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11 The Cauchy Problem for Linear Ordinary Differential Equations
I1
The Cauchy Problem for Linear Ordinary Differential Equations Consider a first-order linear differential equation in an open interval - T < t < T (T > 0) of the real line: du
(11.1)
- - a(t)u
dt
=f.
Suppose that both the coefficient a(t) and the right-hand sidef(t) are continuous functions of t , I t I < 2". We obtain at once a solution of (11.1) by taking (11.2)
u( t ) = eA(',')uO
+
where uo is an arbitrary constant and A(t, t ' )
=
f
Sd
eA("'"f(t'>dt',
1 tl < T ,
a(s) ds,
I'
1 t'l < T.
As a matter of fact, all the solutions of (1 1 .l) are obtained in this manner: They are all of the form (1 1.2). It is clear that u(t) is completely determined by the choice of uo . Observe that
(11.3)
u(0) = 2.40.
In other words, the problem (11.1)<11.3) has a unique solution, given by (1 1.2). Notice, incidentally, that this solution is once continuously dzfferentiable in 3 - T, T [ . We now go to first-order systems of linear ODES: (1 1.4) (11.5)
dU
dt
M(t)u = f,
u(0) 89
= uo
.
90
THE CAUCHY PROBLEM
[Chap. I1
Here u, f, uo are valued in C” and M ( t ) is an m x m matrix with continuous functions as entries. If all these entries were constant functions, i.e., M(t) = M, a complex m x m matrix, the theory of (11.4H11.5) would be a straightforward generalization of the theory of (11.1H11.3). The solution would be given by
+
(11.6)
~ ( t= ) etzHu0
Id
e(t-t’)Mf(r‘)dr‘.
But the situation is radically different when M ( t ) is nonconstant. In the constant coefficients case, set U(t, 1’) = exp((t - t’)M).What lies at the root of formula (1 1.6) is the fact that
dU - -- M U , Ul =Z (the m x m identity matrix). dt t = t’ But this ceases to be true, in general, when M is nonconstant. The thing to do, therefore, is to lay aside the exponential and concentrate on the solution of (1 1.7). It should be noted that problem (1 1.7) is of the same kind as (1 1.4)(11.5): I t is a system of first-order linear ODES. This time there are m2 equations in m 2 unknowns, submitted to rn2 conditions at time t = t’. It follows at once from the fundamental theorem on ordinary differential equations that (11.7) has a unique solution, which furthermore is a C’ function o f t , t’ for Irl < T, It’] < T . The correct generalization of (1 1.2) is then (11.7)
(11.8)
u(t) = U(t, O)u0
+
J-:
U(t, t’)f(t’) dt’.
The function U(t, 1’) is sometimes called the Riemann funcrion of the problem (11.4H11.5).
PROPOSITION 11.1. If
t , t’, t“
(11.9) (11.10)
are three points in rhe interval ] - T, T[, we haue
U(t, t’)U(t’,t “ ) = U(t, r”), U(t, 1) = I ,
U(t, t’)U(t’,r ) = I.
Proof: The first identity in (1 1.10) is part of the definition of U(t, f‘). By differentiating it (with respect to t), we obtain
dU dt
-+-=O
dU dr’
when
t=t‘,
when
t = t’.
whence, by combining this with (11.7), (1 1.1Oa)
dU _ - -M(t’) dr’
Sect. 111
LINEAR ORDINARY DIFFERENTIAL EQUATIONS
91
On the other hand, if we differentiate the first equation in (1 1.7) with respect to t’, we obtain (1 1. lob)
The solution of the Cauchy problem (1 l.lOa)-(ll.lOb) is given by
dU dt, -- - U(t, ow’),
(11.1Oc)
as can be checked directly. Set now V(t, t ’ ) = U(t, t’)U(t‘, t). We have, by (11.7) and (ll.lOc),
dV dU _ -- dU -(t, t’)U(t’, t ) + U(t, t ’ ) - ( f ’ , t) dt’ dt‘ dt = - U(t, t’)M(t‘)U(t’,t )
+ U(t, t’)M(r’)U(t’,r) = 0 ,
hence V(t) = V(0) = I. Now set W(t) = U(t, t‘)U(t’,t ” ) . We have
dW dU - - _ - (t, t‘)U(t’,t ” ) = M(t)U(t, t’)U(f’,t “ ) dt dt
=MW
On the other hand, W = I when t = t“. By the uniqueness of the solution of (1 1.7) we conclude that W = U(t, t“). Q.E.D. Consider now an mth-order differential equation (11.11)
d“u
dm-1U
Lu = - + a,(t) -+ * . . + a,(t)u =f, dt” dt”-’
where the coefficients a j and the right-hand sidef are continuous functions in ] - T, T[. By the standard procedure, already described (see p. 29) we may transform it into an equation of the kind (11.4). This means that, here, M(t) =
(
0
0
-a,(t)
1 0
-~,,,-~(r)
and the components uj of u are given by (11.12)
0 1 -a,-z(t)
... ... * * -
-al(?)
