ADVANCED MATHEMATICS: INTEGRAL CALCULUS AND DIFFERENTIAL EQUATIONS
CONTENTS ORDINARY DIFFERENTIAL EQUATIONS PRELIMINARIES SEPARABLE EQUATIONS EXERCISES EXACT DIFFERENTIAL EQUATIONS INTEGRATING FACTOR LINEAR DIFFERENTIAL EQUATIONS HOMOGENEOUS DIFFERENTIAL EQUATIONS BERNOULLI EQUATION HIGHER-ORDER DIFFERENTIAL EQUATIONS EQUATIONS WITH CONSTANT COEFFICIENTS NONHOMOGENEOUS EQUATIONS 1 10 WRONSKIAN METHOD 1 10 1 EXERCISES 1 11 CAUCHY-EULER EQUATION 1 11 1 EXERCISES PARTIAL DIFFERENTIAL EQUATIONS INTRODUCTION LINEAR EQUATIONS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS EQUATIONS WITH VARIABLE COEFFICIENTS QUASI-LINEAR EQUATIONS BURGER’S EQUATION SHOCK PATH SECOND-ORDER PDES WAVE EQUATION AND D’ALEMBERT’S SOLUTION VIBRATING STRING WITH FIXED ENDS HEAT EQUATION SOLUTION OF THE HEAT EQUATION HEAT EQUATION ON SEMI-INFINITE DOMAIN: DIRICHLET CONDITION HEAT EQUATION ON SEMI-INFINITE DOMAIN: NEUMANN CONDITION WAVE EQUATION ON SEMI-INFINITE DOMAIN MATRICES AND SYSTEMS OF LINEAR EQUATIONS SYSTEMS OF EQUATIONS AND GAUSSIAN ELIMINATION HOMOGENEOUS SYSTEMS MATRICES DETERMINANTS AND INVERSE OF MATRICES APPLICATION TO LEAST SQUARE FITTING VECTOR SPACES EIGENVALUES-EIGENVECTORS INNER PRODUCT SPACES DIAGONALIZATION QUADRATIC FORMS 3 10 FUNCTIONS OF SYMMETRIC MATRICES 3 10 1 EXERCISES CALCULUS OF VARIATIONS EULER-LAGRANGE EQUATION IMPACT OF Y? ON EULER-LAGRANGE EQUATION NECESSARY AND SUFFICIENT CONDITIONS APPLICATIONS GENERALIZATION OF EULER-LAGRANGE EQUATION NATURAL BOUNDARY CONDITIONS IMPACT OF Y?? ON EULER-LAGRANGE EQUATION DISCONTINUITY IN EULER-LAGRANGE EQUATION 4 10 TRANSVERSALITY CONDITION 4 10 1 PROBLEM OF BOLZA 4 10 2 EXERCISES 4 11 CORNERS AND BROKEN EXTREMAL 4 11 1 EXERCISES 4 12 VARIATIONAL PROBLEMS WITH CONSTRAINTS 4 12 1 EXERCISES 4 13 ISOPERIMETRIC PROBLEMS 4 13 1 EXERCISES 4 14 STURM-LIOUVILLE PROBLEM 4 14 1 THE FIRST EIGENVALUE 4 14 2 EXERCISES 4 15 RAYLEIGH RITZ METHOD 4 15 1 EXERCISES 4 16 MULTIPLE INTEGRALS 4 16 1 EXERCISES INTEGRAL EQUATIONS INTRODUCTION AND CLASSIFICATIONS CONNECTION BETWEEN ORDINARY DIFFERENTIAL EQUATIONS AND INTEGRAL EQUATIONS THE GREEN’S FUNCTION FREDHOLM INTEGRAL EQUATIONS AND GREEN’S FUNCTION BEAM PROBLEM FREDHOLM INTEGRAL EQUATIONS WITH SEPARABLE KERNELS SYMMETRIC KERNEL ITERATIVE METHODS AND NEUMANN SERIES APPROXIMATING NON-DEGENERATE KERNELS LAPLACE TRANSFORM AND INTEGRAL EQUATIONS FREQUENTLY USED LAPLACE TRANSFORMS 5 10 ODD BEHAVIOR 5 10 1 EXERCISES IS ONE OF THE ENDPOINTS OF THE INTERVAL I, THEN THE ABOVE DEFINITION OF THE DERIVATIVE BECOMES A ONE-SIDED DERIVATIVE IF F ?(X0) EXISTS AT EVERY POINT X0 ?I, THEN WE SAY F IS DIFFERENTIABLE ON I AND WRITE F ?(X) THE DERIVATIVE OF A FUNCTION F IS AGAIN A FUNCTION F ?; ITS DOMAIN, WHICH IS A SUBSET OF THE DOMAIN OF F, IS THE SET OF ALL POINTS X0 FOR WHICH F IS DIFFERENTIABLE OTHER NOTATIONS FOR THE DERIVATIVE ARE DX F,D F/DX, AND DY/DX, WHERE Y = F(X) THE FUNCTION F ? MAY IN TURN HAVE A DERIVATIVE, DENOTED BY F ??, WHICH IS DEFINED AT ALL POINTS WHERE F ? IS DIFFERENTIABLE F ?? IS CALLED THE SECOND DERIVATIVE OF F FOR HIGHER-ORDER DERIVATIVES, WE USE THE NOTATIONS F ???