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1

This is frequency integration. Hint: RHS = $\int_{s}^{\infty} F(u) du$ = $\int_{s}^{\infty} \int_{0}^{\infty} f(t)e^{-ut} dt du$ If you give up: http://www.eee.hku.hk/~u3500898/LTProperty.pdf I think the justification for the double integral swap is the same as the justification for your swapping of derivative and integral in your attempt.


0

The hitch is at the end : $\mathcal{L(\dfrac{f(t)}{t})}(s)= - \int \mathcal{L(f(t))(s)} ds$ + constant It can be rewritten as: $\mathcal{L(\dfrac{f(t)}{t})}(s)= - \int_{0}^{s} \mathcal{L(f(t))(u)} du +C$ Now, you will have to determine $C$ as $C= \int_{0}^{\infty} \mathcal{L(f(t))(u)} du $


1

Don't you insert your integration limits in a wrong way? Why are you taking the one limit as $0$ instead of $\infty$? Take your intermediate result $$ \frac{d}{du}\mathcal{L(f(t)/t)}(u)=-\mathcal{L(f(t))(u)}$$ then integrate from $k\to\infty$ to $s$ to get $$ \mathcal{L(f(t)/t)}(s)-\lim_{k\to\infty}\mathcal{L(f(t)/t)}(k) = ...


0

Let $u=x+2$ so that $u=0$ when $x=-2$ $$x^2=(u-2)^2$$ $$x^2=4-4u+u^2$$ $x^2$ is linerized as $x^2=4-4u=4-4(x+2)$ which is brought back into the ODE : $$x'=2x^2-8=2(4-4(x+2))-8=8-8(x+2)-8$$ $$x'=-8x-16$$


0

If you build the Taylor-McLaurin series at $x=-2$, you obtain $$2x^2-8=-8 (x+2)+2 (x+2)^2+O\left((x+2)^3\right)$$ Just retain the first term. You get the same if you write $$2x^2-8=2\Big((x+2)-2\Big)^2-8=2\Big((x+2)^2-4(x+2)+4\Big)-8=-8(x+2)+2(x+2)^2$$


0

$f(x) = 2 x^2-8$. We have $f(-2) = 0$ as expected. The linearized system is $y' = f'(-2) y$, which is $y'=-8y$. If you want to express this as a system based around $-2$ rather than zero, let $z=y-2$, or $y=z+2$, which will give the equation: $z'=-8 z -16$.


1

I did the change of variables on the original ODE to see if RRL was right in saying that it would also lead to transforming the spherical ODE to the Cartesian one. Original ODE: $$\frac{d}{dr}\left(r^2\frac{df}{dr}\right)=0$$ and B.C.'s: $f(r_1)=f_1$ and $f(r_2)=f_2$. I did a change of variables $\xi=\frac{r-r_1}{\Delta r}$, where $\Delta r=r_2-r_1$. ...


0

Hint: use the following inequality to show the uniform convergence: $$ (\sum_{k=1}^na_kb_k)^2\le\sum_{k=1}^n|a_k|^2\sum_{k=1}^n|b_k|^2. $$


0

My notation was screwed up and fixing it solves the problem. Basically, my definition above for the infinitesimal rotation tensor was incorrect because the rotation tensor applied to a vector is a matrix. Here's how it works. The infinitesimal rotation tensor is really the skew symmetric part of the Fréchet derivative. Recall, $$ f^\prime(x)= ...


0

Hint: $\dfrac{d}{dt}\dfrac{1}{S(t)} = -\dfrac{dS/dt}{S(t)^2}$. What does the Riccati equation imply about the RHS?


0

I'm making this CW as it's not exactly an answer, but, this is how some physicists define curl and divergence and it does almost immediately connect curl to circulation and divergence to change in flux. Probably semiclassical's comment is most useful towards the path you begin to walk. We usually see definitions for curl and divergence which were based in ...


0

Here is a very subjective list: Arnold, Mathematical Methods of Classical Mechanics This book treats classical mechanics from mathematical point of view and introduces a wealth of various mathematical concepts (such as a differential manifold or differential form) in a right context. The main mathematical tool is ODE and it is really inspiring to see how ...


6

$$ xy''-(1-x)y'+y=0 $$ expanded $$ xy'' -y' +xy' + y = xy'' -y' + \frac{d}{dx}xy $$ now using the fact $$ \frac{d}{dx}xy' = xy'' + y' $$ and manipulating to yield $$ \frac{d}{dx}xy' - 2y' = xy'' - y' $$ we can rewrite the original equation as $$ xy'' -y' +xy' + y = \frac{d}{dx}xy' - 2y'+ \frac{d}{dx}xy = \frac{d}{dx}\left[xy' - 2y + xy\right]=0 $$ I ...


