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[WSG26] Daily Study Group: Introduction to partial differential equations

POSTED BY: Aram Manaselyan
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POSTED BY: Mitchell Sandlin

Hello Mitch,

Thank you for your report. I'll create a ticket per your Lesson 4 report for our team to review and follow up once I have more information.

Best,

Christine Owens Wolfram U Project Manager Wolfram U

POSTED BY: Christine Owens

I would like to suggest a small correction to Exercises_18.nb. In the second exercise, the subscript on the eigenvalue, lambda, should be "nm" instead of "nl."

POSTED BY: Edward Tucholski
POSTED BY: Christine Owens

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?

Posted 4 days ago

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?

POSTED BY: James Cobb
POSTED BY: Christine Owens
Posted 5 days ago

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.

POSTED BY: Glenn Huth

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.

POSTED BY: Aram Manaselyan

Hi;

Are Questions 1 & 2 in quiz 4 correct? In question 1, I get an answer, but it doesn't match any of the given answers - not even close. In question 2, I believe there are some typos in the boundary conditions.

Thanks, Mitch Sandlin

POSTED BY: Mitchell Sandlin

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

Both questions are correct.

Hint: In Problem 2 try to verify the solutions and not to solve the PDE.

POSTED BY: Aram Manaselyan

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.

Hi dott. Haram, with version 14.3, I get a very different solution than you did with version 14.3.1. for the potential along an infinite strip. Do you think it's due to a bug in Mathematica installed on my machine? PS: When I installed version 14.3, I encountered a bug in Graphics3D that caused Mathematica to crash due to incorrect 3DRenderingEngine settings. Now the problem is different, perhaps due to an internal kernel calculation routine, but it seems that Wolfram Mathematica sometimes doesn't give the same results to different users. Can you please help me resolve this issue? I'm infinitely grateful.

Attachments:
POSTED BY: LORIS LORI

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.enter image description here

POSTED BY: Aram Manaselyan

I don't think 14.3.1 is available to the the public yet. Does anybody on this forum have version 14.3.1?

POSTED BY: Carl Hahn

I am sorry, version 14.3.1 is not a released version so you can't update to it.

But I have tested the PDE example from Lesson 16 with released version 14.3.0 from July 8, 2025 and couldn't reproduce the problem (see the screenshot).

Currently I don't know why this example is not working in your case. I will continue to investigate this problem.enter image description here

POSTED BY: Aram Manaselyan

Could you attach this example as a notebook? Maybe I can't see a typo.

POSTED BY: Carl Hahn

I ran the notebook from the post of LORIS LORI on a Mac, (14.3.0 for Mac OS X ARM), and I got the same results that Aram Manaselyan shows in his screenshot.

POSTED BY: Michael Rogers

I am running on a Mac Book Pro with Tahoe 26.2 OS. I got a solution with an inactive integration and when I try to execute Plot3D I just get a blank plot. I tried activating the integration and aborted it after a few minutes.

POSTED BY: Carl Hahn

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?

POSTED BY: Carl Hahn

Some solvers have time constraints. They depend not so much on the OS as on the speed of the system. And only if it's an edge case. Edge cases are very rare. I tried the code several times on my Mac (M4 Max, 36 GB RAM). It always gave me a solution. For me, the PDE does not seem an edge case, although it takes a few seconds to execute.

BTW, I would expect a plot of a formula with an inactivated integral to always be empty. An inactive integral will not evaluate to a number, and plotting functions plot only numbers.

POSTED BY: Michael Rogers
Posted 9 days ago

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

POSTED BY: Glenn Huth

Yes, it is a mistake. The correct function should be u(x,y) and not u(x,t). Thank you for pointing out this issue. I will fix it in the updated version of course web page.

POSTED BY: Aram Manaselyan

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

POSTED BY: Mitchell Sandlin

Both methods are OK. You can use any of them.

POSTED BY: Aram Manaselyan

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:
POSTED BY: LORIS LORI

The fix works for me in version 14.3: enter image description here

POSTED BY: Aram Manaselyan

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

POSTED BY: Carl Hahn

When I take the file that ran correctly in version 14.1 (see previous post) and imported into v 14.3, the solution was different and the plot did not show.

enter image description here

POSTED BY: Stephen Kiss

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:
POSTED BY: LORIS LORI

Using 14.3 - I confirm that solution shown is different.

Using 14.1 - I get the same result as shown in the lesson (see attached)

Attachments:
POSTED BY: Stephen Kiss

screenshot of 14.1 results -

enter image description here

POSTED BY: Stephen Kiss

Thanks for pointing out this issue. Looks like this example in Lesson 16 is not working properly in version 14.3. Here is a quick fix for it: solution = DSolveValue[pde && bcs, u[x, y], {x, y}, Assumptions -> {x, y} \[Element] Reals]

POSTED BY: Aram Manaselyan

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?

Posted 14 days ago

I ran into an issue with DSolveValue when I was working on the Question 1 of Day 5, it's a Laplace eqn in polar.

