In thinking about Sturm-Liouville theory a bit I see I have no actual idea what is going on.  The first issue I have is that my book began with the statement that given  $$L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$  the problem $L[y] \ = \ f$ can be re-cast in the form $L[y] \ = \ \lambda y$.  Now it could be a typo on their part but I see no justification for the way you can just do that!  More importantly though is the motivation for Sturm-Liouville theory in the first place. The story as I know it is as follows:  Given a linear second order ode   $$F(x,y,y',y'') = L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$  it is an exact equation if it is derivable from a differential equation of one order lower, i.e.  $$F(x,y,y',y'') = \frac{d}{dx}g(x,y,y').$$  The equation is exact iff  $$a''(x) - b'(x) + c(x) = 0. $$  If $F$ is not exact it can be made exact on multiplication by a suitable integrating factor $\alpha(x)$.  This equation is exact iff  $$(\alpha(x)a(x))'' - (\alpha(x)b(x))' + \alpha(x)c(x) = 0 $$  If you expand this out you get the Adjoint operator  $$L^*[\alpha(x)] \ = \ (\alpha(x)a(x))'' \ - \ (\alpha(x)b(x))' \ + \ \alpha(x)c(x) \ = 0 $$  If you expand $L^*$ you see that we can satisfy $L \ = \ L^*$ if $a'(x) \ = \ b(x)$ &amp; $a''(x) \ = \ b'(x)$ which then turns $L[y]$ into something of the form  $$L[y] \ = \ \frac{d}{dx}[a(x)y'] \ + \ c(x)y \ = \ f(x).$$  Thus we seek an integratiing factor $\alpha(x)$ so that we can satisfy this &amp; the condition this will hold is that $\alpha(x) \ = \ \frac{1}{a(x)}e^{\int\frac{b(x)}{a(x)}dx}$  Then we're dealing with:  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \alpha(x)f(x)$$  But again, by what my book said they magically re-cast this problem as  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \lambda \alpha(x) y(x)$$  Then calling  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ ( \alpha(x)c(x)y \ - \ \lambda \alpha(x) )y(x) \ = \ 0$$  a Sturm-Liouville problem.  My question is, how can I make sense of everything I wrote above? How can I clean it up &amp; interpret it, like at one stage I thought we were turning our 2nd order ode into something so that it reduces to the derivative of a first order ode so we can easily find first integrals then the next moment we're pulling out eigenvalues &amp; finding full solutions - what's going on? I want to be able to look at $a(x)y'' \ + \ b(x)y' \ + \ c(x)y \ = \ f(x)$ &amp; know how &amp; why we're turning this into a Sturm-Liouville problem in a way that makes sense of exactness &amp; integrating factors, thanks for reading!
122122
In thinking about Sturm-Liouville theory a bit I see I have no actual idea what is going on.  The first issue I have is that my book began with the statement that given  $$L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$  the problem $L[y] \ = \ f$ can be re-cast in the form $L[y] \ = \ \lambda y$.  Now it could be a typo on their part but I see no justification for the way you can just do that!  More importantly though is the motivation for Sturm-Liouville theory in the first place. The story as I know it is as follows:  Given a linear second order ode   $$F(x,y,y',y'') = L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$  it is an exact equation if it is derivable from a differential equation of one order lower, i.e.  $$F(x,y,y',y'') = \frac{d}{dx}g(x,y,y').$$  The equation is exact iff  $$a''(x) - b'(x) + c(x) = 0. $$  If $F$ is not exact it can be made exact on multiplication by a suitable integrating factor $\alpha(x)$.  This equation is exact iff  $$(\alpha(x)a(x))'' - (\alpha(x)b(x))' + \alpha(x)c(x) = 0 $$  If you expand this out you get the Adjoint operator  $$L^*[\alpha(x)] \ = \ (\alpha(x)a(x))'' \ - \ (\alpha(x)b(x))' \ + \ \alpha(x)c(x) \ = 0 $$  If you expand $L^*$ you see that we can satisfy $L \ = \ L^*$ if $a'(x) \ = \ b(x)$ &amp; $a''(x) \ = \ b'(x)$ which then turns $L[y]$ into something of the form  $$L[y] \ = \ \frac{d}{dx}[a(x)y'] \ + \ c(x)y \ = \ f(x).$$  Thus we seek an integratiing factor $\alpha(x)$ so that we can satisfy this &amp; the condition this will hold is that $\alpha(x) \ = \ \frac{1}{a(x)}e^{\int\frac{b(x)}{a(x)}dx}$  Then we're dealing with:  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \alpha(x)f(x)$$  But again, by what my book said they magically re-cast this problem as  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \lambda \alpha(x) y(x)$$  Then calling  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ ( \alpha(x)c(x)y \ - \ \lambda \alpha(x) )y(x) \ = \ 0$$  a Sturm-Liouville problem.  My question is, how can I make sense of everything I wrote above? How can I clean it up &amp; interpret it, like at one stage I thought we were turning our 2nd order ode into something so that it reduces to the derivative of a first order ode so we can easily find first integrals then the next moment we're pulling out eigenvalues &amp; finding full solutions - what's going on? I want to be able to look at $a(x)y'' \ + \ b(x)y' \ + \ c(x)y \ = \ f(x)$ &amp; know how &amp; why we're turning this into a Sturm-Liouville problem in a way that makes sense of exactness &amp; integrating factors, thanks for reading!
