Differential Equations Assignment #1: answers.

213567 solutionsDifferential Equations

Assignment #1: answers.Note.A mathematical text is written in human language; every formula should be part of a

proper sentence. Needless (?) to say, a sentence begins with a capital letter and ends with a punctuation symbol. Otherwise the text is not meant to be read. This elementary principle is often disregarded; however writing lines and lines of formulas is not acceptable. The idea is that your assignments should show your reasoning, not your computation; that someone withlessknowledge than you should be able to see the key ideas, and the names of the methods. By reading the solutions below, you will be impressed by the little amount of symbols. But high-level mathematics proceeds in sentences, not in formulas.Do not focus on the computa- tions, focus on the logic.

Exercise 1.Consider the following equation:

x ?(t) = 2tx(t) +1t exp(t2) defined onR>0, with initial conditionx(1) = 0. 1. W ritethe so-called homogeneous equation (without an yinitial cond ition).Giv ethe general solution (of the homogeneous equation). 2. Giv eone sp ecialsolution of the original equation. Then giv ethe general solution (of the original equation). 3. No wgiv ethe only solution satisfying the initial condition.

Solution.

1.

The asso ciatedhomogeneous equation is:

x ?(t) = 2tx(t) which as we know solves intox(t) = exp(?2tdt) =λet2forλ?R.Notes. (a) Man yof y ouwrote " λ(t)·et2". This is simply wrong -λmust be a constant. (b) Some of y ouha veb eenu singthe so-called "separable v ariablesmetho d",as follo ws: Write x?x = 2t, which integrates aslog(x) =et2+C... This is mathematicallynot rigorousfor two reasons, and naive for one. i. Division b yx: butxcould vanish. Indeed, the mapx(t) = 0is a solution. ii. Use of the logarithm: there is no w ell-definedlogfunction onR. The slightly more correctlog|x|=et2does not solve this issue: again, what aboutx= 0? iii. Ho wdo y ouexp ectto generalise this division to the v ectorcase? In short, nothing beats quick use of the exponential. For desperate (non-linear) scalar situations, separating variables gives some intuition; itsrigoroustreatment is however much heavier than anything we discussed so far. Refrain from using something you cannot justify! 1 solutions2.W euse the metho dcalled "v ariationof the parameter", and lo okfor one sp ecialsolution of the original equation in the formx(t) =λ(t)·et2.

For this to be a solution, we need to have:

?(t)·et2=1t et2 and thereforeλ?(t) =1t This suggests to letx(t) = log(t)·et2, which is well-defined onI=R>0. A quick compu- tation shows that this is a solution of the original equation. Hence the general solution has the form: (λ+ log(t))·et2 for someλ?R.Notes. (a) Do not f orgetto c heckthat the xyou define is a solution indeed. (A necessary condition is not always sufficient, so getting the condition onλis a priori no serious guarantee.) (b) It is mathematically andgrammatically incorrect to write "the one special solu- tion". 3. The linear case of the Cauc hy-Lipschitztheorem p redictsthat there will b eexactly one maximal solution to the equation with initial condition, and that it will be defined onR>0. For an arbitrary solutionx(t) = (λ+ log(t))·et2to satisfyx(1) = 0, one needsλ= 0. One quickly checks thatx(t) = log(t)·et2is a solution of the equation, and satisfies the initial condition; it is even unique as such by the Cauchy-Lipschitz theorem. Exercise 2.We consider the following differential equation: (EH) :x(4)=x(3)+ 7x(2)-13x?+ 6x. 1.

Pro videa basis of the space solution SH.

2. Giv ethe explicit solution with the initial conditions x(0) = 2;x?(0) = 0;x??(0) =-2andx(3)(0) =-4

Solution.

1. W econsider the c haracteristicp olynomialasso ciatedto th eproblem:

P(λ) =λ4-λ3-7λ2+ 13λ-6.

The value1is an obvious root; dividing (there is no need for Euclidean division here, you can do it mentally) one findsP(λ) = (λ-1)(λ3-7λ+ 6). Here again1is a root of the second factor, yieldingP(λ) = (λ-1)2(λ2-λ-6). The latter term has discriminant25, and solutions

1±52

, namely-2and3. Hence the characteristic polynomials has roots (counted with multiplicity):-2,1,1,3. So the functionset,tet,e-2t,e3tare solutions of the differential equation(EH). Because it is linear homogeneous of order4, the Cauchy-Lipschitz theorem predicts that the spaceSHof solutions has dimension4; since the solutions we gave are clearly linearly independent, they form a basis ofSH.

As a conclusion,abasis ofSHis{et,tet,e-2t,e3t}.

