[Jia Liwei university mathematics magazine] 317th Xiamen University 2010 comprehensive foundation I Postgraduate Entrance Exam reference

 

Mathematical Analysis Section ($110 '$)

 

 

1. multiple choice questions ($5 \ times 6' = 30' $)

(1) set the function $ f (x) $ second-order convertible, and satisfy the equation $ \ Bex f'' (x) + 3 [f' (x)] ^ 2 + 2e ^ x f (x) = 0, \ EEx $ set $ x_0 $ to a resident point of $ f (x) $ and meet $ F (x_0) <0 $, then $ F (x_0) $ at the Point $ x_0 $ (B ).

A. obtain the maximum value;

B. Obtain the minimum value;

C. Do not take the extreme values;

D. Not sure.

Answer: $ \ Bex f'' (x_0) =-2E ^ {x_0} f (x_0)> 0. \ EEx $

 

 

(2) function $ \ DPS {f (x) = \ ln x-\ frac {x} {e} + k \ (k> 0 )} $ the number of zero points in the range $ (0, \ infty) $ is (c ).

A. $0 $;

B. $1 $;

C. $2 $;

D. Not sure.

Answer: by $ \ Bex f' (X) =\ frac {1} {x}-\ frac {1} {e }=\ sedd {\ BA {ll}> 0, & 0 <x <e \ <0, & x> E \ EA} \ EEx $ and $ \ Bex F (e) = k> 0, \ quad \ lim _ {x \ to 0} f (x) =-\ infty = \ vlm {x} f (x) \ EEx $ is the conclusion.

 

 

(3) When $ x> 0 $ is known, functions $ e ^ {\ Tan x}-e ^ x $ and $ x ^ N $ are of the same order, then $ n = $ (c ).

A. $1 $;

B. $2 $;

C. $3 $;

D. $4 $.

Answer: $ \ Bex e ^ {\ Tan x}-e ^ x = e ^ \ XI (\ Tan x-x) = e ^ \ Xi \ SEZ {\ frac {1} {3} x ^ 3 + O (x ^ 3 )}. \ EEx $

 

 

(4) which of the following statements is true? ().

A. if $ f (x) $ on $ [a, B] $, and there is an original function $ f (x) $, then $ \ Bex \ int_a ^ B f (x) \ rd x = F (B)-f (a); \ EEx $

B. $ f (x) $ Riann product on $ [a, B] $, then $ \ DPS {\ int_0 ^ x F (t) \ RD t} $ can be exported on $ [a, B] $;

C. if $ f ^ 2 (x) $ can accumulate in Riann on $ [a, B] $, then $ | f (x) | $ is not necessarily in $ [, b] $ on Riann product;

D. if $ | f (x) | $ Riann product on $ [a, B] $, then $ f (x) $ in $ [, b] $ ON THE Riann product.

Answer: This is a good question. proof of a: $ \ beex \ Bea \ int_a ^ B f (x) \ rd x & =\ vlm {n} \ sum _ {k = 1} ^ n f (\ xi_ I) \ lap X_ I \\\\\\ vlm {n} \ sum _ {k = 1} ^ n \ frac {f (x_ I)-f (x _ {I-1 })} {\ lap X_ I} \ cdot \ lap X_ I \\& = F (B)-f (). \ EEA \ eeex $ B counterexample: $ \ Bex f (x) =\sedd {\ BA {ll}-1, &-1 <x <0, \ 0, & x = 0, \ 1, & 0 <x <1. \ EA} \ EEx $ C: $ f ^ 2 \ In \ calr [a, B] \ rA | f | \ In \ calr [a, B] $. in fact, $ f ^ 2 $ is continuous at $ x_0 $ and only when $ | f | $ is continuous at $ x_0 $. they have the same continuous point set. according to the theory of real-time variable functions, the conclusion is known. inverse example of D: $ f (x) $ is the Dirichlet function.

