 Hello, welcome to this lecture of bio mathematics. We have been discussing about diffusion in the last lecture, where we use ideas from vectors and calculus to derive the diffusion equation. We said that just by taking the continuity equation that the current from one side to the other side, when you think about the current, we said that for the concentration to change at a particular place, whatever the current that is the flow that is coming in has to be more different, cannot be equal to the flow that go out. If they are equal, the concentration at this point will not change. And from this just simple common sense argument, we use mathematics which is we called, we said it is like a language. So, we use the, we use mathematics basically to express this idea that the current that flowing in has to be different from the current that is flowing out for the concentration at a particular point to change. And from this simple argument, we derive an equation and that is the diffusion equation. That equation governs the diffusion of particles as we discussed. Today, we will discuss a little more about diffusion. There are very important relations that is related to diffusion. So, we will discuss some very important and interesting results associated or resulting from this equation, which are related to diffusion. So, today's lecture is basically again related to application of calculus and vector algebra in biology and we will again discuss diffusion. So, today again we will discuss the, we will continue discussing diffusion and we will derive some interesting results, interesting expressions related to diffusion based on what we learn so far. These are very important relations, very famous relations, very useful in biology or and in many other fields. So, let us get started. So, we said that when you, so what is shown here is a tube. So, you have a tube, but and this long tube and you introduce some particle at some particular position. Let us call x as the length along the tube. So, this is x equal to 0 and this is x equal to 0 and as we go as x is increasing and x is decreasing. So, let us think of this as x axis and there is in this tube we are introducing, let us say we are putting some ink molecules and there is imagine that there is water in this tube and we are introducing some ink molecules or protein molecules in this tube. So, that is shown here and what is shown at the bottom here is a plot of concentration as a function of the x. So, this is x equal to 0. So, at this moment the concentration is only at x equal to 0. So, what almost like a line up, we may be a little more width, but essentially it is just like a line up. So, we say is that the concentration is only the concentration everywhere else here, here everywhere the concentration is 0 and the concentration is known 0 only at x equal to 0 only here you have particles and here here here here no particles. So, this is what this this picture is written in kind of a in the in the form of a graph here that is what is shown here whatever whatever shown here is written in the form of a graph here. Basically, what you have here is that the the physical picture is that you have a tube and in the middle of the tube you have some for protein molecules. Some some the concentration of the protein molecules is only at x equal to 0 and nowhere else you have this protein molecules. How do we represent this this kind of a thing mathematically? What is what is the what is the equation that you will write for representing things of this fashion in a mathematical form? So, we we said that anything that we speak or any idea that we see or any physical idea we can represent in the form of equations and that is kind of a language in itself. So, how do we how do we say this mathematically the fact that you have only concentration at one point and everywhere else it is 0. It is shown graphically here just a line here and everywhere else this is 0. So, the way to show this mathematically it it turns out that. So, if we have x equal to 0 and you have only at this particular point this is shown by. So, this is a function and it turns out that this function can be written as delta of x. So, this delta this famous function this is called this is mathematics this is called Dirac's delta function. Basically, this represents this represents this this is basically precisely to say what we just said that you have only only something only at some particular point everywhere else this is 0. There are some particular properties for this for the Dirac delta function we would not go into this, but what I am trying to say is that you can represent this mathematically using this delta function. So, the concentration that you see here c of x if you wish can be written as a Dirac delta function c of x can be written as. So, it is a c of x is proportional to this is directly proportional to the delta of x is like Dirac delta function. So, this is a way of telling this mathematically. So, now let us think about let us think physically for the moment let us just keep the mathematics apart for the for a moment and. So, we know that you have a tube here and we have some protein molecules here and there is water in this you can think of this as protein molecules or even ink molecules or any any any molecules here and we call it a time equal to 0 this is what the thing you see. Now, think of this as the time goes what will happen let us say after 10 minutes if you look at this tube what