Solving an DE involving a logarithm
$begingroup$
I've the following DE, describing a physical phenomenon. And the prupose is to solve that DE:
$$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dots$$
The intial conditon is equal to $x(0)=x_0$.
For the constants (because that is maybe important for an approximation):
$r$ can be very large;
$l$ can be very large;
$a$ is round about $0.02526$;
$b$ is very small, round about $300cdot10^{-6}$;
$x_0$ can be very large
I've no idea where to start what so ever. Thanks for any help or ideas
ordinary-differential-equations functions logarithms physics mathematical-physics
asked Dec 30 '18 at 14:33
KlopjasKlopjas
714
$endgroup$
add a comment |
$begingroup$
I've the following DE, describing a physical phenomenon. And the prupose is to solve that DE:
$$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dots$$
The intial conditon is equal to $x(0)=x_0$.
For the constants (because that is maybe important for an approximation):
$r$ can be very large;
$l$ can be very large;
$a$ is round about $0.02526$;
$b$ is very small, round about $300cdot10^{-6}$;
$x_0$ can be very large
I've no idea where to start what so ever. Thanks for any help or ideas
ordinary-differential-equations functions logarithms physics mathematical-physics
asked Dec 30 '18 at 14:33
KlopjasKlopjas
714
$endgroup$
$begingroup$
I think that numerical methods would be required.
$endgroup$
– Claude Leibovici
Dec 30 '18 at 16:36
$begingroup$
@ClaudeLeibovici I've no idea how or why
$endgroup$
– Klopjas
Dec 30 '18 at 16:42
$begingroup$
It's a separable equation, so it's solvable by quadratures in principle, but probably not explicitly in practice since you get a rather nasty integral...
$endgroup$
– Hans Lundmark
Dec 30 '18 at 17:47
add a comment |
$begingroup$
I've the following DE, describing a physical phenomenon. And the prupose is to solve that DE:
$$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dots$$
The intial conditon is equal to $x(0)=x_0$.
For the constants (because that is maybe important for an approximation):
$r$ can be very large;
$l$ can be very large;
$a$ is round about $0.02526$;
$b$ is very small, round about $300cdot10^{-6}$;
$x_0$ can be very large
I've no idea where to start what so ever. Thanks for any help or ideas
ordinary-differential-equations functions logarithms physics mathematical-physics
asked Dec 30 '18 at 14:33
KlopjasKlopjas
714
$endgroup$
I've the following DE, describing a physical phenomenon. And the prupose is to solve that DE:
$$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dots$$
The intial conditon is equal to $x(0)=x_0$.
For the constants (because that is maybe important for an approximation):
$r$ can be very large;
$l$ can be very large;
$a$ is round about $0.02526$;
$b$ is very small, round about $300cdot10^{-6}$;
$x_0$ can be very large
I've no idea where to start what so ever. Thanks for any help or ideas
ordinary-differential-equations functions logarithms physics mathematical-physics
ordinary-differential-equations functions logarithms physics mathematical-physics
asked Dec 30 '18 at 14:33
KlopjasKlopjas
714
asked Dec 30 '18 at 14:33
KlopjasKlopjas
714
asked Dec 30 '18 at 14:33
KlopjasKlopjas
714
asked Dec 30 '18 at 14:33
KlopjasKlopjas
714
asked Dec 30 '18 at 14:33
KlopjasKlopjas
714
714
$begingroup$
I think that numerical methods would be required.
$endgroup$
– Claude Leibovici
Dec 30 '18 at 16:36
$begingroup$
@ClaudeLeibovici I've no idea how or why
$endgroup$
– Klopjas
Dec 30 '18 at 16:42
$begingroup$
It's a separable equation, so it's solvable by quadratures in principle, but probably not explicitly in practice since you get a rather nasty integral...
$endgroup$
– Hans Lundmark
Dec 30 '18 at 17:47
add a comment |
$begingroup$
I think that numerical methods would be required.
$endgroup$
– Claude Leibovici
Dec 30 '18 at 16:36
$begingroup$
@ClaudeLeibovici I've no idea how or why
$endgroup$
– Klopjas
Dec 30 '18 at 16:42
$begingroup$
It's a separable equation, so it's solvable by quadratures in principle, but probably not explicitly in practice since you get a rather nasty integral...
$endgroup$
– Hans Lundmark
Dec 30 '18 at 17:47
$begingroup$
I think that numerical methods would be required.
$endgroup$
– Claude Leibovici
Dec 30 '18 at 16:36
$begingroup$
I think that numerical methods would be required.