2 solutionsNotes. (a) One should not need to write do wna p olynomialdivision here; the factor isation was too simple. (b) Nev erin troducethe discriminan tΔwithout saying the word "discriminant" (this is a typical example of unclear mathematical exposition; as a rule, never use an unexplained notation, however universal you think it is). (c) A v ectorspace almost nev erhas a unique basis. It is therefore mathemati cally incorrect to write "the basis is". 2. W eno wtak ethe initial condition in toacc ount.The ge neralsolution has the form: x(t) =aet+btet+ce-2t+de3t. We can then compute and evaluate the successive derivatives: x(0) =a+c+d; x ?(0) =a+b-2c+ 3d; x ??(0) =a+ 2b+ 4c+ 9d; x ???(0) =a+ 3b-8c+ 27d. Hence the initial condition gives rise to the following system in matrix form, where we let Λbe the vector of coordinates in the given basis: (((1 0 1 1

1 1-2 3

1 2 4 9

1 3-8 27)

(((2 0 -2 -4) This system is quickly handled by introducing an augmented matrix and using Gauß elimination, as follows: (((1 0 1 12

1 1-2 30

1 2 4 9-2

1 3-8 27-4)

)))-------→L

2←L2-L1L3←L3-L1L4←L4-L1(

(((1 0 1 12

0 1-3 2-2

0 2 3 8-4

0 3-9 26-6)

L

3←L3-2L2L4←L4-3L2(

(((1 0 1 12

0 1-3 2-2

0 0 9 40

0 0 0 200

At this stage one clearly hasc=d= 0; finallya= 2andb=-2. So the only solution

with initial condition isx(t) = 2et-2tet= 2(1-t)et.Note.More than ever in this question, what matters is not the mathematical formulas,

but the sentences linking them.

Exercise 3.Consider the following matrix:

A=?0-1

1 0?

Prove that:

exp(A) =?cos(1)-sin(1) sin(1) cos(1)? 3

solutionsSolution.We treatAas a complex-valued matrix. Its characteristic polynomial isλ2+1, which

has roots±i. The2×2matrixAhas two distinct eigenvalues, so it is diagonalisable.

Let us compute the eigenspaces:

ker(A-iI) = ker?-i-1 1-i? = ker? 1-i? =C?i 1? and likewise, ker(A+iI) = ker?i-1 1i? = ker? 1i? =C?1 i? Hence ??i 1? ,?1 i?? is an eigenbasis, which suggests to indtroduce the following coordinate change matrix: P=?i1 1i?

Its inverse is easily computed:

P -1=-12 i-1 -1i? by construction (no computation is required here), P solutionsDifferential Equations

Assignment #1: answers.Note.A mathematical text is written in human language; every formula should be part of a

proper sentence. Needless (?) to say, a sentence begins with a capital letter and ends with a punctuation symbol. Otherwise the text is not meant to be read. This elementary principle is often disregarded; however writing lines and lines of formulas is not acceptable. The idea is that your assignments should show your reasoning, not your computation; that someone withlessknowledge than you should be able to see the key ideas, and the names of the methods. By reading the solutions below, you will be impressed by the little amount of symbols. But high-level mathematics proceeds in sentences, not in formulas.Do not focus on the computa- tions, focus on the logic.

Exercise 1.Consider the following equation:

x ?(t) = 2tx(t) +1t exp(t2) defined onR>0, with initial conditionx(1) = 0. 1. W ritethe so-called homogeneous equation (without an yinitial cond ition).Giv ethe general solution (of the homogeneous equation). 2. Giv eone sp ecialsolution of the original equation. Then giv ethe general solution (of the original equation). 3. No wgiv ethe only solution satisfying the initial condition.

Solution.

1.

The asso ciatedhomogeneous equation is:

x ?(t) = 2tx(t) which as we know solves intox(t) = exp(?2tdt) =λet2forλ?R.Notes. (a) Man yof y ouwrote " λ(t)·et2". This is simply wrong -λmust be a constant. (b) Some of y ouha veb eenu singthe so-called "separable v ariablesmetho d",as follo ws: Write x?x = 2t, which integrates aslog(x) =et2+C... This is mathematicallynot rigorousfor two reasons, and naive for one. i. Division b yx: butxcould vanish. Indeed, the mapx(t) = 0is a solution. ii. Use of the logarithm: there is no w ell-definedlogfunction onR. The slightly more correctlog|x|=et2does not solve this issue: again, what aboutx= 0? iii. Ho wdo y ouexp ectto generalise this division to the v ectorcase? In short, nothing beats quick use of the exponential. For desperate (non-linear) scalar situations, separating variables gives some intuition; itsrigoroustreatment is however much heavier than anything we discussed so far. Refrain from using something you cannot justify! 1 solutions2.W euse the metho dcalled "v ariationof the parameter", and lo okfor one sp ecialsolution of the original equation in the formx(t) =λ(t)·et2.