will you see you will see that this molecules would have diffused this way and this way they just diffuse this way and diffuse this way. So, they would have diffused either along this tube. So, what do we see you will see this look at here they have diffused along this way. So, there are some molecules here some molecules here, but still majority of the molecules here. So, this is what you see after 10 minutes 10 minutes you after 10 minutes you see that there are many molecules still at the center at x equal to 0 and many molecules a few of them are here and few of them here and if you represent this graphically you can represent in this particular way. The concentration if you plot as a function of x what you would get is that at far away the concentration is very less and at the middle the concentration is large. So, forget what the form of this curve what is the equation here, but there is some particular form which is a mathematically this is graphically this concentration can be represented by this particular graph that is it is a it is a peak at the middle that means still majority of the molecules are in the middle very few of them are towards the ends as you go along far away from the center either to the left or to the right you have fewer molecules and there is no particular reason that right is any different from the left because if there are 2 of them 100 or 20 of them diffusing to the right you might as well have 20 of them diffusing to the left I mean there is no symmetry they are symmetric like left and right are symmetric to each other there is no particular reason that left is different from the right on this on this on this particular tube. So, therefore, the concentration on average on the left and the right will be the same there is no particular reason that more of more will diffuse to the left or more will diffuse to the right. So, this function will be symmetric along x equal to 0. So, you have a function which is essentially. So, you either begin to begin with you had this x axis. So, this is x going to if you go if you think of this let us say you are going to plus infinity and this is so this is plus infinity and this is minus infinity. Let us say this is x is going this way to begin with you had just one line as a concentration. So, this was a concentration to begin with. So, this is t equal to 0. Now, as the time went as the time went the concentration. So, if you take this x axis this is minus infinity this is plus infinity this is 0. So, this is 0 here what you would see is that you would see kind of a nice symmetric thing like this. I mean maybe what I do may not be perfect, but I wanted to have the peak at 0 here this is peak at 0 and symmetric about this as it goes to infinity. So, this is the concentration as a function of x as it goes to infinity the concentration will go to 0. So, at you know that in the tube if you look at the tube here far away at infinity you will not find anything after 10 minutes because it would have it would not have yet reached infinity. So, important point see at look at here see at infinity 0 see at minus infinity is also 0. So, that means here and here the concentration will slowly is actually 0 far away the concentration is 0 and at the middle you have maximum concentration. So, you have why after 10 minutes this is what you expect, but the important point to note is that at infinity see at infinity and see at minus infinity has to be 0 because you can imagine after some 10 minutes the concentration it would not have reached infinity. It would not it would not if you look at a distance far away if you look at far away distance it would not have yet reached that far. So, the concentration there has to be 0. So, by knowing just this that it has to be symmetric and that the concentration the molecules would not have reached at infinity or far away we can derive very interesting results. This is some two physical things we have to know that after it will have such a shape and it will have a symmetric shape around x equal to 0 and that concentration will be 0 far away I mean it would not have reached there. No molecules will have reached far away from the initial position if you look at far away from the initial position after 10 minutes molecules would not have reached there. After any time after finite time t it would not reach some infinitely far distance. If you know this much we can derive some interesting. So, what what did we say we said that we have what did we say we have some concentration and see at x equal to infinity is 0. See at x equal to minus infinity it is 0. What does this mean? If you have in a pipe if you had a pipe like this and if you had some molecules here after any finite time they would not have reached far away infinities and minus infinities. Here the concentrations will be 0 here also the concentration will be 0 they will be still diffusing somewhere near the initial point. So, this is finite time at any finite time after 1 minute 2 minutes 3 minutes 4 minutes even after 10 minutes it would not have gone that far it will be still at a finite distance x the concentration. So, knowing this much idea. So, now if you know this it would not have reached here the derivative let us say del c by del x at x equal to infinity this will also be 0. Because, there is no molecules here the c itself is not there. So, c cannot vary there because c does not access there. So, del c by del x will be 0. Here also del c by del x at