Let $y(x,\xi)$ be the value at point $x$ of the maximal solution that satisfies the initial condition $y(0)=\xi$.         Prove that the domain of $y(\cdot, y(s, \xi))$ is $I-s$, where $I$ is the domain of $y(\cdot ,\xi)$.   Prove that for all $s, t$ such that $y(s, \xi)$ and $y(t+s, \xi)$ exist, then $y(t,y(s,\xi))$ also exists and $y(t,y(s,\xi))=y(t+s, \xi)$.   If $y$ is a maximal solution and there exists $T&gt;0$ such that $y(0)=y(t)$ and $f(y(0))\neq 0$, then $y$ is a periodic solution and not constant.      In the above problem, I'm missing only that in 3., $y$ is defined on $\mathbb R$. I asked about this problem . The user Artem has provided some help in the comments regarding this, but I can't transform it into an answer.  I have a theorem that says that if for all $a,b\in I$ such that $y$ is defined on $(a,b)$, there exists $K$ such that $|y(x)|\leq K$ for all $x\in (a,b)$, then $y$ is defined on $I$.  How can I use the above to solve the problem? Why is $y$ bounded?  Otherwise, how I can use the 1. and 2. to prove that $y$ is defined on $\Bbb R$?  Some detail is appreciated, not because I'm lazy but because I feel like I need it.
124247
This isn't homework. I have no idea what theorems I should be looking at to solve this. Guidance, partial and total solutions are all welcomed.     Let $f$ be a locally lipschitz function in an open set $G\subseteq \Bbb R^n$. Consider the autonomous system $y'=f(y)$.      Let $y(x,\xi)$ be the value at point $x$ of the maximal solution that satisfies the initial condition $y(0)=\xi$.         Prove that the domain of $y(\cdot, y(s, \xi))$ is $I-s$, where $I$ is the domain of $y(\cdot ,\xi)$.   Prove that for all $s, t$ such that $y(s, \xi)$ and $y(t+s, \xi)$ exist, then $y(t,y(s,\xi))$ also exists and $y(t,y(s,\xi))=y(t+s, \xi)$.   If $y$ is a maximal solution and there exists $T&gt;0$ such that $y(0)=y(T)$ and $f(y(0))\neq 0$, then $y$ is a periodic solution and not constant.   If $y$ is a solution whose domain is $(a,+\infty)$, if $\eta:=\lim _{x\to +\infty}y(x)$ and $\eta \in G$, then $f(\eta)=0$.      EDIT  I found an alternative solution for 3. Please check my proof and give feedback in comments:  let $u$ be the restriction of $y$ to $[0,T]$, now let $\overline u$ be the periodic extension of $u$ to $\mathbb R$. It is easy to see that $\overline u$ is a solution to the given differential equation. But since $f$ is locally lipschitz, $\overline u$ must coincide with $y$ wherever they are both defined. Since $y$ is a maximal solution, it must be $\overline u$, so $y$ is defined on $\mathbb R$.
112640
Let $X = \{(x,y) \in \mathbb{R^2}: x &gt; 0, y = sin(1/x)\}$. Is this set open, closed, or neither, and what is the boundary, interior, and closure of $X$?  I just want to make sure I am thinking about these problems in a reasonable manner. I am also trying to apply this to other concepts I have learned.  Is $X$ open?  X is open if every $x \in X$ is an interior point. Take an arbitrary $p \in X$ so that $p = (p_x, p_y)$ then $p_x &gt; 0$ and $p_y = sin(1/p_x)$. Define an open ball $B_r(p)$, and take $q \in B_r(p)$ be an arbitrary point. Is $q \in X$? Is it possible for $q$ to not be in X? Since $y = sin(1/x)$ is defined and is continuous for all $x \neq 0$,then there will always exist a $q_y = sin(1/q_x)$, provided $q_x \neq 0$. How do I know $q_x \neq 0$? Otherwise, I propose that X is open.   Alternatively, X is open if whenever $f: \mathbb{R^2 \to R}$ is continuous and $U \subseteq R$ so that if $U$ is open, then $f^{-1}(U)$ is also open. Could I construct such a $U$ to prove that it's open? My only concern is that if $U = (-1,1)$, then $f^{-1}(U) = (-\infty, \infty) = \mathbb{R}$ which is not open nor closed.   Is $X$ closed?   I am not sure about this one. I want to say that is not closed, since $X^c$ is not open. It is not open because $(0,0)$ defines a boundary point which lies in the set $X^c$.  Thanks!
