:

{ u n ( t ) = e 2 t 20 n 2 t { 1 8 ( u n + 1 ( t ) + m = 1 u m ( t ) ) 2 + 1 9 ( m = 1 u m ( t ) ) 1 } u n ( t ) = + e 3 t 18 n 3 t { ( 0 t e ( t + 1 ) s u n ( s ) d s ) 2 + 1 2 ( 0 u n + 2 ( s ) d s ( 1 + t + s ) 2 ) 3 }, 0 < t <, t k ( k = 1, 2, 3, ; n = 1, 2, 3, ), Δ u n | t = k = 5 k 16 n 2 { u 2 n ( k ) + 1 3 ( m = 1 u m ( k ) ) 1 2 } ( k = 1, 2, 3, ; n = 1, 2, 3, ), u n ( ) = 1 2 u n ( 9 2 ) + 6 u n ( 0 ) ( n = 1, 2, 3, ). (85)

Conclusion Infinite system (85) has at least two positive solutions { u n ( t ) } ( n = 1, 2, 3, ) and { u n ( t ) } ( n = 1, 2, 3, ) such that

0 < inf 0 t < n = 1 u n ( t ) sup 0 t < n = 1 u n ( t ) < 1 < sup 0 t < n = 1 u n ( t ), inf 0 t < n = 1 u n ( t ) > 0.

Proof Let E = l 1 = { u = ( u 1,, u n, ) : n = 1 | u n | < } with norm u = n = 1 | u n | and P = { ( u 1,, u n, ) : u n 0, n = 1, 2, 3, }. Then P is a normal cone in E with normal constant N = 1, and infinite system (85) can be regarded as an infinite three-point boundary value problem of form (1). In this situation, u = ( u 1,, u n, ), v = ( v 1,, v n, ), w = ( w 1,, w n, ), t k = k ( k = 1, 2, 3, ), K ( t, s ) = e ( t + 1 ) s, H ( t, s ) = ( 1 + t + s ) 2, η = 9 2, γ = 1 2, β = 6, f = ( f 1,, f n, ) and I k = ( I k 1,, I k n, ), in which

f n ( t, u, v, w ) = e 2 t 20 n 2 t { 1 8 ( u n + 1 + m = 1 u m ) 2 + 1 9 ( m = 1 u m ) 1 } + e 3 t 18 n 3 t ( v n 2 + 1 2 w n + 2 3 ), t J + = ( 0, ), u P + = { u P : u > 0 }, v, w P ( n = 1, 2, 3, ) (86)

and

I k n ( u ) = 5 k 16 n 2 { u 2 n + 1 3 ( m = 1 u m ) 1 2 }, u P + ( k = 1, 2, 3, ; n = 1, 2, 3, ). (87)

It is easy to see that f C [ J + × P + × P × P, P ], I k C [ P +, P ] ( k = 1, 2, 3, ) and condition ( H 1 ) is satisfied and k 1, h 1. We have, by (86),

0 f n ( t, u, v, w ) e 2 t 20 n 2 t { 1 8 ( 2 u ) 2 + 1 9 u 1 } + e 3 t 18 n 3 t ( v 2 + 1 2 w 3 ) e 2 t n 2 t ( 1 40 u 2 + 1 180 u 1 + 1 18 v 2 + 1 36 w 3 ), t J +, u P +, v, w P ( n = 1, 2, 3, ), (88)

so, observing the inequality n = 1 1 n 2 < 2, we get

f ( t, u, v, w ) = n = 1 f n ( t, u, v, w ) e 2 t t ( 1 20 u 2 + 1 90 u 1 + 1 9 v 2 + 1 18 w 3 ), t J +, u P +, v, w P,

which implies that condition ( H 2 ) is satisfied for

a ( t ) = e 2 t t

and

g ( x, y, z ) = 1 20 x 2 + 1 90 x + 1 9 y 2 + 1 18 z 3

with

a = 0 e 2 t t d t < 0 1 d t t + 1 e 2 t d t = 2 + 1 2 e 2 < 29 14.

By (87), we have

0 I k n ( u ) 5 k 16 n 2 ( u + 1 3 u 1 2 ), u P + ( k = 1, 2, 3, ; n = 1, 2, 3, ), (89)

so,

I k ( u ) 1 8 5 k ( u + 1 3 u 1 2 ), u P + ( k = 1, 2, 3, ),

which implies that condition ( H 3 ) is satisfied for γ k = 1 8 5 k ( γ = 1 32 ) and

F ( x ) = x + 1 3 x.

On the other hand, (86) implies

f n ( t, u, v, w ) e 2 t 160 n 2 t u 2, t J +, u P +, v, w P ( n = 1, 2, 3, )

and

f n ( t, u, v, w ) e 2 t 180 n 2 t u 1, t J +, u P +, v, w P ( n = 1, 2, 3, ), (90)

so, we see that condition ( H 5 ) is satisfied for b ( t ) = e 2 t 160 t ( b < 29 2, 240 ), τ ( u ) = u 2 and u 0 = ( 1,, 1 n 2, ) and condition ( H 6 ) is satisfied for c ( t ) = e 2 t 180 t ( c < 29 2, 520 ), σ ( u ) = u 1 and u 1 = ( 1,, 1 n 2, ). In addition, from (90), we have

f ( t, u, v, w ) ( n = 1 1 n 2 ) e 2 t 160 t u 1 > e 2 t 160 t u 1, t J +, u P +, v, w P,

which implies that (3) and (4) hold, i.e., f ( t, u, v, w ) is singular at t = 0 and u = θRBA inflation target faces Labor review
Shadow treasurer Jim Chalmers has flagged that the 2-3 per cent inflation target will be reviewed for the first time in 30 years as part of a deep dive into the Reserve Bank of Australia.
The federal government’s budget and its economic “intersection” with RBA monetary policy would also be considered by the planned review if Labor wins the election, Dr Chalmers revealed in Canberra on Tuesday.
Shadow treasurer Jim Chalmers elaborated on his planned review of the RBA. Alex Ellinghausen
He said a review was timely because of the unusual economic times during the pandemic and extraordinary monetary policy deployed – such as the 0.1 per cent cash rate, $350 billion government bond buying and $188 billion in cheap loans to commercial banks.
Asked after his National Press Club address if the mooted review would consider the RBA’s charter, the inflation target and the size of budget deficits, Dr Chalmers said he would not pre-empt the terms of reference because he wanted to have a proper discussion with governor Philip Lowe and the Treasury Secretary, if Labor won the election.
“But you can imagine, at least a couple of those issues you’ve raised, if not the last one, that that would be a factor,” he said.
“Obviously, crucial, central to the Reserve Bank’s work is the inflation target.”
Dr Lowe has previously dismissed suggestions to reduce the 2-3 per cent inflation target, after undershooting it for six years before the recent spike in inflation due to higher petrol prices and supply chain blockages.
Westpac chief economist Bill Evans, who attended the Canberra speech, said the review was instigated after the bank undershot its inflation target, but “clearly the world has changed significantly”.
“Before we saw this inflation shock I was of the view the inflation target was too high, but clearly now the risk is a long period of high inflation,” Mr Evans said.
“The real issue for central banks globally is what are the implications of the massive expansions of their balance sheets – what were the benefits and what are the ongoing costs of maintaining such a bloated balance sheet?”
The RBA is conducting an internal review of its extraordinary pandemic-era policies, but Mr Evans said it would be good to have an independent outsider take a look.
The Organisation for Economic Co-operation and Development last September said the RBA’s “overly tight”