Systems of differential equations Handout Peyam Tabrizian Friday, November 8th, This handout is meant to give you a couple more example of all the techniques discussed in chapter 9, to counterbalance all the dry theory and complicated applications in the differential equations book! Enjoy! :) Note: Make sure to read this carefully! The methods presented in the book are a bit strange and convoluted, hopefully the ones presented here should be easier to understand! Systems of differential equations Find the general solution to the following system: x (t) = x (t) = x 3(t) = x (t) x (t) + 3x 3 (t) x (t) + x (t) x 3 (t) x (t) x (t) + 3x 3 (t) First re-write the system in matrix form: Where: x = Ax x (t) 3 x = x (t) A = x 3 (t) 3
Now diagonalize A: A = P DP, where: D =, P = Note: To find the eigenvalues, solve det(a λi) =. You should get λ =,,. The diagonal entries of D are λ =,,. Then, for each eigenvalue, find a basis for Nul(A λi). The the columns of P are the eigenvectors you found. Then use the following fact: Fact: For each eigenvalue λ and eigenvector v you found, the corresponding solution is x(t) = e λt v Hence here, the solution is: x(t) = Ae t + Be t + C (Note: Here e t = ). Aside: Why does this work? Suppose you want to solve x = Ax, since A = P DP, this becomes: So: x = P DP x x = P D ( P x ) Now let y = P x, so x = P y (remember Peyam, not Pexam). Then the above becomes:
x = P Dy P x = Dy But P is like a constant, so it gets inside the derivative! Finally, use y = P x, and you get: ( P x ) = Dy y = Dy Now solve the system: y = Dy, which is easier to solve: y (t) = y (t) y (t) = y (t) y 3(t) = Which gives you: Finally, use x = P y to get: y (t) = y (t) = y 3 (t) = Ae t Be t Ce t = C x (t) Ae t x (t) = Be t = Ae t + Be t + C x 3 (t) C Note: The matrix: e t e t X(t) = e e t e t (where you essentially ignore the constants A, B, C) is called a fundamental matrix for the system. 3
Complex eigenvalues. Solve the system x = Ax, where: A = [ ] 8 Eigenvalues of A: λ = ± 4i. From now on, only consider one eigenvalue, say λ = + 4i. [ ] i A corresponding eigenvector is Now use the following fact: Fact: For each eigenvalue λ and eigenvector v you found, the corresponding solution is x(t) = e λt v Hence, one solution is: ] x(t) = e ( +4i)t [ i Now split into real and imaginary parts and multiply everything out and group everything back into real and imaginary parts to get: x(t) = ( e t cos(4t) + ie t sin(4t) ) ([ ] [ ]) + i ( [ ] [ ]) = e t cos(4t) e t sin(4t) + i Hence the general solution is: ( x(t) = A e t cos(4t) [ ] e t sin(4t) ( e t sin(4t) [ ]) ( +B e t sin(4t) [ ] + e t cos(4t) [ ] + e t cos(4t) [ ]) [ ]) 4
3 Undetermined coefficients 3. Solve x = Ax + f, where A = [ ] 3 4 [ ] e f(t) = t + cos(t) 4e t As usual, x(t) = x (t) + x p (t), where: x (t) is the general solution to x = Ax x p (t) is one particular solution to x = Ax + f(t) To find x, use the techniques discussed in. To find x p, use: Undetermined coefficients: First group the terms in f which look alike : [ ] [ ] [ ] e t cos(t) f(t) = 4e t + = e t + 4 [ ] cos(t) Now guess: x p = ae t + b cos(t) + c sin(t) where a = [ a a ], b = [ b b ], c = [ c Now plug in x p into the equation x = Ax + f, and solve for a, b, c c ]. Remarks: 5
) Notice how similar this is to our usual way of doing undetermined coefficients! The only difference here is that a is a vector instead of a number! ) Remember to always put a sin(t) term whenever you see a cos(t) term and vice-versa! 3) The same rule about adding a t or not holds in this case too (but it s very rare). 4 Variation of parameters 4. Find the general solution to x = Ax + f, where: 3 e t A = f(t) = ln(t) 3 tan(t) As usual, x(t) = x (t) + x p (t). We already found x (t) in the first example: e t e t x (t) = A e t + B + C e t e t To find x p (t), use: Variation of Parameters: Suppose e t e t x p (t) = v (t) e t + v (t) + v 3 (t) e t e t Consider the (pre)-wronskian (or fundamental matrix): 6
e t e t W (t) = X(t) = e t e t e t (essentially put all the vectors you found in one matrix) And solve: v (t) e t W (t) v (t) = ln(t) v 3(t) tan(t) That is: v (t) ) e v (t) = ( W t (t) ln(t) v 3(t) tan(t) This gives you v, v, v 3. To get v, v, v 3, integrate the equations you found. And finally, to get x p, use: e t e t x p (t) = v (t) e t + v (t) + v 3 (t) e t e t And hence x(t) = x (t) + x p (t). Note: The following formula might come in handy: [ ] [ ] a b IfA = then A d b = c d ad bc c a 7
5 Matrix exponential 5. Find e At, where: A = 4 Eigenvalues of A: λ =, with multiplicity 3. IMPORTANT: The following technique works only in this case (where we have one eigenvalue with full multiplicity). For all the other cases, use the next example. Then: ( ) e t e At = e t I + (A + I)t + (A + I) t = 4te t e t! te t e t Note: If λ had multiplicity, we would stop at (A + I)t. But if it had multiplicity 4, we would add a (A + I) 3 t3 3! term. General method: 5. Find e At, where: A = [ ] 6 35 6 3 Diagonalize A: A = P DP, where: [ ] D =, P = Then: [ ] 5 7 3 [ ] [ ] [ ] 5 7 e e At = P e Dt P t [ ] 5 7 5e = 3 e t = t 4e t 35e t 35e t 3 6e t 6e t 5e t 4e t 8
Point: The general solution to x = Ax is x(t) = e At c, where c is a constant vector! Here, we get: [ ] [ ] 5e x(t) = A t 4e t 35e 6e t 6e t + B t 35e t 5e t 4e t 9