Mixing
Shifted laplace transform derivative?
$begingroup$
I have a linear ODE that has terms that are shifted, for example
$$frac{d^k f(x-n)}{dx^k}$$
from a general equation
$$f(x)=sum_{k=0}^{n} a_k frac{d^k f(x-n)}{dx^k}$$
where $a_k$ are constants. Since the ODE was linear i assumed a Laplace transform would be easy to apply to this, but a laplace transform for something like $frac{df(t-n)}{dt}$ gives
$$int_{0}^{infty} frac{df(t-n)}{dt} e^{-st}dt=int_{n}^{infty} frac{df(t)}{dt} e^{-s(t+n)}dt$$
$$=e^{-ns}[int_{n}^{infty} frac{d}{dt}(f(t) e^{-st})dt - sint_{n}^{infty}f(t)e^{-st}]$$
$$=e^{-ns}[-f(n^{-})e^{-ns}-s(F(s)-int_{0}^{n} f(t)e^{-st}dt)]$$
Is this correct? What about the last term, is there an evaluation for it or is it computer only? Is there an easier way to solve the above ODE?
ordinary-differential-equations laplace-transform
asked Jan 26 at 8:44
Baklava GainBaklava Gain
14810
$endgroup$
add a comment |
$begingroup$
I have a linear ODE that has terms that are shifted, for example
$$frac{d^k f(x-n)}{dx^k}$$
from a general equation
$$f(x)=sum_{k=0}^{n} a_k frac{d^k f(x-n)}{dx^k}$$
where $a_k$ are constants. Since the ODE was linear i assumed a Laplace transform would be easy to apply to this, but a laplace transform for something like $frac{df(t-n)}{dt}$ gives
$$int_{0}^{infty} frac{df(t-n)}{dt} e^{-st}dt=int_{n}^{infty} frac{df(t)}{dt} e^{-s(t+n)}dt$$
$$=e^{-ns}[int_{n}^{infty} frac{d}{dt}(f(t) e^{-st})dt - sint_{n}^{infty}f(t)e^{-st}]$$
$$=e^{-ns}[-f(n^{-})e^{-ns}-s(F(s)-int_{0}^{n} f(t)e^{-st}dt)]$$
Is this correct? What about the last term, is there an evaluation for it or is it computer only? Is there an easier way to solve the above ODE?
ordinary-differential-equations laplace-transform
asked Jan 26 at 8:44
Baklava GainBaklava Gain
14810
$endgroup$
add a comment |
$begingroup$
I have a linear ODE that has terms that are shifted, for example
$$frac{d^k f(x-n)}{dx^k}$$
from a general equation
$$f(x)=sum_{k=0}^{n} a_k frac{d^k f(x-n)}{dx^k}$$
where $a_k$ are constants. Since the ODE was linear i assumed a Laplace transform would be easy to apply to this, but a laplace transform for something like $frac{df(t-n)}{dt}$ gives
$$int_{0}^{infty} frac{df(t-n)}{dt} e^{-st}dt=int_{n}^{infty} frac{df(t)}{dt} e^{-s(t+n)}dt$$
$$=e^{-ns}[int_{n}^{infty} frac{d}{dt}(f(t) e^{-st})dt - sint_{n}^{infty}f(t)e^{-st}]$$
$$=e^{-ns}[-f(n^{-})e^{-ns}-s(F(s)-int_{0}^{n} f(t)e^{-st}dt)]$$
Is this correct? What about the last term, is there an evaluation for it or is it computer only? Is there an easier way to solve the above ODE?
ordinary-differential-equations laplace-transform
asked Jan 26 at 8:44
Baklava GainBaklava Gain
14810
$endgroup$
I have a linear ODE that has terms that are shifted, for example
$$frac{d^k f(x-n)}{dx^k}$$
from a general equation
$$f(x)=sum_{k=0}^{n} a_k frac{d^k f(x-n)}{dx^k}$$
where $a_k$ are constants. Since the ODE was linear i assumed a Laplace transform would be easy to apply to this, but a laplace transform for something like $frac{df(t-n)}{dt}$ gives
$$int_{0}^{infty} frac{df(t-n)}{dt} e^{-st}dt=int_{n}^{infty} frac{df(t)}{dt} e^{-s(t+n)}dt$$
$$=e^{-ns}[int_{n}^{infty} frac{d}{dt}(f(t) e^{-st})dt - sint_{n}^{infty}f(t)e^{-st}]$$
$$=e^{-ns}[-f(n^{-})e^{-ns}-s(F(s)-int_{0}^{n} f(t)e^{-st}dt)]$$
Is this correct? What about the last term, is there an evaluation for it or is it computer only? Is there an easier way to solve the above ODE?
