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### Messages - Weihan Luo

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##### Final Exam / Alternative solution to the optimization in Problem 2 on the practice final
« on: April 23, 2022, 02:19:38 PM »
Could I have solved the maximization/minimization using Lagrange multipliers? In particular, define $g_1(x,y) = y-x$, $g_2(x,y) = y+x$, and $g_3(x,y) = -(x^2+y^2)+1$. Then, a solution $(x^*,y^*)$ necessarily satisfies $$\nabla{u} + \lambda_1\nabla{g_1} + \lambda_2\nabla{g_2} + \lambda_3\nabla{g_3} = 0$$ and $$\lambda_1{g_1} = 0, \lambda_2{g_2}=0, \lambda_3{g_3}=0$$

for some $\lambda_{i} \geq 0$.

Then, after finding the points $(x^*, y^*)$, I need to verify that $$\nabla^2{u} + \lambda_1\nabla^2{g_1} + \lambda_2\nabla^2{g_2} + \lambda_3\nabla^2{g_3}$$ is positive definite on the tangent space $T_{x^*,y^*}D$.

Would this approach also work?

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##### Test 2 / Sturm Liouville eigenfunctions
« on: March 28, 2022, 11:06:57 PM »
In the practice term test 2 Variant A Problem 1, I had to solve the the following Sturm Liouville problem:

$$X''+\lambda x = 0$$ with boundary conditions $$X'(0) = X'(4\pi) = 0$$ In the answer key, the eigenfunction corresponding to the eigenvalue $\lambda_{0} = 0$ is $X_0 = \frac{1}{2}$. However, if we substituted $\lambda_0 = 0$ into the ODE, we get:

$$X'' = 0$$ which the solution is simply $$X(x) = \alpha + \beta x$$ Plugging into the boundary conditions and we get that $$\beta = 0$$ so The solution is $$X(x) = \alpha$$ where $\alpha \in \mathbb{R}$. How do we get that $X_0 = \frac{1}{2}$?

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##### Chapter 5 / Theorem 3 in chapter 5 question
« on: March 17, 2022, 12:16:07 AM »
In property 3 of Theorem 3 ($g(x) =𝑥𝑓(𝑥)⟹𝑔̂(𝑘)=i\hat{f}'(k)$), for $\hat{f}'(k)$, is the Fourier transform being computed before the derivative, or the other way around? Intuitively, Leibnitz rule says that the order shouldn't matter here, but I just want to get a confirmation. Thank you

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##### Chapter 1 / Re: Classification of PDEs
« on: January 14, 2022, 11:28:53 AM »
In particular, the definition of a linear PDE, from the textbook, is: $au_{x}+bu_{y}+cu-f=0$, where $f= f(x,y)$. However, if we simply move the the $cu$ to the right-hand side, we get: $au_{x}+bu_{y}=f-cu$. Now, define $g(x,y,u) = f(x,y)-cu$, then $au_{x}+bu_{y}=g(x,y,u)$, and the right-hand side now depends on lower-order derivatives, so by definition, it's quasi-linear. Could someone help identify the issue with this argument?