(X), F (4)(X),, F (N)(X), OR DN F DXN FOR N = 1,2,3, H|H| H = 0 IN CONCLUSION, F ?(X) = 2|X| FOR ALL X ?R ? A DIFFERENTIAL EQUATION IS AN EQUATION INVOLVING A FUNCTION AND DERIVATIVES OF THIS FUNCTION DIFFERENTIAL EQUATIONS ARE DIVIDED INTO TWO CLASSES: ORDINARY AND PARTIAL ORDINARY DIFFERENTIAL EQUATIONS CONTAIN ONLY FUNCTIONS OF A SINGLE VARIABLE, CALLED THE INDEPENDENT VARIABLE, AND DERIVATIVES WITH RESPECT TO THAT VARIABLE PARTIAL DIFFERENTIAL EQUATIONS CONTAIN A FUNCTION OF TWO OR MORE VARIABLES AND SOME PARTIAL DERIVATIVES OF THIS FUNCTION THE ORDER OF A DIFFERENTIAL EQUATION IS DEFINED BY THE HIGHEST DERIVATIVE PRESENT IN THE EQUATION AN NTH-ORDER ORDINARY DIFFERENTIAL EQUATION IS A FUNCTIONAL RELATION OF THE FORM F X,Y, DY DX, D2Y DX2 , D3Y DX3 ,, DNY DXN = 0, X ?R, (1 2) BETWEEN THE INDEPENDENT VARIABLE X AND THE DEPENDENT VARIABLE Y, AND ITS DERIVA- TIVES DY DX, D2Y DX2 , D3Y DX3 ,, DNY DXN WE SHALL ALWAYS ASSUME THAT (1 2) CAN BE SOLVED FOR Y(N) AND PUT IN THE FORM Y(N) = F X,Y, Y?, , Y(N?1) (1 3) LOOSELY SPEAKING, BY A SOLUTION OF (1 2) ON AN INTERVAL I, WE MEAN A FUNCTION Y(X) = ?(X) SUCH THAT F X,?(X),??(X),,?(N)(X) PRELIMINARIES IS DEFINED FOR ALL X ?I AND F X,?(X),??(X),,?(N)(X) = F(X,Y+H)?F(X,Y) H MORE OFTEN WE WRITE FX, FY, TO DENOTE ?F ?X AND ?F ?Y , RESPECTIVELY FOR THE (IVP) (1 3) AND (1 4), THE FOLLOWING EXISTENCE AND UNIQUENESS RESULT IS TRUE FOR MORE DISCUSSION ON THE TOPIC AND ON THE PROOF OF THE NEXT THEOREM, WE REFER TO [3], [4], OR [5] THEOREM 1 1 CONSIDER THE (IVP) DEFINED BY (1 3) AND (1 4), WHERE F IS CONTINUOUS ON THE (N+1)-DIMENSIONAL RECTANGLE D OF THE FORM D = {(X, Y0, Y1, , YN?1) : BK?1 < YK?1 < DK?1 AND BN < X < DN, K = 1, 2, , N} IF THE INITIAL CONDITIONS ARE CHOSEN SO THAT THE POINT (X0,A0,A1,,AN?1) IS IN D, THEN THE (IVP) HAS AT LEAST ONE SOLUTION SATISFYING THE INITIAL CONDITIONS IF, IN ADDITION, F HAS CONTINUOUS PARTIAL DERIVATIVES ?F ?Y , ?2 F ?Y? , , ?N?1 F ?YN?1 , IN D, THEN THE SOLUTION IS UNIQUE SUPPOSE F SATISFIES THE HYPOTHESIS OF THEOREM 1 1 IN D THEN A GENERAL SOLUTION OF THE (IVP) IN D IS GIVEN BY THE FORMULA Y = ?(X,C1,C2,,CN) ARE CONTINUOUS IN THE UPPER HALF-PLANE DEFINED BY Y > 0 WE CONCLUDE FROM COROLLARY 1 THAT FOR ANY POINT (X0,Y0), Y0 > 0, THERE IS SOME INTERVAL AROUND X0 ON WHICH THE GIVEN DIFFERENTIAL EQUATION HAS A UNIQUE SOLUTION ? EXAMPLE 1 3 CONSIDER Y?(X) = Y2, Y(0) = 1 HERE, WE HAVE THAT F(X,Y) = Y2 AND ?F ?Y = 2Y ARE CONTINUOUS EVERYWHERE IN THE PLANE AND IN PARTICULAR ON THE RECTANGLE D = {(X,Y) : ?2 < X < 2, 0 < Y < 2} SINCE THE INITIAL POINT (0,1) LIES INSIDE THE RECTANGLE, COROLLARY 1 GUARANTEES A UNIQUE SOLUTION OF THE (IVP) ? IN THE NEXT EXAMPLE, WE ILLUSTRATE THE EXISTENCE OF MORE THAN ONE SOLUTION EXAMPLE 1 4 CONSIDER THE DIFFERENTIAL EQUATION Y?