Initial Conditions

Initial Conditions 1. Introduction Before we try to solve higher order equations with discontinuous or impulsive input we need to think carefully about what happens to the so­ lution at the point of discontinuity. Recall that we have the left and right limits of a function as t → 0: x (0− ) = lim x (t) t ↑0 and x (0+ ) = lim x (t). t ↓0 (Note that we can define these limits as t goes to any value a.) For a con­ tinuous function these two limits are the same, and they are both equal to x (0). For the unit step function we have u(0− ) = 0, u(0+ ) = 1, u(0) is undefined. For the unit impulse function δ(t) we have δ(0− ) = 0, δ(0+ ) = 0, δ(0) = ∞. In this unit our differential equations will always have initial conditions at t = 0. The above examples show that when there is a discontinuity we might need to distinguish between 0− and 0+ . Assuming x is the output, . we will do this by calling x (0− ), x (0− ), . . . the pre-initial conditions and . x (0+ ), x (0+ ), . . . the post-initial condition. Important: Hereafter when we just say initial conditions we will mean the pre-initial conditions. In cases where x (t) is smooth the pre and post-initial conditions are the same and their is no need to distinguish between them. 2. Simple Examples Example 1. Consider the initial value problem . x = u ( t ), x (0− ) = 0. This is a simple calculus problem and has solution � x (t) = 0 t for t < 0 for t > 0. x(t) t Initial Conditions OCW 18.03SC It is easy to see that x (0+ ) = 0, so the post-initial condition is the same as the pre-initial condition. This should not surprise us. Although the rate of input jumps from 0 to 1, it is still only inputting an infinitesimal amount at a time. So, the response x (t) should be continuous. But, note . . that x (0− ) = 0 �= x (0+ ) = 1. Example 2. Consider the initial value problem . x = δ ( t ), x (0− ) = 0. We know how to integrate δ(t) to get x (t) = u(t). 1 x = u(t) t Here the pre-initial condition x (0− ) = 0 does not match the post-initial condition x (0+ ) = 1. The impulse causes a jump in the value of x. Example 3. Consider a second order IVP .. x = u ( t ), . x (0− ) = 0, x (0− ) = 0. � Integrating twice we get x (t) = 0 t2 /2 1 for t < 0 for t > 0. x(t) t . . Again, it’s easy to check that x (0− ) = x (0+ ) and x (0− ) = x (0+ ). That .. is, the pre and post initial conditions are the same. (But, x (0− ) = 0 �= .. + x (0 ) = 1.) Example 4. Consider the initial value problem .. x = δ ( t ), . x (0− ) = 0, x (0− ) = 0. . Integrating once gives x (t) = u(t). Integrating a second time gives � x (t) = 0 t x(t) for t < 0 for t > 0. t Checking the pre and post initial conditions gives x (0− ) = 0 = x (0+ ) . . x (0− ) = 0 � = x (0+ ) = 1 2 Initial Conditions OCW 18.03SC In other words, x (t) itself is continuous, but for the second order equation the input δ(t) caused a jump in the first derivative. If we continued these examples we’d find that for an nth -order equation an input of δ(t) causes a jump in the derivative of order n − 1. 3. Rest Initial Conditions The case where x (t) = 0 for t < 0 is called rest initial conditions. If we have a DE of order n this translates into pre-initial conditions . x (0− ) = 0, x (0− ) = 0, . . . , x (n−1) (0− ) = 0. 4. Conclusion A unit step input u(t) causes a smooth response with matching pre and post-initial conditions. For a unit impulse input δ(t) the pre and post initial conditions match except for the derivative one less than the order of the equation. 3