92
[Chap. I1
THE CAUCHY PROBLEM
We see the meaning here of the initial condition ( I 1.5): Let us denote by uo,o . uo, . . . uo, the components of uo . Condition ( I 1.5) is equivalent. in the present context. to
(11.13)
u‘”(0) = u ~ , ~ . 0
5 k 2 rn
- 1.
Condition ( I I . 13) is thus what generalizes ( I 1.3) to higher order equations. In accordance with the terminology for partial differential equations. w e shall refer to the problem ( 1 1.1 1 )-( 1 1.13) as the Cauchj,problem for the differential operator L. The next question is related to what generalizes formula ( I 1.2) in the higher order case. This can be ansuered by means of formula ( 1 1.8) relative to the equivalent system ( I l.4)+11.5) One must recall that ffr) = f ( r ) e , . We denote by ej(l S j 5 rn) the vector uhose components all vanish. except the jth one. which is equal to one. Observe then that m- 1
u(t,O)uo = k = O uo. u(t.O h , Let us call Uj(r. 1‘) thefirsr component of U ( t . r ‘ ) e j T , In . viebb of (1 1.12) we derive from (1 1.8)
From this identity we derive that U J t . 1 ’ ) is the solution of the Cauchy problem (11.15) (11.16)
LV=O.
at r
= r’.
d’V 10 -dt’ \I
if j # k (Ozjsrn-I). if j = k
In general, it is not easy to compute the functions Uj(r. r’). It is therefore convenient to give an equivalent statement of (11.14) in which all c’, have been eliminated but one. This can be done as follows. Introduce the function (1 1.17)
Clearly iio(t) satisfies the initial conditions (1 1.13). Let us set We have (11.18) (11.19)
r(r) = u ( t ) - ijo(r). LL. =f - LEO.
r‘k’(0)= 0,
0 5 k 5 rn - 1.
Sect. 11 ]
LINEAR ORDINARY DIFFERENTIAL EQUATIONS
93
We may therefore apply (1 1.14) with v instead of u : u(t) =
whence
I‘ 0
U,,r-l(t. r’)( f( t ’ ) - Lu,(t’)) dt‘,
In the first integral, on the right-hand side of (11.20), we may integrate by parts and use the explicit expression of LC,. This leads to a more explicit formula for u(t), involving I Y , , - ~ (t ’~) ,and its derivatives. We leave it to the student to compute it, if he wishes to do so. In the case where the coefficients a j are constant, U m - l ( t .t’) is not difficult to compute; it is the first component of e(r - t ‘ ) M
em
3
We have U m - l ( t ,t’) = U m - , ( t - t ’ , 0 ) = U(t - t ’ ) where U ( x ) is the function so denoted on page 30. With the notation of Sect. 4, we have E = H U where H is the Heaviside function and E is the fundamental solution of L with support in the nonnegative half-line. Formula (1 1.20) in this case can be rewritten as (11.21)
u =Go
+ ( H U ) * { H ( f - LEO)}- ( B U ) * {A(f- LEO)},
where B(r) = H( - t). The support of the first convolution on the right-hand side of (1 1.21) is contained in the half-line t 2 0, the one of the second convolution is contained in the half-line t 5 0. Let us go back, for some final remarks, to the general case of variable coefficients. Concerning the system (1 1.4) and its solution (1 1.8) we may note that if the coefficients M ( t ) and the right-hand side f(t) are N times continuously differentiable functions of t . 1 t / < T , the Riemann function U(t, t’) is an ( N + ])-times continuously differentiable function off, t’ in the same interval. Consequently, u(t) is CNf’in 1- T , T [ . Let us now return to the mth-order equation (I 1.11 ) via the equivalent first-order system ( I 1.4). Assume now that the a j and f are C N functions in ] - T , T[. Because of the relations (1 1.12) we reach the conclusion that u ( ~ - ’is) a CN+’function in 1-T, T [ . We may state
PROPOSITION 11.2. If the coeficients a j (1 5 j 5 m ) and the right-hand side f of Eq. (11.11) are C Nfunctions o f t , I t [ < T , the solution u given by (11.14) is a CNt” function in the interval ] - T, T[. COROLLARY 11. I . I f the coeficients a j and the right-hand side ,f are C“ functions in the interval ] - T , T [ , the same is true of the solution u.
