Consider two concentric spheres of radius r = a and r = b, a < b as shown in the figure. The tem

First, we can rewrite the given differential equation as r*d^2u/dr^2 + du/dr = 0. This is a stan

Consider two concentric spheres of radius r = a and r = b, a...

Transcript

00:01 If the temperature of the region between the two spheres of radius a and radius b, where a is less than b, so if those two circles are going to represent our spheres, well, you have two circles there, and you can see that they have those different radii.

00:22 Now the a again is less than b, so we can see that, and the boundary value, the boundaries value, the boundaries value, values, you for the inner circle or for the inner sphere is u equals to u0 and the boundary for the outer sphere is u equals to u 1.

00:44 Well and u not and u .1 are constant so the quest here is to find show actually that our function u of r is equal to 0x the natural log of r divided by b minus u times the natural log of r divided by a over the natural log of a over b so anything you can see up there the trick here or the if you read between the lines we are really asked to find the solution to this differential equation okay this is a second order linear differential equation okay we are to find a solution to that equation and and basically we're to show that this is equal to this.

01:34 Okay? so we're to show that a solution is going to be equal to that.

01:39 That's really, if you really read between the lines, okay? so let's go ahead.

01:44 I'm going to rewrite that differential equation somewhere here.

01:49 I'm going to rewrite it this way.

01:50 I'm going to rewrite it as r.

01:53 U double prime plus u prime as equal to zero.

01:58 So this is a homogeneous.

02:02 If you can see here, we have a case where we don't, we have the missing, but what we call missing y.

02:09 So let's say, for example, you have your derivatives like this, just a standard second order differential equation where you have a p of x times y prime plus q of x times y equals to r, right? usually this rfx something like that in this case we are missing this term this is the missing term okay so there are techniques to do to solve just that okay so let me just show you what we can it be doing here well the first thing is to get it back to that form and that form is y double prime plus the standard form right y double prime times b of x times derivative first derivative of y plus q of x times y equals to r of x i just wrote it down to a few seconds ago but i'm just writing it we want to write it in this form well you can see here that our what is would technically be our y is the u and that u here is has a uh and a a very what to call the variable coefficients right so we're basically solving second order linear equations with variable coefficients.

03:34 So to do this, like i said again, we can write it in the standard form.

03:37 To do that, we're going to divide through by r.

03:39 And when you do that, we're going to get that this is u prime plus 1 over r to times the derivative of u and that's equal to 0.

03:51 Alright? so what we're going to do here, we're going to do something interesting because usually we use r to represent this, but let's do this.

03:59 Let's say let looking for a variable to use let w be equal to u prime okay so that the so that the so that uh w prime will be equal to u double prime okay so i'm going i'm supposed again i want to do this in red so that i would see that so we say again w equals to u prime and that way w prime will be equal to u double prime so if you want to rewrite this we're going to get that u so w so we can yeah we're going to write it this way w plus one over r times uh u prime but this case is just w equals to zero and so here we can we're going to use the we're going to use our method of uh you know how we saw first of all the differential equations here so let's just write it again in red here okay so this is w prime forgot to mention okay just again yeah w prime forgot to write that down so w prime plus one over our w okay so as we normally would do who see this kind of situation now we've come back to the form of we've come back to this form we've come back to the form of y prime plus p of x times y equals this error so this is the homogenous scenario right here we could use the the pressure that's what do we call it we could use the method of separation of variables i can't talk today i'm really sorry so we can use separation of variables right or we could use the integrating factor okay so we can try using an integrating factor i i want to try using e that one.

06:07 So our integrating factor, okay, so i'm going to call that if, or we can just use mew to do that.

06:17 Our integrating factor, which is a function of r, basically would be equal to e to the integral of p.

06:30 Again, don't forget, we're going to be using r instead of x, as we would normally use dr because this function is a function of r and writing things in terms of r.

06:42 The function of r in terms of r and u.

06:46 So this is our integrating factor.

06:49 So that means that our function w would now be equal to, well first let's solve for what our integrating factor would be, right? and so our integrating factor would be, i can do this right here, our integrating factor would be, just put it here, if would be mu, sorry, e to what is p, so this is going to be pr.

07:21 Pr, p of r is equal to 1 over r, correct? so we're going to have this is going to be integral of 1 over r dr, okay? so that's, that is, that is, our, integrative factor to simplify this and make this look a little better.

