I DO NOT really have time to write this review,but I have to take time to thank you Dr Chris. I do have an exam tomorrow and i have no single idea of whats going on in my pde class, i dont think i can pass the exam until I saw your video, I think I can just ace the exam very easily. Thank you so much Dr Chris, you have saved thousands of students life.Hopefully that I can see your more videos.
For those who do not understand why (x,y)(b,-a) = k, you can easily think of it like this: the equation of a line is given by: m *x* + n *y* + c = 0 where (-n,m) is a *direction vector*, in our case: the direction vector is (a,b) so: n = -a and m = b hence the equation of that line is: b *x* -a *y* + c = 0 or b *x* -a *y* = k where k = -c
Thank you so much! Your video clarified the solution step by step and explained all the methods that were used. It helped me to get a better understanding of the material. Thank you again.
Here are things that initially confused me and later found answer to: 1. For directional derivative to be zero, you can be travelling perpendicular to the gradient too. Why then has Dr. Chris taken only the constant u case? Well, both are one and the same. Gradient is the direction of steepest ascent. Let's assume you are in a point on the mountain and looking towards the steepest line is the gradient. Keep your body fixed, but rotate your head around. You will eventually go from looking up at the steep to down the slope. You will pass through a line that has no slope. And that line is going to have constant u. Also, that line would be perpendicular. Consider this (a,b).(x,y) has some value (considering both are non zero) except for at 90 degrees. 2. Why (x,y).(-b,a)? The equation of a line in cartesian form along (a,b) through (x0,y0) is (x-x0)/a=(y-y0)/b. On solving you get b(x-x0)=a(y-y0). Put b(x0)-(a(y0) as a constant. You can see where that comes from
The right hand side of the PDE is zero. The left hand side of the PDE is a special dot product. Since we have a PDE the left hand and right hand sides must be equal.
I was just watching this other video v=43i-A6lt_MI and one of the comments said that an example, instead of the general case method/demonstration, would be more helpful. Sometimes there's no pleasing everyone so that's why we look in many different places.
Hello, Your are a great professor. I'm learning a lot with your youtube channel. Does exists a PDF for this serie of lesson (on the PDE). Thanks a lot Eric
What if the characteristic lines are not straight lines but some curves? Oh, and does this technique also work for higher-order equations? I think it should, because I know that at least the wave equation can be solved with characteristics. So how can this concept be generalized for higher-order equations? I can't quite see how gradients/directional derivatives apply there...
Thank you so much professor! would be great if you could clear my confusion about Linear PDE: solution of a linear pde can always be represented by a linear combination, is it true?
He's using the normal vector because he's using the property of dot products where if the dot product of two vectors is zero, then those vectors are at right angles (a.k.a. orthogonal) to each other. We are trying to find a family of parallel lines, but for the moment, imagine that the line we are trying to find the equation for is the one that goes through the origin. Let's say that the vector is also located at the origin. You know that the origin is on the line in question and you know that the tip of the vector is on the line, but you do not know any of the other points on the line. We want to find an equation for the line, which is equivalent to finding the set of points on the line. This in turn is equivalent to finding the set of vectors from the origin to a point on the line. We can express the condition "a point (x,y) is on the line" with the equivalent condition "a vector (x,y) is perpendicular to a normal of the line." We can convert this statement into a numerical equation by saying that if v=(x,y) is a vector and n is a vector normal to the line, then the vectors we are interested in are the set that satisfy dot(v,n)=0. Since this is a two dimensional problem, there are two choices for the normal vector, (-b,a) and (b,-a). (These are just negatives of each other and so point in opposite directions) Now that we have a mathematical condition for the one line going through the origin, let's find all the others that are parallel to it. Let's take n=(b,-a). Then dot(v,n)=0 gives bx-ay=0 which gives y=(b/a)x. However, the video doesn't have =0 but =k. If we have dot(v,n)=k we have bx-ay=k which gives ay=bx-k which gives y=(b/a)x-(k/a). If you compare this with the slope intercept form of a linear equation, you'll see that -k/a is the y intercept. So all k does is raise or lower the entire graph. A positive value of k/a lowers it and a negative value of k/a raises it. In this way all the lines parallel to the original line are described.
I have a question what if a and b are variables of x and y not constants? if anyone has an answer to this that would be great because my course has only shown us how to do questions with a and b being constant
The function `f` is ARBITRARY. Which means that it can be ANY function whatsoever (pick your favourite), as long as it is a function of `b·x - a·y` instead of just, let's say, `x`. The particular function is chosen by initial/boundary conditions.
aditya morey I think it is zero not because the actual grad dot (a,b) is zero, but rather to be consistent with the initial pde, the term on the left has to be zero ...idk I could be wrong tho
I think it is zero not because the actual grad dot (a,b) is zero, but rather to be consistent with the initial pde, the term on the left has to be zero ...idk I could be wrong tho
Hi - they are private as there were audio problems with the recordings, but I forgot to remove them from the original playlist. Thanks for reminding me!