$endgroup$
– Claude Leibovici
Dec 30 '18 at 16:36
$begingroup$
@ClaudeLeibovici I've no idea how or why
$endgroup$
– Klopjas
Dec 30 '18 at 16:42
$begingroup$
@ClaudeLeibovici I've no idea how or why
$endgroup$
– Klopjas
Dec 30 '18 at 16:42
$begingroup$
It's a separable equation, so it's solvable by quadratures in principle, but probably not explicitly in practice since you get a rather nasty integral...
$endgroup$
– Hans Lundmark
Dec 30 '18 at 17:47
$begingroup$
It's a separable equation, so it's solvable by quadratures in principle, but probably not explicitly in practice since you get a rather nasty integral...
$endgroup$
– Hans Lundmark
Dec 30 '18 at 17:47
add a comment |
1 Answer
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$begingroup$
$$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dots$$
This is a separable ODE.
$$x'(t)=frac{dx}{dt}=-frac{1}{l}left(rx(t)+alnleft(1+frac{x(t)}{b}right)right)$$
$$dt=-frac{l}{rx+alnleft(1+frac{x}{b}right)}dx$$
$$t=-lintfrac{dx}{rx+alnleft(1+frac{x}{b}right)}$$
With the condition $x(0)=x_0$ :
$$t(x)=-lint_{x_0}^xfrac{dxi}{rxi+alnleft(1+frac{xi}{b}right)}$$
As far as I know there is no closed form of this integral in terms of a finite number of standard functions. The same for the inverse function $x(t)$.
So, the analytic solution is a function defined by an integral. This is very common in practice. One have to proceed with numerical integration.
It is very easy to draw $x(t)$ thanks to usual numerical integration : Draw $t(x)$ from the above integral. Plot the points $(t,x)$ instead of $(x,t)$ , i.e. with $t$ on horizontal axis and $x$ on vertical axis.
To compute the value of $x(t)$ at a given value $t$ , proceed to the numerical integration of $-lint_{x_0}frac{dxi}{rxi+alnleft(1+frac{xi}{b}right)}$ with $xi$ increasing up to reach the specified value $t$ . The value of $xi$ at this point gives the value of $x(t)$ .
answered Dec 30 '18 at 18:08
JJacquelinJJacquelin
44.2k21853
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add a comment |
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$begingroup$
$$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dots$$
This is a separable ODE.
$$x'(t)=frac{dx}{dt}=-frac{1}{l}left(rx(t)+alnleft(1+frac{x(t)}{b}right)right)$$
$$dt=-frac{l}{rx+alnleft(1+frac{x}{b}right)}dx$$
$$t=-lintfrac{dx}{rx+alnleft(1+frac{x}{b}right)}$$
With the condition $x(0)=x_0$ :
$$t(x)=-lint_{x_0}^xfrac{dxi}{rxi+alnleft(1+frac{xi}{b}right)}$$
As far as I know there is no closed form of this integral in terms of a finite number of standard functions. The same for the inverse function $x(t)$.
So, the analytic solution is a function defined by an integral. This is very common in practice. One have to proceed with numerical integration.
It is very easy to draw $x(t)$ thanks to usual numerical integration : Draw $t(x)$ from the above integral. Plot the points $(t,x)$ instead of $(x,t)$ , i.e. with $t$ on horizontal axis and $x$ on vertical axis.
To compute the value of $x(t)$ at a given value $t$ , proceed to the numerical integration of $-lint_{x_0}frac{dxi}{rxi+alnleft(1+frac{xi}{b}right)}$ with $xi$ increasing up to reach the specified value $t$ . The value of $xi$ at this point gives the value of $x(t)$ .
answered Dec 30 '18 at 18:08
JJacquelinJJacquelin
44.2k21853
$endgroup$
add a comment |
$begingroup$
$$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dots$$
This is a separable ODE.
$$x'(t)=frac{dx}{dt}=-frac{1}{l}left(rx(t)+alnleft(1+frac{x(t)}{b}right)right)$$
$$dt=-frac{l}{rx+alnleft(1+frac{x}{b}right)}dx$$
$$t=-lintfrac{dx}{rx+alnleft(1+frac{x}{b}right)}$$
With the condition $x(0)=x_0$ :
$$t(x)=-lint_{x_0}^xfrac{dxi}{rxi+alnleft(1+frac{xi}{b}right)}$$
As far as I know there is no closed form of this integral in terms of a finite number of standard functions. The same for the inverse function $x(t)$.
So, the analytic solution is a function defined by an integral. This is very common in practice. One have to proceed with numerical integration.