minus infinity when x equal to minus infinity this will all be 0 because there is no molecules. So, the change in concentration will be anyway 0. So, if you by knowing this much idea we can get some interesting results. So, let us go ahead and do this. So, now let me do some small let us discuss another interesting fact. So, we know that we have a concentration c of x. So, what is integral c of x d x? So, at every point x you have this concentration c. So, now what you are doing you are doing for summing over all the concentrations along x. So, that is integral c of x d x. So, what is this? This has to be the total concentration c t. This has to be the total concentration c t because this is the now for simplicity for a for a now let let me define as some a new function called c till day of x which is c of x divided by this constant c t. Just for some convenience I am doing this this is just a trick just for convenience I am doing this. So, if you have such a function what will be integral minus infinity to infinity c till day of x d x. This will be integral c of x d x divided by c t. We know that integral c t c integral c of x d x is c t from here. So, this will be c t by c t this will be 1. So, we have this new function let us define and this function called c of c till day of x which is the c of x divided by some constant number which is the total concentration. So, this what is the c till day of x essentially this is the fraction in some sense concentration divided by the total concentration. In some sense you can think of this is a percentage or a fraction or in some senses this is like a fraction, but does not matter what does it mean. It is just a trick we have this concentration we divided by a constant we can always divide by a constant mathematically nothing changes and we have a new function c till day of x. So, we did some tricks we just did realize that integral c of x d x is total concentration and we defined a new function called c till day and we called it c of x by c t such that the integral c till day of x d x is 1. That means the this integral of this becomes 1 we have c till day of x. Now, let us go back to the diffusion equation. So, what is the diffusion equation? Diffusion equation is this equation that we described last time that del c by del t at any point x and t at any point x at any point time at any point x at any time t is equal to d del square c by del x square. So, the change in that how does the concentration change with time is this and how does the concentration change in space x is the distance space variable and this is essentially this is essentially how does the concentration change with space and this how does the space will change with time how does the concentration changes with time. So, this relation is called the diffusion equation. Now, let us look at a few properties that we defined last time we define x average. So, here x average is integral x c till day x d x. So, c till day which is defined it is concentration divided by c t. So, this is some kind of a normalized distance do not worry about this. So, this is the definite let us define some properties called x average and x square average. What is this essentially mean we will discuss this we will discuss what is this essentially mean what is this physically mean, but we can always define x average and x square average in this particular manner. So, if you do this x into c till day of x d x and x square into c till day of x d x we get this x average and x square average. Now, the aim of this is that the aim of this is to derive some interesting results related to the r m s distance of diffusing particle the root mean square distance. So, this is the definition of x square average. So, typically you know you might have seen it many places that you might have heard of r m s velocity r m s root mean square velocity what is root mean square velocity like think about you had you had gas in a box and this gas molecules were freely moving in all directions this ideal gas we had and we could calculate and in such cases you might have studied that v r m s the root mean square velocity this is related to the r t and temperature and all that this is related to temperature we have resigned something about r m s velocity. So, the definition of this is that v square average this is root there is a mean this is mean sorry this is square mean mean square. So, and this. So, this is the definition of r m s velocity you have find the root mean and square root mean square similarly here you have particle in a tube going this way and that way they are diffusing and what we want is x square average root. So, this is you can say x r m s the r m s distance the root mean square distance that this particle will diffuse with time like if you wait 10 minutes what is the root mean square distance this particle will diffuse. So, let us think about just one particle here look at this one particle this one particle will can go little here then it here. So, on an average we also can find out how much what is the average distance and what is x square average this is what we want to find out and this is what we have defined here. So, let us we have defined x average as this and x square average as this. Now, let us calculate this now how do we calculate. So, we had this diffusion equation which is del c by del t is d del square c by del x square I can divide both side by c t. So, c by c t is c till day. So, I can write this as del c