 

 

(5) set $ a> 0 $, $ f (x) $ to meet $ f'' (x)> 0 $ in $ (-a, a) $, and $ | f (x) | \ Leq x ^ 2 $, $ \ DPS {I =\int _ {-A} ^ A f (x) \ RD x }$, then there must be (B ).

A. $ I = 0 $;

B. $ I> 0 $;

C. $ I <0 $;

D. Not sure.

Answer: $ \ beex \ Bea \ frac {1} {2} [f (x) + f (-x)] &> F \ sex {\ frac {x + (-x)} {2} = f (0) = 0, \ quad x> 0; \ I & =\ int _ {-A} ^ A f (x) \ RD x = \ int _ {-A} ^ 0 f (x) \ RD x + \ int_0 ^ A f (x) \ RD x \\\\\int_0 ^ 1 [F (-x) + f (x)] \ RD x \\&> 0. \ EEA \ eeex $

 

2. ($ 10' $) set $ f (x) = A_1 \ SiN x + A_2 \ sin 2x + \ cdots + a_n \ sin NX $, and $ | f (x) | \ Leq | \ SiN x | $, $ a_ I \ (I = 1, 2, \ cdots, n) $ is a real constant. proof: $ \ Bex \ sev {a_1 + 2a_2 + \ cdots + na_n} \ Leq 1. \ EEx $

Proof: at $ \ Bex \ sev {\ frac {f (x )} {\ SiN x }}\ Leq 1 \ EEx $ \ Bex | a_1 + 2a_2 + \ cdots + na_n | \ Leq 1. \ EEx $

 

3. ($15 '$) set the function $ F $ to be consistent and continuous on $ \ BBR $, $ \ ETA> 0 $. on $ \ BBR $, use the following formula to define the function $ G $: $ \ Bex g (x) =\ sup \ sed {| f (y)-f (z) |; y, z \ In (X-\ ETA, x + \ ETA )}. \ EEx $ proof: $ G $ consistent continuity on $ \ BBR $.

Proof: $ F $ consistent continuous knowledge $ \ Bex \ forall \ ve> 0, \ exists \ Delta \ In (0, \ ETA ), \ st | X-x' | <\ Delta \ rA | f (x)-f (x') | <\ ve. \ EEx $ when $ | X-X '| <\ Delta $, the symmetry may be set to $ x <x' $, $ \ beex \ Bea \ SERD {\ BA {RL} x'-\ ETA <Y <X + \ ETA \ x'-\ ETA <z <X + \ ETA \ EA} & \ rA | f (y) -f (z) | <G (x ), \ SERD {\ BA {RL} x'-\ ETA <Y <X + \ ETA \ x + \ ETA \ Leq z <x' + \ ETA \ EA} & \ Ra \ BA {ll} | f (y) -f (z) | & \ Leq | f (y)-f (x + \ ETA) | + | f (x + \ ETA)-f (z) |\& <G (x) + \ ve, \ EA \ SERD {\ BA {RL} X + \ ETA <Y <X '+ \ ETA \ x + \ ETA <z <X' + \ ETA \ EA} & \ rA | f (y) -f (z) | <\ ve. \ EEA \ eeex $ so $ \ Bex g (x') <G (x) + \ ve. \ EEx $ Certificate-like $ \ Bex g (x) <G (x ') + \ ve. \ EEx $ $ \ Bex | G (x)-g (x') | <\ ve. \ EEx $

 

4. ($15 '$) set the function $ \ varphi $ to be continuous in the range $ [0, a] $. The function $ F $ is on the upper order of $ \ BBR $, and $ \ Bex f'' (x) \ geq 0, \ quad \ forall \ x \ In \ BBR. \ EEx $ proof: $ \ Bex f \ sex {\ frac {1} {A} \ int_0 ^ A \ varphi (t) \ RD t} \ Leq \ frac {1} {A} \ int_0 ^ A f [\ varphi (t)] \ rd t. \ EEx $