(X) = 2Y1/3, Y(0) = 0, X ?R HERE, WE HAVE F(X,Y) = 2Y1/3 AND ?F ?Y = 2Y?2/3 SEPARABLE EQUATIONS X3/2, FOR X > SEPARABLE EQUATIONS CONSIDER THE FIRST-ORDER DIFFERENTIAL EQUATION Y? = F(X,Y) (1 5) WHERE F AND ?F ?Y ARE CONTINUOUS FOR ALL (X,Y) IN A RECTANGLE D OF THE FORM D = {(X,Y) : B1 < X < B2 AND D1 < Y < D2} ORDINARY DIFFERENTIAL EQUATIONS IN SOME CASES, WE MAY WANT TO SOLVE (1 5) AND, AT DIFFERENT TIMES, WE MAY WANT TO SOLVE IT SUBJECT TO THE INITIAL CONDITION Y(X0) = Y0, (1 6) WHERE THE POINT (X0,Y0) IS SPECIFIED AND IN D IN SOME INSTANCES, THE FIRST-ORDER DIF- FERENTIAL EQUATION (1 5) CAN BE REARRANGED AND PUT IN THE FORM, G(Y)DY DX +H(X) = G(U)DU, (1 9) WHICH IS EQUIVALENT TO (1 8) SEPARABLE EQUATIONS EXAMPLE 1 5 SOLVE Y? = 5/2?Y/2, Y(0) = 2 THE DIFFERENTIAL EQUATION CAN BE WRIT- TEN AS 5?Y DY DX = 2 USING (1 9), WE OBTAIN X 2DS = Y 5?UDU FOR FINITE X, WE MUST HAVE U < 5; OTHERWISE, THE INTEGRAL ON THE RIGHT DIVERGES THERE- FORE, 5?U > 0 AND INTEGRATION YIELDS X 2 = ?LN(5?Y)+LN(3), OR X 2 = LN OR DY DT = RY 1?Y D , WHERE R = ? ?? AND D = R/? THIS IS THE DIFFERENTIAL LOGISTIC EQUATION NOTE THAT IT IS SEPARABLE AND CAN BE SOLVED EXPLICITLY ? ORDINARY DIFFERENTIAL EQUATIONS EXAMPLE 1 7 THE LAW OF MASS ACTION IS A USEFUL CONCEPT THAT DESCRIBES THE BEHAVIOR OF A SYSTEM THAT CONSISTS OF MANY INTERACTING PARTS, SUCH AS MOLECULES, THAT REACT WITH EACH OTHER, OR VIRUSES THAT ARE PASSED ALONG FROM A POPULATION OF INFECTED INDIVIDUALS TO NON-IMMUNE ONES THE LAW OF MASS ACTION WAS DERIVED FIRST FOR CHEMICAL SYSTEMS BUT SUBSEQUENTLY FOUND WIDE USE IN EPIDEMIOLOGY AND ECOLOGY TO DESCRIBE THE LAW OF MASS ACTION, WE ASSUME M SUBSTANCES S1,S2,,SM TOGETHER FORM A PRODUCT WITH CONCENTRATION P THEN THE LAW OF MASS ACTION STATES THAT DP DT IS PROPORTIONAL TO THE PRODUCT OF THE M CONCENTRATIONS SI,I = 1,,M THAT IS, DP DT = KS1S2 SM SUPPOSE WE HAVE A HOMOGENEOUS POPULATION OF FIXED SIZE, DIVIDED INTO TWO GROUPS THOSE WHO HAVE THE DISEASE ARE CALLED INFECTIVE, AND THOSE WHO DO NOT HAVE THE DIS- EASE ARE CALLED SUSCEPTIBLE LET S = S(T) BE THE SUSCEPTIBLE PORTION OF THE POPULATION AND I = I(T) BE THE INFECTIVE PORTION THEN BY ASSUMPTION, WE MAY NORMALIZE THE POPULATION AND HAVE S+I = 1 WE FURTHER ASSUME THAT THE DYNAMICS OF THIS EPIDEMIC SATISFY THE LAW OF MASS ACTION HENCE, FOR SOME POSITIVE CONSTANT ? WE HAVE THE NONLINEAR DIFFERENTIAL EQUATION I?(T) = ?SI (1 10) LET I(0) = I0, 0 < I(0) < 1 BE A GIVEN INITIAL CONDITION IT FOLLOWS THAT BY SUBSTITUTING S = 1?I INTO (1 10), I?(T) = ?I(1?I), I(0) = I0 (1 11) IF WE CAN SOLVE (1 11) FOR I(T), THEN S(T) CAN BE FOUND FROM THE RELATION I +S = 1 WE SEPARATE THE VARIABLES IN (1 11) AND OBTAIN DI I(1?I) = ?DT USING PARTIAL FRACTIONS ON THE LEFT SIDE OF THE EQUATION AND THEN INTEGRATING BOTH SIDES YIELDS LN(|I|)?LN(|1?I|) = ?T +C, OR FOR SOME POSITIVE CONSTANT C1 WE HAVE I(T) = C1E?T 1+C1E?T APPLYING I(0) = I0 GIVES THE SOLUTION I(T) = I0E?T 1?I0 +I0E?T (1 12) NOW FOR 0 < I(0) < 1, THE SOLUTION GIVEN BY (1 12) IS INCREASING WITH TIME AS EX- PECTED MOREOVER, USING L’HOSPITAL’S RULE, WE HAVE LIM T??I(T) = LIM T?? I0E?T 1?I0 +I0E?T = 1 HENCE, THE INFECTION WILL GROW, AND EVERYONE IN THE POPULATION WILL GET INFECTED EVEN- TUALLY ? SEPARABLE EQUATIONS Y? = ?X?Y, Y(2) = Y? = XLN(Y), Y(1) = Y? = P 1?Y2, Y(0) = 0 EXERCISE 1 8 SHOW THAT THE (IVP) Y? = ? P 4?Y2, Y(0) = Y? +XY = Y, Y(1) = 3 EXACT DIFFERENTIAL EQUATIONS