Mixing
Solving a difficult differential equation
$begingroup$
Well, I've to solve the following DE:
$$y(t)=x(t)cdottext{a}+text{b}cdotlnleft(1+frac{x(t)}{text{c}}right)+int_0^tx(tau)cdot p(t-tau)spacetext{d}tautag1$$
And I've no idea how to start.
Some background information: this DE describes a current in an electric circuit. For the values of $a,b$ and $c$ I know that they are real and positive. For $c$ I know that $10^{-16}le cle10^{-4}$. The function $y(t)=kcdotthetaleft(t-mright)+(n-k)cdotthetaleft(t-vright)$ where all the values of the constants are ral and positive and $theta$ is the Heaviside Theta function. The function $p(t-tau)=mathcal{L}_text{s}^{-1}left{frac{1}{frac{1}{R_2+sl}+frac{s}{sR_3+frac{1}{z}}}right}_{left(t-tauright)}$ where all the constants are again real and positive.
integration ordinary-differential-equations definite-integrals logarithms
asked Jan 17 at 17:38
JanJan
21.9k31240
$endgroup$
add a comment |
$begingroup$
Well, I've to solve the following DE:
$$y(t)=x(t)cdottext{a}+text{b}cdotlnleft(1+frac{x(t)}{text{c}}right)+int_0^tx(tau)cdot p(t-tau)spacetext{d}tautag1$$
And I've no idea how to start.
Some background information: this DE describes a current in an electric circuit. For the values of $a,b$ and $c$ I know that they are real and positive. For $c$ I know that $10^{-16}le cle10^{-4}$. The function $y(t)=kcdotthetaleft(t-mright)+(n-k)cdotthetaleft(t-vright)$ where all the values of the constants are ral and positive and $theta$ is the Heaviside Theta function. The function $p(t-tau)=mathcal{L}_text{s}^{-1}left{frac{1}{frac{1}{R_2+sl}+frac{s}{sR_3+frac{1}{z}}}right}_{left(t-tauright)}$ where all the constants are again real and positive.
integration ordinary-differential-equations definite-integrals logarithms
asked Jan 17 at 17:38
JanJan
21.9k31240
$endgroup$
2
$begingroup$
I see no differential equation, is $y(t)=x'(t)$ or is it given and you seek to find $x(t)$? Then, I would call it integral equation...
$endgroup$
– Alex
Jan 17 at 21:49
add a comment |
$begingroup$
Well, I've to solve the following DE:
$$y(t)=x(t)cdottext{a}+text{b}cdotlnleft(1+frac{x(t)}{text{c}}right)+int_0^tx(tau)cdot p(t-tau)spacetext{d}tautag1$$
And I've no idea how to start.
Some background information: this DE describes a current in an electric circuit. For the values of $a,b$ and $c$ I know that they are real and positive. For $c$ I know that $10^{-16}le cle10^{-4}$. The function $y(t)=kcdotthetaleft(t-mright)+(n-k)cdotthetaleft(t-vright)$ where all the values of the constants are ral and positive and $theta$ is the Heaviside Theta function. The function $p(t-tau)=mathcal{L}_text{s}^{-1}left{frac{1}{frac{1}{R_2+sl}+frac{s}{sR_3+frac{1}{z}}}right}_{left(t-tauright)}$ where all the constants are again real and positive.
integration ordinary-differential-equations definite-integrals logarithms
asked Jan 17 at 17:38
JanJan
21.9k31240
$endgroup$
Well, I've to solve the following DE:
$$y(t)=x(t)cdottext{a}+text{b}cdotlnleft(1+frac{x(t)}{text{c}}right)+int_0^tx(tau)cdot p(t-tau)spacetext{d}tautag1$$
And I've no idea how to start.
