 Hi, I'm Zor. Welcome to Unizor Education. I would like to complete information which I'm giving to you about partial differential equations. My previous lecture was about heat equation as an example of partial differential equations. And well, what good does it actually take to give you an equation and not to discuss how to solve it, right? So today we will solve this equation. Not to the very, very bottom, but basically we will do whatever is necessary to bring us close to a solution as possible. This lecture is part of the course of advanced mathematics for high school students. It's presented on Unizor.com. And if you are watching this lecture from some other source, like from YouTube, for instance, directly, I would rather recommend you to go to this site and find this lecture in the directory, in the menus, because every lecture has very detailed notes to this lecture. And you can read it basically as textbook. There are some exams, not for every topic which I discuss, at least yet, but many topics do have exams. So everything is free, by the way, on this site and there are no advertisements. So I do suggest you to use the Unizor.com to access all these educational materials, including this one. Okay. So heat equation. What is heat equation? Very briefly, we were talking in a previous lecture about thin metal, let's say metal rod, with certain source of heat applied to one end. And then we were talking about how this particular heat is distributed down the road as the time progresses. So basically we are talking about function, temperature of time and distance. So at point A, which is at distance x from the source of the heat, temperature is growing, obviously, as the time goes by. We probably should assume that at time equal to zero, there is some temperature and then we're heating this end and the temperature is spreading down the road. So this is the function which we were talking about, function of temperature and coordinate. So at any given time, any point has certain temperature. Now, what we came up with is so-called a heat equation, which looks like this. Dt, that's from t. So partial derivative by time of this function should be equal to some kind of a constant, positive constant, second partial derivative of the same function by coordinate x. So that's our partial differential equation which we have come up with using certain logical and physical and mathematical considerations. This is the end of the previous lecture. Now, this lecture I would like to devote to solving this particular differential equation. And I will just present to a method which can be used. Now, it doesn't mean that there are no other methods. There are. And it doesn't mean that this particular method is good for any partial differential equation. It's not. It's not. However, in this particular case, it's good. And again, my purpose is just to get acquainted with the concept of partial differential equation and how to solve it. Not going into any more specific details. And obviously, these are subject to higher education. Now, in this particular case, I'm going to suggest one particular way of addressing this problem. So here is how. Let's find a solution as a product of two functions. Now, what does it mean? It means I'm actually hypothesizing. Let's assume that we have a function which looks like this. Now, if I will put this function into this differential equation, is it possible that I will find a particular effort t and g of x, which will satisfy this particular equation? If I will be able to find these functions, well, it means I found a solution. Now, obviously, as in all the cases with differential equations, including partial, you will have some kind of a result which depends on certain constants, because integration always brings some kind of constants which are unknown. And the only way to determine what kind of constants are participating is to know certain initial conditions. And in this particular case, there are two kinds of conditions which we really should know before we finally get the solution. And these conditions are. The first one is, what is the distribution of temperature before we apply this source of heat? Well, let's consider, for instance, that the rod has universal, along the length, temperature of whatever number of degrees. Okay, so we can assume that at time zero, this is equal to, well, some kind of a constant, whatever known constant is. Okay, this is the temperature of the rod before I applied the heat. Now, the second condition, now this is called initial condition. In theory, it can actually be a function of X. That's a more complete representation. But in a practical case, most likely this is a constant function, because the whole rod is probably at one particular temperature before we start our experiment. But again, not necessarily. Another condition, it's called boundary condition. And the boundary condition is, now if we apply this heat, what does it actually mean? Well, for instance, there is some kind of a fire or flame or whatever, which we put at this particular end. It has certain temperature. And regardless of the time, for all different moments of time, this is obviously a fixed temperature. So what does it mean? It means that T of T comma zero, because the coordinate of this end is zero, right? The length from the beginning is zero. So at any given time, this can be some kind of a constant. So that means that at a particular, like a flame, we are putting at this particular end. And it means we are always, at all time, keep this temperature at certain constant level. Now, well, does it mean that that's the general boundary condition? Not necessarily. Actually, it can be dependent on the time as well. It can be some kind of a function. For instance, we put some kind of a hot object at the end. But then this hot object also loses the temperature. We don't maintain the temperature at constant level, as in the previous case, when I put the flame, flame has a certain fixed temperature. But if it's just an object, then obviously the temperature is changing. So it might be some