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The denominator can be simplified using the half-angle formula to 2 ln(b sech(b x + d)), where b=sqrt(a)/2 and d=c/2. This DE has an important application in plasma physics. It comes up when finding equilibrium solutions to the Vlasov-Maxwell equations, where y is the logarithm of the density profile. This particular solution is called a Harris sheet, which applies when the plasma is assumed to be electrically neutral everywhere.
It is a fascinating thing. You can see that it happens very frequently in classical mechanics with the lagrangian/hamiltonian approach, but only under certain conditions (central potentials for example)
@@ingolifs it has to do with the concept of symmetries and conservation laws. For example you can solve the equation for the harmonic oscillator, but it is more interesting see if you can obtain more information of the physical system without solve the equation directly. Then you can try to search for possible symmetries and you find that for example the Harmonic oscillator equation is the same if you make a translation in time. This implies the energy conservation and you can obtain this result without solve the equation. It is a crazy trip, but certainly worth to learn more of this relationships; mathematicians have researched this relationships between differential equations and symmetry groups and related algebras for many years and physicists also make use of this kind of things in order to study specific problems and find properties of physical systems without solve any equation. I find this fascinating and very deep.
@@ingolifssee Pauli’s 1926 article on the spectrum of the hydrogen atom, before the Schrödinger eq. Also: Wigner’s Nobel prize, and ofc Noether’ theorem. And The Eightfold Way. It’s math heavy, though.
You could have done a bit more simplification on the parameter A to get rid of the plus/minus. If we change the A up top to B², then the cosh term becomes cosh(Bx+C), which is a lot nicer. Breaking apart the ln, I think, looks better, and so we get y=ln(B²)-ln(1+cosh(Bx+C))=2ln|B|-ln(1+cosh(Bx+C)).
to solve the first integral 1/sqrt(C-2e^y) you could have used the substitution u=sqrt(C-2e^y) and it’s really easy afterwards since it gives the integral of -2/(C-u^2)
The same differential equation with a parameter p ,that is y'' +p exp[y]=0, supplemented with boundary conditions (say zero b.c.) on an interval e.g. (-d, d) is well studied ,in particular the extension to a two- or threedimensional region , with the Laplacian in the place of the second derivative . It is usually called Gelfand-problem , and it has interesting properties , in particular the dependence on the parameter p. For example , there is no solution if the parameter is larger than a critical value.
Formally there is another solution, when A = 0. If you go all the way, you find: u = -1/2 (x+C)² Now, if we assume y to be a real-valued function on ℝ, u = exp(-y) must be positive, and thus discard this solution, but I think it is worth mentioning it.
I will instead express the solution as: y = ln(A) - ln(1 + cosh(x√A + C)) Because cosh(x) = (e^x + e^-x)/2 is an even function. Also, I will add a constraint such that the solution is valid: A > 0
Oh that is much better indeed.... The constraint however is quite obvious. I mean seriously I've solved so many tough integrals and DEs so far I'm pretty sure people know exactly how the solutions are valid😂
The physical approach may be (and should be) used while considering the DE in form y''+f(y)=0 If the second derivative is separated from the variable and there are no coefficient dependency on x, then the equation describes the movement a particle in potential field, where s.c. "conservation laws" are applicable. In our case, after integration the equation over dy (differential of y) integral(y.''+exp(y), dy) = 0 we'll get the "energy conservation equation" corresponding to initial DE 1/2·y'²+exp(y) = E where E is total energy of the system, K = 1/2·y'² is _kinetic_ energy and P=exp(y) is potential energy.
General a nice solution. However, you where not super rigorous. Technically sqrt(sinh(x)^2) is abs(sinh(x)) and not sinh(x) meaning that you get a sign(x) in the integrand and not a constant. However, because you have that alternating sign outside the integral you can get rid of sign function by matching different signs from the left and right of the discontinuity at 0. So you get the same answer in the end.
the sqrt(a) has to be left as sqrt(a) in the end because it’s not just any constant, it is specifically the constant that is the sqrt of the value in the numerator of the cosh argument, right?
can you do a series of videos on how to solve problems involving floor and ceiling functions. whenever this topic comes up, the methods seem so ad hoc.
Good. Starting with substitution y'=some v and getting the derivatives in terms of v and y would be easier for students to understand (this is a routine trick when the independent var does not appear in the ode). And by the way you have to keep the final root(A) as is, not only because it looks nicer, because it's related to the A in the numerator of the arg of Ln.
Your last comment right at the end that sqrt(A) could just be replaced by another constant is wrong. There is an instance of A in the numerator, so these constants are not independent of each other and you can't freely change sqrt(A) into an independent constant.
