From the series: Differential Equations and Linear Algebra
Gilbert Strang, Massachusetts Institute of Technology (MIT)
With forcing f = cos(ωt), the particular solution is Y*cos(ωt). But if the forcing frequency equals the natural frequency there is resonance.
This is the second video on second order differential equations, constant coefficients, but now we have a right hand side. And the first one was free harmonic motion with a zero, but now I'm making this motion, I'm pushing this motion, but at a frequency omega. This is my forcing term.
So I think I'm having a forcing frequency, omega, and remember that for this one, for the no solution, there was a natural frequency omega n. It's very important are those close, are those well separated? That governs whether the bridge that you're walking over oscillates too much and eventually falls.
Or in the extreme case, are they equal? If omega n is equal to omega that's called resonance. Let me put that word in. Resonance. When omega equals omega n. And we're not going to deal with today, but you should know that always the formula has an omega minus omega n dividing by that. So if that is 0, if omega equals omega n our formula has to change.
Today, this won't happen. No. So what's the formula? What is yp? I'm looking for a particular solution. That's a nice function and also important in practice. So I would like to hope that the particular solution could be some multiple of that cosine omega t.
And in this problem that's possible. Because if I have a cosine, I've got a cosine on the right hand side, and if that cosine comes here, it's on the left side, and the second derivative of the cosine is, again, a cosine, I'm going to have a match of cosine omega t terms. And then I'll just choose the right number capital Y.
I won't be able to do that when there's a first derivative in there, because the first derivative of cosine will bring in signs. I'll have a mixture of cosines and sines and then I better allow for that mixture. But here I don't have to.
There's the forcing function. Response, this is the forced response. I'd like to get used to that word, response, for the solution. Here's the input, the response is the output. So let me just plug that into the equation and find capital Y.
So here I have m, second derivative is going to be a Y, and second derivative will bring out a minus omega squared times the cosine. And here I have kY is Y times the cosine equal the cosine. I could have a constant there, but the whole thing would be no more interesting, no more difficult than with a 1.
So what do I do? The nice thing is here I have all cosines, so I'm just going to have minus omega squared m and a k. So it's k minus m omega squared. Can I write it that way? Times Y. I'm going to cancel the cosines. That's just a 1. On the side is a 1.
I've canceled the cosine, so I've kept kY. I've kept the 1, and I've kept a minus omega squared mY. So that tells me Y right away. It's just like plugging in an exponential and canceling exponentials all the way along. Here, I'm canceling cosines all the way because every term was a cosine.
So I know Y. So I know the answer. So the final answer is Y(t) is Yn. Well, let me put Y particular first plus Yn. So I've just found Y particular. Y particular is this capital Y cosine omega t. So it's cosine omega t times Y and Y is 1 over this.
Here's goes Y. Down below I have k minus m omega squared. Right? That's what we just found, that particular solution. The capital Y, the multiplying constant, was 1 over that constant. And now comes the C1 cosine of omega nt and the C2 sine of omega nt.
Remember, omega n is different from omega. Actually, this is pretty nice here. I could write that another way so you would see the important here. So remember, what is omega n squared? Can I just remember that omega n squared is k over m. Right? Yup.
k is the same as-- I'm going to put that m up here-- k is the same as m omega n squared. k is the same as m omega n squared and here I'm subtracting m omega squared. You'll see the whole point of resonance or near resonance when the bridge is getting forced buy a frequency close to its resonant frequency.
This difference, omega squared, the difference between the two frequencies squared is in the denominator and will be small and then the effect is large. And if we get those too close, the effect is too large. So we'll see this cosine omega t over this is, I would call, the frequency response is this factor. 1 over m omega n squared minus omega squared.
That's the key multiplier for when the forcing term is a pure frequency, that frequency gets exploded. And now, of course, what are capital C1 and capital C2? We find those from the initial condition. At t equals 0, we put in t equals 0, and that tells us what C1 has to be. And we put in t equals 0 again to match the velocity Y prime at 0, and that tells us C2. Are you OK with that?
Just look at the beauty of that solution. This is null part. This is the forced part, the particular part, the cosine divided by that constant. There's one more equation, one more forcing term I'd like often and always and now to discuss. And that is a delta function, an impulse.
So I'm going to add one more example. my double prime plus ky equal the delta function. Delta function. It's called an impulse. So I'd like to solve that equation also. When the forcing term just happens at one second, at the initial second. At t equals 0, the delta function, I'm hitting the spring.
So the spring is sitting or the pendulum is sitting there. Actually, let's set it at rest. Here's my pendulum. I'll try to draw a pendulum. I don't know. That's not much of a pendulum. But it's good enough.
This equation says what happens if I hit it with a point source? At t equals 0, I hit it but I give it a finite velocity. It doesn't move in that instant second. This is where delta functions come in so let me give you the result of what happens and then we'll see them again.
So what am I doing? I want to solve this equation when the forcing function is a delta function. So I'm going to call y the impulse response. It's the solution that comes when the forcing function is an impulse. So y is the impulse response. In fact, it's so important, I'm going to give it its own letter. g. Now, can I turn that y into a g?