kind of a function. Now, another kind of a boundary condition. And maybe it is, maybe it isn't. Maybe if we maintain certain temperature on the other end of the road, or at some point in between. So there are certain boundaries which we really have to know before we start our experiment, certain conditions which will allow us to solve this particular differential equation down to every constant which participates in this solution. All right? So let's put aside all these considerations about initial conditions and boundary conditions. But we will just attempt to find some kind of a solution to this differential equation in basically by looking at the solution in this form. So let's try and we will see. Now, if we will succeed, well, great. If we will not succeed to find such functions, FAT and GFX, which satisfy this differential equation, well, tough luck. It means we just went the wrong way and we probably should look for something else. But obviously I know that this will result in some kind of a solution, otherwise I wouldn't talk to you about this. And this is just one of the methods which can be applied to solve partial differential equations of this type. Okay. Now, using this, what is partial derivative by T? Well, if partial derivative by T and this is the product, so X is basically considered to be constant, so this is like a coefficient. So basically it's just derivative of F of T. So it's derivative of F of T times G of X. Now, what is partial derivative by X, the second partial derivative by X? Again, now F of T is considered to be a constant multiplier and we just have to differentiate twice this, which means it's equal to F of T times second derivative of X. Now, using this, we will just rewrite our equation and it looks like this. F of T times GX equals A square F of T G of X. So that's our differential equation which we would like to solve somehow. Well, we can always rewrite it as F here, G there. So I will have F of T divided by F of T equals A square G second derivative G of X. Now, we have a very strange situation. Let me tell you, I was surprised myself when I saw it. Look, on the left, I have a function of T. On the right, I have a function of X, completely two unrelated arguments. So these are functions and these are functions. T is changing, well, it's a time, right? So it's from zero to infinity. X is distance from the edge of the road, which is also like from zero to infinity. So we have two functions, arguments are completely independent and yet these two functions from two independent arguments must be equal to each other. I mean, at least I'm looking for functions which are supposed to be equal to each other. How is this possible? Well, there is obviously the possibility. What if this function is identically a constant? Is it possible? Yes, there is a function of T. This is a function of T, right? It's a, well, it's combination of function and its derivative and ratio between them, but it's still a function of T. And it might be equal identically to some kind of a constant, as well as this thing might actually be equal to exactly the same constant. So I don't know this constant, but if this and this functions are equal to the same constant, constants identically for all possible values of T and for all possible values of X, then I can say that these two functions of two completely different arguments are equal to each other. So if I will be able to find such a function f of T, such that this thing is identically a, and I will find, I will try to find, to find function g of X when this is identically equal to a, then basically that would satisfy my initial equation. So let's do it. What these are, it's just two ordinary differential equations. This differential equation is kind of easy, right? Because what this means is df of T by dt times f of T is equal to a, right? So I wrote first derivative in the form which is usual to solve differential equations from which dt goes here and df divided by f is actually differential of logarithm of f of T, right? Why? Because the derivative from logarithm is 1 over this function f of T, but then the chain rule gives me a multiplication of derivative of f of T. So that's what it is. So the differential differential of logarithm of f of T is differential of f divided by f of T. On the on the right I will have a times dt. Now I can integrate it, right? And what do I have? Well I will have that logarithm of f of T equals a times T, right? Plus some kind of a constant. Well in theory I should have actually put f absolute value, but it doesn't really matter much because I will explain it in a second. Now let me just simplify it. I will raise e to these powers and what will I have? If I will raise e to this power I will have f of T is equal to e to the power of cc is basically any constant so I will put the constant in front as a multiplier. It's actually e to the power of c, but since c can be any constant I will do this e to the power at. Now whether I will put absolute value or not it doesn't really matter since again c is just any constant. Just simplifies the matter. I just don't want to go into really very very small details because my purpose is just to demonstrate the the avenue you can go to not a little pass which really is very intricate etc. So this is just a general solution to the left part of this. Okay now how about the right part? The right part is something which we already dealt with because what this e is is let me just rewrite it g of x is equal to a so it's equal to a divided by a square g of x. All right if this is equal to a then a square then second derivative is equal to a times g and then we divide by a square. So we have this and I will even put it in a little bit more familiar way like this. We already solved this particular ordinary differential equation when we were talking about spring. Now to again to solve this we have to look for a solution in a specific way and our specific way is look for a solution in this form. It's a linear ordinary differential equation and that's why you can really use this type of a technique