IN THIS SECTION, WE LOOK AT A SPECIAL TYPE OF DIFFERENTIAL EQUATIONS CALLED EXACT DIFFER- ENTIAL EQUATIONS GIVEN A FUNCTION F(X,Y), THE TOTAL DERIVATIVE OF F IS GIVEN BY D F = FXDX+ FYDY DEFINITION 1 1 (EXACT EQUATION) Q(X,Y) DY DX +P(X,Y) = 0 IS AN EXACT EQUATION IF AND ONLY IF THE DIFFERENTIAL FORM Q(X,Y) DY+P(X,Y) DX IS EXACT, I E THERE EXISTS A FUNCTION F(X,Y) FOR WHICH D F = Q(X,Y) DY+P(X,Y) DX IF P(X,Y) DX + Q(X,Y) DY IS AN EXACT DIFFERENTIAL OF F, THEN D F = P(X,Y) DX + Q(X,Y) DY BUT BY THE CHAIN RULE, D F = ?F ?X DX + ?F ?Y DY AND THIS EQUALITY HOLDS FOR ANY DISPLACEMENTS DX,DY SO ?F ?X = P, ?F ?Y = Q FROM THIS, WE HAVE ?2 F ?Y?X = ?P ?Y , ?2 F ?X?Y = ?Q ?X EXACT DIFFERENTIAL EQUATIONS WE KNOW THAT THE TWO MIXED SECOND DERIVATIVES ARE EQUAL SO ?P ?Y = ?Q ?X THE CONVERSE IS NOT NECESSARILY TRUE EVEN IF THIS EQUATION HOLDS, THE DIFFERENTIAL NEED NOT BE EXACT HOWEVER, IT IS TRUE IF THE DOMAIN IS SIMPLY-CONNECTED DEFINITION 1 2 (SIMPLY-CONNECTED DOMAIN) A DOMAIN D IS SIMPLY-CONNECTED IF IT IS CONNECTED AND ANY CLOSED CURVE IN D CAN BE SHRUNK TO A POINT IN D WITHOUT LEAVING D EXAMPLE 1 8 A DISC IN 2D IS SIMPLY-CONNECTED A DISC WITH A “HOLE” IN THE MIDDLE IS NOT SIMPLY-CONNECTED BECAUSE A LOOP AROUND THE HOLE CANNOT BE SHRUNK INTO A POINT SIMILARLY, A SPHERE IN 3D IS SIMPLY-CONNECTED BUT A TORUS IS NOT ? THEOREM 1 2 IF ?P ?Y = ?Q ?X THROUGHOUT A SIMPLY-CONNECTED DOMAIN D, THEN P(X,Y) DX+Q(X,Y) DY = DX+ XCOS(XY)+EY DY = 0 WE HAVE P = YCOS(XY)+1, Q = XCOS(XY)+EY THEN ?P ?Y = ?Q ?X = COS(XY)?XYSIN(XY) HENCE, WE ARE DEALING WITH AN EXACT EQUATION TO FIND THE FUNCTION F WE INTEGRATE EITHER ?F ?X = YCOS(XY)+1, OR ?F ?Y = XCOS(XY)+EY INTEGRATING THE FIRST EQUATION GIVES F(X,Y) = SIN(XY)+X+G(Y) NOTE THAT SINCE IT WAS A PARTIAL DERIVATIVE WITH RESPECT TO X HOLDING Y CONSTANT, THE “CONSTANT” TERM CAN BE ANY FUNCTION OF Y DIFFERENTIATING THE DERIVED F WITH RESPECT TO Y, WE HAVE ?F ?Y = XCOS(XY)+G?(Y) = XCOS(XY)+EY THUS G?(Y) = EY AND G(Y) = EY THE CONSTANT OF INTEGRATION NEED NOT BE INCLUDED IN THE PRECEDING LINE SINCE THE SOLUTION IS F(X,Y) = C THIS IS DUE TO THE FACT THAT D F(X,Y) = IMPLIES THAT F(X,Y) = CONSTANT THUS, THE FINAL SOLUTION IS SIN(XY)+X+EY = C IF WE ARE GIVEN AN INITIAL CONDITION, SAY Y(?/2) = 1, THEN C = 1+?/2+E ? INTEGRATING FACTOR IN SOME CASES, IT IS POSSIBLE TO CONVERT A FIRST-ORDER DIFFERENTIAL EQUATION OF THE FORM (1 13) THAT IS NOT EXACT INTO AN EXACT ONE IN SUCH A SCENARIO, IT IS POSSIBLE TO MULTIPLY (1 13) BY A FUNCTION µ(X,Y), CALLED THE INTEGRATING FACTOR THUS WE MAKE THE FOLLOWING DEFINITION DEFINITION 1 3 AN INTEGRATING FACTOR µ(X,Y) OF (1 13) IS A FUNCTION THAT MAKES µ(X,Y)P(X,Y) DX+ µ(X,Y)Q(X,Y) DY = (1 14) AN EXACT DIFFERENTIAL EQUATION THAT IS ?µ(X,Y)P ?Y = ?µ(X,Y)Q ?X WE HAVE THE FOLLOWING THEOREM THEOREM 1 3 EQUATION (1 13) HAS (A) AN INTEGRATING FACTOR OF X ALONE, µ(X) = E F(X) IF AND ONLY IF Q PY ?QX = F(X), (1 15) (B) AN INTEGRATING FACTOR OF Y ALONE, µ(Y) = EG(Y) IF AND ONLY IF P QX ?PY = G(Y) (1 16) PROOF SUPPOSE (1 15) HOLDS WITH µ(X) = EF(X) THEN ? µQ ?X = ?µ ?X Q+ µ ?Q ?X = µ Q PY ?QX Q+ µ ?Q ?X SINCE µ?