Integro-differential equation including a convolution of the first derivative.
$begingroup$
I am having difficulty finding the right approach to solving the following differential equation,
$$
y''(t)+int_t^Tg(s-t)y'(s),ds=f(t),
$$
with the boundary conditions,
$$y(0)=y_0,,quad y(T)=0.$$
I considered representing $y(t)$ using a Fourier series, but term-wise differentiation is not guaranteed because the periodic extension of $y(t)$ isn't
differentiable at the boundaries $t=0,T$.
What approach is the smartest to solve this equation? Any help will be appreciated.
ordinary-differential-equations convolution boundary-value-problem integro-differential-equations
edited Jan 9 at 15:36
Adrian Keister
4,82761933
asked Jan 9 at 15:22
honey.mustardhoney.mustard
8619
$endgroup$
add a comment |
$begingroup$
I am having difficulty finding the right approach to solving the following differential equation,
$$
y''(t)+int_t^Tg(s-t)y'(s),ds=f(t),
$$
with the boundary conditions,
$$y(0)=y_0,,quad y(T)=0.$$
I considered representing $y(t)$ using a Fourier series, but term-wise differentiation is not guaranteed because the periodic extension of $y(t)$ isn't
differentiable at the boundaries $t=0,T$.
What approach is the smartest to solve this equation? Any help will be appreciated.
ordinary-differential-equations convolution boundary-value-problem integro-differential-equations
edited Jan 9 at 15:36
Adrian Keister
4,82761933
asked Jan 9 at 15:22
honey.mustardhoney.mustard
8619
$endgroup$
$begingroup$
Just a thought, but can you subtract off the boundary by writing $y(t) = u(t) + frac{y_0}{T}(T-t)$? Then due to linearity, $u(t)$ would satisfy a very similar equation
$endgroup$
– Dylan
Jan 9 at 18:42
$begingroup$
@Dylan I considered that too. In that case, the first derivative of the Fourier series would indeed converge to $y'(t)$, but $y'(0) = y'(T)$ is not guaranteed so the second derivative of the Fourier series would NOT converge to $y''(t)$.
$endgroup$
– honey.mustard
Jan 10 at 1:21
add a comment |
$begingroup$
I am having difficulty finding the right approach to solving the following differential equation,
$$
y''(t)+int_t^Tg(s-t)y'(s),ds=f(t),
$$
with the boundary conditions,
$$y(0)=y_0,,quad y(T)=0.$$
I considered representing $y(t)$ using a Fourier series, but term-wise differentiation is not guaranteed because the periodic extension of $y(t)$ isn't
differentiable at the boundaries $t=0,T$.
What approach is the smartest to solve this equation? Any help will be appreciated.
ordinary-differential-equations convolution boundary-value-problem integro-differential-equations
edited Jan 9 at 15:36
Adrian Keister
4,82761933
asked Jan 9 at 15:22
honey.mustardhoney.mustard
8619
$endgroup$
I am having difficulty finding the right approach to solving the following differential equation,
$$
y''(t)+int_t^Tg(s-t)y'(s),ds=f(t),
$$
with the boundary conditions,
$$y(0)=y_0,,quad y(T)=0.$$
I considered representing $y(t)$ using a Fourier series, but term-wise differentiation is not guaranteed because the periodic extension of $y(t)$ isn't
differentiable at the boundaries $t=0,T$.
What approach is the smartest to solve this equation? Any help will be appreciated.
ordinary-differential-equations convolution boundary-value-problem integro-differential-equations
ordinary-differential-equations convolution boundary-value-problem integro-differential-equations
edited Jan 9 at 15:36
Adrian Keister
4,82761933
asked Jan 9 at 15:22
honey.mustardhoney.mustard
8619
edited Jan 9 at 15:36
Adrian Keister
4,82761933
asked Jan 9 at 15:22
honey.mustardhoney.mustard
8619
edited Jan 9 at 15:36
Adrian Keister
4,82761933
edited Jan 9 at 15:36
Adrian Keister
4,82761933
edited Jan 9 at 15:36
Adrian Keister
4,82761933
4,82761933
asked Jan 9 at 15:22
honey.mustardhoney.mustard
8619
asked Jan 9 at 15:22
honey.mustardhoney.mustard
8619
asked Jan 9 at 15:22
honey.mustardhoney.mustard
8619
8619
$begingroup$
Just a thought, but can you subtract off the boundary by writing $y(t) = u(t) + frac{y_0}{T}(T-t)$? Then due to linearity, $u(t)$ would satisfy a very similar equation
$endgroup$
– Dylan
Jan 9 at 18:42
$begingroup$
@Dylan I considered that too. In that case, the first derivative of the Fourier series would indeed converge to $y'(t)$, but $y'(0) = y'(T)$ is not guaranteed so the second derivative of the Fourier series would NOT converge to $y''(t)$.
