Fakultät für Mathematik und Informatik 11. April 2013 TU Bergakademie Freiberg
Prof. Dr. O. Rheinbach/Dr. M. Helm
Numerical Analysis of Differential Equations Initial Value Problems (I)
Picard-Lindelöf theorem
Assume the function f : [a, b]×Rn→Rn is continuous, and Lipschitz continuous in its second argument, i. e.
kf(x,y2)−f(x,y1)k ≤Lky2−y1k for some constantL >0. Then the initial value problem
y0(x) =f(x,y(x)), y(a) =ya
has a unique solution on[a, b].
Exercise 1
Consider the initial value problem
y0(t) =tsin(2y), y(0) =y0.
a) Show that the initial value problem has a unique solution on each interval[0, te].
(Hint: Apply the Picard-Lindelöf theorem. For the proof of Lipschitz continuity remember the mean value theorem.)
b) Solve the initial value problem by separation of variables.
c) Take Matlab or your pocket calculator to draw some pictures of the solution for different initial values y0. Choose the time interval [0, te] in a way, that you get a feeling for the behavior of the solution.
Exercise 2
Reactivate your prior knowledge about systems of linear differential equations of first order with constant coefficients. Solve the following initial value problems.
a)
y0= y10(t)
y20(t)
=Ay=
3 0 0 −2
y1(t) y2(t)
,
y1(0) y2(0)
= 4
7
b)
y0= y10(t)
y20(t)
=Ay= 1 2
4 3
y1(t) y2(t)
,
y1(0) y2(0)
= 1
0
Hints:
– Determine the eigenvalues and corresponding eigenvectors of the matrixAfirst.
– Diagonalize AtoD=V−1AV, whereV is generated by the eigenvectors of A.
– The substitutionz=V−1yleads to the systemz0=Dzof differential equations. Solve this system and determineyusing the backward substitutiony=Vz.
Initial Value Problems 2
Exercise 3
Determine the exact solution of the following linear initial value problems of first order.
a)
y01(t) y02(t)
= 1 2
4 3
y1(t) y2(t)
+ 2
0
,
y1(0) y2(0)
= 1
0
b)
y10(t) y20(t)
=
0 1
−1 2
y1(t) y2(t)
+ e2t
0
,
y1(0) y2(0)
= 3
−1
Exercise 4
Rewrite the following higher order initial value problems as first order systems.
(a) y00−2y0+y=tet−t , 0≤t≤1 y(0) = 0, y0(0) = 0.
(b) y000= 1
2y2, t≥1
y(1) = 1, y0(1) =−1, y00(1) = 2.
(c) t2y00(t)−2ty0(t) + 2y(t) =t3lnt , t≥1 y(1) = 1, y0(1) = 0.