4

Apply $\exp$ to both, express trig functions in terms of exponentials, and take the difference. I get $$ \left( \dfrac12\,{{\rm e}^{C_{{3}}+iC_{{4}}}}-4\,{{\rm e}^{C_{{2}}}}C_{{1} } \right) {r}^{2}+{\frac {\dfrac12\,{{\rm e}^{C_{{3}}-iC_{{4}}}}-{{\rm e}^{ C_{{2}}}}}{{r}^{2}}} $$ For this to be $0$ for all $r$, you need both coefficients to be $0$. This ...


2

The two functions $kx^2+.5$ and $\ln(x)$ are increasing for $x>0, k>0$. The first is concave up and the second is concave down. So if their graphs intersect once they will intersect again. Thus the only way for there to be one solution is if the graphs are tangent (double solution) at the point of intersection. Then you can solve the derivative ...


3

A common method for fitting an analytical scalar field $F(\textbf{x})$ (in any dimension) to a vector field of (potentially noisy) derivatives is to expand the field in a radial basis function $$ F(\textbf{x}) = \sum_i a_ie^{-|\textbf{x}_i-\textbf{x}|^2/2\sigma^2} $$ where you pick a grid of centres $\textbf{x}_i$. Then you can find the $a_i$ and $\sigma$ ...


0

First rewrite the equation by multiplying both sides by $r^2$: $$\frac{1}{r}\left(\frac{d}{dr}\left(r\frac{dw}{dr}\right)\right)-\frac{\lambda^2}{r^2}w=0\\ \implies r\left(\frac{d}{dr}\left(r\frac{dw}{dr}\right)\right)-\lambda^2w=0.$$ Make the substiution $r=e^x$ and define $w(r)=w(e^x)=:y(x)$. Then, $x=\log{r}$, $\frac{dx}{dr}=\frac{1}{r}$, and ...


1

By the Product Rule, $\dfrac{d}{dr}\left(r\dfrac{dw}{dr}\right) = r\dfrac{d^2w}{dr^2}+\dfrac{dw}{dr}$. Thus, the differential equation becomes $\dfrac{d^2w}{dr^2}+\dfrac{1}{r}\dfrac{dw}{dr}-\dfrac{\lambda^2}{r^2}w = 0$. Now, try something in the form $w = r^{\alpha}$. Then, $\dfrac{dw}{dr} = \alpha r^{\alpha-1}$ and $\dfrac{d^2w}{dr^2} = \alpha(\alpha-1) ...


0

I am also an enthusiast of this technique solve a differential equation (See this link). With the operator D, we find the general solution of non-linear equations homogenêneas with f being a polynomial, trigonometric functions, exponential, or combinations of these. You can also define a differential operator to study the differential equations of ...


1

Your idea is correct. As the thickness of the shell diminishes -- the effect of the curvature of the domain becomes negligible. The solution in rectangular coordinates becomes a close approximation to the curvilinear solution. As you suggested, let $\Delta = r_2 - r_1$ and $\xi = r - r_1$. The spherical solution is $$f(r) = ...


0

Each root from the indical equations may not always only find one group of the linearly independent solutions, sometimes we can find more than one group of the linearly independent solutions at the same time. For the examples see Find the form of a second linear independent solution when the two roots of indicial equation are different by a integer and ...


1

It is essentially the same, if you expand the formula for the cosine or sine of a sum. Then $A,B$ would be different, but since without initial values conditions they are arbitrary, then you encompass all solutions anyway.


0

From where you are, I'd factor. $$v(yv'+v)=0$$ so either $v=0$ or $(yv)'=0$. I think the easiest way to solve it from the start, though would be to recognize the left side as being the result of the product rule, reducing the equation to $(yy')'=0$


1

$v=\frac{dy}{dx}=y'$ so $y''=\frac{dv}{dx}=\frac{dv}{dy}\cdot\frac{dy}{dx}=v'v$ $y\cdot y'' + (y')^2 = 0$ so $yv'v+v^2=0$ then $yv'v=-v^2$ and $yv'=-v$ $\frac{v'}{v}=-\frac{1}{y}$ then $y'=\frac{dy}{dx}=v=\frac{1}{y}$ and $\frac{1}{2}y^2=x$


0

Hint: Let $V = \dfrac{1}{2}y^2$. Then, $\dfrac{dV}{dt} = yy'$ and $\dfrac{d^2V}{dt^2} = yy''+(y')^2 = 0$ This gives you an easy differential equation to solve for $V$. Use that to get $y$.


