Ordinary Differential Equations

Try $y=x^2:2x^2-4x^3+4x^3-2x^2=0$. So $\phi (x)=x^2$, giving $\phi (4)=16$ and $\phi (5)=25$.

$f\in C^1\Rightarrow$ locally Lipschitz ⇒ unique local solution (Picard). |x′| ≤ sup|f| $<\infty \Rightarrow$ the solution cannot blow up, so it is global, and it is Lipschitz with constant sup|f|.

$y^{1/3}$ is not Lipschitz at 0; y ≡ 0 and y = $(2x/3{)}^{3/2}$ (for x ≥ 0) both work, and so do translates.

Both y ≡ 0 and $y=x^3$ solve it (for the latter, $3(x^3{)}^{2/3}$ $=3x^2)$, and gluing — y = 0 up to any c ≥ 0, then $(x-c{)}^3$ — produces a continuum of solutions. No contradiction with Picard–Lindelöf, because (4) fails: $|3y^{2/3}$|/|y| = $3|y{|}^{-1/3}$ $\to \infty$ as y → 0, so no Lipschitz constant exists at 0.

Eigenvalues of A are 6 and 1 (trace 7, det 6), both positive: every non-zero solution grows like $e^{6t}$ or $e^t$. So (1) fails, (2) fails along the 6-eigendirection, (3) fails along the 1-eigendirection, and (4) holds since 10 > 6.

The integrating factor eˣ gives (eˣy)′ = 1, so eˣy = x and y = $x{e}^{-x}$. Hence y(1) = 1/e — the resonance case where the forcing solves the homogeneous equation, producing the factor x.

W = eˣ·$(-e^{-x})$ − $e^{-x}$·(eˣ) = −1 − 1 = −2, a nonzero constant — consistent with Abel's identity for y″ − y = 0, where p ≡ 0 makes W constant, and with the independence of the two solutions.

Abel's identity gives W′ = −p(x)W, so $W(x)=W(x_0)\exp (-\int p).(1)$ If W vanishes once it vanishes identically. (2) Combining (1) with the two-dimensionality of the solution space: independence forces W nowhere zero, and W nowhere zero forces independence. (3) Fails unless p ≡ 0 — take y″ + y′ = 0, where W is a non-constant exponential. (4) Fails if $y_1,y_2$ happen to be dependent, $e.g.y_2=2y_1$, in which case they do not span the solution space.

The characteristic polynomial is ${\lambda }^2+1$. The flow is rotation: $(x,y)(t)=(x_0\cos t+y_0\sin t,-x_0\sin t+y_0\cos t)$, so every orbit is a circle traversed with period $2\pi$ and ‖x(t)‖ is constant. Purely imaginary eigenvalues give a centre; spiralling requires a nonzero real part, so (4) fails.

The operator is self-adjoint with respect to the inner product ⟨f, g⟩ $=\int fgw$, which gives (1) and (2) by the usual argument: for eigenpairs $({\lambda }_1,y_1)$ and $({\lambda }_2,y_2)$, integrating by parts twice yields $({\lambda }_1-{\lambda }_2)$⟨$y_1,y_2$⟩ = 0. (3) and (4) both fail: a regular Sturm–Liouville problem has a countably infinite sequence of simple eigenvalues ${\lambda }_1<{\lambda }_2<\cdots$ with ${\lambda }_n\to +\infty$. The model case $-y^{\prime \prime }=\lambda y$ on $[0,\pi ]$ with $y(0)=y(\pi )=0$ has ${\lambda }_n=n^2$.

The boundary conditions force y = sin(nx) with $\lambda =n^2,n\ge 1$. Orthogonality: ${\int }_0^{\pi }\sin ($mx)sin(nx) dx = 0 for m ≠ n. For $\lambda =0$ the general solution ax + b dies under both conditions, so 0 is not an eigenvalue. Simplicity holds in every regular Sturm–Liouville problem with separated conditions: two independent eigenfunctions for one $\lambda$ would exhaust the solution space of the second-order ODE, contradicting the boundary conditions.

Trace = −4 < 0, det = 3 > 0, discriminant 16 − 12 = 4 > 0: real negative eigenvalues (−1, −3).

(c) fails (a centre or unstable node can have det > 0); (d) fails — [[−1, 5], [5, −1]] has eigenvalues 4 and −6.

$\det A={\lambda }_1{\lambda }_2<0$ forces real eigenvalues of opposite sign — a complex pair would give the positive product |$\lambda$|${}^2$. Opposite signs is precisely a saddle, which is always unstable, so (3) fails. The invariant lines are the two eigendirections: decay along one, growth along the other, with all other trajectories sweeping between them.

Complex eigenvalues with nonzero imaginary part give rotation (a spiral/focus); the real part −1 < 0 makes every trajectory decay like $e^{-t}$. Centre would need real part exactly 0, saddle needs real eigenvalues of opposite sign.