94
THE CAUCHY PROBLEM
[Chap. I1
In this corollary one may replace C" everywhere by analytic. This follows at once from the fact that if the coefficient M ( r ) in ( I 1.4) is analytic with respect to t , the Riemann function U(t. r') is analytic with respect to t. r' (in the interval 1- T. TD. Exercises 11.1. Suppose that for any two points 1. t' in the interval 1- 7.T [ the m x rn matrices M(r) and M ( r ' ) commute. In this case. give a simple expression of the Riemann function C(t. t ' ) of the problem ( 1 1.4H11.5). 11.2. Suppose that. in (1 1.4), the matrix M IS constant. Suppose furthermore that M is negative definite. i.e.. (Mv. v) < 0 for all v E C". v # 0 [we have denoted by ( . ) the Hermitian inner product In Cm].Let U(r) denote the solution of the problem (1 1.7). Shorn that there is a constant c > 0 such that the matrix norm of U ( t ) decreases at least as fast as e-" when t + + x. 11.3 Using Exercise 4.10. gike a complete description of the solution U of problem (1 1.7) when M is an arbitrary m x m matrix. 11.4. Give a complete description of the function U,,-l(t. t ' ) in ( I 1.14) in the case where the coefficients a, of L are constant (cf. Exercises 4.10 and 11.3). 11.5. Let L be the differential operator defined in (11.11). Suppose that the coefficients a,(r) and the right-hand side f ( t ) are C" functions in the interval 1- T, T [ .Suppose furthermore that f ( t ) = 0 if t < 0 and that I( is a distribution in ] - T. T [ of the form
with g a continuous function of t. t < T, vanishing when t < 0. Use the fact that Lu = j t o prove that u is a C" function in 1- T. T [ . 11.6. We assume that the coefficients uj (.j = 1. . . . . rn) of the operator L and the second member f in (1 1.1 1) are (complex-valued) C" functions in the real line. It follows at once from formula (11.14) that the problem (1 1.1 1)(1 1.13) has a unique solution I( which is also a C" function in R' (the uniqueness follows from the fundamental theorem about ODES). Let then u'(t) be the function equal to u ( t ) for r 2 0 and to zero for t < 0. Compute the distribution Lu'. In particular, compute the distributions Lu' when u = Uk(t,0). k = 0, .. . , m - 1 [see (11.14) to (11.16)]. 11.7. Suppose that the differential operator L in (11.11) has constant coefficients and let u' denote the function defined in Exercise 11.6. Prove that if c is any distribution in the real line. vanishing for t < 0, i.e., L' E P ;
Sect. 111
95
LTNEAR ORDINARY DIFFERENTIAL EQUATIONS
(cf. Exercise 4.1 1) such that LD= Lii, then necessarily we must have u = E. Is this fact still true when 1. has variable (but Cm)coefficients? 11.8. Give an example of a homogeneous first-order ODE with analytic coefficients (in the whole real line) whose solutions all vanish of infinite order at the origin. 11.9. Analyze completely the possibility of solving the “initial value problem ”
(1 1.22)
xu”
+
U l 24’
+ ut xu = 0,
(1 1.23)
u(0) = uo.
Show that, in general (this is to be made precise), the Cauchy problem, obtained by adjoining the condition (11.24)
u’(0) = 241,
cannot be solved. Is it true that, when it can be solved, its solution is unique? 11.10. Let a, A and u be three nonnegative continuous functions on the closed interval [0, TI, uo a nonnegative constant. By comparing u to a solution of (1 l . l ) , prove that the inequality (11.25)
v ( t ) 5 uo
+
f(s) ds
+ Iza(s)u(r) ds, 0
0 5 t 5 T,
implies the Gronwall inequality (11.26)
u(t)s
(u,+Jb’f(s)ds)exp(~~u(s)ds), O s t j T .
What inequality, bearing on u, would have followed from (11.27)
Iu‘(f)l
5 u 1 + f(s) ds
assuming that a(s) is monotone?
