Mixing
Find the resolvent kernel of particular form of volterra integral equation of second kind
$begingroup$
Find the resolvent kernel for the integral equation
$$u(x)= 30 +6x+∫_0^x(5t-6t^2 )u(t)dt$$
I try to solve but its lengthy
$$k(x,t)=(5t-6t^2)$$
$$k_2(x,t)=∫_ξ^x (5t-6t^2)(5ξ-6ξ^2)dt$$
its very lengthy how i find out $$k_3 $$ and$$k_n+1$$
Suppose that the kernel (x,t ) is a polynomial of degree (n − 1) in x, which can always
be represented in the form
$$a_0(x)+a_1(x)(x-t)$$
then how we take the values if a_0 and a_1
integration definite-integrals
asked Oct 15 '14 at 17:37
nooraynooray
65
$endgroup$
add a comment |
$begingroup$
Find the resolvent kernel for the integral equation
$$u(x)= 30 +6x+∫_0^x(5t-6t^2 )u(t)dt$$
I try to solve but its lengthy
$$k(x,t)=(5t-6t^2)$$
$$k_2(x,t)=∫_ξ^x (5t-6t^2)(5ξ-6ξ^2)dt$$
its very lengthy how i find out $$k_3 $$ and$$k_n+1$$
Suppose that the kernel (x,t ) is a polynomial of degree (n − 1) in x, which can always
be represented in the form
$$a_0(x)+a_1(x)(x-t)$$
then how we take the values if a_0 and a_1
integration definite-integrals
asked Oct 15 '14 at 17:37
nooraynooray
65
$endgroup$
add a comment |
$begingroup$
Find the resolvent kernel for the integral equation
$$u(x)= 30 +6x+∫_0^x(5t-6t^2 )u(t)dt$$
I try to solve but its lengthy
$$k(x,t)=(5t-6t^2)$$
$$k_2(x,t)=∫_ξ^x (5t-6t^2)(5ξ-6ξ^2)dt$$
its very lengthy how i find out $$k_3 $$ and$$k_n+1$$
Suppose that the kernel (x,t ) is a polynomial of degree (n − 1) in x, which can always
be represented in the form
$$a_0(x)+a_1(x)(x-t)$$
then how we take the values if a_0 and a_1
integration definite-integrals
asked Oct 15 '14 at 17:37
nooraynooray
65
$endgroup$
Find the resolvent kernel for the integral equation
$$u(x)= 30 +6x+∫_0^x(5t-6t^2 )u(t)dt$$
I try to solve but its lengthy
$$k(x,t)=(5t-6t^2)$$
$$k_2(x,t)=∫_ξ^x (5t-6t^2)(5ξ-6ξ^2)dt$$
its very lengthy how i find out $$k_3 $$ and$$k_n+1$$
Suppose that the kernel (x,t ) is a polynomial of degree (n − 1) in x, which can always
be represented in the form
$$a_0(x)+a_1(x)(x-t)$$
then how we take the values if a_0 and a_1
integration definite-integrals
integration definite-integrals
asked Oct 15 '14 at 17:37
nooraynooray
65
asked Oct 15 '14 at 17:37
nooraynooray
65
edited Oct 16 '14 at 19:17
nooray
asked Oct 15 '14 at 17:37
nooraynooray
65
asked Oct 15 '14 at 17:37
nooraynooray
65
asked Oct 15 '14 at 17:37
nooraynooray
65
65
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By using the Laplace transform and the convolution theorem, examples found here,
it is seen that for
begin{align}
mathcal{L}{ f(t) } = int_{0}^{infty} e^{- s t} f(t) , dt = f(s)
end{align}
then the integral equation
begin{align}
u(t)= 30 + 6t + int_{0}^{t} (5y-6y^2 )u(y)dy
end{align}
becomes
begin{align}
u(s)= frac{30}{s} + frac{6}{s^2} + left(frac{5}{s} - frac{12}{s^3} right) u(s).
end{align}
By solving for $u(s)$ the result becomes
begin{align}
u(s) = frac{6 s + 30 s^2}{s^3 - 5 s + 12}.
end{align}
Inversion of this yields
begin{align}
u(t) = frac{6}{77} , e^{-3 t} left[ 46 sqrt{7} , e^{frac{9 t}{2}} sinleft(frac{sqrt{7} t}{2} right) + 238 , e^{frac{9 t}{2}} cosleft(frac{sqrt{7} t}{2}right) + 147 right] .
end{align}
answered Oct 15 '14 at 18:51
LeucippusLeucippus
19.7k102871
$endgroup$
$begingroup$
we need to find out kernel $$k_1$$ to $$k_n$$
$endgroup$
– nooray
Oct 16 '14 at 11:30
$begingroup$
Any one solve it with alternative method
$endgroup$
– nooray
Oct 16 '14 at 19:16
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
By using the Laplace transform and the convolution theorem, examples found here,
it is seen that for
begin{align}
mathcal{L}{ f(t) } = int_{0}^{infty} e^{- s t} f(t) , dt = f(s)
end{align}
then the integral equation
begin{align}
u(t)= 30 + 6t + int_{0}^{t} (5y-6y^2 )u(y)dy
end{align}
becomes
begin{align}
u(s)= frac{30}{s} + frac{6}{s^2} + left(frac{5}{s} - frac{12}{s^3} right) u(s).
end{align}
By solving for $u(s)$ the result becomes
begin{align}
u(s) = frac{6 s + 30 s^2}{s^3 - 5 s + 12}.
end{align}
Inversion of this yields
begin{align}
u(t) = frac{6}{77} , e^{-3 t} left[ 46 sqrt{7} , e^{frac{9 t}{2}} sinleft(frac{sqrt{7} t}{2} right) + 238 , e^{frac{9 t}{2}} cosleft(frac{sqrt{7} t}{2}right) + 147 right] .