(X) = µ(X)F(X) = µ ?P ?Y = ? µP ?Y THE REST OF THE PROOF WILL BE IN THE EXERCISES EXAMPLE 1 10 CONSIDER THE DIFFERENTIAL EQUATION 6XYDX+(4Y+9X2)DY = 0, Y > WHICH IS NOT EXACT NOW Q PY ?QX = 6X?18X , EXACT DIFFERENTIAL EQUATIONS WHICH IS NOT A FUNCTION OF ONE VARIABLE ALONE HOWEVER, P QX ?PY = 6XY 18X?6X = Y AND SO BY (B) OF THEOREM 1 3, THE EQUATION HAS THE INTEGRATING FACTOR µ(Y) = E Y DY = Y2 HENCE THE NEW DIFFERENTIAL EQUATION Y2H 6XYDX+(4Y+9X2)DY I = (Y2 COS(X)?3X2Y?4X)DX+(2YSIN(X)?X3 +LN(Y)+1)DY = 0, Y(0) = E EXERCISE 1 23 SUPPOSE µ(X) IS A FUNCTION OF ONLY X AND IS AN INTEGRATING FACTOR OF P(X,Y)DX+Q(X,Y)DY = 0 SHOW THAT Q PY ?QX IS A FUNCTION OF X ALONE EXERCISE 1 24 PROVE PART (B) OF THEOREM 1 3 G(S,Y(S))E? X S A(U)DUDS, X ?X0 (1 18) IT CAN BE EASILY SHOWN THAT IF Y(X) SATISFIES (1 18), THEN IT SATISFIES (1 17) EXPRESSION (1 18) IS KNOWN AS THE VARIATION OF PARAMETERS FORMULA WE NOTE THAT (1 18) IS A FUNCTIONAL EQUATION IN Y SINCE THE INTEGRAND IS A FUNCTION OF Y IF WE REPLACE THE LINEAR DIFFERENTIAL EQUATIONS E?A(X?S)H(S)DS, X ?X0 (1 20) IT IS EASY TO COMPUTE, USING (1 20) THAT THE DIFFERENTIAL EQUATION Y?(X)?3Y = E2X, Y(0) = EXERCISES EXERCISE 1 27 SOLVE EACH OF THE GIVEN DIFFERENTIAL EQUATION (A) XDY DX +2Y = 5XY, Y(1) = 0 (B) DY DX +2Y = EX, Y(1) = 0 (C) (X+1)DY DX +(X+2)Y = 2XE?X (D) DY DX + EX 1+X2 = Y, Y(1) = 0 (E) XDY DX = YLN(X), Y(1) = 2 EXERCISE 1 28 FIND A CONTINUOUS SOLUTION SATISFYING Y?(X)+Y(X) = F(X), Y(0) = 0, X > 1 IS THE SOLUTION DIFFERENTIABLE AT X = 1? ORDINARY DIFFERENTIAL EQUATIONS EXERCISE 1 29 FIND A CONTINUOUS SOLUTION SATISFYING Y?(X)+2XY(X) = F(X), Y(0) = WHERE F(X) = ? ? ? X, 0 ?X < HOMOGENEOUS DIFFERENTIAL EQUATIONS IN THIS SECTION WE LOOK AT FIRST-ORDER DIFFERENTIAL EQUATIONS THAT CAN BE WRITTEN IN THE FORM P(X,Y) DX+Q(X,Y) DY = 0 (1 21) WE BEGIN WITH THE FOLLOWING DEFINITION DEFINITION 1 4 A FUNCTION G(X,Y) OF TWO VARIABLES IS CALLED HOMOGENEOUS OF DEGREE ? IF FOR ALL X,Y WHERE G IS DEFINED AND FOR ALL POSITIVE CONSTANTS ? WE HAVE G(?X,?Y) = ? ?G(X,Y) (1 22) FOR EXAMPLE THE FUNCTION H(X,Y) = XY + Y2 IS HOMOGENEOUS OF DEGREE ? = 2, SINCE H(?X,?Y) = ? 2(XY+Y2) = ? 2H(X,Y) ON THE OTHER HAND, H(X,Y) = XY + Y2 + 1 IS NOT HOMOGENEOUS OF ANY DEGREE ? SINCE H(?X,?Y) = ? 2(XY+Y2)+1 ?= ? ?H(X,Y), FOR ANY ? DEFINITION 1 5 THE DIFFERENTIAL EQUATION (1 21) IS CALLED A HOMOGENEOUS DIFFERENTIAL EQUATION IF BOTH P AND Q SATISFY CONDITION (1 22) FOR THE SAME ? METHOD OF FINDING SOLUTIONS SUPPOSE (1 21) IS A HOMOGENEOUS DIFFERENTIAL EQUATION THEN YOU MAY USE EITHER TRANSFORMATION (A) Y = UX OR (B) X = VY WHERE U AND V ARE CONTINUOUS AND ARE FUNCTIONS OF THE VARIABLE X THE CHOICE OF EITHER (A) OR (B) DEPENDS ON THE NUMBER OF TERMS THAT MULTIPLY P AND Q FOR EXAMPLE, IF P IS MULTIPLIED BY FEWER TERMS, THEN GO WITH (B) SIMILARLY, IF Q IS MULTIPLIED BY FEWER HOMOGENEOUS DIFFERENTIAL EQUATIONS TERMS, THEN GO WITH (A) YOU MAY USE EITHER (A) OR (B) IF THEY ARE MULTIPLIED BY THE SAME NUMBER OF TERMS SAY WE GO WITH Y = UX THEN COMPUTE DY DX = XDU DX +U MULTIPLYING BOTH SIDES BY DX WE GET DY = X DU+U DX (1 23) NEXT