From the series: Differential Equations and Linear Algebra
Gilbert Strang, Massachusetts Institute of Technology (MIT)
Resonance occurs when the natural frequency matches the forcing frequency — equal exponents from inside and outside.
Well, you see spring has finally come to Boston. My sweater is gone and it's April the 16th, I think. It's getting late spring. So today, this video is the nice case, constant coefficient, linear equations, and the right hand side is an exponential. Those are the best.
And we've seen that before. In fact, let me extend, we saw it for first order equations, here it is for second order equations, and it could be an nth order equation. We could have the nth derivative and all lower derivatives. The first derivative, the function itself, 0-th derivative with coefficients, constant coefficients, equalling e to the st. That's what makes it easy.
And what do we do when the right hand side is e to the st? We look for a solution, a multiple of e to the st. Capital Y times e to the st, that's going to work. We just plug into the equation to find that transfer function capital Y. Can I just do that?
I'll do it for the nth degree. Why not? So will I plug this in for y, every derivative brings an s. Capital Y is still there, the exponential is going to be still there, and then there are all the s's that come down from the derivatives, and s's from the nth derivative. One s from the first derivative, no s from the constant term.
Do you see that equation is exactly like what we had before with as squared plus Bs plus C. We have quadratic equation, the most important case. Now, I'm including that with any degree equation, nth degree equation. And what's the solution for Y?
Because e to the st cancels e to the st. That whole thing equals 1. I divide by this and I get Y equal 1 over that key polynomial. It's a nth degree polynomial. And the 1 over it is called the transfer function. And that transfer function transfers the input-- e to the st-- to the output-- Ye to the st. It gives the exponential response.
Very nice formula. Couldn't be better. And you remember for second degree equations, our most important case is as squared Bs plus C. That's the solution almost every time. But one thing can go wrong.
One thing can go wrong. Suppose for the particular s, the particular exponent, in the forcing function, suppose that s in the forcing function is also one of the s's in the no solutions. You remember the no solutions, there are two s's-- s1 and s2-- that make this 0, that make that 0. Those are the s1 and s2 that go into the no solutions.
Now, if the forcing s is one of those no solution s's, we have a problem. Because this is 1 over 0 and we haven't got an answer yet. 1 over 0 has no meaning. So I have to, this is called resonance. Resonance is when the forcing exponent is one of the no exponents that make this 0.
And there are two of those for second order equations and there will be n different s's-- s1, s2, up to sn-- for nth degree equations. Those special s's, I could also call them poles of the transfer function. The transfer function has this in the denominator and when this is 0, that identifies a pole. So the s1, s2, to sn are the poles and we hope that this s is not one of those, but it could be.
And if it is, we need a new formula. So that's the only remaining case. This is a completely nice picture. We just need this last case with some resonance when s equals say s1. I'll just pick s1 and when A, I know that As 1 squared plus Bs 1 plus C is 0. So Y would be 1 over 0. And we can't live with that.
I've written here for the second degree equation same possibility for the nth degree, An s1 to the nth plus A0 equals 0. That would be a problem of resonance. In the nth degree equation, this gives us resonance, you see, because remember, the no solutions were e to the s1t was a no solution.
If I plug-in e to the s1t, the left side will give 0. So I can't get for equal to a forcing term on the right side. I need a new solution. I need a new y of t.
Can I tell you what it is? It's a typical case of L'Hopital's rule from calculus when we approach this bad situation, and we are getting a 1 over 0. Well, you'll see a 0 over 0 and that's L'Hopital's aim for. So where do we start?
A Y particular solution is this e to the st over this As squared plus Bs plus C. Right. That's our particular solution. If it works. Resonance is the case when this doesn't work because that's 0.
Now, that's a particular solution. I'll subtract off a no solution. I could do that. I still have a solution. So I subtract off e to the s1t. So S1 is, e to the s1t is a no solution. This is what I would call a very particular solution.
It's very particular because a t equals 0, it's 0. Do you see that, you see what's happening here. The question, resonance happens when s approaches s1. Resonance is s equal to, resonance itself is at the thing.
Now, we let s sneak up on s1 and we ask what happens to that formula. You see, we're sneaking up on resonance. At resonance, when s equals s1, that will be 0 and that will be 0. That's our problem. So approach it and you end up with the derivative of this divided by the derivative of this.
Do you remember L'Hopital? It was a crazy rule in calculus, but here it's actually needed. So as s goes to s1 this goes to 0 over 0, so I have to take derivative over derivative. So let me write the answer. Y resonant.
Can I call this the resonant solution when s equals s1. And what does it equal? Well, I take the s derivative of this and divide it by the s derivative of that. Derivative over derivative. The s derivative of that is te to the st. And the s derivative of this is 2As plus B.
And now, I have derivative over derivative, I can let s go to s1. So s goes to s1, this goes to s1, and I get an answer. The right answer. This is the correct solution and you notice everybody spots this t factor. That t factor is a signal to everyone that we're in a special case when two things happen to be equal.
Here the two things are the s and the s1. So that will work. So do you see the general picture? It's always this te to the st, t above. And down below we have the derivative of this polynomial at s equal s1. You know it's theoretically possible that we could have double resonance. We would have double resonance if that thing is 0.
If s1 was a double root, if s1 was a double root, then, well, that's just absurd, but it could happen. Then not only is that denominator 0, but after one use of L'Hopital, so we have to drag L'Hopital back from the hospital and say do it again. So we would have a second derivative. I won't write down that solution because it's pretty rare.
So what has this video done? Simply put on record the simplest case possible with a forcing function, e to the st. And above all, we've identified this transfer function. And let me just anticipate that if we need another way to solve these equations, instead of in the t domain, we could go to the Laplace transform in the s domain. We could solve it in the s domain.
And this is exactly what we'll meet when we take the Laplace transform. That will be the Laplace transform, which we have to deal with. So that transfer function is a fundamental, this polynomial tells us, its roots tell us the frequencies, s1, s2, and the no solutions. And then 1 over that tells us the right multiplier in the force solution.
So constant coefficients, exponential forcing, the best case possible. 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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