So that g is g of t is the impulse response. If I can solve that equation. You might say, not so easy. With a delta function, it's not even a genuine function. It's a little bit crazy. It all happens in one second. I'm sorry, in one instant. Not over one second, but one moment.
But I can solve it. I can solve it for this reason. I can think of it as an impulse here or I have an option, another way which clearly I can think of it as solving it with no force mg double prime. Same problem, same solution is 0.
But I start from rest. Nothing's happening. y of 0 is 0. And it starts from an initial velocity, y prime of 0. The impulse starts it out like a golf ball. Just go. And there's a 1 over m there. I'll discuss that another time.
What I want to see now is that I have either this somewhat mysterious equation or this totally normal equation, even a no equation starting from y of 0 equals 0. But with an initial velocity that the impulse gave to the system. And I should be calling this g. This is the g. We'll see impulse responses again, but let's see it this time by solving this equation.
So I plan to solve that equation and actually we solved it last time. You remember the solution to this one? When it starts from 0, there's no cosine. But when the initial velocity is 1 over m, there is a sign. So I'm going to just write down the g of t, which is just sine of omega nt.
And why is it the natural frequency? Because I'm solving the no. I'm looking for a no solution. But the previous video on no solutions gets me this. Only I have to divide by, get 1 over m as the initial velocity. You'll see that that will solve the no equation.
This is what happens to the pendulum or the golf ball. Well, pendulum much better. Actually, golf ball is poor example. Sorry about that. Golf balls don't swing back and forth. They tend to go.
I'm looking at pendulums, springs going up and down. So the spring starts out, has an initial velocity of 1 over m and then after that nothing happens. So that is the impulse response. The response to an impulse. And why do I like that? First of all, its beautiful. Simple answer.
Secondly, every forcing function, and the output comes from this one. We'll see that point. So we've introduced forcing functions, cos omega t, where the particular solution was a multiple of cos omega t. And now, we've introduced a forcing function delta, the delta function where the response is a sine function. Thank you.
1.1: Overview of Differential Equations Linear equations include dy/dt = y, dy/dt = –y, dy/dt = 2ty . The equation dy/dt = y *y is nonlinear.
1.2: The Calculus You Need The sum rule, product rule, and chain rule produce new derivatives from the derivatives of xn , sin(x ) and ex . The Fundamental Theorem of Calculus says that the integral inverts the derivative.
1.4b: Response to Exponential Input, exp(s*t) With exponential input, est , from outside and exponential growth, eat , from inside, the solution, y(t), is a combination of two exponentials.
1.4c: Response to Oscillating Input, cos(w*t) An oscillating input cos(ωt ) produces an oscillating output with the same frequency ω (and a phase shift).
1.4d: Solution for Any Input, q(t) To solve a linear first order equation, multiply each input q(s) by its growth factor and integrate those outputs.
1.4e: Step Function and Delta Function A unit step function jumps from 0 to 1. Its slope is a delta function: zero everywhere except infinite at the jump.
1.5: Response to Complex Exponential, exp(i*w*t) = cos(w*t)+i*sin(w*t) For linear equations, the solution for f = cos(ωt ) is the real part of the solution for f = eiωt . That complex solution has magnitude G (the gain).
1.6: Integrating Factor for a Constant Rate, a The integrating factor e-at multiplies the differential equation, y’=ay+q, to give the derivative of e-at y: ready for integration.
1.6b: Integrating Factor for a Varying Rate, a(t) The integral of a varying interest rate provides the exponent in the growing solution (the bank balance).
1.7: The Logistic Equation When –by2 slows down growth and makes the equation nonlinear, the solution approaches a steady state y( ∞) = a/b.
1.7c: The Stability and Instability of Steady States Steady state solutions can be stable or unstable – a simple test decides.
2.1: Second Order Equations For the oscillation equation with no damping and no forcing, all solutions share the same natural frequency.
2.1b: Forced Harmonic Motion With forcing f = cos(ωt ), the particular solution is Y *cos(ωt ). But if the forcing frequency equals the natural frequency there is resonance.
2.3: Unforced Damped Motion With constant coefficients in a differential equation, the basic solutions are exponentials est . The exponent s solves a simple equation such as As2 + Bs + C = 0 .
2.3c: Impulse Response and Step Response The impulse response g is the solution when the force is an impulse (a delta function). This also solves a null equation (no force) with a nonzero initial condition.
2.4: Exponential Response - Possible Resonance Resonance occurs when the natural frequency matches the forcing frequency — equal exponents from inside and outside.
2.4b: Second Order Equations With Damping A damped forced equation has a particular solution y = G cos(ωt – α). The damping ratio provides insight into the null solutions.
2.5: Electrical Networks: Voltages and Currents Current flowing around an RLC loop solves a linear equation with coefficients L (inductance), R (resistance), and 1/C (C = capacitance).
2.6: Methods of Undetermined Coefficients With constant coefficients and special forcing terms (powers of t , cosines/sines, exponentials), a particular solution has this same form.