DSolveValue[Laplacian[f[r,\[Theta]],{r,\[Theta]},"Polar"]==0,f[r,\[Theta]],{r,\[Theta]}]

It appears DSolveValue doesn't do anything about it. Can someone please adivse on what's going on?

POSTED BY: Xiangyang Zhou

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.

POSTED BY: Aram Manaselyan
Posted 12 days ago

Thank you, that solved the problem.

POSTED BY: Xiangyang Zhou

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.

POSTED BY: LORIS LORI
Posted 14 days ago

There is an error, a small typo, in example 11.2 of notebook 11. In the equation, it should be the time derivative of u that is multiplied by gamma. This typo is in the "English" statement and the equations below it though are correct.

POSTED BY: George Schils

Can you post a screenshot showing the error and correction ?

POSTED BY: Stephen Kiss
Posted 14 days ago

The red arrow in the graphic below (taken from notebook lesson 11) shows the derivative with respect to x, and it should be the derivative with respect to t. Note it is correct in the pde equation, below, as input to WL.

enter image description here

POSTED BY: George Schils

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

POSTED BY: Aram Manaselyan

Some might find this output from Copilot interesting as it describes "Nabla" and it "Square" in relativity. I learned Nabla as Del. I asked copilot to answer as a mathematician/physicist professor.

Excellent. Let’s step into that professorial voice and treat these operators the way a mathematician–physicist would in a graduate seminar: historically aware, structurally precise, and attentive to why these symbols matter in the architecture of modern theory.


The Nabla and the D’Alembertian — A Professor’s Explanation

1. The Nabla (∇): Origins and Adoption

If you look at ∇ and think “inverted delta,” you’re not wrong — but that’s not where the name comes from.
The term nabla is a 19th‑century scholarly joke that escaped into the wild.

Historical origin

  • The operator itself was introduced by William Rowan Hamilton in the 1830s in his quaternionic calculus.
  • Hamilton originally wrote it sideways, as a kind of differential arrow.
  • Peter Guthrie Tait, Hamilton’s student and evangelist, rotated it upright and used it extensively.

The name

  • The word nabla comes from the ancient Greek name for a Phoenician harp shaped like a small triangular lyre.
  • The resemblance between the harp’s frame and the ∇ symbol was pointed out by William Robertson Smith, a Semitic philologist.
  • Tait shared the term with James Clerk Maxwell, who found it amusing enough to write parody poetry about it.
  • The name eventually stuck, partly because Victorian scientists enjoyed classical references.

Why it matters

Once vector calculus crystallized (largely through Gibbs and Heaviside), ∇ became the unifying operator from which the gradient, divergence, curl, and Laplacian all emerge. Its geometric meaning is now foundational across physics.


2. The D’Alembertian (□): Introduction and Popularization

The square operator, □, is the natural relativistic analogue of the Laplacian. Its history is a two‑stage story.

Stage 1: d’Alembert’s original operator (1747)

In the mid‑18th century, Jean le Rond d’Alembert introduced the wave equation for vibrating strings:

[ \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}. ]

This operator — a second derivative in time minus a second derivative in space — is the conceptual ancestor of the modern □.

Stage 2: The four‑dimensional operator (1905–1910)

The relativistic d’Alembertian did not exist until physics discovered spacetime:

[ \Box = \partial_\mu \partial^\mu. ]

This form becomes meaningful only after: - Einstein’s special relativity (1905)
- Minkowski’s spacetime formalism (1908)

Once spacetime was understood as a unified 4‑dimensional manifold, the operator d’Alembert had introduced in 1‑D was recognized as the unique Lorentz‑invariant second‑order differential operator.

When it gained popularity

The □ operator became ubiquitous in the 1920s and 1930s, when it appeared in: - the Klein–Gordon equation (1926)
- the covariant form of Maxwell’s equations
- the structure of the Dirac equation (1928)
- early quantum field theory

By that point, □ had become the natural language of relativistic wave phenomena.


A professor’s closing remark

Mathematical symbols often look timeless, but they have biographies.
The ∇ and □ operators are excellent examples: one named after a harp, the other after a French Enlightenment polymath, both now indispensable in the language of modern physics.

If you’d like, I can continue in this voice and compare ∇² and □ as Laplacians on Euclidean vs. Minkowski manifolds — a very instructive contrast.

POSTED BY: Deuard Worthen

Thank you for the interesting information.

POSTED BY: Aram Manaselyan
Posted 15 days ago

I have noticed that the method of separation of variables seems to be particularly effective when dealing with IVBVPs. Is this a standard situation, i.e. is there some reason why separation of variables is so effective in such cases, or is it just for convenience?

POSTED BY: Michal Stano

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.

