1Basics of Ordinary Differential Basics of Ordinary Differential EquationsEquationsLarry CarettoMechanical Engineering 501ABSeminar in Engineering AnalysisSeptember 21, 20042Outline• Review last week• Basics of differential equations• First order equations– Separable solutions– General solution for linear equation• Introduction to second order equations– Problems considered– Basis of solutions– Constant-coefficient, homogenous case3Review Numerical Solutions• Gauss elimination is basic approach• Need pivoting strategies to reduce round-off error in solution• Modifications of Gauss elimination– Gauss-Jordan sometimes used for finding inverse of matrix– LU method generally preferred• Does most of the elimination work without knowing the right-hand-side (b) vector• 1D integer vector required for pivoting4Basic Differential Equations• A differential equation is an equation that contains derivatives of a dependent variable, e.g., y(x) or u(x,y)• Differential equation solution gives y(x) or u(x,y) as a function of independent variable(s)– Ordinary differential equations (ODE) have one independent variable– Partial differential equations (PDE) have more than one independent variable5Definitions and Terms• Differential equations have boundary conditions or initial conditions• A general solution to the differential equation is one which can fit any boundary or initial condition by adjusting “constants” in the solution• A solution that satisfies the differential equation and the boundary or initial conditions is called a particular solution6More Definitions and Terms• The order of a differential equation is the order of the highest derivative in the equation• A linear differential equation is one in which the dependent variable and its derivatives all appear in linear terms• A homogenous differential equation is one in which all terms involve the dependent variable and its derivatives27Examples of ODEs• Third-order, linear, homogenous0)sin(22=+ ydxyd• Second-order, non-linear, homogenous• Second-order, linear, non-homogenous0)sin(233=−+ yxdxdyxdxyd133=+dxdyydxyd)cos(22xeydxydx=+• Third-order, non-linear, non-homogenous8Applications• First order differential equations are often used to model rate processes– Newton’s cooling dT/dt = -k(T - T∞)– chemical reactions, dci/dt = f(c,T)• Newton’s second law, F = ma leads to second order equations for mechanical systems• Deflection, y, of rectangular beam oriented in x direction EId4y/dx4= f(x)iiFdtydm =229Separable Forms• Simple differential equations can be written as integrals– Even if numerical quadrature is required this is more accurate than numerical solution of ODECuuhduxdxxyhdxdyCdxxfdyygygxfdxdyCdxxfyxfdxdy+−=⇒⎟⎠⎞⎜⎝⎛=+=⇒=+=⇒=∫∫∫∫∫)()()()()()()(10P(x,y)dx + Q(x,y)dy = 0• Is P(x,y)dx + Q(x,y)dy = 0 an exact form?• From differential of a function of two variables, f(x,y), see if P and Q satisfy partial derivative relation• If df = 0, f = Cdyyfdxxfdf∂∂+∂∂=xfyyfxyxfyxf∂∂∂∂=∂∂∂∂⇒∂∂∂=∂∂∂22xfyxP∂∂≡),(yfyxQ∂∂≡),(QdyPdxdf +=yPxQ∂∂=∂∂11Exact Form• If , P(x,y)dx +Q(x,y)dy = df• We may not know (or care) what f is, but we use df = P(x,y)dx +Q(x,y)dy to solve the differential equation• We also know that P(x,y)dx +Q(x,y)dy = 0 means that df = 0 or f = a constant• We also know that P and Q are derivatives of this mysterious f functionyPxQ∂∂=∂∂yPxQifonlyyfyxQandxfyxP∂∂=∂∂∂∂=∂∂= ),(),(12Exact Forms II• Integrate df = P(x,y)dx + Q(x,y)dy for constant y (dy = 0)• f = constant, C, because df = 0dyydgdxyxPyyfyxQconsty)(),(),( +⎥⎥⎦⎤⎢⎢⎣⎡∂∂=∂∂=∫=CygdxyxPdffconsty=+==∫∫=)(),()(),(),()(yhdxyxPyyxQdyydgconsty=⎥⎥⎦⎤⎢⎢⎣⎡∂∂−=∫=313Exact Forms III• Final equation must be a function of y only• Integrate this equation for g(y))(),(),()(yhdxyxPyyxQdyydgconsty=⎥⎥⎦⎤⎢⎢⎣⎡∂∂−=∫=CygdxyxPdffconsty++==∫∫=)(),(CdydxyxPyyxQdgygconsty+⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡∂∂−==∫∫∫=),(),()(• Substitute g(y) into equation for f14Exact Forms IV• Combine constants into a single constant• Obtain implicit relationship between y and x1)(),( CygdxyxPdffconsty=+==∫∫=2),(),()( CdydxyxPyyxQygconsty+⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡∂∂−=∫∫=12),(),(),( CCdydxyxPyyxQdxyxPconstyconsty=+⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡∂∂−+∫∫∫==15Solving Exact Pdx + Qdy = 0• Step 1 – Integrate P(x,y)dx with y constantCdydxyxPyyxQdxyxPconstyconsty=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡∂∂−+∫∫∫==),(),(),(• Step 2 – Take the y derivative of the step 1 result and subtract it from Q(x,y)• Step 3 – Integrate result of step 2, that will be a function of y only, over y• Step 4 – Add results of steps 1 and 316Integrating Factors• Used to integrate P(x,y)dx + Q(x,y)dy = 0 if P and Q are not exact• Basic idea is to find a factor, F, that multiplies the original equation: FPdx + FQdy = 0• Find the F factor so that FP and FQ are exactyFPxFQ∂∂=∂∂• Use trial and error or process outlined in Kreyszig to find F17First-order Equations• First order rate equation where rate is proportional to amount dy/dt = -ky•y = y0e-k(t-t0)• General linear first order equation for y(x): dy/dx + f(x)y = g(x) has closed form solution shown below• C is found from initial condition[]∫∫+==−dxxgeCeydxxfppp)()(18Existence and Uniqueness• Important because we can try numerical solution of an ODE with no solution• Examine dy/dx = f(x,y) with y(x0) = y0in a region |x – x0| < a and |y – y0| < b• Derivate is bounded: |f(x,y)| ≤ K• Equation has a solution in region |x – x0| < min(a, b/K) • Uniqueness requires |∂f/∂y| ≤ M419Existence and Uniqueness• Example: xy’ = 4, y(0) = 0• Here we have dy/dx = f(x,y) =4/x with y(x0= 0) = y0= 0• Region is |x – 0| < a and |y – 0| < b• Derivate is not bounded at x = x0= 0• There there are no solutions : |f(x,y)| ≤ K• Attempted solution is y = 4ln(x) + C, but we cannot apply this at x = x0= 020Second Order Equations• First look at homogenous linear equations• Then consider nonhomogenousequations• Most nonlinear equations require numerical solution0)()(22=++