113359
Let $V = \{p(x) \in F[x] \ | \ \deg(p(x)) \le n\}$. Let $T : V \to V$ be given by differentiation, in essence $$T(p(x)) = p'(x)$$ It seems to me that the only eigenvalue that can exist is $\lambda =0$, since the given operator lowers the degree of any given polynomial and hence makes it impossible for a constant multiple of a polynomial to be equal to the polynomial's derivative. Is my claim correct?
108789
We consider the differential equation $y'=x+y^2$ and for any $\eta \geq 0$ we denote by $\phi_\eta:(\alpha_\eta,\beta_\eta)\rightarrow\mathbb{R}$ the unique saturated solution of the Cauchy Problem $y'=x+y^2$, $y(0)=\eta$. I am required to do the change of variable $z(x)=y(-x)$ and I succeded here, obtaining $y'(-x)=y^2(-x)-x$ and so $z'(x)=x-z^2(x)$, then I was required to represent the $0-isocline$ and it was also simple as I obtained a parabolla and also I was required to write the equation of the tangent in the point $(0,\eta)$ which was also pretty simple as I obtained $\phi'_\eta(0)=\eta^2$ and the equation of the tangent is $y=\eta+\eta^{2}x$. Now, I am required to find $\phi^{(5)}_0(0)$( that $5$ indicates the number of differentiations), which puzzles me as I don't know the form of the solution and if it can be found out and also I am told to show that $\frac{1}{\eta}\geq\beta_\eta$ What I did not prove after the first edit : and that for $\eta &gt; 0$, $\alpha_\eta=-\infty$ and for the Cauchy problem $z'=x-z^2,z(0)=\eta$ the saturated solution is defined up to $+\infty$. At this last ones I am lost. Any help, please? Edit: I have passed beyond finding $\phi^{(5)}_0(0)$ and I think I did show that $\frac{1}{\eta}\geq\beta_\eta$, now it remains to prove the last questions.
19753
I'm wondering whether there any nice identities (or relationships) that can simplify or possibly compact the following expressions:  $$\operatorname{Li}_2(\beta e^{\alpha x}) - \operatorname{Li}_2(\beta e^{-\alpha x}) $$  and  $$\operatorname{Li}_2(pe^{\alpha x}) + \operatorname{Li}_2(qe^{\alpha x}) $$
13459
So I was reading this:  and while the answers for the first question are good, nobody gave much attention to the second one. In fact, people only said that it can be proved without giving any proof or any.  So, if I have a Lagrangian and ADD an arbitrary function of $\dot{q}$, $q$ and $t$ in such a way that the equations of motion are the same, does this extra function MUST be a total time derivative?   EDIT Ok, I changed my question a little bit:  Question: If I have a function that obeys the Euler-Lagrange equation off-shell, this implies that my function is a time derivative? (This was used in Qmechanic's answer of this other question: , equation 14.)  Also, why people only talk about things that change the Lagrangian only by a total derivative? If this is not always the case that keeps the equation of motion the same, so why is it so important? And why in the two questions I posted about the same statement on Landau&amp;Lifshitz's mechanics book only consider this kind of change in the Lagrangian?