Dr Chris Tisdell If we suppose (x,y) is a point on the line, it is then parallel to (a,b) right? So why then is (x,y).(b,-a) = constant and not zero. I think i am missing something.
Josh Winfield I see it the same way as you. It's the case the right hand side of the PDE is a constant (not 0) .. Then all the lines where u=Const are at an constant angle to the vector (a,b).
www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/gradient-and-directional-derivatives/v/directional-derivative All of these videos help immensely in describing a directional derivative. It is similar to a derivative in two dimensions where it describes the slope at a point. In two dimensions, it shows small (usually positive) change in x leads to a small change in y. However, in three dimensions, we now have x and y as a domain. Rather than just having a positive small change in x that we look at, we can have changes going in any direction in any direction in two dimensions. The directional derivative asks, what is our change in our third dimension (here, it is u) when we change along the direction in the x, y plane. Hard to explain in a comment, but I hope I helped some.
I DO NOT really have time to write this review,but I have to take time to thank you Dr Chris. I do have an exam tomorrow and i have no single idea of whats going on in my pde class, i dont think i can pass the exam until I saw your video, I think I can just ace the exam very easily. Thank you so much Dr Chris, you have saved thousands of students life.Hopefully that I can see your more videos.
I love math. I will watch these videos when I'm done watching the ODE videos. Thanks for these videos.
For those who do not understand why (x,y)(b,-a) = k, you can easily think of it like this: the equation of a line is given by: m *x* + n *y* + c = 0 where (-n,m) is a *direction vector*, in our case: the direction vector is (a,b) so: n = -a and m = b hence the equation of that line is: b *x* -a *y* + c = 0 or b *x* -a *y* = k where k = -c
Why is a negative? How do we know solutions to this PDE are linear equations with a negative slope?
Thank you so much! Your video clarified the solution step by step and explained all the methods that were used. It helped me to get a better understanding of the material. Thank you again.
Here are things that initially confused me and later found answer to:
1. For directional derivative to be zero, you can be travelling perpendicular to the gradient too. Why then has Dr. Chris taken only the constant u case?
Well, both are one and the same. Gradient is the direction of steepest ascent. Let's assume you are in a point on the mountain and looking towards the steepest line is the gradient. Keep your body fixed, but rotate your head around. You will eventually go from looking up at the steep to down the slope. You will pass through a line that has no slope. And that line is going to have constant u. Also, that line would be perpendicular. Consider this (a,b).(x,y) has some value (considering both are non zero) except for at 90 degrees.
2. Why (x,y).(-b,a)?
The equation of a line in cartesian form along (a,b) through (x0,y0) is (x-x0)/a=(y-y0)/b. On solving you get b(x-x0)=a(y-y0). Put b(x0)-(a(y0) as a constant. You can see where that comes from
This is something that I plan to do later this year when I teach a full course on PDE.
Awesomeeeee....Now I understand the concept not just the procedure....thanks Chris
Yes, this is an interesting generalization. You can find the method for that problem in my PDE playlist.
Thank you so much! This cleared up a lot of issues I was having.
you understand it clearly. good
The right hand side of the PDE is zero. The left hand side of the PDE is a special dot product. Since we have a PDE the left hand and right hand sides must be equal.
Hi Dr.Chris, If possible can you tell why the points on u=constant satisfy the equation bx-ay=k?
I was just watching this other video v=43i-A6lt_MI and one of the comments said that an example, instead of the general case method/demonstration, would be more helpful.
Sometimes there's no pleasing everyone so that's why we look in many different places.
Where is this general solution? Can't find it.
Thank you for all your work!
thanks alot sir.. very helpful.. do make more vids..
sophomore grad IIT kanpur , India
Hello,
Your are a great professor. I'm learning a lot with your youtube channel.
Does exists a PDF for this serie of lesson (on the PDE).
Thanks a lot
Eric
Thanks. I am keeping going!
What if the characteristic lines are not straight lines but some curves?
Oh, and does this technique also work for higher-order equations? I think it should, because I know that at least the wave equation can be solved with characteristics. So how can this concept be generalized for higher-order equations? I can't quite see how gradients/directional derivatives apply there...