48060
Let   $$y''-q(x)y=0$$be a differential equation with initial conditions  on $0\leq x&lt;\infty,$ as $y(0)=1,y'(0)=1$ where $q(x)$ is a positive monotonically increasing continuous function. Then which of the following are true?   $y(x)\rightarrow\infty$ as $x\rightarrow\infty$. $y'(x)\rightarrow\infty$ as $x\rightarrow\infty$. $y(x)$ has finitely many zeros in $[0,\infty)$. $y(x)$ has infinitely many zeros in $[0,\infty)$.   Please don't mind, actually I am new in differential equation. I only know that by Picard's theorem the above differential equation has the unique solution, but I don't know what is the solution as I tried by direct hit and trial method. According to me the solution of above differential equation will be some thing in exponential form, so according to me its answer will be $a$, $b$, $c$. But I don't know the exact method. Please help me to solve the above problem. Thanks in advance.
72136
I don't know if it's late or I'm tired, but for some reason my following argument for proving every field is an integral domain seems wrong to me:     Let $k$ be a field and $x,y \in k$. Assume that   $xy=0$. Then $$ x^{-1}xy=y=0=x=xyy^{-1}.$$   The reason I think my argument is wrong is that the conclusion I arrive to is $x=0$ and $y=0$, whereas the proper approach to proving this by the contrapositive is obtaining $x=0$ or $y=0$. Could someone highlight my mistake?
57911
Assume the IVP $$ x'=x^2, \quad x(0)=1 $$ By separating the variables and using the initial condition $$ x(t)=(1-t)^{-1} $$ Then, my book states that since $x_0=1&gt;0$ then the solution can only be extended to the left with maximum existence interval $\, I=(-\infty,1)$.   My question is, since the solution $x(t)$ we found is defined for every $t\neq1$, why isn't the maximum interval of its existence $I^*=(-\infty,1)\cup(1,+\infty)$? Why does it depend on the sign of $x_0$?
92477
Let $f$ be holomorphic on $D(0,1)\cup \{1\}$, and $f(0)=0$, $f(1)=1$, $f(D(0,1))\subset D(0,1)$. Prove that $|f'(1)|\geq 1$.  I have no idea. Maybe $f'(\xi)=[f(1)-f(0)]/(1-0)$ for some $\xi \in (0,1)$, and the maximal modules principle applies? This is not exact.
61357
Let $f:\mathbb R^n\to\mathbb R$ differntiable and $f(0)=0$. Prove exist $g_i$ s.t for $x=(x_1,\dots,x_n):f(x)=\sum\limits_{i=1}^nx_i\cdot g_i(x)$. hint:$f(x)=\int\limits_0^1f\prime(tx)dt$.   I dont have any idea how to solve it. maybe we can build $g_i=f(x_i\cdot x)$ but im not sure it may help since I cannot take one time $\int f(x_1\cdot x)dx_1$ and in another $\int f(x_n\cdot x)dx_n$).  how can I build the sequence of functions and furthermore, what is the intuition for this type of questions involves building functions $\varphi:\mathbb R^n\to\mathbb R^m$ based on given condition with sums or diffrentiablity or partial derivatives?
14606
$f(S \cup T) = f(S) \cup f(T)$  $f(S)$ encompasses all $x$ that is in $S$ $f(T)$ encompasses all $x$ that is in $T$  Thus the domain being the same, both the LHS and RHS map to the same $y$, since the function $f$ is the same for both.  Can you post the solution?