+ Izu(s)u(s) ds, 0
0
t
5 T,
The Cauchy Problem for Linear Ordinary Differential Equations Consider a first-order linear differential equation in an open interval - T < t < T (T > 0) of the real line: du
(11.1)
- - a(t)u
dt
=f.
Suppose that both the coefficient a(t) and the right-hand sidef(t) are continuous functions of t , I t I < 2". We obtain at once a solution of (11.1) by taking (11.2)
u( t ) = eA(',')uO
+
where uo is an arbitrary constant and A(t, t ' )
=
f
Sd
eA("'"f(t'>dt',
1 tl < T ,
a(s) ds,
I'
1 t'l < T.
As a matter of fact, all the solutions of (1 1 .l) are obtained in this manner: They are all of the form (1 1.2). It is clear that u(t) is completely determined by the choice of uo . Observe that
(11.3)
u(0) = 2.40.
In other words, the problem (11.1)<11.3) has a unique solution, given by (1 1.2). Notice, incidentally, that this solution is once continuously dzfferentiable in 3 - T, T [ . We now go to first-order systems of linear ODES: (1 1.4) (11.5)
dU
dt
M(t)u = f,
u(0) 89
= uo
.
90
THE CAUCHY PROBLEM
[Chap. I1
Here u, f, uo are valued in C” and M ( t ) is an m x m matrix with continuous functions as entries. If all these entries were constant functions, i.e., M(t) = M, a complex m x m matrix, the theory of (11.4H11.5) would be a straightforward generalization of the theory of (11.1H11.3). The solution would be given by
+
(11.6)
~ ( t= ) etzHu0
Id
e(t-t’)Mf(r‘)dr‘.
But the situation is radically different when M ( t ) is nonconstant. In the constant coefficients case, set U(t, 1’) = exp((t - t’)M).What lies at the root of formula (1 1.6) is the fact that
dU - -- M U , Ul =Z (the m x m identity matrix). dt t = t’ But this ceases to be true, in general, when M is nonconstant. The thing to do, therefore, is to lay aside the exponential and concentrate on the solution of (1 1.7). It should be noted that problem (1 1.7) is of the same kind as (1 1.4)(11.5): I t is a system of first-order linear ODES. This time there are m2 equations in m 2 unknowns, submitted to rn2 conditions at time t = t’. It follows at once from the fundamental theorem on ordinary differential equations that (11.7) has a unique solution, which furthermore is a C’ function o f t , t’ for Irl < T, It’] < T . The correct generalization of (1 1.2) is then (11.7)
(11.8)
u(t) = U(t, O)u0
+
J-:
U(t, t’)f(t’) dt’.
The function U(t, 1’) is sometimes called the Riemann funcrion of the problem (11.4H11.5).
PROPOSITION 11.1. If
t , t’, t“
(11.9) (11.10)
are three points in rhe interval ] - T, T[, we haue
U(t, t’)U(t’,t “ ) = U(t, r”), U(t, 1) = I ,
U(t, t’)U(t’,r ) = I.
Proof: The first identity in (1 1.10) is part of the definition of U(t, f‘). By differentiating it (with respect to t), we obtain
dU dt
-+-=O
dU dr’
when
t=t‘,
when
t = t’.
whence, by combining this with (11.7), (1 1.1Oa)
dU _ - -M(t’) dr’
Sect. 111
LINEAR ORDINARY DIFFERENTIAL EQUATIONS
91
On the other hand, if we differentiate the first equation in (1 1.7) with respect to t’, we obtain (1 1. lob)
The solution of the Cauchy problem (1 l.lOa)-(ll.lOb) is given by
dU dt, -- - U(t, ow’),
(11.1Oc)
as can be checked directly. Set now V(t, t ’ ) = U(t, t’)U(t‘, t). We have, by (11.7) and (ll.lOc),
dV dU _ -- dU -(t, t’)U(t’, t ) + U(t, t ’ ) - ( f ’ , t) dt’ dt‘ dt = - U(t, t’)M(t‘)U(t’,t )
+ U(t, t’)M(r’)U(t’,r) = 0 ,
hence V(t) = V(0) = I. Now set W(t) = U(t, t‘)U(t’,t ” ) . We have
dW dU - - _ - (t, t‘)U(t’,t ” ) = M(t)U(t, t’)U(f’,t “ ) dt dt
=MW
On the other hand, W = I when t = t“. By the uniqueness of the solution of (1 1.7) we conclude that W = U(t, t“). Q.E.D. Consider now an mth-order differential equation (11.11)
d“u
dm-1U
Lu = - + a,(t) -+ * . . + a,(t)u =f, dt” dt”-’
where the coefficients a j and the right-hand sidef are continuous functions in ] - T, T[. By the standard procedure, already described (see p. 29) we may transform it into an equation of the kind (11.4). This means that, here, M(t) =
(
0
0
-a,(t)
1 0
-~,,,-~(r)
and the components uj of u are given by (11.12)
0 1 -a,-z(t)
... ... * * -
-al(?)
