of single variable. This equation look like Fredholm equation of the first type. But it can't be represent in form [math]f(y)=\int_a^b K(x,y)\varphi(x)dx[/math] where [math]K(x,t),f(x) [/math] is known functions and need to find [math] \varphi[/math] In my case the [math]C(yW(x))=K(x,y)[/math] is not known and need to find. I can replace variable [math]\xi=yW(x) => F(y)=\frac{1}{y^4} \int_0^y C(\xi) \frac{\xi^3}{W'(x)} d\xi [/math], but it means that I need to find relation of [math] x[/math] as function of variable ([math]\xi,y[/math] ) [math]x=\Phi(\xi,y)[/math] it is impossible (look to definition of [math] W(x)[/math]) Soo we get Volterra equation of first kind [math]F(y)y^4=\int_0^y C(\xi) K(\xi,y) d\xi[/math] where [math]K(\xi,y)=\frac{\xi^3}{W'(\Phi(y,\xi))}, x=\Phi(\xi,y) [/math]- solution of equation [math]yW(x)=\xi [/math] Edited July 19, 2016 by Airat
Country Boy Posted August 20, 2016 Posted August 20, 2016 I don't understand. How can F(y) be a function? The integral (if it exists, which I haven't checked) is a definite integral and therefore a pure number and therefore a constant, not a function at all. and what is y anyway? The integration is with respect to x. F is a function of the variable y.
studiot Posted August 20, 2016 Posted August 20, 2016 The integration is with respect to x. F is a function of the variable y. Many thanks for your explanation. I had understood that the integration was with respect to x. I think I missed the first y.
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