WEBVTT mathematics/differential-equations/murray
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Hi and welcome back to the differential equations lecture here on educator.com.
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My name is Will Murray , and today we are studying partial differential equations.
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Meet some of the most common partial differential equations , today focus on the heat equation we are to look at boundary value problems in some of the associated conditions are often given as part of a package with differential with partial differential equations.
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To learn how to interpret each 1 of those conditions in terms of physical setting so, let us get started.
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So remember, a partial differential equation is an equation that relates a function of 2 variables which were to call U of X and T X and T are the variables and U is the function.
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So, partial differential equation relates U with 1 or more of its partial derivatives.
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So, we did a whole lecture reviewing partial derivatives so, I hope U are comfortable with partial derivatives by now, if you are not comfortable with partial derivatives maybe go back and watch the previous lecture here E differential equations lecture series.
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Got a long review of partial derivatives, if that lecture is not enough then you might want to go back and watch the videos from the multivariable calculus where they really do partial derivatives in more detail.
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So, the most common partial differential equations have the second partial derivative of U with respect to X, the U of X-X and then I gets related to either the partial derivative with respect to T or the second partial derivative with respect to T .
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So, we will meet a couple of those are common partial differential equations in just a moment let me tell you about at boundary value problem .
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A boundary value problem is a partial differential equation which comes together in a package with initial conditions and boundary conditions.
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Often, the boundary conditions or are 0 so, we show let me show you what I mean by that.
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The initial condition is this first condition here the idea there is that the plug-in time = 0 that is why it is called the initial condition is because se it takes place at time 0.
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So, we plugged in T = 0 and so, we just get a function of X so, this would be given to you all these context of the problem all of these equations would be given to you.
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The boundary conditions tell you what happens when X = 0, when X = some Extreme value which were usually going to call L so, we plugged in X = 0 or X = L which is a constant.
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And the boundary conditions tells you what happens at different times at the edges of your interval at 0 on one side and it fell on the other.
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So, these collectively are boundary conditions and will see how those get interpreted physically but the important thing to know, right now, is that L is a constant and all of these of this initial condition and these boundary conditions these would all be given to you any package along with the partial differential equation.
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The goal is to find a function U of XT satisfies all of those.
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We will see how to how to solve that over the course of the next few lectures in the meantime I just want to study these and this initial condition these boundary conditions little more carefully and see how we interpret them physically.
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So, let us introduce the most common partial differential equations.
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For 3 very common 1s are known as the heat equation and the wave equation and Laplace’s equation each 1 takes a quite a long time to really study and solve.
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So, you kind of an studying the same equations over and over again once you learn each 1 then you really have a good grip of partial differential equations.
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So, the heat equation says U of T remember that is the partial derivative of U with respect to T = α ² x U of XX.
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What that α ² is the constant and I will say what that measures in just a moment U XX is the second partial derivative of U with respect to X and what this equation describes is a function U of XT which tells you the temperature of a solid rod of rod could be of different of different materials it could be a metal rod it could be a wooden rod we often think of it as being a metal rod.
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And it tells you the temperature of a rod at position X at time T and X here is the position along the rod E left hand of the rod is being X = 0 the right-hander the rod is X = L E rod is L units long and so, were trying to solve this discover what happens for all positions X between 0 and how and for all time T.
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And the ideas usually that you know, the initial distribution of heating the rod and then gradually the heat diffuses throughout the rod in a cool down and you are trying to find a description of what temperature each place along the rod is that each time.
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So, that is the heat equation very common partial differential equation that is the 1 were actually going to study and solved in the next few lectures here.
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The wave equation is another very common 1 and it looks very similar U TT = C ² x U XX and again C is a constant.
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I forgot to mention back were talking about the heat equation that α ² is a constant and it is a constant of thermal diffusivity which means it is a constant that depends on the material in the rod.
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If the material in the rod conducts heat very easily than you have a higher value for α ².
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If the material in the rod conducts heat poorly then U have a lower value for α ².