92
[Chap. I1
THE CAUCHY PROBLEM
We see the meaning here of the initial condition ( I 1.5): Let us denote by uo,o . uo, . . . uo, the components of uo . Condition ( I 1.5) is equivalent. in the present context. to
(11.13)
u‘”(0) = u ~ , ~ . 0
5 k 2 rn
- 1.
Condition ( I I . 13) is thus what generalizes ( I 1.3) to higher order equations. In accordance with the terminology for partial differential equations. w e shall refer to the problem ( 1 1.1 1 )-( 1 1.13) as the Cauchj,problem for the differential operator L. The next question is related to what generalizes formula ( I 1.2) in the higher order case. This can be ansuered by means of formula ( 1 1.8) relative to the equivalent system ( I l.4)+11.5) One must recall that ffr) = f ( r ) e , . We denote by ej(l S j 5 rn) the vector uhose components all vanish. except the jth one. which is equal to one. Observe then that m- 1
u(t,O)uo = k = O uo. u(t.O h , Let us call Uj(r. 1‘) thefirsr component of U ( t . r ‘ ) e j T , In . viebb of (1 1.12) we derive from (1 1.8)
From this identity we derive that U J t . 1 ’ ) is the solution of the Cauchy problem (11.15) (11.16)
LV=O.
at r
= r’.
d’V 10 -dt’ \I
if j # k (Ozjsrn-I). if j = k
In general, it is not easy to compute the functions Uj(r. r’). It is therefore convenient to give an equivalent statement of (11.14) in which all c’, have been eliminated but one. This can be done as follows. Introduce the function (1 1.17)
Clearly iio(t) satisfies the initial conditions (1 1.13). Let us set We have (11.18) (11.19)
r(r) = u ( t ) - ijo(r). LL. =f - LEO.
r‘k’(0)= 0,
0 5 k 5 rn - 1.
Sect. 11 ]
LINEAR ORDINARY DIFFERENTIAL EQUATIONS
93
We may therefore apply (1 1.14) with v instead of u : u(t) =
whence
I‘ 0
U,,r-l(t. r’)( f( t ’ ) - Lu,(t’)) dt‘,
In the first integral, on the right-hand side of (11.20), we may integrate by parts and use the explicit expression of LC,. This leads to a more explicit formula for u(t), involving I Y , , - ~ (t ’~) ,and its derivatives. We leave it to the student to compute it, if he wishes to do so. In the case where the coefficients a j are constant, U m - l ( t .t’) is not difficult to compute; it is the first component of e(r - t ‘ ) M
em
3
We have U m - l ( t ,t’) = U m - , ( t - t ’ , 0 ) = U(t - t ’ ) where U ( x ) is the function so denoted on page 30. With the notation of Sect. 4, we have E = H U where H is the Heaviside function and E is the fundamental solution of L with support in the nonnegative half-line. Formula (1 1.20) in this case can be rewritten as (11.21)
u =Go
+ ( H U ) * { H ( f - LEO)}- ( B U ) * {A(f- LEO)},
where B(r) = H( - t). The support of the first convolution on the right-hand side of (1 1.21) is contained in the half-line t 2 0, the one of the second convolution is contained in the half-line t 5 0. Let us go back, for some final remarks, to the general case of variable coefficients. Concerning the system (1 1.4) and its solution (1 1.8) we may note that if the coefficients M ( t ) and the right-hand side f(t) are N times continuously differentiable functions of t . 1 t / < T , the Riemann function U(t, t’) is an ( N + ])-times continuously differentiable function off, t’ in the same interval. Consequently, u(t) is CNf’in 1- T , T [ . Let us now return to the mth-order equation (I 1.11 ) via the equivalent first-order system ( I 1.4). Assume now that the a j and f are C N functions in ] - T , T[. Because of the relations (1 1.12) we reach the conclusion that u ( ~ - ’is) a CN+’function in 1-T, T [ . We may state
PROPOSITION 11.2. If the coeficients a j (1 5 j 5 m ) and the right-hand side f of Eq. (11.11) are C Nfunctions o f t , I t [ < T , the solution u given by (11.14) is a CNt” function in the interval ] - T, T[. COROLLARY 11. I . I f the coeficients a j and the right-hand side ,f are C“ functions in the interval ] - T , T [ , the same is true of the solution u.
