Thank you for pointing out this issue. You are right — this problem inadvertently has two correct answers. I apologize for the inconvenience. I will contact the Wolfram U team to remove this question from the list of final-exam questions or to slightly modify one of the answers.

Hello Christine,

thank you for the information. However, I am not sure why I should delete any files, since I do not aim to take the exam again.

On the other hand, I thought I earned a certificate, but I did not receive one. As a participant in the daily study group, I thought it was enough to complete the final exam by February 6 to earn it. Did I misunderstand anything?

Best regards

VR

Hello Václav,

I wanted to clarify that, in order to earn Completion certification via the Daily Study Group, you need to have passed all of the quizzes. It looks like you passed the exam (Feb 6) but did not take the quizzes. Can you please confirm? You can also follow up via wolfram-u@wolfram.com.

Thank you,
Christine

Hello Mitch,

Thank you for letting us know. I'll put in a ticket with our team for review.

Best,

Christine Owens

Thanks for the interesting question.

The sphere presented in the lesson represents spherical coordinates and describes physical space. In contrast, the Bloch sphere represents abstract quantum states. It does not correspond to physical space, but rather provides a geometric representation of the state of a two-level quantum system. Thus, although the two pictures look similar, their meanings are completely different.

Laplacian equations are central to quantum mechanics and play an important role in the design and modeling of quantum devices, but they are not responsible for the Bloch-sphere representation itself. Instead, Laplacians and related partial differential equations are used to model real physical systems, such as nanostructures, electromagnetic fields, and superconducting devices that form the basis of modern quantum technologies.

The case of graphene is slightly different. In graphene, the effective equation governing low-energy electronic states is not the Schrödinger equation with a Laplacian, but a two-dimensional Dirac equation involving first-order derivatives. Nevertheless, PDE methods and boundary-value problems remain absolutely central in the theoretical description of such systems.

Hi;

In working on Exercise 18, Problem 2, I would appreciate some help in determining where the parameter values for the DEigenvalues [] function came from - I have attached a notebook containing the problem below.

From the given problem, it appears that the first two values of the given equation were reversed and prefixed with negative sighs and the given equation was terminated (Lamba + u = 0).

Additionally, I do not understand how you are getting the boundary condition using the function DirichletCondition[u[x, y] == 0, True].

Next it appears that the function u[x,y] is being defined but then we have these other numbers that I don't understand their origin or use - {x, 0, 2}, {y, 0, 2}.

The rest of the function I understand.

Please bear with me since some of these techniques are very new to me and still playing catch-up on some of the information given in this study group. I am completely blown away that Eigenvalues & Eigenvectors can be used in separation of values calculations and would appreciate any additional information on the technique.

Thanks,

Mitch Sandlin

Attachments:

Ex18Prob2DEigen.nb

Hi;

In Lesson 4 when trying to use the Mathematica function Discriminant [] to calculate the Discriminates, I get the following messages - see attached notebook. Additionally, I tried to use the Derivative [] function as opposed to the u with subscripts but could not find any examples in the documentation. In my opinion, Mathematica should put some examples of the Derivative [] function in the documentation, since it seems to be a really powerful function. In any event, would someone please show me how to use the Mathematica function to calculate these Discriminant's.

Thanks,

Mitch Sandlin

Attachments:

Lesson4Discrimin...nb

In Wolfram Language, the Discriminant function is designed to work only with polynomials, so it cannot be used with PDEs.

The Derivative function can be used to define initial and boundary conditions that contain partial derivatives of a multivariable function. For example, Derivative[0,1][u][x,y] returns the first-order partial derivative of the function u[x,y] with respect to y, and Derivative[0,1][u][x,0] returns the first-order partial derivative of u[x,y] at the points (x,0). You can find more examples in the Scope section of the reference page for the Derivative function in the Wolfram Documentation.

Hello Richard,

Thank you for your feedback. I'll create a ticket per your report for our team to review and follow up once I have confirmation.

Best,

Christine Owens Wolfram U Project Manager Wolfram U

I finished the seven quizzes for this corse. The "Track My Progress" page shows an inactive grey button to access the Course Completion Certificate. How do I get the certificate?

Fix for the notebook Exercises 15, Solving the Laplace Equation: Exercises 1 and 3 repeat the text "Use DSolveValue to solve this BVP:" for both the solution and the plot. I'm sure you intend to have the Plot3D function with its own introductory statement, as in Exercises 2 and 4. I'm mentioning this not as an error, but as a fix. Your notebooks are so spectacular that I'm sure you want them to be perfect.

Thank you for the valuable comment. I agree, the statement in this poll question is not very accurate. I will rephrase this question in the updated version of the course.

[Comment deleted. I was looking at a different quiz.]

QUESTION: How do we submit challenges to the automated grading of free-response questions in the Framework quizzes? I created a variable name "discrm" to represent a particular quantity to be computed, then set that variable equal to an expression using terms presented in the question. The automatic grader approved my answer but gave me only 60% credit. The only indication of anything wrong was that it highlighted my variable "discrm" with red wavy underscores, as though it couldn't find it in a dictionary. Why should it require free-response variable names to come from a dictionary? REFERENCE: Quiz 1.

It looks like there was an integration bug in version 14.3.0, which has been fixed in version 14.3.1. Due to this bug DSolve returns an unevaluated integral as a solution for this PDE. This integral can be easily calculated in versions 14.3.1 and 14.2 and gives the correct solution of the PDE. So my suggestion is - update to latest version 14.3.1.

Then I ran it with the condition of x and y being real and got a different solution altogether. It plots and its not the same thing at all. All very odd... Why would it care about the OS?