Proof: by the mean point theorem, $ \ Bex \ exists \ Xi \ In (0, ), \ ST \ frac {1} {A} \ int_0 ^ A \ varphi (t) \ rd t = \ varphi (\ XI ). \ EEx $ expand $ f (x) $ at $ \ varphi (\ xi) $ Taylor with $ \ beex \ Bea F (\ varphi (t )) & = f (\ varphi (\ XI) + f' (\ varphi (\ XI) [\ varphi (t)-\ varphi (\ xi)] + \ frac {f'' (\ ETA)} {2} [\ varphi (t)-\ varphi (\ xi)] ^ 2 \ & \ geq F (\ varphi (\ XI) + f' (\ varphi (\ XI) [\ varphi (t) -\ varphi (\ xi)]. \ EEA \ eeex $ credits on $ [0, a] $ \ beex \ Bea \ int_0 ^ a f (\ varphi (t )) \ rd t & \ geq a f (\ varphi (\ XI) + f' (\ varphi (\ XI) \ SEZ {\ int_0 ^ A \ varphi (t) \ rd t-A \ varphi (\ xi) }\\& = a f (\ varphi (\ XI )). \ EEA \ eeex $

 

5. ($10 '$) set $ f_0 (x) $ to be continuous in the range $ [0, a] $, so $ \ Bex f_n (X) = \ int_0 ^ x F _ {n-1} (t) \ RD t \ Quad (n = 1, 2, \ cdots ). \ EEx $ proof: $ \ sed {f_n (x)} $ in the range $ [0, a] $, the consistent convergence on $0 $.

Proof: If $ | f_0 | \ Leq M $ is set, it is easy to know by mathematical induction $ \ Bex \ Max _ {x \ in [0, a]} | f_n (X) | \ Leq \ frac {Ma ^ n} {n !}. \ EEx $

 

6. ($15 '$) evaluate the function $ f (x, y) = AX ^ 2 + 2bxy + cy ^ 2 $ maximum and minimum values on $ x ^ 2 + y ^ 2 \ Leq 1 $.

Answer: Set $ \ Bex a =\sex {\ BA {ll} A & B \ B & C \ EA }, \ EEx $ is easy to obtain its feature value $ \ Bex \ lm _ {1} =\ frac {A + C} {2} \ PM \ frac {\ SQRT {( a-C) ^ 2 + B ^ 2 }}{ 2 }. \ EEx $ there is an orthogonal array $ p $ to make $ \ Bex a = P ^ t \ diag \ sed {\ lm_1, \ lm_2} p. \ EEx $ note $ \ Bex \ sex {u \ atop v} = p \ sex {x \ atop y }, \ EEx $ the original question is $ \ Bex g (u, v) = \ lm_1 U_1 ^ 2 + \ lm_2u_2 ^ 2 \ EEx $ the maximum and minimum values under $ U ^ 2 + V ^ 2 \ Leq 1 $. obviously, $ \ Bex \ max F = \ Max G = \ frac {A + C} {2} + \ frac {\ SQRT {(a-c) ^ 2 + B ^ 2 }}{ 2 }, \ quad \ min F = \ min G = \ frac {A + C} {2}-\ frac {\ SQRT {(a-c) ^ 2 + B ^ 2 }}{ 2 }. \ EEx $

 

7. ($ 15' $) set $ z = z (x, y) $ by equation $ \ Bex f \ sex {x + \ frac {z} {y }, Y + \ frac {z} {x }}= 0 \ EEx $ is determined. $ F $ has a continuous partial derivative, which proves: $ \ Bex xz_x + yz_y = z-XY. \ EEx $

Proof: returns the offset of $ \ Bex F_1 '\ sex {1 + \ frac {z_x} {y} + f_2' \ sex {-\ frac {z} {x ^ 2.} + \ frac {z_x} {y }}= 0, \ quad F_1 '\ sex {-\ frac {z} {y ^ 2} + \ frac {z_y} {y} + f_2' \ sex {1 + \ frac {z_y} {x }}= 0. \ EEx $ then $ \ Bex z_x = \ frac {-F_1 '+ \ frac {z} {x ^ 2} F_2' }{\ frac {F_1 '} {Y} + \ frac {F_2 '} {x }}, \ quad z_y = \ frac {z} {y ^ 2} F_1 '-F_2'} {\ frac {F_1 '} {y} + \ frac {F_2 '} {x }}, \ EEx $ \ beex \ Bea xz_x + yz_y & =\ frac {-xf_1 '+ \ frac {z} {x} F_2' + \ frac {z} {Y} F_1 '-yf_2'} {\ frac {F_1 '} {y} + \ frac {F_2'} {x }}\\& = z-XY. \ EEA \ eeex $

