Toronto Math Forum

Wrong step in the correct direction meaning I shouldn't use limits but I should eventually integrate by parts?

Should I use local integration?

Sorry, I am still confused. Why do we assume $\varphi(0)\ne 0$?

Would I be starting in the right direction with this instead?

$f'(\varphi) = f(\varphi ') = \int f(x) \varphi'(x)\,dx = \int \mathrm{ln}|x| \varphi'(x)\,dx$

c. and d. Correction to Hamiltonian:

First express $\dot{q}$ in terms of generalized momenta $p$.

We have $p_i = \frac{\partial L}{\partial \dot q_i} =  m\dot q_{i} + eA(q)_{i}$

so

$\dot{q}_i = \frac{p_i}{m} - \frac{eA(q)_i}{m}$

For the Hamiltonian to really be a Hamiltonian, we must replace the $\dot{q}$ term:

$H = \frac{m}{2}{\dot{q}}^2 + V(q)$

Now replace $\dot{q}_i$:

$H = \frac{m}{2}\sum_{i=1}^{n}( \frac{p_i}{m} - \frac{eA(q)_i}{m})^2 + V(q)$

I suppose $A$ affects the Hamiltonian because it affects generalized momenta.

e.

Find $\ddot{q}$ due to $A$.

The Euler-Lagrange equation is as follows:

$eA'(q) - V'(q) = m\ddot{q} + eA'(q)\dot q + \frac{\partial A}{\partial t}$

Solving for $\ddot{q}$:

$\ddot{q} = \frac{1}{m}(eA'(q) - V'(q) - A'(q)\dot{q} + \frac{\partial A}{\partial t})$

Part (a):

To check that $u = e^{\frac{1}{2}ax^2}$ satisfies $u'(x) = -ax u(x)$, differentiate $u$. I am not sure if we can just treat $u$ like an exponential or if we have to expand $u = e^{\frac{1}{2}\mathrm{Re}(a)x^2}e^{\frac{1}{2}\mathrm{Imaginary}(a)x^2}$ using euler's formula first.

Next, use $u'(x) = -ax u(x)$ to show $i\omega{u}(\omega) = -ia\hat{u}'(\omega)$.

First, use fourier transform property $g(x) = xf(x) \rightarrow \hat{g}(k) = i\hat{f}'(k)$.
Using $u'(x) = g(x)$ and $-au(x) = f(x)$, as well as the definition of a fourier transform, we get $\int_{\infty}^{\infty} u'(x)e^{-i\omega x} = -ia\hat{u}'(\omega)$.

Now use fourier transform property $g'(x) = f(x) \rightarrow \hat{g}(k) = ik\hat{f}(k)$. Using $-axu(x) = f(x)$ and $u'(x) = g'(x)$ we get $\int_{\infty}^{\infty} -axu(x)e^{-i\omega x} = i\omega \hat{u} (\omega)$. Replacing $-axu(x)$ with $u'(x)$ we get $\int_{\infty}^{\infty} u'(x)e^{-i\omega x} = i\omega \hat{u} (\omega)$. Apply this equality to the previously shown equation $\int_{\infty}^{\infty} u'(x)e^{-i\omega x} = -ia\hat{u}'(\omega)$ and get $i\omega \hat{u} (\omega) = -ia\hat{u} ' (\omega)$.

Finally, find $\hat{u}$. Consider that $i\omega \hat{u} (\omega) = -ia\hat{u} ' (\omega) \rightarrow  -\frac{\omega}{a} \hat{u} (\omega) = \hat{u} ' (\omega)$. This is an ODE with an exponential solution, $\hat{u} (\omega) = Ce^{-\frac{\omega ^2}{2a}}$.

The second part of the question asks us to prove (2), starting with the premises (1), (3), and (4).