ordinary-differential-equations laplace-transform
ordinary-differential-equations laplace-transform
asked Jan 26 at 8:44
Baklava GainBaklava Gain
14810
asked Jan 26 at 8:44
Baklava GainBaklava Gain
14810
edited Jan 28 at 17:11
Baklava Gain
asked Jan 26 at 8:44
Baklava GainBaklava Gain
14810
asked Jan 26 at 8:44
Baklava GainBaklava Gain
14810
asked Jan 26 at 8:44
Baklava GainBaklava Gain
14810
14810
add a comment |
add a comment |
1 Answer
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$begingroup$
The Laplace transforms applies to functions that are assumed to be zero on $t < 0$. The shifted function $f(t-t_0)$ would be zero on $t < t_0$. The time-shifting property says
$$ int_0^infty f^{(n)}(t-t_0)e^{-st}dt = int_{t_0}^infty f^{(n)}(t)e^{-s(t+t_0)}dt = e^{-t_0s} int_0^infty f^{(n)}(t)e^{-st}dt $$
where the last integral is just the Laplace transform of $f^{(n)}(t)$
answered Jan 27 at 7:11
DylanDylan
14.1k31127
$endgroup$
$begingroup$
Is there a way to solve the above ODE using another method, like Fourier transforms then or some other method and does it still give a partial integral? Technically, the solution of my problem is not defined for $x leq 0$, though values could be obtained through interpolation.
$endgroup$
– Baklava Gain
Jan 28 at 16:59
$begingroup$
What is the ODE? I think you should ask a separate question for that.
$endgroup$
– Dylan
Jan 28 at 17:03
$begingroup$
Ok i've edited the question to include the full equation
$endgroup$
– Baklava Gain
Jan 28 at 17:12
$begingroup$
This is called a delay differential equation. It's better to write it as $$ sum_{k=0}^n a_k f^{(k)}(x) = f(x+x_0) $$ You're free to look up different methods of solving it, or ask another question about it. I'm not going to answer it here, as it's now a completely different problem.
$endgroup$
– Dylan
Jan 28 at 18:25
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The Laplace transforms applies to functions that are assumed to be zero on $t < 0$. The shifted function $f(t-t_0)$ would be zero on $t < t_0$. The time-shifting property says
$$ int_0^infty f^{(n)}(t-t_0)e^{-st}dt = int_{t_0}^infty f^{(n)}(t)e^{-s(t+t_0)}dt = e^{-t_0s} int_0^infty f^{(n)}(t)e^{-st}dt $$
where the last integral is just the Laplace transform of $f^{(n)}(t)$
answered Jan 27 at 7:11
DylanDylan
14.1k31127
$endgroup$
$begingroup$
Is there a way to solve the above ODE using another method, like Fourier transforms then or some other method and does it still give a partial integral? Technically, the solution of my problem is not defined for $x leq 0$, though values could be obtained through interpolation.
$endgroup$
– Baklava Gain
Jan 28 at 16:59
$begingroup$
What is the ODE? I think you should ask a separate question for that.
$endgroup$
– Dylan
Jan 28 at 17:03
$begingroup$
Ok i've edited the question to include the full equation
$endgroup$
– Baklava Gain
Jan 28 at 17:12
$begingroup$
This is called a delay differential equation. It's better to write it as $$ sum_{k=0}^n a_k f^{(k)}(x) = f(x+x_0) $$ You're free to look up different methods of solving it, or ask another question about it. I'm not going to answer it here, as it's now a completely different problem.
$endgroup$
– Dylan
Jan 28 at 18:25
add a comment |
$begingroup$
The Laplace transforms applies to functions that are assumed to be zero on $t < 0$. The shifted function $f(t-t_0)$ would be zero on $t < t_0$. The time-shifting property says
$$ int_0^infty f^{(n)}(t-t_0)e^{-st}dt = int_{t_0}^infty f^{(n)}(t)e^{-s(t+t_0)}dt = e^{-t_0s} int_0^infty f^{(n)}(t)e^{-st}dt $$
where the last integral is just the Laplace transform of $f^{(n)}(t)$
answered Jan 27 at 7:11
DylanDylan
14.1k31127
$endgroup$
$begingroup$
Is there a way to solve the above ODE using another method, like Fourier transforms then or some other method and does it still give a partial integral? Technically, the solution of my problem is not defined for $x leq 0$, though values could be obtained through interpolation.
$endgroup$
– Baklava Gain
Jan 28 at 16:59
$begingroup$
What is the ODE? I think you should ask a separate question for that.
$endgroup$
– Dylan
Jan 28 at 17:03
$begingroup$
Ok i've edited the question to include the full equation
$endgroup$
– Baklava Gain
Jan 28 at 17:12
$begingroup$
This is called a delay differential equation. It's better to write it as $$ sum_{k=0}^n a_k f^{(k)}(x) = f(x+x_0) $$ You're free to look up different methods of solving it, or ask another question about it. I'm not going to answer it here, as it's now a completely different problem.