SUBSTITUTE Y = UX AND (1 23) INTO (1 21) AND THE RESULTING DIFFERENTIAL EQUATION IS SEPARABLE IN X AND U AND CAN BE EASILY SOLVED IF WE GO WITH (B), THEN USE DX = Y DV+V DY (1 24) AND THEN SUBSTITUTE BACK INTO (1 21) TO OBTAIN A SEPARABLE EQUATION IN TERMS OF V AND Y EXAMPLE 1 11 CONSIDER XDY DX +X?3Y = 0 (1 25) THEN THE EQUATION (1 25) TAKES THE FORM XDY+(X?3Y)DX = 0, WHICH IS HOMOGENEOUS OF DEGREE 1 SINCE DY IS MULTIPLIED BY ONE TERM ONLY, WE USE Y = UX USING THE ABOVE PROCEDURE, THE DIFFERENTIAL EQUATION REDUCES TO X(UDX+XDU)+(X?3XU)DX = 0 SIMPLIFYING BY X AND THEN REGROUPING WE ARRIVE AT THE SEPARABLE EQUATION DX X = DU 2U?1 INTEGRATING BOTH SIDES AND THEN SUBSTITUTING U = Y/X WE ARRIVE AT THE SOLUTION OF THE ORIGINAL PROBLEM LN|X| = +C, FOR SOME CONSTANT C ? NOTICE THAT (1 25) CAN BE WRITTEN AS DY DX = 1+3Y X = F(1, Y X), AND THIS HINTS AT ANOTHER WAY OF DEFINING HOMOGENEOUS DIFFERENTIAL EQUATIONS THUS WE MAKE ANOTHER ALTERNATE DEFINITION DEFINITION 1 6 A DIFFERENTIAL EQUATION DY DX = G(X,Y) (1 26) IS CALLED A HOMOGENEOUS DIFFERENTIAL EQUATION IF G IS A HOMOGENEOUS FUNCTION OF DE- GREE 0, THAT IS G(?X,?Y) = ? 0G(X,Y) = G(X,Y) ORDINARY DIFFERENTIAL EQUATIONS IF WE LET ? = X, THEN G(?X,?Y) = ? 0G(X,Y) = G(X,Y) = G(1, Y X) HENCE, IF WE LET F( Y X) = G(1, Y X), THEN (1 26) CAN BE PUT IN THE FORM DY DX = F(Y X) (1 27) THIS SUGGESTS MAKING THE SUBSTITUTION U = Y X IN THIS CASE THE DIFFERENTIAL EQUATION GIVEN BY (1 27) IS REDUCED TO DU DX = F(U)?U X , WHICH IS SEPARABLE ANOTHER TYPE OF FIRST-ORDER DIFFERENTIAL EQUATION THAT REQUIRES TRANSFORMATIONS IN BOTH THE DEPENDENT AND INDEPENDENT VARIABLES IS OF THE FORM DY DX = AX+BY+C DX+EY+G, (1 28) WITH AE ?= BD TO SOLVE (1 28) WE PROPOSE THE SUBSTITUTIONS X = T ?P, Y = W?K, WHERE THE CONSTANTS P AND K MUST BE CAREFULLY CHOSEN SO THAT THE RESULTING EQUATION IS HOMOGENEOUS USING THE CHAIN RULE, WE SEE THAT DY DX = D DX(W?K) = DW DX = DW DT DT DX = DW DT (1) = DW DT MOREOVER, AX+BY+C = AT +BW+(C?AP?BK) SIMILARLY, DX+EY+G = DT +EW+(G?DP?EK) IN ORDER FOR THE RESULTING EQUATION TO BE HOMOGENEOUS WE REQUIRE C?AP?BK = 0, G?DP?EK = 0, WHICH HAS A UNIQUE SOLUTION P AND K SINCE AE ?= BD THIS RESULTS IN THE NEW DIFFERENTIAL EQUATION DW DT = AT +BW DT +EW, (1 29) WHICH IS HOMOGENEOUS IN W AND T HOMOGENEOUS DIFFERENTIAL EQUATIONS ?2X?Y+4 IT IS CLEAR THAT AE ?= BD WE USE THE SUBSTITUTIONS X = T ?P, Y = W?K AND SOLVE FOR THE UNIQUE SOLUTION OF ?5?P?2K = 0, 4+2P+K = XDY DX = Y+ P X2 ?Y2, Y(1) = 0 BERNOULLI EQUATION IN THE PREVIOUS SECTION WE USED SUBSTITUTIONS TO SOLVE FIRST-ORDER DIFFERENTIAL EQUA- TIONS THE SAME INTUITION CAN BE APPLIED TO THE BERNOULLI DIFFERENTIAL EQUATION Y? +P(X)Y = Q(X)YN (1 30) NOTE THAT IF N = 0, THEN (1 30) IS A LINEAR DIFFERENTIAL EQUATION AND IT CAN BE SOLVED BY THE METHOD OF SECTION 1 4 SIMILARLY, IF N = 1, THEN (1 30) IS A SEPARABLE DIFFERENTIAL EQUATION THAT CAN BE SOLVED BY THE METHOD OF SECTION 1 2 THUS, WE CONSIDER (1 30) ONLY FOR N ?