94
THE CAUCHY PROBLEM
[Chap. I1
In this corollary one may replace C" everywhere by analytic. This follows at once from the fact that if the coefficient M ( r ) in ( I 1.4) is analytic with respect to t , the Riemann function U(t. r') is analytic with respect to t. r' (in the interval 1- T. TD. Exercises 11.1. Suppose that for any two points 1. t' in the interval 1- 7.T [ the m x rn matrices M(r) and M ( r ' ) commute. In this case. give a simple expression of the Riemann function C(t. t ' ) of the problem ( 1 1.4H11.5). 11.2. Suppose that. in (1 1.4), the matrix M IS constant. Suppose furthermore that M is negative definite. i.e.. (Mv. v) < 0 for all v E C". v # 0 [we have denoted by ( . ) the Hermitian inner product In Cm].Let U(r) denote the solution of the problem (1 1.7). Shorn that there is a constant c > 0 such that the matrix norm of U ( t ) decreases at least as fast as e-" when t + + x. 11.3 Using Exercise 4.10. gike a complete description of the solution U of problem (1 1.7) when M is an arbitrary m x m matrix. 11.4. Give a complete description of the function U,,-l(t. t ' ) in ( I 1.14) in the case where the coefficients a, of L are constant (cf. Exercises 4.10 and 11.3). 11.5. Let L be the differential operator defined in (11.11). Suppose that the coefficients a,(r) and the right-hand side f ( t ) are C" functions in the interval 1- T, T [ .Suppose furthermore that f ( t ) = 0 if t < 0 and that I( is a distribution in ] - T. T [ of the form
with g a continuous function of t. t < T, vanishing when t < 0. Use the fact that Lu = j t o prove that u is a C" function in 1- T. T [ . 11.6. We assume that the coefficients uj (.j = 1. . . . . rn) of the operator L and the second member f in (1 1.1 1) are (complex-valued) C" functions in the real line. It follows at once from formula (11.14) that the problem (1 1.1 1)(1 1.13) has a unique solution I( which is also a C" function in R' (the uniqueness follows from the fundamental theorem about ODES). Let then u'(t) be the function equal to u ( t ) for r 2 0 and to zero for t < 0. Compute the distribution Lu'. In particular, compute the distributions Lu' when u = Uk(t,0). k = 0, .. . , m - 1 [see (11.14) to (11.16)]. 11.7. Suppose that the differential operator L in (11.11) has constant coefficients and let u' denote the function defined in Exercise 11.6. Prove that if c is any distribution in the real line. vanishing for t < 0, i.e., L' E P ;
Sect. 111
95
LTNEAR ORDINARY DIFFERENTIAL EQUATIONS
(cf. Exercise 4.1 1) such that LD= Lii, then necessarily we must have u = E. Is this fact still true when 1. has variable (but Cm)coefficients? 11.8. Give an example of a homogeneous first-order ODE with analytic coefficients (in the whole real line) whose solutions all vanish of infinite order at the origin. 11.9. Analyze completely the possibility of solving the “initial value problem ”
(1 1.22)
xu”
+
U l 24’
+ ut xu = 0,
(1 1.23)
u(0) = uo.
Show that, in general (this is to be made precise), the Cauchy problem, obtained by adjoining the condition (11.24)
u’(0) = 241,
cannot be solved. Is it true that, when it can be solved, its solution is unique? 11.10. Let a, A and u be three nonnegative continuous functions on the closed interval [0, TI, uo a nonnegative constant. By comparing u to a solution of (1 l . l ) , prove that the inequality (11.25)
v ( t ) 5 uo
+
f(s) ds
+ Iza(s)u(r) ds, 0
0 5 t 5 T,
implies the Gronwall inequality (11.26)
u(t)s
(u,+Jb’f(s)ds)exp(~~u(s)ds), O s t j T .
What inequality, bearing on u, would have followed from (11.27)
Iu‘(f)l
5 u 1 + f(s) ds
assuming that a(s) is monotone?
+ Izu(s)u(s) ds, 0
0
t
5 T,