7737
I'm having this question in my homework assignment in Linear Algebra and diffrential equation class, and trying to find the general solution for this second ODE.  $$y''y^3 = 1$$  Using substitution I said $p = y'$ and $p' = y''  \rightarrow  \frac{dp}{dx}= \frac{dp}{dy} \times \frac{dy}{dx}$  Then we get $y^3 \frac{dp}{dx}$ now what's the next step should I integrate or differentiate the equation? What's the trick on these type of ODEs?  Thanks in advance!
22015
The standard treatment of the one-dimensional  (SHO) using the raising and lowering operators arrives at the countable basis of eigenstates $\{\vert n \rangle\}_{n = 0}^{\infty}$ each with corresponding eigenvalue $E_n = \omega \left(n + \frac{1}{2}\right)$. Refer to this construction as the abstract solution.  How does the abstract solution also prove uniqueness? Why is there only one unique sequence of countable eigenstates? In particular, can one prove the state $\vert 0\rangle$ is the unique ground state without resorting to coordinate representation? (It would then follow that the set $\{\vert n \rangle\}_{n = 0}^{\infty}$ is also unique.)  The uniqueness condition is obvious if one solves the problem in coordinate representation since then one works in the realm of differential equations where uniqueness theorems abound. Most textbooks ignore this detail (especially since they often solve the problem both in coordinate representation and abstractly), however I have found two exceptions:   Shankar appeals to a theorem which proves one-dimensional systems are non-degenerate, however this is unsatisfactory for two reasons:   Not every one-dimensional system is non-degenerate, however a general result can be proven for a large class of potentials (the SHO potential is in such a class). The proof requires a departure from the abstract solution since it classifies the potentials according to their functional properties.  Griffiths addresses this concern in a footnote stating that the equation $a \vert 0\rangle = 0$ uniquely determines the state $\vert 0\rangle$. Perhaps this follows from the abstract solution, however I do not see how.
5422
What is the reason for the observation that across the board fields in physics are generally governed by second order (partial) differential equations?    If someone on the street would flat out ask me that question, then I'd probably mumble something about physicists wanting to be able to use the Lagrangian approach. And to allow a positive rotation and translation invariant energy term, which allows for local propagation, you need something like $-\phi\Delta\phi$.  I assume the answer goes in this direction, but I can't really justify why more complex terms in the Lagrangian are not allowed or why higher orders are a physical problem. Even if these require more initial data, I don't see the a priori problem.  Furthermore you could come up with quantities in the spirit of $F\wedge F$ and $F \wedge *F$ and okay yes... maybe any made up scalar just doesn't describe physics or misses valuable symmetries. On there other hand in the whole renormalization business, they seem to be allowed to use lots and lots of terms in their Lagrangians. And if I understand correctly, supersymmetry theory is basically a method of introducing new Lagrangian densities too.   Do we know the limit for making up these objects? What is the fundamental justification for order two?
107151
Show that there exists a unique $v_0 \in H^1(0,1)$ such that $u(0)=\int_0^1(u'v_0'+uv_0), \forall u \in H^1(0,1)$. Further Show that $v_0$ is the solution of some differential equation with appropriate boundary conditions. Compute $v_0$ explicitly.   Let $f: H^1(0,1) \to \mathbb{R}$ be defined by $f(u)=u(0)$. Then I showed that $f$ is linear and continuous. Hence there exists $v_0 \in H^1(0,1)$ such that $u(0)=\int_{0}^1u'v_0'+\int_{0}^1uv_0$  The differential equation is given by $f=-u''+u$ with the conditions $u'(0)=u'(1)=0$.   How do I compute $v_0$ explicitly??
73334
In a book I am reading on differential equations, the author writes about the solution to a homogenous, linear, second order differential equation with constant coefficients. The author says something like, "Let us suppose that the solution is of the form $y=e^{mx}$ " . After this the author introduces the characteristic equation of a differential equation of the form mentioned above, and proceeds to describe how to solve it w/ undetermined coefficients, variation of parameters, etc.   How did mathematicians first come up with this "assumption" that the solution was of the form $y=e^{mx}$? And, are there any other forms of solutions for these types of equations?
