From the series: Differential Equations and Linear Algebra
Gilbert Strang, Massachusetts Institute of Technology (MIT)
A damped forced equation has a particular solution y = G cos(ωt – α). The damping ratio provides insight into the null solutions.
I'm coming back to the number one example, but not the easiest example, of a second order equation with an oscillating forcing term, cosine omega t. We have to know the answer to this problem. And it's a little messy, but the method is not messy. The method is straightforward.
So let me begin by looking for the rectangular form. I call this the rectangular form. It separates the cosine with its amplitude and the sine with its amplitude into two separate pieces. So if I'm looking for that solution, and m and n are the numbers I want to find, how do I proceed?
It's a case of undetermined coefficients, M and N. And the way to determine them is substitute this into the equation and match the cosine term and find M and N. And the way we find M and N, we need two equations for two quantities, M and N.
And imagine this substituted in there. I'll get some cosines. So the cosines on one side will match the cosine on the other side. And also from the derivative, I'll get some sines and they should match 0 because I have no sine omega t on the right hand side.
So I have two equations, matching the sines, matching the cosines. And I solve those. Two equations, two unknowns. And I just write the answer down.
M involves a C minus omega squared. M is coming from the cosines. And we get cosines from that term and that term. Divided by some number, D, that I'll write down.
And N is just B omega divided by that same D. And now I'll write down D. That's C minus A omega squared squared plus B omega squared. This is what comes out from the two equations for M and N.
I just solve those equations. This D here is the two by two determinant if we think about the linear algebra behind two equations. And that's what it is. And so the answer now is in terms of A, C, B, and D, which is a mixture of all of A, B, and C. That's the solution.
Only I always want to show you a different form of the solution. And in this case, a better form. Because the most important physical quantity is the magnitude. How large does y get? What is the amplitude of this?
This is a sinusoid. And we remember that every sinusoid can be written in a polar form. Says that y of t is some amplitude of G, the gain, times a cosine of omega t with a shift, with a lag, with an angle alpha. So I have two numbers now.
That's the gain. And this is the phase shift alpha. And that's an attractive form because it has only one term. The two numbers, G and alpha, get put into a single term where we can see the magnitude of the oscillation.
And what does that come out to be? I won't go through all the steps. I'll just write down what G turns out to be. G turns out to be-- it comes from there-- and it's 1 over the square root of D. Well, G is the square root of M squared plus N squared.
The square root of M squared plus N squared. And if I put M squared and N squared, then I have D over D squared. I get that answer. That's the gain.
Let me write that word, gain, again. Because you got it there. Here it is again. And as always, the tangent of alpha is the N over the M, which is just B omega over C minus A omega squared. I like that polar form.
And I feel I should just do an example. I didn't do any of the algebra in this video. But you know where the algebra came from. It came from substituting the form we expect for the solution. And of course, that form that we expect is the form we get provided omega, the driving frequency, is different from omega N.
Well, no. I guess we're all right even if omega is omega N, because we have a damping term. So that's the answer.
So an example. Why not an example? y double prime plus y prime plus 2y equals cosine of t. That's a simple example. I took omega to be 1, you see. And there is omega. And then A is 1, B is 1, C is 2.
We can evaluate everything. In fact, I think M and N are 1/2. D, by the way, will be 1 squared plus 1 squared. That's 2 square root. Sorry. D will be 2. 1 squared plus 1 squared.
So what do I know? Do I know the rectangular form? Yes. Rectangular form is 1/2. 1/2 for both the cosine and the sine. 1/2 of cosine t plus sine t. That's the rectangular form.
Two simple things, but I have to add them. And in my mind, I don't necessarily see how the cosine adds to the sine. But the sinusoidal identity, the polar form, gives it to me. So what is it in polar form?
So G, the gain, is going to be 1 over the square root of 2. At the highest point, the cosine and the sine are the same. They're both 1 over the square root of 2. I have two of them. So I get 1 over the square root of 2 cosine of t minus pi over 4 is the angle, the phase lag.
When I add the cosine and the sine, I get a sinusoid that's sitting over pi over 4, 45 degrees. So those are the two forms. So in a nice example, we certainly got a nice answer. We certainly did. Yes.
So that is the-- worked out, more or less worked out, in principle, worked out-- is the solution to what I think of as the most important application when the forcing term is a cosine. So it gives oscillating motion. It gives a phase shift. And it gives these formulas.
The only thing I would add is that I need to comment on better notation. So I have used in these formulas A, B, and C. But those have meaning as mass, damping constant, spring constant. M, B, and K.
And it's combinations of those that come in. So let me just take this moment to say better notation. Or maybe I should say engineering notation instead of A, B, C, which are mass, damping, spring constant.
Well, that's already better to use letters that have a meaning. But the small but very important point is that two combinations of A, B, C, M, B, K are especially good. One is the natural frequency that we've seen already, square root of C over A.
Square root of K over M. So that gives us one important combination of A and C. And the other one is the damping ratio. And it's called zeta. And that damping ratio is B over the square root of 4ac.
Ha! You'll say, where does that come from? Or I can use these letters, B over the square root of 4mk. That damping ratio is, so to speak, it's the right dimensionless quantity. The dimensions of this ratio are just numbers.
Those two quantities have the same dimension. And we can see that because in the quadratic formula comes-- you remember that in a quadratic formula comes the square root of b squared minus 4ac? Now if you see a formula that has b squared minus 4ac in it, you know that these must have the same units.
Otherwise, subtraction would be a crime. So they have the same ratio and the same units and therefore the ratio is dimensionless. Let me write that word. Dimensionless.
So conclusion. I could rewrite the answer in terms of these quantities omega n and zeta. I won't do that here. That can wait for another time.
But just to say since we've found a solution to the most important application with cosine omega t there, since we found the solution, appropriate to comment that we could write the answer in terms of omega n, the natural frequency, and z, zeta, the damping ratio. 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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