1

Assuming that $u(x)$ is at least $C^2(\mathbb{R})$ and that your domain is $\mathbb{R}$ (connected) you have: $$u'(x)=1-u(x)$$ So $u'(x)=0 \iff u(x)=1$. Also $u'(0)=1$ so we just have to prove that $u'(x) \geq 0 \quad \forall x$ But $u'(x)$ is continuous so $u'(x) \geq 0 \quad \forall x$ (it can't go to a negative value because it should pass by zero and ...


2

Building on your own answer, one sees that the crucial part is getting some lower bound on $f(t_{max}-\delta)$. The differential equation yields $$ \frac{df}{dt} = f-f^3-\alpha f(t-\delta) \leq (1+\alpha)f_{\max} $$ By Taylor expansion, $\exists \tau,\, f(t_{\max})-f(t_{max}-\delta) =\delta f'(\tau)$. Thus, using your global bound, ...


0

$\partial_tP(x,y,t)=x\partial_xP(x,y,t)+(y-1)\partial_yP(x,y,t)+2P(x,y,t)$ $\partial_tP(x,y,t)-x\partial_xP(x,y,t)+(1-y)\partial_yP(x,y,t)=2P(x,y,t)$ Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$ $\dfrac{dx}{ds}=-x$ , letting $x(0)=x_0$ , we have ...


1

$\nabla f = A(n+1)+\lambda \mathrm{diag}(e^{u_1},e^{u_2},...,e^{u_n})$. ($\mathrm{diag}$ is building a diagonal matrix, with the diagonal elements given in the brackets)


0

a couple of remarks: 1/ It is not obvious to me that you necessarily get $t$ such that $f'(t)=0$. It looks like this is the case in your numerical examples, but analytically, I'm not sure how you would prove this. 2/ If indeed you have oscillatory solutions, then I agree with you, you need to solve the equations above. But as you have a parameter $\delta$ ...


0

There are no explicit solutions for most of the nonlinear systems. Therefore, they are linearized near the fixed points. The eigenvalues near the fixed points mostly give idea about the phase portrait solutions of such systems. For this system fixed points are $(0,0)$ and $(p/3,p/12)$. The eigenvalues near the $(p/3,p/12)$ are: ...


0

First solve the homogeneous equation which gives the following fundamental set of solutions $$ \left\{ e^{-\frac{s}{c}x}, e^{\frac{s}{c}x} \right\}. $$ Since $ c,s > 0 $ you should pick up the solution $U_h= \alpha \, e^{-\frac{s}{c}x}$, for some $\alpha$ to be determined later using the initial condition, to satisfy your condition of boundedness. ...


-1

Solutions to these types of equations .....(dx/dt) =F1(x,y.z) ;(dy/dt) =F2(x,y,z);(dz/dt) =F3(x,y,z) , ....which are called State Space Equations , are found in Non-linear Control System Engg. and Advanced Control System Engg. books.


1

Drawing a phase diagram would probably have shown you right away that the nonnegative quadrant $Q=(X\geqslant0,Y\geqslant0)$ is stable by the dynamics, that the equilibria in the quadrant $Q$ are the point $(X,Y)=(0,0)$ and the segment $S=(X\geqslant0,Y\geqslant0,X+Y=N)$, and that $(0,0)$ is an unstable node. At every point $(X_0,Y_0)$ on $S$, an eigenvalue ...


0

Hint: Do $U(x) = e^{\lambda x}$ for homogeneous equations.


0

It depends on what notation the author has decided to use, but $| \cdot |$ is usually reserved for the absolute value of a real number, or the modulus of a complex number, whereas $\| \cdot\|$ is reserved for the norm of a vector.


2

What you have done seems correct to me - and also quite simple. You got the multiply-by-$\dot{x}$ trick which is the 'standard' way to solve this. I think it will be hard finding a much easier route. To complete the solution, you only need to fix the value of $C$ and integrate up once more. This is also fairly simple since from $$\dot{x} = ...


0

Here you variables are separated, so it is not a big problem. Rewrite the equation as $x(y)' = - \frac{1}{2} \times \frac{\frac{d \sin \omega y}{dy}}{\sin \omega y}$ or $x = x_0 -\frac{1}{2}\ln(\sin(\omega y))$ If you want the more general case you need to look at total differential site here and especially the Schwarz conditions.