It is very easy to draw $x(t)$ thanks to usual numerical integration : Draw $t(x)$ from the above integral. Plot the points $(t,x)$ instead of $(x,t)$ , i.e. with $t$ on horizontal axis and $x$ on vertical axis.
To compute the value of $x(t)$ at a given value $t$ , proceed to the numerical integration of $-lint_{x_0}frac{dxi}{rxi+alnleft(1+frac{xi}{b}right)}$ with $xi$ increasing up to reach the specified value $t$ . The value of $xi$ at this point gives the value of $x(t)$ .
answered Dec 30 '18 at 18:08
JJacquelinJJacquelin
44.2k21853
$endgroup$
add a comment |
$begingroup$
$$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dots$$
This is a separable ODE.
$$x'(t)=frac{dx}{dt}=-frac{1}{l}left(rx(t)+alnleft(1+frac{x(t)}{b}right)right)$$
$$dt=-frac{l}{rx+alnleft(1+frac{x}{b}right)}dx$$
$$t=-lintfrac{dx}{rx+alnleft(1+frac{x}{b}right)}$$
With the condition $x(0)=x_0$ :
$$t(x)=-lint_{x_0}^xfrac{dxi}{rxi+alnleft(1+frac{xi}{b}right)}$$
As far as I know there is no closed form of this integral in terms of a finite number of standard functions. The same for the inverse function $x(t)$.
So, the analytic solution is a function defined by an integral. This is very common in practice. One have to proceed with numerical integration.
It is very easy to draw $x(t)$ thanks to usual numerical integration : Draw $t(x)$ from the above integral. Plot the points $(t,x)$ instead of $(x,t)$ , i.e. with $t$ on horizontal axis and $x$ on vertical axis.
To compute the value of $x(t)$ at a given value $t$ , proceed to the numerical integration of $-lint_{x_0}frac{dxi}{rxi+alnleft(1+frac{xi}{b}right)}$ with $xi$ increasing up to reach the specified value $t$ . The value of $xi$ at this point gives the value of $x(t)$ .
answered Dec 30 '18 at 18:08
JJacquelinJJacquelin
44.2k21853
$endgroup$
$$x(t)cdot r+x'(t)cdot l+acdotlnleft(1+frac{x(t)}{b}right)=0spaceLongleftrightarrowspace x(t)=dots$$
This is a separable ODE.
$$x'(t)=frac{dx}{dt}=-frac{1}{l}left(rx(t)+alnleft(1+frac{x(t)}{b}right)right)$$
$$dt=-frac{l}{rx+alnleft(1+frac{x}{b}right)}dx$$
$$t=-lintfrac{dx}{rx+alnleft(1+frac{x}{b}right)}$$
With the condition $x(0)=x_0$ :
$$t(x)=-lint_{x_0}^xfrac{dxi}{rxi+alnleft(1+frac{xi}{b}right)}$$
As far as I know there is no closed form of this integral in terms of a finite number of standard functions. The same for the inverse function $x(t)$.
So, the analytic solution is a function defined by an integral. This is very common in practice. One have to proceed with numerical integration.
It is very easy to draw $x(t)$ thanks to usual numerical integration : Draw $t(x)$ from the above integral. Plot the points $(t,x)$ instead of $(x,t)$ , i.e. with $t$ on horizontal axis and $x$ on vertical axis.
To compute the value of $x(t)$ at a given value $t$ , proceed to the numerical integration of $-lint_{x_0}frac{dxi}{rxi+alnleft(1+frac{xi}{b}right)}$ with $xi$ increasing up to reach the specified value $t$ . The value of $xi$ at this point gives the value of $x(t)$ .
answered Dec 30 '18 at 18:08
JJacquelinJJacquelin
44.2k21853
answered Dec 30 '18 at 18:08
JJacquelinJJacquelin
44.2k21853
answered Dec 30 '18 at 18:08
JJacquelinJJacquelin
44.2k21853
answered Dec 30 '18 at 18:08
JJacquelinJJacquelin
44.2k21853
44.2k21853
add a comment |
add a comment |
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$begingroup$
I think that numerical methods would be required.
$endgroup$
– Claude Leibovici
Dec 30 '18 at 16:36
$begingroup$
@ClaudeLeibovici I've no idea how or why
$endgroup$
– Klopjas
Dec 30 '18 at 16:42
$begingroup$
It's a separable equation, so it's solvable by quadratures in principle, but probably not explicitly in practice since you get a rather nasty integral...
$endgroup$
– Hans Lundmark
Dec 30 '18 at 17:47