For this to be a solution, we need to have:

?(t)·et2=1t et2 and thereforeλ?(t) =1t This suggests to letx(t) = log(t)·et2, which is well-defined onI=R>0. A quick compu- tation shows that this is a solution of the original equation. Hence the general solution has the form: (λ+ log(t))·et2 for someλ?R.Notes. (a) Do not f orgetto c heckthat the xyou define is a solution indeed. (A necessary condition is not always sufficient, so getting the condition onλis a priori no serious guarantee.) (b) It is mathematically andgrammatically incorrect to write "the one special solu- tion". 3. The linear case of the Cauc hy-Lipschitztheorem p redictsthat there will b eexactly one maximal solution to the equation with initial condition, and that it will be defined onR>0. For an arbitrary solutionx(t) = (λ+ log(t))·et2to satisfyx(1) = 0, one needsλ= 0. One quickly checks thatx(t) = log(t)·et2is a solution of the equation, and satisfies the initial condition; it is even unique as such by the Cauchy-Lipschitz theorem. Exercise 2.We consider the following differential equation: (EH) :x(4)=x(3)+ 7x(2)-13x?+ 6x. 1.

Pro videa basis of the space solution SH.

2. Giv ethe explicit solution with the initial conditions x(0) = 2;x?(0) = 0;x??(0) =-2andx(3)(0) =-4

Solution.

1. W econsider the c haracteristicp olynomialasso ciatedto th eproblem:

P(λ) =λ4-λ3-7λ2+ 13λ-6.

The value1is an obvious root; dividing (there is no need for Euclidean division here, you can do it mentally) one findsP(λ) = (λ-1)(λ3-7λ+ 6). Here again1is a root of the second factor, yieldingP(λ) = (λ-1)2(λ2-λ-6). The latter term has discriminant25, and solutions

1±52

, namely-2and3. Hence the characteristic polynomials has roots (counted with multiplicity):-2,1,1,3. So the functionset,tet,e-2t,e3tare solutions of the differential equation(EH). Because it is linear homogeneous of order4, the Cauchy-Lipschitz theorem predicts that the spaceSHof solutions has dimension4; since the solutions we gave are clearly linearly independent, they form a basis ofSH.

As a conclusion,abasis ofSHis{et,tet,e-2t,e3t}.

2 solutionsNotes. (a) One should not need to write do wna p olynomialdivision here; the factor isation was too simple. (b) Nev erin troducethe discriminan tΔwithout saying the word "discriminant" (this is a typical example of unclear mathematical exposition; as a rule, never use an unexplained notation, however universal you think it is). (c) A v ectorspace almost nev erhas a unique basis. It is therefore mathemati cally incorrect to write "the basis is". 2. W eno wtak ethe initial condition in toacc ount.The ge neralsolution has the form: x(t) =aet+btet+ce-2t+de3t. We can then compute and evaluate the successive derivatives: x(0) =a+c+d; x ?(0) =a+b-2c+ 3d; x ??(0) =a+ 2b+ 4c+ 9d; x ???(0) =a+ 3b-8c+ 27d. Hence the initial condition gives rise to the following system in matrix form, where we let Λbe the vector of coordinates in the given basis: (((1 0 1 1

1 1-2 3

1 2 4 9

1 3-8 27)

(((2 0 -2 -4) This system is quickly handled by introducing an augmented matrix and using Gauß elimination, as follows: (((1 0 1 12

1 1-2 30

1 2 4 9-2

1 3-8 27-4)

)))-------→L

2←L2-L1L3←L3-L1L4←L4-L1(

(((1 0 1 12

0 1-3 2-2

0 2 3 8-4

0 3-9 26-6)

L

3←L3-2L2L4←L4-3L2(

(((1 0 1 12

0 1-3 2-2

0 0 9 40

0 0 0 200

At this stage one clearly hasc=d= 0; finallya= 2andb=-2. So the only solution

with initial condition isx(t) = 2et-2tet= 2(1-t)et.Note.More than ever in this question, what matters is not the mathematical formulas,

but the sentences linking them.

Exercise 3.Consider the following matrix:

A=?0-1

1 0?

Prove that:

exp(A) =?cos(1)-sin(1) sin(1) cos(1)? 3

solutionsSolution.We treatAas a complex-valued matrix. Its characteristic polynomial isλ2+1, which

has roots±i. The2×2matrixAhas two distinct eigenvalues, so it is diagonalisable.

Let us compute the eigenspaces:

ker(A-iI) = ker?-i-1 1-i? = ker? 1-i? =C?i 1? and likewise, ker(A+iI) = ker?i-1 1i? = ker? 1i? =C?1 i? Hence ??i 1? ,?1 i?? is an eigenbasis, which suggests to indtroduce the following coordinate change matrix: P=?i1 1i?

Its inverse is easily computed:

P -1=-12 i-1 -1i? by construction (no computation is required here), P