= 0,1 WE MAKE THE SUBSTITUTION W = Y1?N, SO THAT Y? = N?1W N 1?NW ? SUBSTITUTING INTO (1 30) WE ARRIVE AT THE LINEAR DIFFERENTIAL EQUATION IN W AND X W ? +(1?N)P(X)W = (1?N)Q(X) (1 31) THAT CAN BE NOW SOLVED BY THE METHOD OF SECTION 1 4 EXAMPLE 1 13 CONSIDER XDY DX ?Y = EXY3, X > 0 HIGHER-ORDER DIFFERENTIAL EQUATIONS X, AND Q(X) = EX X THUS, BY (1 31) WE HAVE THE NEW LINEAR DIFFERENTIAL EQUATION W ? + HIGHER-ORDER DIFFERENTIAL EQUATIONS NOW IS THE TIME TO CONSIDER HIGHER-ORDER LINEAR DIFFERENTIAL EQUATIONS IN PARTICULAR, WE CONSIDER THE GENERAL NTH ORDER LINEAR DIFFERENTIAL EQUATION AN(X)DNY DXN +AN?1(X)DN?1Y DXN?1 ++A1(X)DY DX +A0(X)Y = F(X) (1 32) UNLESS OTHERWISE NOTED, WE ALWAYS ASSUME THAT THE COEFFICIENTS AI(X), I = 1,2,,N AND THE FUNCTION F(X) ARE CONTINUOUS ON SOME OPEN INTERVAL I THE INTERVAL I MAY BE UNBOUNDED IF THE FUNCTION F(X) VANISHES FOR ALL X ?I, THEN WE CALL (1 32) A HOMO- GENEOUS LINEAR EQUATION; OTHERWISE, IT IS NONHOMOGENEOUS THUS THE HOMOGENEOUS ORDINARY DIFFERENTIAL EQUATIONS LINEAR EQUATION ASSOCIATED WITH (1 32) IS AN(X)DNY DXN +AN?1(X)DN?1Y DXN?1 ++A1(X)DY DX +A0(X)Y = 0 (1 33) DEFINITION 1 7 LET F1, F2, , FN BE SET OF FUNCTIONS DEFINED ON AN INTERVAL I WE SAY THE SET { F1, F2, , FN} IS LINEARLY DEPENDENT ON I IF THERE EXISTS CONSTANTS C1, C2, , CN NOT ALL ZERO, SUCH THAT C1 F1 +C2 F2 ++CN FN = FOR EVERY X ?I IF THE SET OF FUNCTIONS IS NOT LINEARLY DEPENDENT ON THE INTERVAL I, IT IS SAID TO BE LINEARLY INDEPENDENT DEFINITION 1 8 (FUNDAMENTAL SET OF SOLUTIONS) A SET OF N SOLUTIONS OF THE LINEAR DIF- FERENTIAL SYSTEM (1 33) ALL DEFINED ON THE SAME OPEN INTERVAL I, IS CALLED A FUNDAMENTAL SET OF SOLUTIONS ON I IF THE SOLUTIONS ARE LINEARLY INDEPENDENT FUNCTIONS ON I WE HAVE THE FOLLOWING COROLLARY COROLLARY 2 LET THE COEFFICIENTS AI(X), I = 1,2,,N BE CONTINUOUS ON AN INTERVAL I IF {?1(X), ?2(X), , ?N(X)} FORM A FUNDAMENTAL SET OF SOLUTIONS ON I, THEN THE GENERAL SOLUTION OF (1 33) IS GIVEN BY Y(X) = C1?1(X)+C2?2(X)++CN?N(X), FOR CONSTANTS CI, I = 1,2,,N EXAMPLE 1 14 THE SECOND-ORDER DIFFERENTIAL EQUATION Y??(X)+Y?(X)?2Y(X) = 0, (1 34) HAS THE TWO SOLUTIONS ?1(X) = EX AND ?2(X) = E?2X AND THEY ARE LINEARLY INDEPENDENT ON I = (??,?) AND HENCE THE FUNDAMENTAL SOLUTION IS Y(X) = C1?1(X)+C2?2(X) = C1EX +C2E?2X, FOR CONSTANTS C1 AND C2 ? TO COMPLETELY DESCRIBE THE SOLUTION OF EITHER (1 32) OR (1 33), WE IMPOSE THE INITIAL CONDITIONS Y(X0) = D0, Y?(X0) = D1, , Y(N?1)(X0) = DN?1, (1 35) FOR AN INITIAL POINT X0, AND CONSTANTS DI, I = 0,1,2,,N?1, THEOREM 1 4 [EXISTENCE AND UNIQUENESS] CONSIDER THE (IVP) DEFINED BY (1 32) AND (1 35), WHERE F AND AI(X), I = 1,2,,N ARE CONTINUOUS ON SOME OPEN INTERVAL I AS- SUME AN(X) ?= 0 FOR ALL X ?I FOR X0 ?I AND GIVEN THE CONSTANTS DI, I = 0,1,2,,N?1, THE DIFFERENTIAL EQUATION (1 32) HAS A UNIQUE SOLUTION ON THE ENTIRE INTERVAL I SATISFY- ING THE INITIAL CONDITION (1 35) HIGHER-ORDER DIFFERENTIAL EQUATIONS FN F ? F ? F ? N F (N?1) F (N?1) AND Y2 ARE LINEARLY INDEPENDENT IF AND ONLY IF W(Y1(X),Y2(X)) ?