0

You have $$\frac{dx}{dy} + \frac{w\cos wy}{2\sin wy}=0$$so the integrating factor is $$e^{\int \frac{w\cos wy}{2\sin wy}}=e^{\frac12\ln(\sin wy)} = (\sin wy)^{\frac12}$$Indeed $$(\sin wy)^\frac12\ \frac{dx}{dy} + \frac w 2\cos wy (\sin wy)^{-\frac12} = \frac{d}{dy}\left(x(\sin wy)^\frac12\right)$$


0

If you have $e^{At}$ and you want to find $A$, note that having $e^{At}$, in whatever form you've got it, and there are many different formulas, sometimes depending on $A$, such as $e^{Jt} = (\cos t) I + (\sin t) J \tag{1}$ for matrices satisfying $J^2 = -I; \tag{2}$ the point is, you might have $e^{At}$ in a form which is easier to ...


2

Given any exponential matrix $e^{A}$, first find a matrix $P$ such that $P e^A P^{-1}$ is in Jordan normal form. $$P A P^{-1} = J \stackrel{def}{=} \begin{bmatrix} J_{m_1}(\lambda_1) & 0 & \ldots & 0 \\ 0 & J_{m_2}(\lambda_2) & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \ldots & ...


1

Okay, we are going to use the fact that if $A = P^{-1}DP$ we also have $e^A = P^{-1}e^DP$ And $e^D$ is simply the diagonal matrix $e^{d_{ i,i}}$. So if you find the $P$ matrix by finding the eigenvectors of $e^{Dt}$ you can use the associated $D$ matrix of eigenvalues (need to take log of each value and divide by $t$, if the result is still a function of ...


1

I think the easiest way to see that $x(t) = a \cos \omega t + b\sin \omega t \tag{1}$ is the general solution of $\dfrac{\partial^2 x(t)}{\partial t^2}  + \omega^2 x(t) = 0 \tag{2}$ is to exploit the good ol' plug and grind method, to wit:  we have, from (1) $\dfrac{\partial x(t)}{\partial t} = -a \omega \sin \omega t + b \omega \cos \omega t, \tag{3}$ ...


1

Two thirds of the way: Forget about $K$ for the moment. Assume we are given a continuous $f:\ [0,1]\to{\mathbb R}$, and we want to construct a $g$ fulfilling the given conditions. From $g'''=f$ we conclude that $$g''(x)=\int_0^x f(t)\>dt +c\ ,$$ where $c$ is as yet undetermined. This leads to $$g'(x)=g'(0)+\int_0^x g''(t)\>dt=\int_0^x\left(\int_0^t ...


0

I have kind of same issue .. Can this method can be useful for me too? I have an inhomogenous ODE. The main issue here is variables are matrices. It is bit of matrix calculus. A solution would be highly appreciated interms of x . I guess we can use same methods for solving ODEs but have to be careful because these are matrices $R'(x)-(C_1 +C_2 x) ...


1

A hint as to what the problem wants you to do: You want to show that the first equation implies the second. Try working backwards---does the second imply the first? (How do you go from an equation with one time derivative to one with two?) Second hint: You want to find some function which, when squared and added to the square of its derivative, is equal to ...


0

Given $x''+\omega^2 x = 0$, we have the auxiliary equation $m^2+1=0$, after substituting $x=e^{mt}$. Roots of the auxiliary equation are imaginary: $m_1=\omega i,m_2=-\omega i$. The solution takes the form $$x(t)=a e^{m_1 t}+b e^{m_2 t}.$$ Can you plug in the roots of $m_1=\omega i$ and $m_2=-\omega i$ into your $x(t)$ expression? After some derivation, you ...


1

By `solution' you mean solution to the initial value problem (IVP) $$ \frac{dy}{dx} = y,\qquad y(0) =c,\qquad x\in\ \mathbb{R}.$$ It follows from the Picard-Lindeloff theorem there exists a solution to the above IVP and that solution is unique. So if $y(x) = ce^x$ is a solution, then it is the only one. For the second 'solution', $\lim_{x\rightarrow 0}y $ ...


0

Let $y = 1 + \frac{x}{t}$. Then $$\frac{dy}{dt} = \frac{\frac{dx}{dt}t - x}{t^2}$$ Substituting $\frac{dx}{dt} = 1 + \frac{x}{t}$ into the expression above, you get $$\frac{dy}{dt} = \frac{1}{t}$$. The rest should be clear.


0

Here's a quick way to do the same thing without an integrating factor. Use an invariant Lie group, a simple one: $t'=\lambda t$, $x'=\lambda^\beta x$. $$ \frac{dx'}{dt'}=1+\frac{x}{t} $$ $$ \frac{\lambda^\beta dx}{\lambda dt}=\lambda^0 1+\frac{\lambda^\beta x}{\lambda t} $$ For invariance, $\beta =1$. Stabilizers for this group are ...



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