$endgroup$
– honey.mustard
Jan 10 at 1:21
add a comment |
$begingroup$
Just a thought, but can you subtract off the boundary by writing $y(t) = u(t) + frac{y_0}{T}(T-t)$? Then due to linearity, $u(t)$ would satisfy a very similar equation
$endgroup$
– Dylan
Jan 9 at 18:42
$begingroup$
@Dylan I considered that too. In that case, the first derivative of the Fourier series would indeed converge to $y'(t)$, but $y'(0) = y'(T)$ is not guaranteed so the second derivative of the Fourier series would NOT converge to $y''(t)$.
$endgroup$
– honey.mustard
Jan 10 at 1:21
$begingroup$
Just a thought, but can you subtract off the boundary by writing $y(t) = u(t) + frac{y_0}{T}(T-t)$? Then due to linearity, $u(t)$ would satisfy a very similar equation
$endgroup$
– Dylan
Jan 9 at 18:42
$begingroup$
Just a thought, but can you subtract off the boundary by writing $y(t) = u(t) + frac{y_0}{T}(T-t)$? Then due to linearity, $u(t)$ would satisfy a very similar equation
$endgroup$
– Dylan
Jan 9 at 18:42
$begingroup$
@Dylan I considered that too. In that case, the first derivative of the Fourier series would indeed converge to $y'(t)$, but $y'(0) = y'(T)$ is not guaranteed so the second derivative of the Fourier series would NOT converge to $y''(t)$.
$endgroup$
– honey.mustard
Jan 10 at 1:21
$begingroup$
@Dylan I considered that too. In that case, the first derivative of the Fourier series would indeed converge to $y'(t)$, but $y'(0) = y'(T)$ is not guaranteed so the second derivative of the Fourier series would NOT converge to $y''(t)$.
$endgroup$
– honey.mustard
Jan 10 at 1:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Set $v(t) = y'(t)$. Thus $y(t) = - int_t^T v(s) , ds$.
Now solve the first order integrodifferential equation
$$
v'(t)+int_t^Tg(s-t)v(s),ds=f(t), ; v(T) = alpha
$$
for general $alpha in mathbb{R}$. The solution may be expressed in the form
$$
v(t) = alpha r(t) + int_t^T r(s-t)f(s), ds = alpha r(t) + tilde f(t)
$$
where $r$ is known as differential resolvent (see monographs on integral equations, e.g. Gripenberg - Londen - Staffans, CUP 1990, Theorem 3.1). Therefore
$$
y(t) = - int_t^T v(s) ds = - alpha int_t^T r(s) ds - int_t^T tilde f(s) , ds
$$
Set $t = 0$ to find the right value of $alpha$. This gives a unique solution if $int_0^T r(t) , dt ne 0$. If $int_0^T r(t) , dt = 0$, there either are infinitely many solutions or there is no solution, depending on whether $int_0^T tilde f(s) + y_0 = 0$ or not.
answered Jan 10 at 3:11
Hans EnglerHans Engler
10.4k11836
$endgroup$
add a comment |
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$begingroup$
Set $v(t) = y'(t)$. Thus $y(t) = - int_t^T v(s) , ds$.
Now solve the first order integrodifferential equation
$$
v'(t)+int_t^Tg(s-t)v(s),ds=f(t), ; v(T) = alpha
$$
for general $alpha in mathbb{R}$. The solution may be expressed in the form
$$
v(t) = alpha r(t) + int_t^T r(s-t)f(s), ds = alpha r(t) + tilde f(t)
$$
where $r$ is known as differential resolvent (see monographs on integral equations, e.g. Gripenberg - Londen - Staffans, CUP 1990, Theorem 3.1). Therefore
$$
y(t) = - int_t^T v(s) ds = - alpha int_t^T r(s) ds - int_t^T tilde f(s) , ds
$$
Set $t = 0$ to find the right value of $alpha$. This gives a unique solution if $int_0^T r(t) , dt ne 0$. If $int_0^T r(t) , dt = 0$, there either are infinitely many solutions or there is no solution, depending on whether $int_0^T tilde f(s) + y_0 = 0$ or not.
answered Jan 10 at 3:11
Hans EnglerHans Engler
10.4k11836
$endgroup$
add a comment |
$begingroup$
Set $v(t) = y'(t)$. Thus $y(t) = - int_t^T v(s) , ds$.