Well ofcourse if I were to change sart(A) into another constant I would've transformed the A in the numerator accordingly. But again, that wouldn't look as cool as the solution I got at the end. Like I say my friend: math is alot easier than English 😂....and apparently alot faster too....
Your very first step has a nice physical interpretation. Whatever equation of the type y'' = f(y) can be reinterpreted as Newton's third law of motion, F = m*a. The step multiplying by y' and recognizing the derivatives is akin to inventing the potential and kinetic energy, and you get KE + PE = total energy = const. The kinetic energy is 1/2 y'^2 = 1/2 v^2 (mass = 1), the potential energy is - int^y f(y2) dy2. This always works out in 1D, as "all forces are potential in a single dimension" (you can't take different paths between a and b), whereas in 2+D, even if your equation reads y'' = f(y) for some vector field f only dependent on position y, this only works if the potential, i.e. the integral U(y) = -int^y f(y2) \cdot dy2 is independent of the path taken between the reference point to the final point y. This is of course traditionally checked by taking the curl of the potential field (there's a connection to the path integral independence via Stokes theorem) :P The force is then calculated as - grad U Typical forces of the form f(r) * rhat are all "potential" (zero curl/line integral independent of the path taken, can be expressed as a grad of a scalar potential), so as far as all the two body problems and springs go, we're dandy. P.S.: physicists aren't crazy, the minus sign in the definition of the potential is so that visually, valleys of a potential naturally correspond to regions where kinetic energy must be higher, thus an object "speeds up" in a potential valley and vice versa with peaks.
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Nice resolution. Do you have a link to the original video? I can't find it, but I did find another video that also mentions the original one.
The denominator can be simplified using the half-angle formula to 2 ln(b sech(b x + d)), where b=sqrt(a)/2 and d=c/2.
This DE has an important application in plasma physics. It comes up when finding equilibrium solutions to the Vlasov-Maxwell equations, where y is the logarithm of the density profile. This particular solution is called a Harris sheet, which applies when the plasma is assumed to be electrically neutral everywhere.
It's fascinating to see that solving odes always boils down to either finding Lie groups or conservation laws 😂😅
It is a fascinating thing. You can see that it happens very frequently in classical mechanics with the lagrangian/hamiltonian approach, but only under certain conditions (central potentials for example)
I don't know enough maths to understand what this means. Can I get an explanation for this please?
@@ingolifs it has to do with the concept of symmetries and conservation laws. For example you can solve the equation for the harmonic oscillator, but it is more interesting see if you can obtain more information of the physical system without solve the equation directly. Then you can try to search for possible symmetries and you find that for example the Harmonic oscillator equation is the same if you make a translation in time. This implies the energy conservation and you can obtain this result without solve the equation. It is a crazy trip, but certainly worth to learn more of this relationships; mathematicians have researched this relationships between differential equations and symmetry groups and related algebras for many years and physicists also make use of this kind of things in order to study specific problems and find properties of physical systems without solve any equation. I find this fascinating and very deep.
@@ingolifssee Pauli’s 1926 article on the spectrum of the hydrogen atom, before the Schrödinger eq. Also: Wigner’s Nobel prize, and ofc Noether’ theorem. And The Eightfold Way. It’s math heavy, though.
You could have done a bit more simplification on the parameter A to get rid of the plus/minus. If we change the A up top to B², then the cosh term becomes cosh(Bx+C), which is a lot nicer. Breaking apart the ln, I think, looks better, and so we get y=ln(B²)-ln(1+cosh(Bx+C))=2ln|B|-ln(1+cosh(Bx+C)).
Agreed
to solve the first integral 1/sqrt(C-2e^y) you could have used the substitution u=sqrt(C-2e^y) and it’s really easy afterwards since it gives the integral of -2/(C-u^2)
The same differential equation with a parameter p ,that is y'' +p exp[y]=0, supplemented with boundary conditions (say zero b.c.) on an interval e.g. (-d, d) is well studied ,in particular the extension to a two- or threedimensional region , with the Laplacian in the place of the second derivative . It is usually called Gelfand-problem , and it has interesting properties , in particular the dependence on the parameter p.
For example , there is no solution if the parameter is larger than a critical value.
Formally there is another solution, when A = 0. If you go all the way, you find: u = -1/2 (x+C)²
Now, if we assume y to be a real-valued function on ℝ, u = exp(-y) must be positive, and thus discard this solution, but I think it is worth mentioning it.
Nice. I used the same technique for Hooke's Law : y ' ' =- (k^2)y
Multiply both sides by 2y' .