= 0, FOR ALL X ?I PROOF LET Y1 AND Y2 BE LINEARLY INDEPENDENT THEN, C1Y1(X)+C2Y2(X) = 0, FOR ALL X ?I, IS TRUE ONLY WHEN C1 = C2 = 0 THEN A SIMPLE DIFFERENTIATION YIELDS C1Y1(X)+C2Y2(X) = 0 AND C1Y? 1(X)+C2Y? 2(X) = (1 37) ORDINARY DIFFERENTIAL EQUATIONS FOR ALL X ?I NOW (1 37) CAN ONLY BE TRUE FOR SOME C1, AND C2 NOT BOTH ZERO IF AND ONLY IF W(Y1(X),Y2(X)) = 0, FOR ALL X ?I IF (1 37) HOLDS FOR SOME POINT X0 ?I, THEN THE FUNCTION Y = C1Y1 + C2Y2 IS A SOLUTION (1 36) AND SATISFIES THE INITIAL CONDITIONS, Y(X0) = Y?(X0) = 0 ON THE OTHER HAND THE ZERO FUNCTION; Y = 0 IS ALSO A SOLUTION AND SATISFIES THE INITIAL CONDITIONS THIS VIOLATES THE UNIQUENESS OF THE SOLUTION UNLESS Y AND THE ZERO SOLUTION, Y = 0 ARE THE SAME NOW Y = 0 IMPLIES (1 37) IS TRUE FOR ALL X ?I THIS SHOWS THAT W(Y1(X),Y2(X)) = 0, FOR ALL X ?I IF AND ONLY IF W(Y1(X),Y2(X)) = 0, FOR AT LEAST ONE X0 ?I THIS COMPLETES THE PROOF NEXT, WE DEFINE THE GENERAL SOLUTION OF THE NONHOMOGENEOUS DIFFERENTIAL EQUATION (1 32) THE PROOF OF THE NEXT THEOREM WILL BE LEFT AS AN EXERCISE THEOREM 1 6 LET THE COEFFICIENTS AI(X), I = 1,2,,N AND THE FUNCTION F(X) BE CON- TINUOUS ON AN INTERVAL I SUPPOSE {?1(X), ?2(X), , ?N(X)} FORM A FUNDAMENTAL SET OF SOLUTIONS ON I OF THE HOMOGENEOUS DIFFERENTIAL EQUATION (1 33), DENOTE SUCH SOLU- TION WITH YH(X) = C1?1(X)+C2?2(X)++CN?N(X), FOR CONSTANTS CI, I = 1,2,,N LET YP(X) BE A PARTICULAR SOLUTION OF THE NONHOMOGE- NEOUS DIFFERENTIAL EQUATION (1 32) THEN THE GENERAL SOLUTION OF (1 32) ON I IS GIVEN BY Y(X) = YH(X)+YP(X) EXAMPLE 1 15 CONSIDER THE SECOND-ORDER DIFFERENTIAL EQUATION Y?? +3Y? +2Y = 6, Y(0) = 1, Y?(0) = ?3 (1 38) CLEARLY, EACH OF THE FUNCTIONS ?1(X) = E?X, AND ?2(X) = E?2X IS A SOLUTION OF THE HOMOGENEOUS EQUATION Y?? +3Y? +2Y = 0 ALSO, THEY ARE LINEARLY INDEPENDENT SINCE W(?1,?2) = E?X E?2X ?E?X ?2E?2X = ?3E?3X ?= 0, FOR ALL X ?(??,?) THUS, THE HOMOGENEOUS SOLUTION OF Y?? +3Y? +2Y = 0 IS GIVEN BY YH(X) = C1E?X +C2E?2X, FOR CONSTANTS C1 AND C2 MOREOVER, YP(X) = 3 IS A PARTICULAR SOLUTION OF (1 38), SINCE Y?? P +3Y? P +2YP = 6 THUS, THE GENERAL SOLUTION OF (1 38) IS Y(X) = YH(X)+YP(X) = C1E?X +C2E?2X +3 APPLYING THE INITIAL CONDITIONS WE ARRIVE AT C1 = ?7, AND C2 = 5 LATER ON IN THE CHAPTER WE WILL LOOK AT DIFFERENT TECHNIQUES FOR FINDING THE PARTICULAR SOLUTION YP(X) ? EQUATIONS WITH CONSTANT COEFFICIENTS EQUATIONS WITH CONSTANT COEFFICIENTS WE CONSIDER THE NTH ORDER DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS ANY(N)(X)+AN?1Y(N?1)(X)++A2Y??(X)+A1Y?(X)+A0Y(X) = = 0 SINCE, ERX ?= 0 FOR ANY FINITE R OR X, WE MUST HAVE THAT ANRN +AN?1RN?1 ++A2R2 +A1R +A0 = 0 (1 41) EQUATION (1 41) IS REFERRED TO AS THE CHARACTERISTIC EQUATION OR THE AUXILIARY EQUA- TION D DX +A0 (1 43) EQUATIONS WITH CONSTANT COEFFICIENTS KY = 0 (1 46) EVERY SOLUTION OF THE KTH ORDER DIFFERENTIAL EQUATION IN (1 46) WILL ALSO BE A SOLUTION OF THE ORIGINAL DIFFERENTIAL EQUATION L Y = 0 SINCE ER1X IS ALREADY A KNOWN SOLUTION, WE SEARCH FOR OTHER SOLUTIONS OF THE FORM Y(X) = U(X)ER1X, (1 47) WHERE THE FUNCTION U IS TO BE DETERMINED USING THE PRODUCT RULE WE OBTAIN (D?R1) U(X)ER1X = (DU(X))ER1X +R1U(X)ER1X ?R1U(X)ER1X = (DU(X))ER1X OR, (D?R1) U(X)ER1X = (DU(X))ER1X APPLYING U(X)ER1X TO (1 46) AND BY AN INDUCTION ARGUMENT ON K (SEE EXERCISE 1 43) IT CAN BE SHOWN THAT (D?R1)K U(X)ER1X = (DKU(X))ER1X