117220
$S := \{x \in \Bbb R^3: ||x||_2 = 1 \} \subset (\Bbb R^3, || \cdot ||_2 )$ and $T:S^2 \to (\Bbb R, |\cdot |)$ is  a continuous function.  I've already shown that $$T_{\mathrm{max}} := \mathrm{sup}\{ T(x) : x \in S^2\}  \: \: \mathrm{ and }\: \: T_{\mathrm{min}}:= \mathrm{inf}\{ T(x) : x \in S^2\}$$ are reached by $T$ with the argument that $S^2$ is closed and bounded and therefore after the Heine-Borel-Theorem compact, and therefore since T is continuous it assumes its max and min in $S^2$.   Now i have answer the following:     1) Is there a point $x_0 \in S^2$ with $T(x_0)=T(-x_0)$ ?      2) Is there a value $T \in(T_{\mathrm{min}}, T_{\mathrm{max}})$ which is only assumed at exactly one point of $S^2$ ?   I wanted to show that $T$ is injective to get that 1) is false and 2) is true but i didn't get anywhere. Any ideas? Thanks in advance!
106543
I don't know how to prove this example. Let $y(t):\mathbb{R}^+\to \mathbb{R}$ be a nonincreasing function. Assume that there is a constant $A&gt;0$ such that \begin{equation} \int_{s}^{+\infty}y(t)dt \leq Ay(s), \quad 0\leq s&lt;+\infty \end{equation} Then $y(t)\leq y(0)e^{1-\frac{t}{A}}$, for all $t&gt;0$.
114779
Suppose $f(x)$ is continuous in $[0,1]$,  derivable in $(0,1)$, $f(0)=0, f(1)=\frac 12$, prove    $\exists \xi, \eta \in (0,1)$, such that   $$  f(\xi)+f'(\eta)=\xi+\eta. $$   I have tried Cauchy mean value theorem:  let $F(x)=\int_0^x f(t)-t \,dt, G(x)=\frac{1}{2}x^2-f(x)$, then $\exists \xi \in (0,1), \eta \in (0,1)$, such that  $$ \frac{F(1)-F(0)}{1-0} = F'(\xi)  \Leftrightarrow f(\xi)-\xi = \int_0^1 f(t) \,dt - \frac 12. $$ $$ \frac{G(1)-G(0)}{1-0} = G'(\eta)  \Leftrightarrow \eta-f'(\eta) = \frac 12 - f(1) = 0. $$  but failed.  Any hints are appreciated, thanks for your help.    The conditions of this question are wrong, but if i rectify them, the question will contradict the anwser of Mr. Robert  Z to some extent, so please be sure to see the anwser after skimming this question. I am sorry for the inconvenience.
36751
I have to show that the domain defined by the half unit circle: $\left \{ \left ( x,y \right )\in \mathbb{R}^2: x^2+y^2\leq 1, x\geq 0 \right \}$ for the function $f:D\rightarrow \mathbb{R}$ is closed (the function is $f(x,y)=2xy^2-2x^2$), so that I can argue the function has a min/max value. I have shown it is bounded and continuous.
42672
This is from Putnam:     If a function $f: \mathbb{R} \to [-1, 1]$ is such that $f''(x)$ exists for all $x \in \mathbb{R}$ and $\{f(0)\}^{2} + \{f'(0)\}^{2} = 4$ then prove that there is a point $c$ such that $f(c) + f''(c) = 0$.   Now the condition concerning values of $f(0), f'(0)$ (and the fact that range of $f$ is subset of $[-1, 1]$) implies that $f'(0) \neq 0$ so that function is not a constant. If we take $g(x) = \{f(x)\}^{2} + \{f'(x)\}^{2}$ then we can see that $g' = 2f'(f + f'')$. One hope of solving the problem is to show that $g$ attains local extrema at some point different from an extrema of $f$. This would ensure that $g'$ vanishes without making $f'$ vanish and that will lead to $f + f'' = 0$. However the constraint on $f(0), f'(0)$ does not help in analyzing the extrema of $f$ or $g$.  Another line of thought which could be helpful here is to consider $F(x) = f'(x)\sin x - f(x)\cos x$ so that $F'(x) = \sin x\{f(x) + f''(x)\}$ but in this case I am not able to see how to use the condition on values $f(0), f'(0)$ to impose some constraint on $F(x)$.  Perhaps both the approaches which came to my mind are not in right direction (or may be they are, but I am unable to see). Please provide any hints.  Update: Based on the answer by Sandeep Thilakan, I have dropped the requirement for continuity of $f''(x)$.