 

 

Real-time function ($ 40' $)

 

 

1. Fill in blank questions ($3 \ times 4' = 12' $)

(1) The closure of rational number set $ \ BBQ $ is ($ \ BBR $ ).

(2) If $ \ sed {g _ \ al; \ Al \ In I} $ is an open set family in $ \ BBR ^ N $, the Union of $ \ DPS {g =\cup _ {\ Al \ In I} g _ \ Al} $ is (open set ).

(3) If $ e \ subset \ BBR ^ N $ is the union of multiple closed sets, $ e $ is the $ F _ \ Sigma $ set; if $ e \ subset \ BBR ^ N $ is the union of several open sets, $ e $ is the $ g _ \ Delta $ set, then, the complementary set of $ F _ \ Sigma $ is ($ g _ \ Delta $ set ). if $ F $ is a real-value function defined on the open set $ g \ subset \ BBR ^ N $, then $ f (x) the continuous set of $ is ($ g _ \ Delta $ set ).

Answer: $ \ Bex \ sed {X; f \ mbox {In} X \ mbox {in a row }}=\ sed {X; \ omega_f (X) = 0 }=\ CaP _ {n = 1} ^ \ infty \ sed {X; \ omega_f (x) <\ frac {1} {n }}, \ EEx $ \ Bex \ omega_f (x) = \ lim _ {\ Delta \ to 0} \ sup \ sed {| f (x ') -F (x') |; \ x', x' \ In U (x_0, \ delta)}, \ EEx $ \ Bex \ sed {X; \ omega_f (x) <\ frac {1} {n }}\ mbox {open set }. \ EEx $

 

2. Question ($4 \ Times 3' = 12' $)

(1) If $ f (x) $ is set to L product on $ [a, B] $, its indefinite points $ f (x) $ are not necessarily absolute continuous functions. ()

(2) set $ f (x) $ to the bounded variance function on $ [a, B] $, then $ f (x) $ at $ [, b] $ almost everywhere there is a finite derivative $ f' (x) $; and $ f' (x) $ can be accumulated on $ [a, B] $. ()

(3) The $ \ Sigma $ algebra generated by an open set family composed of $ \ BBR ^ N $ is called the Borel algebra, and its element is called the Borel set, then, the sum, intersection, and upper (lower) limit set of the Borel set are all Borel sets. ()

(4) If $ f \ subset \ BBR ^ N $ is a non-empty set and $ x_0 \ In \ BBR ^ N $, $ y_0 \ In F $, make $ | x_0-y_0 | = \ RD (x_0, f) $. ()

Answer:

(1) error. It should be ''''.

(2) pair.

(3) pair.

(4) error. If $ F $ is not empty.

 

3. the short answer ($ 8' $) describes the function columns $ \ sed {f_k (x)} $ which almost converge to the definition of $ f (x) $ everywhere; description: $ \ sed {f_k (x)} $ converges on $ e $ according to the definition of $ f (x) $, and briefly describes the relationship between the two.

Answer:

(1) $ f_k \ to F, \ AE $ e $ indicates: $ \ Bex \ exists \ e_0 \ subset E, \ me_0 = 0, \ st x \ In e \ BS e_0 \ rA f_k (x) \ to f (x), \ K \ To \ infty. \ EEx $

(2) $ f_k \ rA F $ indicates: $ \ Bex \ forall \ Delta> 0, \ vlm {k} m e [| f_k-f | \ geq \ Delta] = 0. \ EEx $

(3) If $ f_k \ to F, \ AE $ is at $ e $, and $ mE <\ infty $, then $ f_k \ rA F $.