Quiz 1, Problem 5 has an error. The solution should be of the form u(x,y) and not u(x,t)

Is there a preferred method to entering the parameters "pde", "ics" and "bcs" into the function DSolve[]. In the examples, I see both commas(,) and && used and was wondering if it makes a difference whether commas or && is used. DSolve[{pde,ics,bcs},...} or DSolve[pde&&ics&&bcs,...]

Thanks,

Mitch Sandlin

Hi Dr. Aram, even the quick solution you proposed, in which you impose x, y belonging to real numbers, does not work in the latest version 14.3. Mathematica version 3 returns a solution in the form of a definite integral between minus infinity and plus infinity.

Attachments:

potential along ...nb

In 14.3.0 I get the error message Syntax::sntxf: "(" cannot be followed by "x,y)" when I execute the line solution = DSolveValue[pde && bcs, u[x, y], {x, y}, Assumptions -> (x, y) [Element] Reals] You are running 14.3.1? Does that matter?

Carl

Hello Dr. Aram Manaselyan, I tried to repeat your example of the potential along an infinite strip using my version 14.3 of Mathematica. I'm asking because DSolveValue returns a different result than the one shown in the example in lesson 16 on applications of Laplace's equation. Furthermore the 3D plot of the solution returns nothing.

Attachments:

potential along ...nb

screenshot of 14.1 results -

I tried the following code pde=∇{x,y}2 u[x,y]==0; bcsx=u[0,y]==y(3-y)&&u[2,y]==0; bcsy=u[x,0]==x(2-x)&&u[x,1]==0; solution = DSolveValue[pde && bcsinx && bcsiny, u[x, y], {x, y}] in Mathematica, but I experienced a problem with the Laplacian notation formula ∇^2 as it is in the study material, which is not recognised by the Mathematica interpreter. Could you suggest, please any change?

This is because DSolveValue can't solve the Laplace equation in polar coordinates without boundary conditions. Instead, in poll question 1 for day 5 you should substitute the solutions into Laplace equation to verify them.

In order to solve the equation you should specify some boundary conditions. See e.g. Example 15.4 in Lesson 15.

First question: why in my notebook, identical to the example of the potential along an infinite strip, the solution is returned as an infinite integral while in lesson 16 it is given in the form of a harmonic function? Second question: why do I have to activate the integral calculation to obtain the same harmonic function that I called sol ? This step takes 176 seconds to calculate on my PC. I'm wondering if this is normal or if it's my old laptop's fault. Third question: even calculating the solution on my laptop takes a long time, about 62 seconds, while in the webinar the solution is calculated in a few seconds. I wonder if this is also the fault of my PC, which only has 4 cores.

Attachments:

potentialinfinitestrip.nb

Can you post a screenshot showing the error and correction ?

Thanks, for pointing out this issue. You are right, there is a typo here. I will fix it.

Thank you for the interesting information.

Thank you for very interesting question.

You are right, The method of separation of variables works well for many IVBVPs of linear second order PDEs, like heat and wave equations, with homogeneous boundary conditions. This is because when one tries u(x,t)=X(x)T(t), the PDE splits into a spatial eigenvalue problem with boundary conditions and a temporal ODE controlled by the eigenvalues obtained from the spatial equation. This converts the PDE into a spectral problem. The separated eigenfunctions form a basis, and any reasonable initial condition can be expanded in it, and time evolution is easy.

Thus initial conditions give coefficients like Fourier series coefficients. Boundary conditions determine the eigenbasis. Together, they make the PDE well-posed and solvable.

When Aram was discussing the Heat Equation in Lesson 8, I asked, “I assume that rho could be a function of position. Can it also be a function of time?” Juan Ortiz answered, “rho is the density of the material, it is mostly considered to be constant.” I replied, “Can standard solution techniques handle the case where rho varies with position (even if that is not the case in real-world problems)? If not, can they be extended to handle such a problem?” And Juan answered, “I do not think so. All the solutions here are for simpler examples but please post this question in the community forum so that Aram can discuss it better.” So here it is. Meanwhile, I asked Copilot, “The heat equation contains a term rho = the density of the medium. This is generally assumed to be constant, but are there real-world situations in which rho could be a function of position?” Then further, “The heat equation also contains a constant C = the heat capacity. Could that also be variable rather than constant?” I thought its responses might be of interest, so I will attach these.

Attachments:

Rho and C in the...pdf

Thanks for the interesting question.

In many macroscopic solid-state heat transfer problems the mass density and heat capacity are material properties and can be considered as constants. For example for gold the density is 19.3g/cm^3, for iron 7.8g/cm^3 etc. For the case of gases and liquids, where convective effects can play an important role, both the density and heat capacity can be coordinate and time dependent. But the problem of heat transfer in such complicated systems usually requires numerical methods such as FEM.

You can find some interesting examples in the Heat Transfer section of the "PDE Models Overview" tutorial in Wolfram Documentations:

https://reference.wolfram.com/language/PDEModels/tutorial/HeatTransfer/HeatTransfer.html#509267359

Lesson 14 is missing.

Are there any external resources for PDES please that you can recommend? In particular, I'm interested in the method of characteristics, the solutions method and the transformations of PDES to ODEs

Thank you very much for your response

I am sorry to hear about that. I will forward your message to our tech support team. Meanwhile you can try to disable Dynamic Updating from the Evaluation menue of your Wolfram App. I hope this will help.

I think the discrimanent is missing “4” on the “a*c” term

https://www.bigmarker.com/wolfram-u/wsg74-fundamentals-of-pdes

this link worked for me

Here is the link for study group materials: https://wolfr.am/1BiZEQ06B

Is there any way of looking at a recording? I missed today's but wanted to catch up with the recording but can't find it...

I would like, if possible, to address the topic of transparent boundary conditions for confined systems and where the function escapes the domain.

Thank you so much! I'll check them all!