where lambda can be anything including complex number and if you remember we were using complex numbers to to have solution which is oscillation based on cosine or sine when we were talking about the spring and obviously when lambda is imaginary this thing is using the formula of Euler converts into the power of imaginary number into cosine plus i sine. Okay so this is the general form we are looking for a solution. Now second derivative of this would be of the first one is lambda e to the power lambda x and the second one is lambda square times e to the power of lambda x right. So whenever we will substitute it e to the power lambda x will cancel out right and what will remain the remain will be as a quadratic equation lambda square mean minus a equals to zero right from which we can see that lambda equals plus minus square root of a divided by a. Okay by the way what is this a this basically characterizes the rod itself there are certain constants related to physics of the process of heat distribution how fast for instance silver is a more conducive to the heat than steel for instance so this that's what this a actually is all right from which we see that there are two solutions and since i have two solutions then any linear combination of these two solutions is also a solution so general solution is this c1 times e to the power square root of a divided by a x plus c2 e to the power minus square root of a divided by a x. So this is the general solution we found it it depends on two constants and if we will substitute this by the way into this you obviously will have identity like zero equals to zero. So we have both functions so what's my solution to original equation c of time and x is equal to product c e to the power of a t times c1 e to the power square root of a over a x plus c2 e to the power square root of a a x minus where we have no idea what c c1 c2 or a is so all these must be somehow determined based on initial conditions which we are imposing on the road and we have plenty of conditions obviously as i was saying we have condition of maintaining the road ages at certain temperature or initial temperature of the every point on the road etc etc so at this point i would like actually to stop because what it means is first of all a might be negative in which case i will have imaginary number here and that means i will have sine or cosine as a result if a is positive which i hope it is actually then this is just a regular real function so they don't have to worry about this so i hope that a is a positive constant and this actually describes how the heat is is distributing now in in your experience let's put it this way in your experience what what can you say about physics of that thing so you have some kind of a rod and this is the source of heat so let's say a rod had some temperature initially uniformly like zero or whatever and heat is also applied always at the same temperature like a flame for instance we just put at the edge of the of the road then obviously you understand that at any point x as the time goes by temperature will increase it will probably increase to a certain degree which is basically to accept certain temperature which is no greater than initial temperature right so as the time goes by the temperature will probably from the initial will go something like this where this is the temperature i call it t zero for instance of the edge of the flame which we are applying so that's how it should be right that's from experience temperature from the initial point as the time goes by goes there for any point for any fixed point for any fixed x so if this is x and the x is fixed right this is as we see some kind of a exponent so it's it's it's growing whereas whether it's growing according to you know certain our understanding can it will asymptotically finish at certain point is a big big question but in any case this is just one of the solutions which probably can can be applied now is the temperature going you see this is in infinitely growing function right that's an exponential function does it mean that our exponential function will be our temperature would grow exponentially well this is a very interesting point you see we if we are dealing with their real life metal rod it will not grow higher than this temperature primarily because the heat would just be distributed to the air and stuff like this now if this is let's say in the vacuum and you have a permanent source of heat then the temperature will grow indefinitely because there is no place this particular heat should go I mean yes eventually it will melt the rod and our experiment will obviously stop on this but in theory if there is no way to distribute the heat outside of the rod then the temperature will grow exponentially so that does make sense but in case you have certain other elements of this experiment like you are losing heat at certain at certain speed then this is much more complicated job and the equation would be much more complicated in this case so this equation is in assumption that there is no way to heat to escape the rod and that's why we have this exponent exponent now as far as this is concerned again obviously you understand that the further we go the slower the lower temperature will be right so maybe this component c1 should be zero in this case and this should really signify how the temperature will decrease with increasing x because this is again exponential function and providing c2 is positive this function is going down right so it's no wonder that this component actually would specify how the temperature is decreasing as we are further and further from the source because the source of the heat is further and the heat will take longer to reach that particular point right so that's also kind of a natural from our understanding case well that's it again i wanted to give you just a taste of partial differential equation not a complete theory or anything like this and that would complete this part of the course so i suggest you to read the notes for this lecture on unizor.com and good luck thank you