Now solve the first order integrodifferential equation
$$
v'(t)+int_t^Tg(s-t)v(s),ds=f(t), ; v(T) = alpha
$$
for general $alpha in mathbb{R}$. The solution may be expressed in the form
$$
v(t) = alpha r(t) + int_t^T r(s-t)f(s), ds = alpha r(t) + tilde f(t)
$$
where $r$ is known as differential resolvent (see monographs on integral equations, e.g. Gripenberg - Londen - Staffans, CUP 1990, Theorem 3.1). Therefore
$$
y(t) = - int_t^T v(s) ds = - alpha int_t^T r(s) ds - int_t^T tilde f(s) , ds
$$
Set $t = 0$ to find the right value of $alpha$. This gives a unique solution if $int_0^T r(t) , dt ne 0$. If $int_0^T r(t) , dt = 0$, there either are infinitely many solutions or there is no solution, depending on whether $int_0^T tilde f(s) + y_0 = 0$ or not.
answered Jan 10 at 3:11
Hans EnglerHans Engler
10.4k11836
$endgroup$
add a comment |
$begingroup$
Set $v(t) = y'(t)$. Thus $y(t) = - int_t^T v(s) , ds$.
Now solve the first order integrodifferential equation
$$
v'(t)+int_t^Tg(s-t)v(s),ds=f(t), ; v(T) = alpha
$$
for general $alpha in mathbb{R}$. The solution may be expressed in the form
$$
v(t) = alpha r(t) + int_t^T r(s-t)f(s), ds = alpha r(t) + tilde f(t)
$$
where $r$ is known as differential resolvent (see monographs on integral equations, e.g. Gripenberg - Londen - Staffans, CUP 1990, Theorem 3.1). Therefore
$$
y(t) = - int_t^T v(s) ds = - alpha int_t^T r(s) ds - int_t^T tilde f(s) , ds
$$
Set $t = 0$ to find the right value of $alpha$. This gives a unique solution if $int_0^T r(t) , dt ne 0$. If $int_0^T r(t) , dt = 0$, there either are infinitely many solutions or there is no solution, depending on whether $int_0^T tilde f(s) + y_0 = 0$ or not.
answered Jan 10 at 3:11
Hans EnglerHans Engler
10.4k11836
$endgroup$
Set $v(t) = y'(t)$. Thus $y(t) = - int_t^T v(s) , ds$.
Now solve the first order integrodifferential equation
$$
v'(t)+int_t^Tg(s-t)v(s),ds=f(t), ; v(T) = alpha
$$
for general $alpha in mathbb{R}$. The solution may be expressed in the form
$$
v(t) = alpha r(t) + int_t^T r(s-t)f(s), ds = alpha r(t) + tilde f(t)
$$
where $r$ is known as differential resolvent (see monographs on integral equations, e.g. Gripenberg - Londen - Staffans, CUP 1990, Theorem 3.1). Therefore
$$
y(t) = - int_t^T v(s) ds = - alpha int_t^T r(s) ds - int_t^T tilde f(s) , ds
$$
Set $t = 0$ to find the right value of $alpha$. This gives a unique solution if $int_0^T r(t) , dt ne 0$. If $int_0^T r(t) , dt = 0$, there either are infinitely many solutions or there is no solution, depending on whether $int_0^T tilde f(s) + y_0 = 0$ or not.
answered Jan 10 at 3:11
Hans EnglerHans Engler
10.4k11836
answered Jan 10 at 3:11
Hans EnglerHans Engler
10.4k11836
answered Jan 10 at 3:11
Hans EnglerHans Engler
10.4k11836
answered Jan 10 at 3:11
Hans EnglerHans Engler
10.4k11836
10.4k11836
add a comment |
add a comment |
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$begingroup$
Just a thought, but can you subtract off the boundary by writing $y(t) = u(t) + frac{y_0}{T}(T-t)$? Then due to linearity, $u(t)$ would satisfy a very similar equation
$endgroup$
– Dylan
Jan 9 at 18:42
$begingroup$
@Dylan I considered that too. In that case, the first derivative of the Fourier series would indeed converge to $y'(t)$, but $y'(0) = y'(T)$ is not guaranteed so the second derivative of the Fourier series would NOT converge to $y''(t)$.
$endgroup$
– honey.mustard
Jan 10 at 1:21