(4) If $ f_k \ rA F $, $ \ sed {f_n} $ has subcolumns $ \ sed {F _ {n_k} $ almost always converge to $ F $.

 

4. proof question ($8 '$) use the control convergence theorem to prove the derivation theorem under the integral number: Set $ f (x, y) $ to be defined in $ e \ times (A, B) $ functions, as $ x $ functions, are product-ready on $ e $, and $ y $ functions are micro-scalable on $ (a, B) $, if $ f \ In L (E) $ exists, make $ \ Bex \ sev {\ frac {\ RD} {\ RD y} f (x, y )} \ Leq f (x,), \ Quad (x, y) \ In e \ times (a, B ), \ EEx $ \ Bex \ frac {\ RD} {\ RD y} \ int_e f (x, y) \ RD x = \ int_e \ frac {\ RD} {\ RD y} f (x, y) \ rd x. \ EEx $

Proof: For $ n \ In \ BBN $, note $ \ Bex g_n (y) = \ frac {\ int_ef \ sex {X, Y + \ frac {1} {n }}\ rd x-\ int_ef (x, y) \ RD x }{\ frac {1} {n }}=\ int_e \ frac {f \ sex {X, Y + \ frac {1} {n }}- f (x, y) }{\ frac {1} {n }}\ rd x, \ EEx $ is composed of $ \ Bex \ frac {f \ sex {X, Y + \ frac {1} {n}-f (x, y )} {\ frac {1} {n }}\ to \ frac {\ RD} {\ RD y} f (x, y), \ quad n \ To \ infty; \ EEx $ \ Bex \ sev {\ frac {f \ sex {X, Y + \ frac {1} {n}-f (x, y )} {\ frac {1} {n }}=\ sev {\ frac {\ RD} {\ RD y} f (x, y + \ xi_n )} \ Leq f (x) \ EEx $ and Lebesgue control convergence theorem knowledge $ \ Bex \ frac {\ RD} {\ RD y} \ int_ef (x, y) \ RD x = \ vlm {n} g_n (x) = \ int_e \ frac {\ RD} {\ RD y} f (x, y) \ rd x. \ EEx $

 

 

Ordinary Differential Equations ($40 '$)

 

 

1. calculation question ($3 \ times 8' = 24' $)

(1) $ (2x + 1) y' = 4x + 2y $;

(2) $ Y \ RD x + x \ RD y + y ^ 2 (x \ rd y-Y \ rd x) = 0 $;

(3) $ (2xy-x ^ 2y-y ^ 3) \ rd x-(x ^ 2 + y ^ 2-x ^ 3-xy ^ 2) \ RD y = 0 $.

Answer:

(1) $ \ BEX (4x + 2y) \ rd x-(2x + 1) \ rd y = 0. \ EEx $ note $ \ Bex M = 4x + 2y, \ quad n =-(2x + 1 ), \ EEx $ then $ \ Bex M_y-N_x = 2-(-2) = 4 \ Ra \ frac {M_y-N_x} {n} =-\ frac {4} {2x + 1 }, \ EEx $ and the original ODE has the credit factor $ \ Bex e ^ {-4 \ int \ frac {1} {2x + 1} \ RD x} = e ^ {- 2 \ Ln (2x + 1 )} = \ frac {1} {(2x + 1) ^ 2 }. \ EEx $ then $ \ Bex \ rd u =\frac {4x + 2y)} {(2x + 1) ^ 2} \ rd x-\ frac {1} {2x + 1} \ rd y = 0. \ EEx $ by $ \ Bex u_y =-\ frac {1} {2x + 1} \ rA u =-\ frac {y} {2x + 1} + \ Phi (x) \ EEx $ Zhi $ \ Bex \ frac {4x + 2y} {(2x + 1) ^ 2} = u_x = \ frac {2y} {(2x + 1) ^ 2} + \ Phi '(x), \ EEx $ \ Bex \ Phi' (x) = \ frac {4x} {(2x + 1) ^ 2} \ Ra \ PHI (x) = \ ln | 2x + 1 | + \ frac {1} {2x + 1} + C. \ EEx $ the original ode call is $ \ Bex-\ frac {y} {2x + 1} + \ ln | 2x + 1 | + \ frac {1} {2x + 1} + c = 0, \ EEx $ \ Bex y = C (2x + 1) + 1 + (2x + 1) \ ln | 2x + 1 |. \ EEx $