93368
For context, this equation is used in the following (from Schwartz's QFT 3.44)  $$\partial_{\mu} \frac{\partial(\partial_{\beta}A_{\gamma})^2}{\partial(\partial_{\mu}A_{\nu})} = \partial_{\mu}\left[2(\partial_{\alpha}A_{\alpha})\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} g_{\beta\gamma}\right] = \partial_{\mu}[2(\partial_{\alpha}A_{\alpha})g_{\beta\mu}g_{\gamma\nu}g_{\beta\gamma}]=2\partial_{\nu}(\partial_{\alpha}A_{\alpha}).\tag{3.44}$$  But doesn't $$\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} = g_{\beta\mu}g_{\gamma\nu}$$ imply, for instance, that  $$\frac{\partial(\partial_0 A_3)}{\partial(\partial_0 A_3)} = g_{00}g_{33} = -1$$ instead of $=1$? Shouldn't we instead have $\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} = \delta_{\beta\mu}\delta_{\gamma\nu}$?
35924
In my analysis class, my professor (and the textbook) reference the fact that $\mathbf{D}T(\bar{x}) = T$ for any $\bar{x} \in \mathbb{R}^n$ where $T \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^n) $. I'm wondering about the precise meaning of this statement. Is it that the differential of any linear transformation is also linear? Or is that the differential of any linear transformation is itself (and therefore also linear)? The second seems confusing to me because the derivative of a function is often not itself. For example, $f(x) = 2x$ is a linear transformation but $f'(x) = 2 \ne 2x$ for all $x \in \mathbb{R}$.
39021
The question is to investigate the number of zeroes of $y''(x)+e^{x^2}y(x)=0$ in $[0,3π]$.  Solving this ODE would not be an easy task as one has to use the power series solution and then investigating the zeroes of the solution will require more analysis. I thought it to compare this ODE with the standard $y''(x)+y(x)=0$ whose solution has three or four zeroes in the interval $[0,3π]$.  Since the coefficient of $y(x)$ is $e^{x^2}\ge 1$ for $x\in [0,3π]$ so the solution of given ODE must have atleast three zeroes in $[0,3π]$. However, what  I thought was in the lights of Sturm-Comparison theorem so I am not sure.  Am I correct to interpret this?
42094
Just had a quick question regarding the uniqueness of the general solution of first-order differential equations:  For $x'(t)=Ax(t)$ where $A$ is an $n\times n$ matrix. and initial condition $x(0)=x_0$, the unique solution is $x(t)=e^{tA}x_0$.  The proof for showing that the solution is unique involves differentiating the matrix product $e^{-tA}x(t)$.  Could someone please explain why we always multiply by $e^{-tA}$.   Many thanks, Best.
103425
How should I approach this problem? Should I put ln logarithm on both sides and move factor of variable of lnx on right side in front of the ln, or do somehing else? I understand general solving and types of solving these differential equations of first order, but I have no idea what my next step should be.
86756
Suppose $f:[0,1]\to\mathbb{R}$ is twice differentiable with  $$\lim_{x\to0^+}\frac{f(x)}{x}=1,\lim_{x\to 1^-}\frac{f(x)}{x-1}=2$$  Show that there is $\eta\in(0,1)$ s.t. $f''(\eta)=f(\eta)$.  It is equivalent to prove that $f''(x)-f(x)$ has a root in $(0,1)$. Then I tried to construct function. First I noted that the ODE  $$y''-y=0$$  has solution of the form $\alpha e^x+\beta e^{-x}$, then I consider the following function :  $$(f-f')e^{x}$$  and apply mean value theorem to is. But thing does not behave that nice, and I was stuck.
15353
The question is to find an approximate solution, up to the order of epsilon of following problem.  $$y'' + y+\epsilon y^3 = 0$$ $$y(0) = a$$ $$y'(0) = 0$$  I tried to solve the given problem using perturbation theory.  $$y(t) = y_0(t) + \epsilon y_1(t) + \epsilon^2 y_2(t) + \cdots$$ $$t = w(\epsilon)t = (1 + \epsilon w_1 + \epsilon^2 w_2 + \cdots )t$$  However, i failed to find an appropriate approximate solution... help me!!