till day by del t is del square c till day by del x square where c till day it defined as c by c t you just divide here by a constant here also by this constant c t both sides. So, I divide by c t here I divide by c t here. So, I divide by c t here I divide by c t here since it is a constant I can get this equation. So, now we have this equation. So, now that we have this equation del c till day by del t is equal to d del square c till day by del x square I can multiply both sides. So, first let us first calculate x square average I can multiply both sides with x square and integrate. So, let us do this. So, there is del by del t I multiply with x square and there is c till day and is equal to d del I have x square I multiplied and I have del square c by del x square and I integrate both from minus infinity to infinity d x d x. So, what did I do I just multiplied both sides with x square and integrated both sides nothing will change I multiplied both sides with x square I integrated both sides with respect to x. Now, here the derivative is with respect to time and this is a partial derivative. So, does not matter whether I take the derivative outside and the integral is with respect to x. So, if the integral is with respect to x you can take this del by del t outside because the result of this is anyway has nothing to do with time sorry this is this is only depends on x. So, again whether I write x square del by del t inside or outside it does not matter because the integral is only with respect to x and this is a partial derivative. So, I have this kind of an equation. So, what do we have I multiply what did I do I had this equation I multiplied both sides with x square and I integrated with g x this is what I did. So, let us write it down once more to be clearly. So, we had del by del t of integral minus infinity to infinity x square c till day of x d x is equal to d into integral minus infinity to infinity x square del c till day by del x square d x this is what we have and this is what precisely we have it here. Now, what is this part what is x square c till day of what is this we just learn the we just define this as x square average. So, this quantity which is just defined this quantity as x square average this quantity what I put in this around which I drew this line I have we just define this as x square average. So, the left hand side is del by del t of x square average. So, now what is on the right hand side we have this now you have some function x square some other function del c till day by del x square. So, you have two functions which depends on x and we have to find out the integral of this. So, we have minus infinity plus function x square and another function del c till day by del x square d x. So, now we have to find the integral of this. So, something that we learned in calculus we have to make use of this to get this. So, what did we learn in calculus. So, there is an interesting relation that we learned in calculus that we can make use here and what is that. So, we defined we learned that if you have two function u and v the we have u of x and v of x. We learned that the derivative we learn something called product rule the derivative of this we learned that d by d x of u v is u into d v by d x plus v into d u by d x. We learned this in calculus this is called we call product rule. So, that is derivative of product of two function is first function with derivative of the second function plus second function into the derivative of the first function. This is what we learned now we can integrate this just like this. So, we can just multiply with this and integrate this and we can write this d x here. So, we can just integrate all throughout and get a relation we can say u into d v by d x is equal to. So, we can write. So, what can I write. So, look at here. So, I can write u into integral d v by d x d x is equal to. So, this is equal to integral of d by d x of u v d by d x of u v d x minus I take this side. So, I just rearrange this term I write this is equal to this minus this I can take this term this side. So, u integral this minus this term which is integral v d u by d x d x d u by d x. So, some we are just making use of something that we learned in calculus that we use the product rule and using this product rule using this product rule look at here using this product rule we rewrote it in terms of integrals and is a integral u d v by d x is this. So, in other words I can rewrite this thing I can rewrite this thing. So, you know d v by d x into d x you can write this basically. So, let us write this little more carefully u into integral d v by sorry I just made a small mistake when I wrote here. This integral has to be here integral u d v by d x because this what we had here integral u d v by d x is equal to d by d x of u v d x minus v d u by d x. So, when I write little more carefully I have write. So, what do I write integral u d v by d x is equal to d by d x of u v d x. So, derivative of function integral of a derivative is a function itself because you know integral v set is a anti derivative. So, integral of d by d x of u v is u v itself. So, this is essentially the u v in the limits. So, if you have this integral from minus infinity to infinity all this integral will be from minus infinity to infinity. If this integral is from minus infinity to infinity what you would have is in the limits minus infinity to infinity think about this. This is what you will get just carefully