Some background information: this DE describes a current in an electric circuit. For the values of $a,b$ and $c$ I know that they are real and positive. For $c$ I know that $10^{-16}le cle10^{-4}$. The function $y(t)=kcdotthetaleft(t-mright)+(n-k)cdotthetaleft(t-vright)$ where all the values of the constants are ral and positive and $theta$ is the Heaviside Theta function. The function $p(t-tau)=mathcal{L}_text{s}^{-1}left{frac{1}{frac{1}{R_2+sl}+frac{s}{sR_3+frac{1}{z}}}right}_{left(t-tauright)}$ where all the constants are again real and positive.
integration ordinary-differential-equations definite-integrals logarithms
integration ordinary-differential-equations definite-integrals logarithms
asked Jan 17 at 17:38
JanJan
21.9k31240
asked Jan 17 at 17:38
JanJan
21.9k31240
asked Jan 17 at 17:38
JanJan
21.9k31240
asked Jan 17 at 17:38
JanJan
21.9k31240
asked Jan 17 at 17:38
JanJan
21.9k31240
21.9k31240
2
$begingroup$
I see no differential equation, is $y(t)=x'(t)$ or is it given and you seek to find $x(t)$? Then, I would call it integral equation...
$endgroup$
– Alex
Jan 17 at 21:49
add a comment |
2
$begingroup$
I see no differential equation, is $y(t)=x'(t)$ or is it given and you seek to find $x(t)$? Then, I would call it integral equation...
$endgroup$
– Alex
Jan 17 at 21:49
2
2
$begingroup$
I see no differential equation, is $y(t)=x'(t)$ or is it given and you seek to find $x(t)$? Then, I would call it integral equation...
$endgroup$
– Alex
Jan 17 at 21:49
$begingroup$
I see no differential equation, is $y(t)=x'(t)$ or is it given and you seek to find $x(t)$? Then, I would call it integral equation...
$endgroup$
– Alex
Jan 17 at 21:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I almost certainly sure, that this integral equation has no analytic solution, since log makes it nonlinear. However, numerically speaking it's not that hard to solve.
First, we need to reverse non-linear part of the equation $G(x) = ax+bln(1+x/c)$. Since both terms are monotonously increasing, there's a unique inverse $G^{-1}(x)$.
So no we can, for example, find $x(0) = G^{-1}(y(0))$. It is an important step to use in solving the integral equation as Cauchy equation.
We take derivative from both sides. I will use $y'(t)=0$, since it's true almost everywhere, except for jumps in $t=m,v$, which I discuss later.
$$
0=ax'(t)+frac{bx'(t)}{c+x(t)}+x(t)p(0)+int_0^tx(tau)p'(t-tau)dtau,\
x'(t)=-left(a+frac{b}{c+x(t)}right)^{-1}left(x(t)p(0)+int_0^tx(tau)p'(t-tau)dtauright).
$$
We expressed the derivative $x'(t)$ as a function of $t,x(t)$, so we can use any method for solving ODE (it would be longer than usual, since integral, but still very doable on modern machines). You can precalculate
$$p'(t-tau)=mathcal L^{-1}_sleft{mathcal L_s p'(t-tau) right}_{t-tau}=
mathcal L^{-1}_sleft{mathcal s P(s)-p(0^{+}) right}_{t-tau}=\
mathcal L^{-1}_sleft{frac{s}{frac{1}{R_2+ldots}+ldots} right}_{t-tau} - p(0^+)delta(t-tau)$$
and since $0<tau<t$, we can neglect Dirac delta.
The only thing left to discuss is what to do with jumps in $y(t)$. I would assume, $m<v$. Say we have found the solution for $x(t)$ from $0$ to $m^{-}$ (right before the jump). Then we can calculate the value $x(m^+)$ (after the jump), using the same technique, we used to calculate $x(0)$:
$$
y(m^+) = G(x(m^+)) + int_0^m x(tau)p(m-tau)dtau,\
x(m^+) = G^{-1}left( y(m^+) - int_0^m x(tau)p(m-tau)dtau right)
$$
So it goes like this: you calculate $x_0=x(0)$, then solve ODE from $0$ to $m$, then calculate value $x_1=x(m^+)$, then solve ODE from $m$ to $v$, then calculate value $x_2=x(v+)$, and finally solve ODE from $v$ to whatever you need.