118331
Show that the system $x^{'}=x-y-x^{3}$,$y^{'}=x+y-y^{3}$ has a periodic solution.   I converted to polar:  $r r^{'}=x^{'}x+y^{'}y.$  Thus  $r r^{'}=x^{2}-x^{4}+y^{2}-y^{4}.$  Collecting squares:  $rr^{'}=x^{2}+y^{2}-(x^{2}+y^{2})^{2} +2x^{2}y^{2}.$  Substitute:  $r^{'}=r-r^{3}(1-(\sin^{2}(2\theta))/2)$  $r^{'}=r-r^{3}(1+\cos^{2}(2\theta))/2.$  Setting $r^{'}=0$ yields  $r^{2}=2/(1+\cos^{2}(2\theta).$  Thus $r_{max}= 2^{(1/2)}$ and $r_{min}=1.$  I suck at trig and this doesn't look right. Clearly $(0,0)$ is the only fixed point, so I have a bounded region. I just don't think I have found correct values for $r_{min}$ and $r_{max}$ to find my limit cycle.
9368
Theorem: If $R$ is an integral domain, then $R[x]$ is an integral domain as well.  Proof: Let $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$ and $g(x)=b_0+b_1x+b_2x^2+\cdots+b_mx^m$ be two elements of $R[x]$. Let $0\leq i\leq n$ and $0\leq j\leq m$ such that $i$ & $j$ are the smalles numbers such that $a_i\ne 0$ & $b_j\ne 0$. $f(x)\cdot g(x)=\sum_{i=0}^{n+m}{c_ix^i}.$ $c_{i+j}=\overbrace{a_{i+j}b_0+a_{i+j-1}b_1+\cdots +a_{i+1}b_{j-1}}^{=0} +\overbrace{a_ib_j}^{\ne 0}+\overbrace{a_{i-1}b_{j+1}+\cdots+a_0b_{i+j}}^{=0}$ $=a_i+b_j\ne 0\Longrightarrow f(x)g(x)\ne 0$    I don't know why $a_{i+j}b_0+a_{i+j-1}b_1+\cdots +a_{i+1}b_{j-1}=0$ and $a_{i-1}b_{j+1}+\cdots+a_0b_{i+j}=0$.
2791
Let $y$ be a nontrivial solution of the boundary value problem  $$\frac{d^2y}{dx^2}+xy=0,\ x \in [a,b]$$  $$y(a)=y(b)=0$$ Then:   $b&gt;0$ $y$ is monotone in $(a,0)$ if $a&lt;0&lt;b$. $y'(a)=0$. $y$ has infinitely many zeros in $[a,b]$   I have no idea how to solve it and which options is/ are correct. please help me.
124029
So, for the ODE \begin{align} x' = -x^{3} + \sin t, \end{align} we can show that there exists a $2\pi$ periodic solution.  To do this, we denote by \begin{align} x(t,\alpha) \end{align} The solution $x(t)$ of the ODE such that $x(0) = \alpha$.  Then, let $\alpha \in [-2,2]$.  Consider the function $f(\alpha) = x(2\pi, \alpha)$ \begin{align} x &gt; 1 &amp;\Rightarrow x' &lt; 0 \\ x &lt; 1 &amp;\Rightarrow x' &gt; 0. \end{align}  Thus, solutions beginning in $[-2,2]$ stay there and we can use Brouwer's fixed point theorem to show that there exists a fixed point of $f$.  Therefore there is some $\alpha^{*}$ such that $f(\alpha^{*}) = \alpha^{*}$, which represents the periodic solution $x(t,\alpha)$.  Now, my question: Is this $2\pi$- periodic solution unique? Are there other $2\pi$ periodic solutions of this ODE?  I have started by trying to subtract two periodic solutions from one another but this hasn't taken me anywhere productive.