POSTED BY: Aram Manaselyan
Posted 16 days ago

On Day 1, Devendra said in the Q&A, “Sneddon, Fritz John, and Evans are all very good introductions to PDEs.” —

  1. Ian N. Sneddon, Elements of Partial Differential Equations, Dover reprint of the New York: NY, McGraw-Hill, 1957 edition (352 pages)
  2. Fitz John, Partial Differential Equations, Fourth Edition, New York: NY, Springer-Verlag, 1982 (261 pages)
  3. Lawrence C. Evans, Partial Differential Equations, Second Edition, Providence: RI, American Mathematical Society, 2010 (749 pages)

In his introduction, Devendra also mentioned “Courant and Hilbert … and the mighty four-volume book by Lars Hörmander on linear PDEs.” —

  1. Richard Courant and David Hilbert, Methods of Mathematical Physics, Volume 2: Partial Differential Equations, New York: NY, John Wiley & Sons, 1962 (Wiley Classics edition 1989, 852 pages, original German edition 1937)
  2. Lars Hörmander, The Analysis of Linear Partial Differential Operators, New York: NY, Springer — I: Distribution Theory and Fourier Analysis, Second Edition, 2003 (451 pages); II: Differential Operators with Constant Coefficients, 2005 (406 pages); III: Pseudo-Differential Operators, 2007 (533 pages); IV: Fourier Integral Operators, 2009 (359 pages)
POSTED BY: Gerald Oberg
Posted 16 days ago

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:
POSTED BY: Gerald Oberg

In real life, mass density varies all the time (both figuratively and literally). Look at convection, in which warmer material is less dense and rises while cooler material is more dense and falls.

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

POSTED BY: Aram Manaselyan

Lesson 14 is missing from the download.

enter image description here

enter image description here

POSTED BY: Stephen Kiss

Lesson 14 is missing.

POSTED BY: Stephen Kiss

This is fixed now. Thanks

POSTED BY: Aram Manaselyan

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

I recommend following books about PDEs:

  1. W. A. Strauss, Partial Differential Equations: An Introduction, Hoboken, NJ: John Wiley & Sons, 2008.
  2. R. L. Herman, Introduction to Partial Differential Equations, 2015.
  3. I. P. Stavroulakis and S. A. Tersian, Partial Differential Equations: An Introduction with Mathematica and MAPLE, River Edge, NJ: World Scientific, 2004.
  4. P. Kythe, P. Puri and M. R. Schaferkotter, Partial Differential Equations and Mathematica, Boca Raton, FL: CRC Press, 1997.
  5. E. Kreyszig, Advanced Engineering Mathematics, 10th ed., Hoboken, NJ: John Wiley & Sons, 2011.
POSTED BY: Aram Manaselyan

Thank you very much for your response

Posted 17 days ago

Having a strange issue with the lessons. Starting with Lesson 5, when I scroll to example 5.4 (not evaluating each cell), there is a warning about dynamic content, and then Mathematica crashes. I tried it several times, and several Lessons (5, 6, 7, 21 & 23 for sure). Any thoughts on how to view the material and not crash the software? I am using version 14.0.0 Personal Edition, Windows 11 Pro with 24GB RAM. Thank you!

POSTED BY: Perry Holman

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.

POSTED BY: Aram Manaselyan
Posted 15 days ago

Aram, I tried disabling dynamic evaluation, no change in behavior. Also, my Mathematica version is 14.3.0.0. Thank you

POSTED BY: Perry Holman
Posted 17 days ago

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

POSTED BY: Robert Lyons

It depends how you define the PDE. If it is defined as au_{xx}+bu{xy}+cu{yy}+...=0, then the discriminant will be b^2-4ac. But if we write au{xx}+2bu_{xy}+cu{yy}+...=0, where the factor 2 near second term is written for convenience, then the discriminant can be defined as b^2-a*c.

POSTED BY: Aram Manaselyan
Posted 15 days ago

Thank you for your clarification on the discriminant. All my other experiences with PDEs had the form of the PDE without the factor of 2 on the second element. Bob

POSTED BY: Robert Lyons

Will there be a recording? The link did not work.

POSTED BY: Charles Glover
Posted 18 days ago

The link to download did not work. Please send us the new link? Download Study Group materials: https://wolfr.am/WSG

POSTED BY: Taiboo Song

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

POSTED BY: Aram Manaselyan

Thank you!

Posted 18 days ago

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...

POSTED BY: John Drummond

Yes, all recordings will be available. Here is the link for the recording of the first day: https://www.bigmarker.com/wolfram-u/wsg74-fundamentals-of-pdes

POSTED BY: Aram Manaselyan

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

POSTED BY: Alejandro Puga

Thanks for the question. The problem of PDEs with transparent boundary conditions is not covered in the course and study group. But you may find some useful information about PDEs with open boundaries using following links to Wolfram Documentations:

https://reference.wolfram.com/language/ref/NeumannValue.html#435728308

https://reference.wolfram.com/language/ref/AcousticAbsorbingValue.html

https://reference.wolfram.com/language/PDEModels/tutorial/Acoustics/AcousticsTimeDomain.html#587721845

https://reference.wolfram.com/language/PDEModels/tutorial/Acoustics/AcousticsFrequencyDomain.html#82427960

POSTED BY: Aram Manaselyan

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

POSTED BY: Alejandro Puga

This is a reminder that the PDE Study Group begins on Monday, January 12.

Aram has worked with single-minded dedication to develop this introduction to the amazing world of PDEs in just 24 lessons.

I strongly recommend the Study Group to everyone!

POSTED BY: Devendra Kapadia
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