(2) $ \ beex \ Bea 0 & = Y \ RD x + x \ RD y + y ^ 2 (x \ rd y-Y \ rd x) \\&=\ RD (xy) + (x ^ 2y ^ 2) \ RD \ frac {y} {x }, \ 0 & =\ frac {1} {(xy) ^ 2} \ RD (xy) + \ RD \ frac {y} {x} \\\&=\ RD \ sex {-\ frac {1} {XY} + \ frac {y} {x }}. \ EEA \ eeex $. Therefore, the original ode call is $ \ Bex \ frac {y} {x}-\ frac {1} {XY} = C. \ EEx $

(3) $ \ Bex \ frac {\ RD y} {\ RD x }=\ frac {2xy-x ^ 2y-y ^ 3} {x ^ 2 + y ^ 2-x ^ 3-xy ^ 2 }. \ EEx $ order $ Y = UX $, then $ \ beex \ Bea x \ frac {\ RD u} {\ RD x} + u & =\ frac {2x ^ 2u-x ^ 3u-x ^ 3u ^ 3} {x ^ 2 + u ^ 2x ^ 2-x ^ 3-x ^ 3u ^ 2 }\\&=\ frac {2u-xu-xu ^ 3} {1 + u ^ 2-x-xu ^ 2 }, \ x \ frac {\ RD u} {\ RD x} & =\ frac {(2u-xu-xu ^ 3)-(U + u ^ 3-ux-xu ^ 3 )} {1 + u ^ 2-x-xu ^ 2 }\\&=\ frac {u-u ^ 3} {1 + u ^ 2-x (1 + u ^ 2 )} \\&=\ frac {u (1-u ^ 2) }{( 1 + u ^ 2 )}, \\\ frac {1 + u ^ 2} {u (1-u ^ 2)} \ rd u & =\ frac {1} {x ()} \ rd x, \ sex {\ frac {1} {u}-\ frac {1} {u + 1}-\ frac {1} {U-1} \ rd u & = \ sex {\ frac {1} {x}-\ frac {1} {X-1} \ rd x, \ ln \ sev {\ frac {u} {u ^ 2-1 }}&=\ ln \ frac {x} {X-1} + C_1, \ frac {u} {u ^ 2-1} & = C \ frac {x} {X-1} \ quad \ sex {C> 0 }, \ frac {XY} {y ^ 2-x ^ 2} & =\ frac {cx} {X-1 }, \ frac {y} {y ^ 2-x ^ 2} & = C (x-1), \ y (x-1) & = C (y ^ 2-x ^ 2 ), \ Quad (C> 0 ). \ EEA \ eeex $

 

2. proof question ($8 '$) set $ y' = f (x, y) $, where $ \ Bex f (x, y) = \ sedd {\ BA {ll} \ frac {4x ^ 3Y} {x ^ 4 + y ^ 2}, & x ^ 2 + y ^ 2> 0, \ 0, & X = Y = 0. \ EA} \ EEx $

(1) it is proved that the Lipschitz condition is not met in any region containing the origin;

(2) it is proved that there are infinite solutions for the above equation satisfying the initial condition $ Y (0) = 0 $.