First prove the hint:
$$
v = \int_{0}^{t} u(x,t')\,dt'.
$$
Define:
$$
g(x') = \frac{d}{dx'}G(x');
$$
We already defined:
$$
u(x,t) = \frac{1}{2}(g(x+ct) + g(x-ct)
$$
from (1) and
$$
v(x,t) = \frac{1}{2c} \int_{x-ct}^{x+ct}g(x')\,dt'
$$
from (3). Then:
\begin{gather*}
\int_{0}^{t} u(x,t')dt' = \frac{1}{2} \Bigl(\int_{0}^{t} g(x + ct')dt' + \int_{0}^{t} g(x - ct')\,dt'\Bigr)\\
= \frac{1}{2} \Bigl(\frac{1}{c}G(x+ct) - \frac{1}{c}G(x) + \frac{1}{-c}G(x-ct) - \frac{1}{-c}G(x)\Bigr)\\
= \frac{1}{2c} \Bigl(G(x + ct) - G(x-ct)\Bigr) = \frac{1}{2c} \int_{x-ct}^{x+ct}g(x')dt' = v(x,t)
\end{gather*}
Now I have proved the hint and I will implement it to help prove $(2)$. From the hint: $v_t = u$.

Thus $v_{tt} = u_{t}$. We know from (4):

$v_{tt} = c^{2}v_{xx}$

Thus $u_{t} = c^{2}v_{xx}$

Thus $u_{tt} = c^{2}v_{xxt}$

Let us find $v_{xxt}$. (We hope to prove the first condition of $(2)$.)

Previously we have shown $v(x,t) = \frac{1}{2c}(G(x+ct) - G(x-ct))$

Thus $v_x = \frac{1}{2c}(G'(x+ct) - G'(x-ct)) = \frac{1}{2c}(g(x+ct) - g(x-ct))$

Thus $v_{xx} = \frac{1}{2c}(g'(x+ct) - g'(x-ct))$

Thus $v_{xxt} = \frac{1}{2}g''(x+ct) + g''(x-ct)$

Let us find $u_{xx}$ in hopes that it is equal to $v_{xxt}$. This will help complete our proof.

$u_{x} = \frac{1}{2}(g'(x+ct) + g'(x-ct)$

Thus $u_{xx} = \frac{1}{2}g''(x+ct) + g''(x-ct) = v_{xxt}$.

Thus $u_{tt} = c^{2}u_{xx}$.

So $u_{tt} - c^{2}u_{xx} = 0$, proving that the first condition of $(2)$ must follow from $(1)$, $(3)$, and $(4)$.

To prove that the rest of $(2)$ follows, we don't even need $(3)$ or $(4)$. Plug $t=0$ into $u(x,t)$ and its partial derivative $u_t$ to prove those conditions.

Tristan,

I don't think you are making too many assumptions if your goal is to find one possible solution, but it is valid to be concerned if it there are leaps in logic for a general solution.

Your solutions (I) and (II) are interpreted to be the general solutions to u_xx=0 and u_yy=0, respectively. I believe your solution steps for u_x and u_y for respective constraints to be correct. However, I have modified the final steps for the general solutions:

(I) : u(x,y) =  xϕ(y)+ψ(y)+bx+C
(II) : u(x,y) = yα(x)+β(x)+cy+C

Note that if you take the derivative of (I) with respect to x, b gets absorbed into C. Analogous conditions apply to (II).

When constrained to variables x and y, C is a constant (and so are b and c).

Finally, note that the general solution which obeys both u_xx=0 and u_yy=0 is equal to both solution (I) and solution (II).

Thus, solve: xϕ(y)+ψ(y)+bx+C = yα(x)+β(x)+cy+C

Let's assume none of the terms are equal to 0. This will allow us to impose constraints that will deny trivial solutions. If any of ϕ(y), ψ(y), α(x), and β(x) are equal to 0, the general solution should still be valid.

Thus, xϕ(y) = yα(x).

Thus, ϕ(y)=ay and α(x)=ax. (This is pretty much what you said, except I imposed the extra constraint that the functions share the same coefficient). Check that xϕ(y) = yα(x) = axy.

So now we have axy + ψ(y) + bx + C = axy + β(x) + cy + C

Thus, ψ(y) = cy and  β(x) = bx.

Thus, the general solution is axy + bx + cy + C.