$endgroup$
– Dylan
Jan 28 at 18:25
add a comment |
$begingroup$
The Laplace transforms applies to functions that are assumed to be zero on $t < 0$. The shifted function $f(t-t_0)$ would be zero on $t < t_0$. The time-shifting property says
$$ int_0^infty f^{(n)}(t-t_0)e^{-st}dt = int_{t_0}^infty f^{(n)}(t)e^{-s(t+t_0)}dt = e^{-t_0s} int_0^infty f^{(n)}(t)e^{-st}dt $$
where the last integral is just the Laplace transform of $f^{(n)}(t)$
answered Jan 27 at 7:11
DylanDylan
14.1k31127
$endgroup$
The Laplace transforms applies to functions that are assumed to be zero on $t < 0$. The shifted function $f(t-t_0)$ would be zero on $t < t_0$. The time-shifting property says
$$ int_0^infty f^{(n)}(t-t_0)e^{-st}dt = int_{t_0}^infty f^{(n)}(t)e^{-s(t+t_0)}dt = e^{-t_0s} int_0^infty f^{(n)}(t)e^{-st}dt $$
where the last integral is just the Laplace transform of $f^{(n)}(t)$
answered Jan 27 at 7:11
DylanDylan
14.1k31127
answered Jan 27 at 7:11
DylanDylan
14.1k31127
answered Jan 27 at 7:11
DylanDylan
14.1k31127
answered Jan 27 at 7:11
DylanDylan
14.1k31127
14.1k31127
$begingroup$
Is there a way to solve the above ODE using another method, like Fourier transforms then or some other method and does it still give a partial integral? Technically, the solution of my problem is not defined for $x leq 0$, though values could be obtained through interpolation.
$endgroup$
– Baklava Gain
Jan 28 at 16:59
$begingroup$
What is the ODE? I think you should ask a separate question for that.
$endgroup$
– Dylan
Jan 28 at 17:03
$begingroup$
Ok i've edited the question to include the full equation
$endgroup$
– Baklava Gain
Jan 28 at 17:12
$begingroup$
This is called a delay differential equation. It's better to write it as $$ sum_{k=0}^n a_k f^{(k)}(x) = f(x+x_0) $$ You're free to look up different methods of solving it, or ask another question about it. I'm not going to answer it here, as it's now a completely different problem.
$endgroup$
– Dylan
Jan 28 at 18:25
add a comment |
$begingroup$
Is there a way to solve the above ODE using another method, like Fourier transforms then or some other method and does it still give a partial integral? Technically, the solution of my problem is not defined for $x leq 0$, though values could be obtained through interpolation.
$endgroup$
– Baklava Gain
Jan 28 at 16:59
$begingroup$
What is the ODE? I think you should ask a separate question for that.
$endgroup$
– Dylan
Jan 28 at 17:03
$begingroup$
Ok i've edited the question to include the full equation
$endgroup$
– Baklava Gain
Jan 28 at 17:12
$begingroup$
This is called a delay differential equation. It's better to write it as $$ sum_{k=0}^n a_k f^{(k)}(x) = f(x+x_0) $$ You're free to look up different methods of solving it, or ask another question about it. I'm not going to answer it here, as it's now a completely different problem.
$endgroup$
– Dylan
Jan 28 at 18:25
$begingroup$
Is there a way to solve the above ODE using another method, like Fourier transforms then or some other method and does it still give a partial integral? Technically, the solution of my problem is not defined for $x leq 0$, though values could be obtained through interpolation.
$endgroup$
– Baklava Gain
Jan 28 at 16:59
$begingroup$
Is there a way to solve the above ODE using another method, like Fourier transforms then or some other method and does it still give a partial integral? Technically, the solution of my problem is not defined for $x leq 0$, though values could be obtained through interpolation.
$endgroup$
– Baklava Gain
Jan 28 at 16:59
$begingroup$
What is the ODE? I think you should ask a separate question for that.
$endgroup$
– Dylan
Jan 28 at 17:03
$begingroup$
What is the ODE? I think you should ask a separate question for that.
$endgroup$
– Dylan
Jan 28 at 17:03
$begingroup$
Ok i've edited the question to include the full equation
$endgroup$
– Baklava Gain
Jan 28 at 17:12
$begingroup$
Ok i've edited the question to include the full equation
$endgroup$
– Baklava Gain
Jan 28 at 17:12
$begingroup$
This is called a delay differential equation. It's better to write it as $$ sum_{k=0}^n a_k f^{(k)}(x) = f(x+x_0) $$ You're free to look up different methods of solving it, or ask another question about it. I'm not going to answer it here, as it's now a completely different problem.
$endgroup$
– Dylan
Jan 28 at 18:25
$begingroup$
This is called a delay differential equation. It's better to write it as $$ sum_{k=0}^n a_k f^{(k)}(x) = f(x+x_0) $$ You're free to look up different methods of solving it, or ask another question about it. I'm not going to answer it here, as it's now a completely different problem.
$endgroup$
– Dylan
Jan 28 at 18:25
add a comment |
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