Proof:

(1) $ \ Bex \ sev {\ frac {f (x, x ^ 4)-f (x, 0 )} {x ^ 4 }}=\ frac {1} {x ^ 4} \ sev {\ frac {4x ^ 3x ^ 4} {x ^ 4 + (x ^ 4) ^ 2 }}=\ frac {4 }{| x + x ^ 5 |}\ to + \ infty, \ quad x \ to 0. \ EEx $

(2) please first find $ \ Bex \ frac {\ RD y} {\ RD x }=\ frac {4x ^ 3Y} {x ^ 4 + y ^ 2} \ LRA 4x ^ 3Y \ rd X-(x ^ 4 + y ^ 2) \ rd y = 0. \ EEx $ order $ \ Bex M = 4x ^ 3Y, \ quad n =-(x ^ 4 + y ^ 2 ), \ EEx $ then $ \ Bex M_y-N_x = 4x ^ 3 + 4x ^ 3 = 8x ^ 3 \ Ra \ frac {M_y-N_x} {-m} =-\ frac {2} {y }, \ EEx $ and the original ODE has the credit factor $ \ Bex e ^ {-\ int \ frac {2} {y} \ RD y }=\ frac {1} {Y ^ 2 }. \ EEx $ then $ \ beex \ Bea 0 & =\ frac {4x ^ 3} {y} \ rd x-\ frac {x ^ 4} {y ^ 2} \ RD y-\ RD y \\\&=\ RD \ sex {\ frac {x ^ 4} {y}-y }, \\\ frac {x ^ 4} {y}-Y & = C, \\ x ^ 4-y ^ 2 & = cy. \ EEA \ eeex $ obtain the classless Inter-Domain code $ \ Bex y =\frac {1} {2} \ sex {-C ++ \ SQRT {C ^ 2 + 4x ^ 4 }}\ Quad (C> 0 ), \ EEx $ suitable for $ Y (0) = 0 $.

 

3. Proof question ($8 '$) proves the gronwall inequality in the differential form.

(1) set $ \ ETA (\ cdot) $ to a non-negative, absolutely continuous function on $ [0, T] $, for all $ T $ conditions of inequality $ \ bee \ label {317_ode_3_eq1} \ ETA '(t) \ Leq \ varphi (t) \ ETA (t) + \ varpsi (t), \ EEE $ here $ \ varphi (t) $ and $ \ varpsi (t) $ are in $ [0, t] $ non-negative sum, for all $0 \ Leq t \ Leq T $ \ bee \ label {317_ode_3_eq2} \ ETA (t) \ Leq e ^ {\ int_0 ^ t \ varphi (s) \ RD s} \ SEZ {\ ETA (0) + \ int_0 ^ t \ varpsi (s) \ RD s}; \ EEE $

(2) In particular, if on $ [0, T] $, $ \ Bex \ ETA '(t) \ Leq \ varphi (t) \ ETA (t ), \ mbox {And} \ ETA (0) = 0, \ EEx $ on $ [0, T] $ \ ETA \ equiv 0 $.

Proof:

(1) $ \ beex \ Bea \ ETA '(t) & \ Leq \ varphi (t) \ ETA (t) + \ varpsi (t ), \\\ SEZ {e ^ {-\ int_0 ^ t \ varphi (s) \ RD s} \ ETA (t )} '& \ Leq e ^ {-\ int_0 ^ t \ varphi (s) \ RD s} \ varpsi (t) \ Leq \ varpsi (t ), \ e ^ {-\ int_0 ^ t \ varphi (s) \ RD s} \ ETA (t)-\ ETA (0) & \ Leq \ int_0 ^ t \ varpsi (s) \ rd s, \ ETA (t) & \ Leq e ^ {\ int_0 ^ t \ varphi (s) \ RD s} \ SEZ {\ ETA (0) + \ int_0 ^ t \ varpsi (s) \ RD s }. \ EEA \ eeex $

(2) by (1), $ \ Bex \ ETA (t) \ Leq \ ETA (0) e ^ {\ int_0 ^ t \ varphi (s) \ RD s }. \ EEx $ $ \ Bex \ ETA (0) = 0 \ Ra \ ETA \ equiv 0. \ EEx $

[Jia Liwei university mathematics magazine] 317th Xiamen University 2010 comprehensive foundation I Postgraduate Entrance Exam reference