do this integral u v d v by d x into d x will be equal to u v minus integral v d u by d x into d x. So, this is call integrate by pass you have any function u d v by d x you can write in this particular form. . Look at this equation look at this equation what do we have here what do we have here we have in this side a function let us call this. So, what did what did we have here we had we had an equation which is del by del by del t of x average is equal to integral minus infinity to infinity d times x square del c by del square c tilde by del x square d x this is what we had. So, we can always define this as u x square as u and this as d v d x. So, let us take x square as u and del square c by del x square as d v d x. So, what is this this will be integral u d v d x d x. So, you can use this formula now to do this integration of this. So, if you look at this carefully let us write down only the right hand side the right hand side if you write down the right hand side only what do we have is that d into integral x square del square c tilde by del x square d x as integral I call this x square as u and this as d v by d x into d x and we said that this is just nothing but integral u v in the limits minus integral v d u v d u by d x minus infinity to infinity. Now here u is x square this we take as u and d v by d x is del square c tilde by del x square. So, what is v if this is d v by d x v will be del c tilde by del x. So, the derivative of this will be del square c tilde by del x square. So, you will get this. So, if you apply this formula and rewrite this integral what you would get is the following. So, look at here. So, you will get u which is x square into v in the limits minus integral v which is del c tilde by x into d x u v d u derivative of x square which is 2 x. So, if you apply that formula that is integrate by parts do that carefully you will get exactly this. You apply that formula take x square as u and del c tilde by del x as v you will get exactly this formula. Now what is the derivative at infinity and minus infinity we said that the derivative at infinity and minus infinity there is not anything. So, this derivative at both the places will go to 0. So, this term will go to 0 because there is no concentration at infinity and minus infinity there is no derivative also because there is no change in concentration either because there is not anything at all. So, essentially or in other words it is symmetric whatever on the left side has to be equal to the right side. So, this term will go to 0 and this term will go to 0 and this. So, what remains is just this what remains is just this. So, del by del t of x square average has to be minus I take this 2 outside 2 D minus infinity to infinity del c tilde by del x into x d x. So, you have this function now. Now you can do the same way integrate by parts once more. So, now let us see what we have in hand what we have in hand is that look at what we have in hand we have in hand is del by del t of x square average is minus 2 D into minus infinity to infinity x del c tilde by del x d x. Again we can do this integrate by parts we can take this x as u and this as d v by d x or del v by del x if you want and you can integrate this by parts by the same formula that we used and what do we get what you would get is that again let us say x is u x is u. So, u into v in the limits minus v d u v is c tilde and d u is d x which is 1 sorry which is the derivative of x is what d x by d x is 1. So, what you will get is 2 D into integral c of x d x again using symmetry arguments and that the concentration anyway at infinity will be 0 you can show that this term is 0 why is this 0 because the concentration at infinity and the concentration minus infinity is anyway 0. So, this term has to be 0. Now this term is integral 2 D 2 D integral c of x c tilde of x d x. So, what we have essentially at the end of the day if you say that this if you equate this term to 0 because the concentration at infinity and minus infinity is 0. So, if you equate this term to 0 what you have is that del by del t of x square average is equal to 2 D integral c tilde of x d x minus infinity to infinity. At the beginning we said that this is 1 that is the definition of c tilde of we define c tilde of x such a way that integral c tilde of x d x is 1. .. So, now what we have del by del t of x square average is 2 D this is what we have. So, what do we have del by del t of x square average is 2 D. So, now let us integrate both sides and see what we get. So, what we get is that. So, if we have since this look at here since this is integral c tilde of x d x is 1 we have del by del t of x square average is 2 D. So, we have x square average is equal to 2 D t. So, we have x square average is 2 D t. This is a very interesting and very important relation as far as diffusion is concerned very widely used in many many context in biology and in many places. So, what did we get? We get x square average is 2 D t. We can say x r m s is square root of x square average is square root of 2 D t. What is that tell you? What does this tell you? So, this has lot of significance. So, let us think about for a minute about the significance of this result. So, look at here this once more is the result. Look at this x square average is 2 D t. What is x square average means? The r m s or in the r m s distance means the average distance that this would go to either side. That is r m s distance this molecule will diffuse in