end{align}
answered Oct 15 '14 at 18:51
LeucippusLeucippus
19.7k102871
$endgroup$
$begingroup$
we need to find out kernel $$k_1$$ to $$k_n$$
$endgroup$
– nooray
Oct 16 '14 at 11:30
$begingroup$
Any one solve it with alternative method
$endgroup$
– nooray
Oct 16 '14 at 19:16
add a comment |
$begingroup$
By using the Laplace transform and the convolution theorem, examples found here,
it is seen that for
begin{align}
mathcal{L}{ f(t) } = int_{0}^{infty} e^{- s t} f(t) , dt = f(s)
end{align}
then the integral equation
begin{align}
u(t)= 30 + 6t + int_{0}^{t} (5y-6y^2 )u(y)dy
end{align}
becomes
begin{align}
u(s)= frac{30}{s} + frac{6}{s^2} + left(frac{5}{s} - frac{12}{s^3} right) u(s).
end{align}
By solving for $u(s)$ the result becomes
begin{align}
u(s) = frac{6 s + 30 s^2}{s^3 - 5 s + 12}.
end{align}
Inversion of this yields
begin{align}
u(t) = frac{6}{77} , e^{-3 t} left[ 46 sqrt{7} , e^{frac{9 t}{2}} sinleft(frac{sqrt{7} t}{2} right) + 238 , e^{frac{9 t}{2}} cosleft(frac{sqrt{7} t}{2}right) + 147 right] .
end{align}
answered Oct 15 '14 at 18:51
LeucippusLeucippus
19.7k102871
$endgroup$
$begingroup$
we need to find out kernel $$k_1$$ to $$k_n$$
$endgroup$
– nooray
Oct 16 '14 at 11:30
$begingroup$
Any one solve it with alternative method
$endgroup$
– nooray
Oct 16 '14 at 19:16
add a comment |
$begingroup$
By using the Laplace transform and the convolution theorem, examples found here,
it is seen that for
begin{align}
mathcal{L}{ f(t) } = int_{0}^{infty} e^{- s t} f(t) , dt = f(s)
end{align}
then the integral equation
begin{align}
u(t)= 30 + 6t + int_{0}^{t} (5y-6y^2 )u(y)dy
end{align}
becomes
begin{align}
u(s)= frac{30}{s} + frac{6}{s^2} + left(frac{5}{s} - frac{12}{s^3} right) u(s).
end{align}
By solving for $u(s)$ the result becomes
begin{align}
u(s) = frac{6 s + 30 s^2}{s^3 - 5 s + 12}.
end{align}
Inversion of this yields
begin{align}
u(t) = frac{6}{77} , e^{-3 t} left[ 46 sqrt{7} , e^{frac{9 t}{2}} sinleft(frac{sqrt{7} t}{2} right) + 238 , e^{frac{9 t}{2}} cosleft(frac{sqrt{7} t}{2}right) + 147 right] .
end{align}
answered Oct 15 '14 at 18:51
LeucippusLeucippus
19.7k102871
$endgroup$
By using the Laplace transform and the convolution theorem, examples found here,
it is seen that for
begin{align}
mathcal{L}{ f(t) } = int_{0}^{infty} e^{- s t} f(t) , dt = f(s)
end{align}
then the integral equation
begin{align}
u(t)= 30 + 6t + int_{0}^{t} (5y-6y^2 )u(y)dy
end{align}
becomes
begin{align}
u(s)= frac{30}{s} + frac{6}{s^2} + left(frac{5}{s} - frac{12}{s^3} right) u(s).
end{align}
By solving for $u(s)$ the result becomes
begin{align}
u(s) = frac{6 s + 30 s^2}{s^3 - 5 s + 12}.
end{align}
Inversion of this yields
begin{align}
u(t) = frac{6}{77} , e^{-3 t} left[ 46 sqrt{7} , e^{frac{9 t}{2}} sinleft(frac{sqrt{7} t}{2} right) + 238 , e^{frac{9 t}{2}} cosleft(frac{sqrt{7} t}{2}right) + 147 right] .
end{align}
answered Oct 15 '14 at 18:51
LeucippusLeucippus
19.7k102871
answered Oct 15 '14 at 18:51
LeucippusLeucippus
19.7k102871
answered Oct 15 '14 at 18:51
LeucippusLeucippus
19.7k102871
answered Oct 15 '14 at 18:51
LeucippusLeucippus
19.7k102871
19.7k102871
$begingroup$
we need to find out kernel $$k_1$$ to $$k_n$$
$endgroup$
– nooray
Oct 16 '14 at 11:30
$begingroup$
Any one solve it with alternative method
$endgroup$
– nooray
Oct 16 '14 at 19:16
add a comment |
$begingroup$
we need to find out kernel $$k_1$$ to $$k_n$$
$endgroup$
– nooray
Oct 16 '14 at 11:30
$begingroup$
Any one solve it with alternative method
$endgroup$
– nooray
Oct 16 '14 at 19:16
$begingroup$
we need to find out kernel $$k_1$$ to $$k_n$$
$endgroup$
– nooray
Oct 16 '14 at 11:30
$begingroup$
we need to find out kernel $$k_1$$ to $$k_n$$
$endgroup$
– nooray
Oct 16 '14 at 11:30
$begingroup$
Any one solve it with alternative method
$endgroup$
– nooray
Oct 16 '14 at 19:16
$begingroup$
Any one solve it with alternative method
$endgroup$
– nooray
Oct 16 '14 at 19:16
add a comment |
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