07:40 We're going to have that this is going to be a to the natural log of r so well essentially when you have this e to the natural log of r well this is going to evaluate to r correct.

08:01 Okay, so we're going to have that this is what w here would be have to be very clear that i'm using m here is going to be one over mu, okay.

08:16 And i guess i should be careful, just use the view there.

08:19 It will separate the u from the mu.

08:21 So it's not going to be confusing.

08:25 But this is going to be over, that's up times the integral of our rx here.

08:33 So what's rx? rx was this function here.

08:38 Okay.

08:39 So our rx, all right.

08:44 So that is times our rx.

08:49 Or sometimes it's called q of x.

08:51 I'm going to use, or precase rx.

08:54 The variables here are really messy.

08:58 So this is times me of x.

09:02 The integrating factor, why do i call it rx? ok, so let me just make sure that i'm not going to get you confused.

09:12 So remember that when you want to find this, you want to solve this thing to use the integrating factor, what you're going to use, what you're going to say is that that function, right, was going to be equal to one over your integrating factor, mu, times the integral of mu times some gfx, right, dx plus c.

09:36 So that's what the solution is using the integrating factor, that's what the particular solution was going to be.

09:42 So let's just apply that here.

09:44 So this is going to be one over mu, we got that mu, we got that mu, is.

09:47 R okay muo is r all right and this is going to be times the integral of well mu again is r sorry view again is going to be r but this is going to be times our g of x and our gfx in this case is zero all right so times zero okay and this is d r and all of this is going to be plus c all right this is going to be plus c all right okay uh let me put that in the bracket supposed to be in the brackets here um so just to make that a little complete just to do this just want to show you that's the case here all right so i'm going to just clean this up just to make sure we're not getting confused and then we're going to go ahead well if you clearly see this part is going to go to zero, right? this whole bit side is going to go to zero.

11:06 Leaving us with the fact that w would be equal to, well, that's going to be c over r.

11:16 We could call that c1 for now.

11:19 You can call that c1.

11:21 So that c1 over r.

11:27 Now, but remember, i'm going to just push this up a little bit.

11:35 So remember that we said that w was what? w was basically u prime.

11:45 That was the case here.

11:47 W was u prime.

11:50 So that means that u prime is equal to c1 over r.

11:59 And so that also means that u would be equal to the integral of c1 over r.

12:10 Dr.

12:15 Let's work that out here.

12:20 I can bring this one down again.

12:22 Move this, just a torch here.

12:24 Alright? yep, i'm going to go ahead and move that this way.

12:34 So we have that u is equal to the integral, just separating the board here so you can see.

12:47 The mu equals the integral of c1 divided by r plus so times the r okay um well of course we can bring this out if you see that this is going to be c1 all right we're going to bring out the c1 and we are looking exactly for the integral of 1 over r d r okay well that is just going to be u is equal to c1 and this is going to be natural log of r plus c2 okay so that if that is the case then we could i guess we could proceed now i want you to see we can do this also again don't forget we can say this is actually a function of r so there's u .r equals to c1 we can do this as c1 times the natural log of r plus c1 times c2 so that c1 and that will be times c2 okay so this is just a case of product of constant and when you multiply constant you're going to get a constant okay so this would be we can call that constant c3 and we move along so this is still c1 times the natural log of r.

14:58 But now we are at a place where we need to find out what c1 and c3 are.

15:04 Because we already have a, what do you call it, we already have the boundary values.

15:10 We have the values at ua and ub.

15:14 And so what are those? so we can say this is at, at, or not just say four, let's say four, for u a okay which is equal to u not we have that u a well that's equal to c1 times um times natural log of r plus c3 well that is so natural log of a because we have replaced r now with a and forget this is the case here so this is natural log of a, all right? so that's a natural log of a plus c3.

16:05 And that we're told is equal to u -not.

16:12 Okay, all right.

16:21 And the same thing happens here for ub.

16:30 So this is a little tedious now.

16:34 I'm writing this, but just want to make sure you're following up.

16:36 So for ub we have that this is equal to u1.

16:42 So that we have that u at b equals to c1 natural log of u at b at c3 and that is equal to u1.

16:58 Okay, so i'm just going to fill in this guys here.

17:03 Alright, so we're almost there.

17:05 So we're just going to bring those two equations and we'll solve them as systems of linear equations.

17:12 So this is our first equation and this is our second equation, correct? that's equation one, equation two.

17:21 All right.