Noice
Nice one, at 2:55 I substituted u=sqrt(2e^y/A) then t=sqrt(1-u^2). Same result but without hyperbolic function
Noice
I will instead express the solution as:
y = ln(A) - ln(1 + cosh(x√A + C))
Because cosh(x) = (e^x + e^-x)/2 is an even function.
Also, I will add a constraint such that the solution is valid: A > 0
Oh that is much better indeed....
The constraint however is quite obvious. I mean seriously I've solved so many tough integrals and DEs so far I'm pretty sure people know exactly how the solutions are valid😂
@@maths_505 Just a habit to include a constraint. Nice to see you solving hard integrals and differential equations.
A very cool solution - great👍
The physical approach may be (and should be) used while considering the DE in form y''+f(y)=0
If the second derivative is separated from the variable and there are no coefficient dependency on x, then the equation describes the movement a particle in potential field, where s.c. "conservation laws" are applicable.
In our case, after integration the equation over dy (differential of y)
integral(y.''+exp(y), dy) = 0
we'll get the "energy conservation equation" corresponding to initial DE
1/2·y'²+exp(y) = E
where E is total energy of the system, K = 1/2·y'² is _kinetic_ energy and P=exp(y) is potential energy.
Very interesting. Thank you.
I loved the structure here because it's very similar to second order DEs in physics representative of conservation of energy
General a nice solution. However, you where not super rigorous. Technically sqrt(sinh(x)^2) is abs(sinh(x)) and not sinh(x) meaning that you get a sign(x) in the integrand and not a constant. However, because you have that alternating sign outside the integral you can get rid of sign function by matching different signs from the left and right of the discontinuity at 0. So you get the same answer in the end.
the sqrt(a) has to be left as sqrt(a) in the end because it’s not just any constant, it is specifically the constant that is the sqrt of the value in the numerator of the cosh argument, right?
can you do a series of videos on how to solve problems involving floor and ceiling functions. whenever this topic comes up, the methods seem so ad hoc.
I've done a couple in the past. I'll add more problems along the way.
what is ad hoc?
Good. Starting with substitution y'=some v and getting the derivatives in terms of v and y would be easier for students to understand (this is a routine trick when the independent var does not appear in the ode). And by the way you have to keep the final root(A) as is, not only because it looks nicer, because it's related to the A in the numerator of the arg of Ln.
Just an idea: would it work to substitute u=exp(y) and integrate twice?
Your last comment right at the end that sqrt(A) could just be replaced by another constant is wrong. There is an instance of A in the numerator, so these constants are not independent of each other and you can't freely change sqrt(A) into an independent constant.
Well ofcourse if I were to change sart(A) into another constant I would've transformed the A in the numerator accordingly. But again, that wouldn't look as cool as the solution I got at the end. Like I say my friend: math is alot easier than English 😂....and apparently alot faster too....
I'm glad it was enough to lure you into commenting 😂
@@maths_505 lol! Algorithm bait 😜
@@zunaidparker 😂😂😂
beautiful
Your very first step has a nice physical interpretation. Whatever equation of the type y'' = f(y) can be reinterpreted as Newton's third law of motion, F = m*a. The step multiplying by y' and recognizing the derivatives is akin to inventing the potential and kinetic energy, and you get KE + PE = total energy = const. The kinetic energy is 1/2 y'^2 = 1/2 v^2 (mass = 1), the potential energy is - int^y f(y2) dy2.
This always works out in 1D, as "all forces are potential in a single dimension" (you can't take different paths between a and b), whereas in 2+D, even if your equation reads y'' = f(y) for some vector field f only dependent on position y, this only works if the potential, i.e. the integral U(y) = -int^y f(y2) \cdot dy2 is independent of the path taken between the reference point to the final point y. This is of course traditionally checked by taking the curl of the potential field (there's a connection to the path integral independence via Stokes theorem) :P The force is then calculated as - grad U
Typical forces of the form f(r) * rhat are all "potential" (zero curl/line integral independent of the path taken, can be expressed as a grad of a scalar potential), so as far as all the two body problems and springs go, we're dandy.
P.S.: physicists aren't crazy, the minus sign in the definition of the potential is so that visually, valleys of a potential naturally correspond to regions where kinetic energy must be higher, thus an object "speeds up" in a potential valley and vice versa with peaks.
Ladies and gentlemen,we've found the son of Einstein.
2:52 I would substitute u^2 = A-2exp(y)
Would a secant trig sub be easier?
Depends on taste....
That's not intresting but awesome equation ❤❤😊
y'' + e^y=0 y"=-e^y y'=1^(-e^y)+1 y=1^1^(-e^y)+2
first you have Y'''=Y' Y'' and then let Y'=W and solve for W. Much easier and quicker than the solution given here.
No it’s not