Edit. I just noticed that your function $p(t-tau)$ has Dirac delta component $R_3delta(t-tau)$, which lies on the end of integration interval. Since you use Laplace transform, I believe the convolution arises from the similar matters. In that case, I believe $int_{0}^t$ is really $int_{0^-}^{t^+}$, so domain of delta function lies inside the integration domain. Thus, you need to add manually $x(t)R_3$ to integrals.
answered Jan 18 at 12:31
Vasily MitchVasily Mitch
2,3141311
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077274%2fsolving-a-difficult-differential-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I almost certainly sure, that this integral equation has no analytic solution, since log makes it nonlinear. However, numerically speaking it's not that hard to solve.
First, we need to reverse non-linear part of the equation $G(x) = ax+bln(1+x/c)$. Since both terms are monotonously increasing, there's a unique inverse $G^{-1}(x)$.
So no we can, for example, find $x(0) = G^{-1}(y(0))$. It is an important step to use in solving the integral equation as Cauchy equation.
We take derivative from both sides. I will use $y'(t)=0$, since it's true almost everywhere, except for jumps in $t=m,v$, which I discuss later.
$$
0=ax'(t)+frac{bx'(t)}{c+x(t)}+x(t)p(0)+int_0^tx(tau)p'(t-tau)dtau,\
x'(t)=-left(a+frac{b}{c+x(t)}right)^{-1}left(x(t)p(0)+int_0^tx(tau)p'(t-tau)dtauright).
$$
We expressed the derivative $x'(t)$ as a function of $t,x(t)$, so we can use any method for solving ODE (it would be longer than usual, since integral, but still very doable on modern machines). You can precalculate
$$p'(t-tau)=mathcal L^{-1}_sleft{mathcal L_s p'(t-tau) right}_{t-tau}=
mathcal L^{-1}_sleft{mathcal s P(s)-p(0^{+}) right}_{t-tau}=\
mathcal L^{-1}_sleft{frac{s}{frac{1}{R_2+ldots}+ldots} right}_{t-tau} - p(0^+)delta(t-tau)$$
and since $0<tau<t$, we can neglect Dirac delta.
The only thing left to discuss is what to do with jumps in $y(t)$. I would assume, $m<v$. Say we have found the solution for $x(t)$ from $0$ to $m^{-}$ (right before the jump). Then we can calculate the value $x(m^+)$ (after the jump), using the same technique, we used to calculate $x(0)$:
$$
y(m^+) = G(x(m^+)) + int_0^m x(tau)p(m-tau)dtau,\
x(m^+) = G^{-1}left( y(m^+) - int_0^m x(tau)p(m-tau)dtau right)
$$
So it goes like this: you calculate $x_0=x(0)$, then solve ODE from $0$ to $m$, then calculate value $x_1=x(m^+)$, then solve ODE from $m$ to $v$, then calculate value $x_2=x(v+)$, and finally solve ODE from $v$ to whatever you need.
Edit. I just noticed that your function $p(t-tau)$ has Dirac delta component $R_3delta(t-tau)$, which lies on the end of integration interval. Since you use Laplace transform, I believe the convolution arises from the similar matters. In that case, I believe $int_{0}^t$ is really $int_{0^-}^{t^+}$, so domain of delta function lies inside the integration domain. Thus, you need to add manually $x(t)R_3$ to integrals.
answered Jan 18 at 12:31
Vasily MitchVasily Mitch
2,3141311
$endgroup$
add a comment |
$begingroup$
I almost certainly sure, that this integral equation has no analytic solution, since log makes it nonlinear. However, numerically speaking it's not that hard to solve.
First, we need to reverse non-linear part of the equation $G(x) = ax+bln(1+x/c)$. Since both terms are monotonously increasing, there's a unique inverse $G^{-1}(x)$.
So no we can, for example, find $x(0) = G^{-1}(y(0))$. It is an important step to use in solving the integral equation as Cauchy equation.