Thank you so much professor!
would be great if you could clear my confusion about Linear PDE: solution of a linear pde can always be represented by a linear combination, is it true?
Thank you so much Doc
Can someone please explain why is he using normal vector (to the line) when he is searching for characteristics
He's using the normal vector because he's using the property of dot products where if the dot product of two vectors is zero, then those vectors are at right angles (a.k.a. orthogonal) to each other.
We are trying to find a family of parallel lines, but for the moment, imagine that the line we are trying to find the equation for is the one that goes through the origin. Let's say that the vector is also located at the origin. You know that the origin is on the line in question and you know that the tip of the vector is on the line, but you do not know any of the other points on the line. We want to find an equation for the line, which is equivalent to finding the set of points on the line. This in turn is equivalent to finding the set of vectors from the origin to a point on the line. We can express the condition "a point (x,y) is on the line" with the equivalent condition "a vector (x,y) is perpendicular to a normal of the line." We can convert this statement into a numerical equation by saying that if v=(x,y) is a vector and n is a vector normal to the line, then the vectors we are interested in are the set that satisfy dot(v,n)=0.
Since this is a two dimensional problem, there are two choices for the normal vector, (-b,a) and (b,-a). (These are just negatives of each other and so point in opposite directions)
Now that we have a mathematical condition for the one line going through the origin, let's find all the others that are parallel to it. Let's take n=(b,-a). Then dot(v,n)=0 gives bx-ay=0 which gives y=(b/a)x. However, the video doesn't have =0 but =k. If we have dot(v,n)=k we have bx-ay=k which gives ay=bx-k which gives y=(b/a)x-(k/a). If you compare this with the slope intercept form of a linear equation, you'll see that -k/a is the y intercept. So all k does is raise or lower the entire graph. A positive value of k/a lowers it and a negative value of k/a raises it. In this way all the lines parallel to the original line are described.
So, basically, delU has to be either (0,0) or (b, -a)
I have a question what if a and b are variables of x and y not constants? if anyone has an answer to this that would be great because my course has only shown us how to do questions with a and b being constant
hello,Sir i need to ask do you have some PDE's lectures about Sobolev spaces ?because your lectures are really very helpful.Thanks
Hi - thanks. I plan to speak about Sobolev spaces but not in the near future as I am teaching different courses now.
Would you please recommend some textbooks of PDEs, professsor?
(By the way, I have learned linear algebra, calculus, ODEs)
GIVE HINT, INTRO TO PDE
Can someone give an example of a function that satisfies the PDE. Thank you.
Hi, for example: cos(b*x-a*y) .
The function `f` is ARBITRARY. Which means that it can be ANY function whatsoever (pick your favourite), as long as it is a function of `b·x - a·y` instead of just, let's say, `x`.
The particular function is chosen by initial/boundary conditions.
how is the dot product 0 ... please explain
aditya morey I think it is zero not because the actual grad dot (a,b) is zero, but rather to be consistent with the initial pde, the term on the left has to be zero ...idk I could be wrong tho
I think it is zero not because the actual grad dot (a,b) is zero, but rather to be consistent with the initial pde, the term on the left has to be zero ...idk I could be wrong tho
Why Video 40-45 are private?? do i have to buy them??
Hi - they are private as there were audio problems with the recordings, but I forgot to remove them from the original playlist. Thanks for reminding me!
Okay ?
8:00 : doesn't the constant K have to be zero for the vectors to be normal to each other?
Hi - I'm not saying the two vectors are normal. What I am doing at 08:00 is forming an equation of a line.
Dr Chris Tisdell I see, thanks!
Dr Chris Tisdell If we suppose (x,y) is a point on the line, it is then parallel to (a,b) right? So why then is (x,y).(b,-a) = constant and not zero. I think i am missing something.
Josh Winfield Oh i think i got it now. It is only parallel to (a,b) if there is no constant in the equation for a line e.g y = x for (a=1,b=1).
Josh Winfield I see it the same way as you. It's the case the right hand side of the PDE is a constant (not 0) .. Then all the lines where u=Const are at an constant angle to the vector (a,b).
what is a directional derivative
www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/gradient-and-directional-derivatives/v/directional-derivative
All of these videos help immensely in describing a directional derivative. It is similar to a derivative in two dimensions where it describes the slope at a point. In two dimensions, it shows small (usually positive) change in x leads to a small change in y.
However, in three dimensions, we now have x and y as a domain. Rather than just having a positive small change in x that we look at, we can have changes going in any direction in any direction in two dimensions. The directional derivative asks, what is our change in our third dimension (here, it is u) when we change along the direction in the x, y plane.
Hard to explain in a comment, but I hope I helped some.