Method of Characteristics: How to solve PDE

[donaldgornto]

Thank you Dr Tisdell. Your videos helped me earn an A in PDE after 30 years after my undergrad.

[DrChrisTisdell]

It is my pleasure and I wish the very very best with your studies.

[DaleIsWigging]

My whole time at UNSW I've been hoping to have you for the third year PDEs class, Ah well missed my chance at least I have these videos.
I must say I am truly thankful of you and Wildberger for continuing to support free education. You have both inspired me to one day do the same.

[irumbacolin]

Actually I had almost given up on method of characteristics but your tutorials have tremendously blessed me thanks Dr.

[Bharathimohan-tkyn]

It was an amazing video Dr. Tisdell. It made things crystal clear. Thank you SO much!

[justinstadlbauer]

Just wanted to say thank you for this video. I am teaching myself PDEs and this was very helpful!

[ajayanagar]

Better Understanding !! I can now get the actual idea behind the steps for solving pde.

[clarejoyce]

Definitely, I think I should take this opportunity to thank you for your videos over the years. I have found them very useful for my distance education degree in maths!!! KUDOS to you!

[jameshuang]

2 years after graduating from unsw and I'm still seeking for your help. You are the true hero Chris

[mmaween]

Thanks a lot for your effort. I understand from your illustration more than I usually understand from my lecturer's in college. I have a Final exam tomorrow morning in PDEs. Wish me luck.. ^_^

[philliposromeo]

and also you are good when it comes to explain please keep it like that

[GoldsmithsStats]

I have now thought some more about this. Apologies for the length of this comment, but I was struggling with the ideas for while and this at last made sense of it, for me at least.

Let a solution to the PDE (the “solution surface”) be a surface in (xyz)-space described by z = u(x, y). This can be written u(x, y) – z = 0, and the normal to this is the gradient vector grad(u(x, y) – z) = (∂u/∂x, ∂u/∂y, -1) ...(*).

The PDE itself tells us that the vector (a, b, f) is normal to (*). So it lies in the tangent plane to the surface. This is true at every point so if you join the vector field elements (a, b, f) to make a family of curves in (xyz)-space, the solution surface is comprised entirely of these curves. These are the “characteristic curves”.

How do you find the equation of a particular characteristic curve? By definition, along the curve, dx/a = dy/b = dz/f. Solving the first two on their own, gives a whole family of characteristic curves which together comprise a surface in (xyz)-space. It will contain an arbitrary constant of integration c1. The equation will be of form
g(x, y, z) = c1. The taking the second and third, or the first and the third, will give another surface formed of characteristic curves, something of the form
h(h, y, z) = c2.

If you take a fixed value for the c’s, and find the intersection of the two surfaces formed by g and h, you will get a single characteristic curve (in general). If you now vary the c’s, you can get all the characteristic curves in turn. Each one is the intersection of the g and h surfaces for a particular pair of values of the c’s. In fact the c’s act like a pair of coordinates for the whole set of characteristic curves.

We know that a given solution surface is a set of characteristic curves. In general, this will be a one-parameter set out of the whole two-parameter set of characteristic curves, as indexed by the c’s. How do you describe a one-parameter set of points in a two-dimensional space? In general, a function k of the form c2 = k(c1) will do it. This is where the arbitrary function comes in. Adding the Cauchy condition determines k uniquely, in general.

Prof Tisdell's video is brilliant, but it left me with the odd gap in understanding which maybe this comment will help fill for others, too.

[GoldsmithsStats]

Excellent video. Can I suggest just one other thought which may help to understand this, at least it helped me. But I'm not sure if I have got this right. Perhaps someone can comment.

The characteristic curves with constants c1 and c2 can be seen as providing a way of coordinatizing the surface along which the solution u travels ('travels', as a function of t, that is). The set of points comprising the positions of u as time proceeds, forms a curve on this surface.

If c1 and c2 are looked on as coordinates of the points u, then the points have 'coordinates' c1 and c2 which together form a curve in this two-dimensional space. One way of describing a curve in two-dimensional space is as the graph of a function, which can be written in the form c2 = g(c1), which is the result Dr Tisdell arrives at as the solution. (a) is that right, and (b) does it help to understand it?

[luchaojin]

Thank you so much for this lecture. I was trapped by this MOC when solving practical problem, but I am totally clear now.

[ankurc]

Thank you sir! My teacher didn't explain this fully like how we got the actual formula and all!

[emmanuelo.oketch]

Thank you very much Dr. Chris. I have really enjoyed the video.

[alexrosellverges]

Best video out there on the topic, thanks a lot!!

[haoqiangqi]

Thanks a lot for helping me understand this method😁

[rishabkumar]

A lot of thanks sir, you made my life simpler

[kollisomeshrao]

Respected Dr Chris
Please make a video on method of characteristics for quasi linear problem. I was going thru elements of pde by Ian N Sneddon pg no 62 method of characteristics, but didn't understood the method.
I will be thankful to you if you can explain that

Regards