We take derivative from both sides. I will use $y'(t)=0$, since it's true almost everywhere, except for jumps in $t=m,v$, which I discuss later.
$$
0=ax'(t)+frac{bx'(t)}{c+x(t)}+x(t)p(0)+int_0^tx(tau)p'(t-tau)dtau,\
x'(t)=-left(a+frac{b}{c+x(t)}right)^{-1}left(x(t)p(0)+int_0^tx(tau)p'(t-tau)dtauright).
$$
We expressed the derivative $x'(t)$ as a function of $t,x(t)$, so we can use any method for solving ODE (it would be longer than usual, since integral, but still very doable on modern machines). You can precalculate
$$p'(t-tau)=mathcal L^{-1}_sleft{mathcal L_s p'(t-tau) right}_{t-tau}=
mathcal L^{-1}_sleft{mathcal s P(s)-p(0^{+}) right}_{t-tau}=\
mathcal L^{-1}_sleft{frac{s}{frac{1}{R_2+ldots}+ldots} right}_{t-tau} - p(0^+)delta(t-tau)$$
and since $0<tau<t$, we can neglect Dirac delta.
The only thing left to discuss is what to do with jumps in $y(t)$. I would assume, $m<v$. Say we have found the solution for $x(t)$ from $0$ to $m^{-}$ (right before the jump). Then we can calculate the value $x(m^+)$ (after the jump), using the same technique, we used to calculate $x(0)$:
$$
y(m^+) = G(x(m^+)) + int_0^m x(tau)p(m-tau)dtau,\
x(m^+) = G^{-1}left( y(m^+) - int_0^m x(tau)p(m-tau)dtau right)
$$
So it goes like this: you calculate $x_0=x(0)$, then solve ODE from $0$ to $m$, then calculate value $x_1=x(m^+)$, then solve ODE from $m$ to $v$, then calculate value $x_2=x(v+)$, and finally solve ODE from $v$ to whatever you need.
Edit. I just noticed that your function $p(t-tau)$ has Dirac delta component $R_3delta(t-tau)$, which lies on the end of integration interval. Since you use Laplace transform, I believe the convolution arises from the similar matters. In that case, I believe $int_{0}^t$ is really $int_{0^-}^{t^+}$, so domain of delta function lies inside the integration domain. Thus, you need to add manually $x(t)R_3$ to integrals.
answered Jan 18 at 12:31
Vasily MitchVasily Mitch
2,3141311
$endgroup$
add a comment |
$begingroup$
I almost certainly sure, that this integral equation has no analytic solution, since log makes it nonlinear. However, numerically speaking it's not that hard to solve.
First, we need to reverse non-linear part of the equation $G(x) = ax+bln(1+x/c)$. Since both terms are monotonously increasing, there's a unique inverse $G^{-1}(x)$.
So no we can, for example, find $x(0) = G^{-1}(y(0))$. It is an important step to use in solving the integral equation as Cauchy equation.
We take derivative from both sides. I will use $y'(t)=0$, since it's true almost everywhere, except for jumps in $t=m,v$, which I discuss later.
$$
0=ax'(t)+frac{bx'(t)}{c+x(t)}+x(t)p(0)+int_0^tx(tau)p'(t-tau)dtau,\
x'(t)=-left(a+frac{b}{c+x(t)}right)^{-1}left(x(t)p(0)+int_0^tx(tau)p'(t-tau)dtauright).
$$
We expressed the derivative $x'(t)$ as a function of $t,x(t)$, so we can use any method for solving ODE (it would be longer than usual, since integral, but still very doable on modern machines). You can precalculate
$$p'(t-tau)=mathcal L^{-1}_sleft{mathcal L_s p'(t-tau) right}_{t-tau}=
mathcal L^{-1}_sleft{mathcal s P(s)-p(0^{+}) right}_{t-tau}=\
mathcal L^{-1}_sleft{frac{s}{frac{1}{R_2+ldots}+ldots} right}_{t-tau} - p(0^+)delta(t-tau)$$
and since $0<tau<t$, we can neglect Dirac delta.
The only thing left to discuss is what to do with jumps in $y(t)$. I would assume, $m<v$. Say we have found the solution for $x(t)$ from $0$ to $m^{-}$ (right before the jump). Then we can calculate the value $x(m^+)$ (after the jump), using the same technique, we used to calculate $x(0)$:
$$
y(m^+) = G(x(m^+)) + int_0^m x(tau)p(m-tau)dtau,\
x(m^+) = G^{-1}left( y(m^+) - int_0^m x(tau)p(m-tau)dtau right)
$$
So it goes like this: you calculate $x_0=x(0)$, then solve ODE from $0$ to $m$, then calculate value $x_1=x(m^+)$, then solve ODE from $m$ to $v$, then calculate value $x_2=x(v+)$, and finally solve ODE from $v$ to whatever you need.