2.6b: An Example of Method of Undetermined Coefficients This method is also successful for forces and solutions such as (at2 + bt +c) est : substitute into the equation to find a, b, c .
2.6c: Variations of Parameters Combine null solutions y1 and y2 with coefficients c1(t) and c2(t) to find a particular solution for any f(t).
2.7: Laplace Transform: First Order Equation Transform each term in the linear differential equation to create an algebra problem. You can then transform the algebra solution back to the ODE solution, y(t) .
2.7b: Laplace Transform: Second Order Equation The second derivative transforms to s2Y and the algebra problem involves the transfer function 1/ (As2 + Bs +C).
3.1: Pictures of the Solutions The direction field for dy/dt = f(t,y) has an arrow with slope f at each point t, y . Arrows with the same slope lie along an isocline.
3.2: Phase Plane Pictures: Source, Sink Saddle Solutions to second order equations can approach infinity or zero. Saddle points contain a positive and also a negative exponent or eigenvalue.
3.2b: Phase Plane Pictures: Spirals and Centers Imaginary exponents with pure oscillation provide a “center” in the phase plane. The point (y, dy/dt) travels forever around an ellipse.
3.2c: Two First Order Equations: Stability A second order equation gives two first order equations for y and dy/dt . The matrix becomes a companion matrix.
3.3: Linearization at Critical Points A critical point is a constant solution Y to the differential equation y’ = f(y) . Near that Y , the sign of df/dy decides stability or instability.
3.3b: Linearization of y'=f(y,z) and z'=g(y,z) With two equations, a critical point has f(Y,Z) = 0 and g(Y,Z) = 0. Near those constant solutions, the two linearized equations use the 2 by 2 matrix of partial derivatives of f and g .
3.3c: Eigenvalues and Stability: 2 by 2 Matrix, A Two equations y’ = Ay are stable (solutions approach zero) when the trace of A is negative and the determinant is positive.
5.1: The Column Space of a Matrix, A An m by n matrix A has n columns each in R m . Capturing all combinations Av of these columns gives the column space – a subspace of R m .
5.4: Independence, Basis, and Dimension Vectors v 1 to v d are a basis for a subspace if their combinations span the whole subspace and are independent: no basis vector is a combination of the others. Dimension d = number of basis vectors.
5.5: The Big Picture of Linear Algebra A matrix produces four subspaces – column space, row space (same dimension), the space of vectors perpendicular to all rows (the nullspace), and the space of vectors perpendicular to all columns.
5.6: Graphs A graph has n nodes connected by m edges (other edges can be missing). This is a useful model for the Internet, the brain, pipeline systems, and much more.
6.1: Eigenvalues and Eigenvectors The eigenvectors x remain in the same direction when multiplied by the matrix (A x = λx ). An n x n matrix has n eigenvalues.
6.2: Diagonalizing a Matrix A matrix can be diagonalized if it has n independent eigenvectors. The diagonal matrix Λis the eigenvalue matrix.
6.3: Solving Linear Systems d y /dt = A y contains solutions y = eλt x where λ and x are an eigenvalue / eigenvector pair for A .
6.4: The Matrix Exponential, exp(A*t) The shortest form of the solution uses the matrix exponential y = eAt y (0) . The matrix eAt has eigenvalues eλt and the eigenvectors of A.
6.4b: Similar Matrices, A and B=M^(-1)*A*M A and B are “similar” if B = M-1AM for some matrix M . B then has the same eigenvalues as A .
6.5: Symmetric Matrices, Real Eigenvalues, Orthogonal Eigenvectors Symmetric matrices have n perpendicular eigenvectors and n real eigenvalues.
7.2: Positive Definite Matrices, S=A'*A A positive definite matrix S has positive eigenvalues, positive pivots, positive determinants, and positive energy vT Sv for every vector v. S = AT A is always positive definite if A has independent columns.
7.2b: Singular Value Decomposition, SVD The SVD factors each matrix A into an orthogonal matrix U times a diagonal matrix Σ (the singular value) times another orthogonal matrix VT : rotation times stretch times rotation.
7.3: Boundary Conditions Replace Initial Conditions A second order equation can change its initial conditions on y(0) and dy/dt(0) to boundary conditions on y(0) and y(1) .
8.1: Fourier Series A Fourier series separates a periodic function F(x) into a combination (infinite) of all basis functions cos(nx) and sin(nx) .
8.1b: Examples of Fourier Series Even functions use only cosines (F(–x) = F(x) ) and odd functions use only sines. The coefficients an and bn come from integrals of F(x) cos(nx ) and F(x) sin(nx ).
8.1c: Fourier Series Solution of Laplace's Equation Inside a circle, the solution u (r , θ) combines rn cos(n θ) and rn sin(n θ). The boundary solution combines all entries in a Fourier series to match the boundary conditions.
8.3: Heat Equation The heat equation ∂u /∂t = ∂2u /∂x2 starts from a temperature distribution u at t = 0 and follows it for t > 0 as it quickly becomes smooth.
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