a time t is x r m s. So, what does this mean? If you wait for 100 seconds. So, let us say. So, let us d is a constant t is time. So, let us let we know that let us take for example, let us take for example. So, let us take this x r m s is equal to root of 2 D t. This is a very famous relation. Now, let us see what is it? Let us take d is equal to. So, what is the unit of d here? d will have a unit which is. So, let us take d is 1. Let us say it is meter square per second. This will have a unit of meter square per second. Diffusion coefficient has a minute of meter square per second. We will come into this. We will we will see how this is coming. But for the moment let us take d as some number which is 1 meter square per second given to you. Now, the question is let us say when t is equal to 10 second. How far this would have diffused? So, x r m s is equal to root of 2 into diffusion is 1 into 10. So, this is a diffuse the distance of root 20 meter root 20 meter. Now, let us say after 100 seconds when t is equal to 100 seconds. How much this will have diffused? When you take t equal to 100 second what do you get? x r m s is equal to root of 2 into d is 1 itself. This is 100. So, you know this is 200. So, what is this? This is square root of 2 into 100. So, square root of 100 is 10. So, this is essentially what we get? Root of 200 meter. So, this is root of 20 and this is root of 200. So, now we can write root of 200 as 10 into root 2. So, this you can write root 10 into root 2 and this is 10 into root 2. So, we see in this particular way we can calculate how much distance it will travel in time t. If it is 100 seconds it will have gone. So, you know root of 20 what is root of 20? So, this is like root of 16 is 4, root of 25 is 5. So, root of 20 is somewhere in between 4 and 5. So, let us look at here. So, first case first when t is equal to 10 second we get x r m s is equal to root of 20. We know that this has to be 4 point something because root of 25 is 5, root of 16 is 4. So, this is 4 point something. Calculated yourself what is root 20 and what is in 100 seconds that is the time change by 10 times. If you wait 10 times longer it would have diffused a distance root of 200. What is root of 100? Root of 100 is 10. So, what root of 100 is 10? So, this is little bit above 10. So, root of 15. So, this is some quantity much larger than that. So, this is basically what you get you can write it as root of 200 can be written as root of 2 into root of sorry, this is wrong. So, what will be this number? This number will be like something between very close to 15 because we know that root of 225 is root of 200 is. So, this is something close to 15. So, this is some number which is very little less than 15. So, calculate this number yourself. So, even though the time increase 10 times here it did not travel 10 times it only traveled much less than 10 times. So, this is what essentially it says even though the time increases by 10 times the distance traveled will be much less than 10 times. So, let us take another example. Let us take. So, what we have is X r m s is equal to root of 2 d t and let us now calculate let us ask the question. So, let us do in another way let us add t is equal to x square average. So, this is what we had 2 d t is x square average. So, let us say t is equal to x square average by 2 d. So, now let us ask the question if we can even ask the question again do it yourself ask this question. If you want something to diffuse let us say if it is diffuse 1 micron let us take the example of some more protein molecule let us say actin molecules and ask the question what is the time it takes to diffuse 1 micron and ask the question how far how long it will take how much time how long it will take to diffuse twice the distances 2 micron. So, such kind of things can be calculated from what we learn so far. So, just like what we did now we can also calculate x average which we will discuss later, but this is a very important relation which we used many many different places many different occasions both in biology chemistry physics chemical engineering any field you take pretty much this relation is used that is the x square average is 2 d t if and this is a very important consequence. And anything that will diffuse if you wait 10 times longer it will not go 10 times further it will only go much less than the 10 times much less than 10 times. So, that is the whole message we just saw that if you we just saw that if you wait 10 times more time the distance that will be travel will be much less than the 10 times. So, essentially what to summarize the lecture what we learn today is a very important relation. So, this is our summary just one equation x square average is 2 d t. So, this is our summary for today because this is a very important relation in biology in physics if you have learned spectroscopy or any many many different areas you might you might come across this relation. So, just realize that without solving the diffusion equation we started from this equation without solving this equation by doing a trick by just realizing that the at infinities the concentration will be 0 we found that x square average is 2 d t. So, with this important relation we will stop today's lecture we will continue little more about diffusion in the coming lecture very and there is another important relation we will discuss in that in the next lecture. So, for this lecture this lecture let us stop it here now by.