Edit. I just noticed that your function $p(t-tau)$ has Dirac delta component $R_3delta(t-tau)$, which lies on the end of integration interval. Since you use Laplace transform, I believe the convolution arises from the similar matters. In that case, I believe $int_{0}^t$ is really $int_{0^-}^{t^+}$, so domain of delta function lies inside the integration domain. Thus, you need to add manually $x(t)R_3$ to integrals.
answered Jan 18 at 12:31
Vasily MitchVasily Mitch
2,3141311
$endgroup$
I almost certainly sure, that this integral equation has no analytic solution, since log makes it nonlinear. However, numerically speaking it's not that hard to solve.
First, we need to reverse non-linear part of the equation $G(x) = ax+bln(1+x/c)$. Since both terms are monotonously increasing, there's a unique inverse $G^{-1}(x)$.
So no we can, for example, find $x(0) = G^{-1}(y(0))$. It is an important step to use in solving the integral equation as Cauchy equation.
We take derivative from both sides. I will use $y'(t)=0$, since it's true almost everywhere, except for jumps in $t=m,v$, which I discuss later.
$$
0=ax'(t)+frac{bx'(t)}{c+x(t)}+x(t)p(0)+int_0^tx(tau)p'(t-tau)dtau,\
x'(t)=-left(a+frac{b}{c+x(t)}right)^{-1}left(x(t)p(0)+int_0^tx(tau)p'(t-tau)dtauright).
$$
We expressed the derivative $x'(t)$ as a function of $t,x(t)$, so we can use any method for solving ODE (it would be longer than usual, since integral, but still very doable on modern machines). You can precalculate
$$p'(t-tau)=mathcal L^{-1}_sleft{mathcal L_s p'(t-tau) right}_{t-tau}=
mathcal L^{-1}_sleft{mathcal s P(s)-p(0^{+}) right}_{t-tau}=\
mathcal L^{-1}_sleft{frac{s}{frac{1}{R_2+ldots}+ldots} right}_{t-tau} - p(0^+)delta(t-tau)$$
and since $0<tau<t$, we can neglect Dirac delta.
The only thing left to discuss is what to do with jumps in $y(t)$. I would assume, $m<v$. Say we have found the solution for $x(t)$ from $0$ to $m^{-}$ (right before the jump). Then we can calculate the value $x(m^+)$ (after the jump), using the same technique, we used to calculate $x(0)$:
$$
y(m^+) = G(x(m^+)) + int_0^m x(tau)p(m-tau)dtau,\
x(m^+) = G^{-1}left( y(m^+) - int_0^m x(tau)p(m-tau)dtau right)
$$
So it goes like this: you calculate $x_0=x(0)$, then solve ODE from $0$ to $m$, then calculate value $x_1=x(m^+)$, then solve ODE from $m$ to $v$, then calculate value $x_2=x(v+)$, and finally solve ODE from $v$ to whatever you need.
Edit. I just noticed that your function $p(t-tau)$ has Dirac delta component $R_3delta(t-tau)$, which lies on the end of integration interval. Since you use Laplace transform, I believe the convolution arises from the similar matters. In that case, I believe $int_{0}^t$ is really $int_{0^-}^{t^+}$, so domain of delta function lies inside the integration domain. Thus, you need to add manually $x(t)R_3$ to integrals.
answered Jan 18 at 12:31
Vasily MitchVasily Mitch
2,3141311
edited Jan 18 at 12:40
answered Jan 18 at 12:31
Vasily MitchVasily Mitch
2,3141311
answered Jan 18 at 12:31
Vasily MitchVasily Mitch
2,3141311
answered Jan 18 at 12:31
Vasily MitchVasily Mitch
2,3141311
2,3141311
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077274%2fsolving-a-difficult-differential-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
I see no differential equation, is $y(t)=x'(t)$ or is it given and you seek to find $x(t)$? Then, I would call it integral equation...
$endgroup$
– Alex
Jan 17 at 21:49