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So, this let me interpret the wave equation for you the idea of the wave equation is you have a string that is attached at 2 ends so, again we have X going from 0 to L and this string what you do is you pluck it and so, it vibrates back and forth .
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So, the string vibrates back and forth as time goes by and so, what you have is the string has a particular position particular amount of displacement at any given time.
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So, this displacement depends on where you are on the string and on what time it is and then U of XT describes the displacement of a string from 0 at time T.
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Now, that C ² that constant is a factor of how elastic the string is of course if the string were so, completely rigid object if it were just a metal rod then it would not vibrated all.
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If the string were something very very elastic like a rubber band for example it is going to vibrate a lot and so, that C ² is a measure of how elastic the material in your string is.
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The last equation that is very common when you study partial differential equations is Laplace's equation which says that U XX + UTT = 0 and this 1 comes up a lot in complex analysis.
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I am not going to interpret that 1 physically for you because se it is not 1 that were going to be studying later we are to spend most of our time studying the equation later.
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I just want to show you Laplace's equation because these really are sort of the big 3 partial differential equations that people study the heat equation the wave equation and Laplace's equation.
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So, let us see how the initial conditions and boundary conditions affect the physical situation here.
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So, let us jump in some examples the first example this is give us a physical interpretation for each element of the boundary value problem associated with the heat equation than what we have here is a partial differential equation and then several boundary and initial conditions.
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So, let me show you how to interpret each of these each of these equations here I told her that α² is a constant that depends on how easily the material rod conducts heat so, it depends on the thermal diffusivity of the rod diffusivity of the material in the rod.
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Just to give you an idea if the material is very conducted then you get a higher value there for example Silver has α ² = about 1.7 whereas if it is lower if it is a material with lower conductivity than you have a lower value of α iron for example does not conduct heat as readily as silver does.
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So, iron has α ² = 0.12 so, you get different values of α depending on depending on how conducted the material in Ur rod is.
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Now, let us look at these other conditions the initial conditions in boundary conditions let me draw rod here and you want to think of X going from 0 left hand side to L on the right-hand side now, U of X-0 that means we plugged in 0 for T time 0 = FX .
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So, what that really tells you is the distribution of heat in rod when we begin experiment so, that is something that an experimenter data a physicist or scientist would have to go and measure physically before starting the experiment.
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So, that is the initial temperature of the rod of U of 0T that means we plugged in X = 0 which means were looking at the left end of the rod here we were being told is a U 00T = 0 for all time.
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So, that means the left end of the rod is being kept at temperature 0 for all time so, you want to think of that as the left end is being To temperature 0 so, it is being artificially forced to stay temperature 0 perhaps by being packed in ice.
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So, no matter how much heat might travel to that left rod say from the middle of the rod were forcing the left end of the rod to stay at temperature 0.
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U of LT that means we plugged in X = L which corresponds to the right end of the rod if that is also, forced to stay at 0 that would mean that the right end of the rod is also, packed in ice.
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So, all of these conditions would have to be given to you at the beginning as part of a package which is known as a boundary value problem given all of these conditions then you could start solving at solving the partial differential equation trying to match the initial conditions the boundary value conditions and we have not learned how to do that.
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That is going to be the content of the next several lectures here on educator.com so, this point were just kind of learning how to interpret each 1 of these conditions what they need physically and learn how to solve the next time.
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So, let me recap there what we said was this UT = α ² UX X the partial differential equation and the only thing that ever changes there is the α ² and even that never changes very much because se that is always a constant it just depends on the material of that your rod is made out of the U of X-0 that means U plugged in T = 0 so, that means you are looking at the initial temperature of the rod .
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Again this would all be given to YOU so, FX would be a function that the experimenter has determined by going through in measuring the initial temperature of the rod and then U of 0T and U of LT that means U plugged in X = 0 and X = out which means you are looking at the left-hand end of the rod.
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E right-hand end of the rod and if for keeping them = to 0 that would suggest that leaves the artificially forced both ends to stay at 0 perhaps by submerging both ends of the rod and ice bath and so, that is why how we know, the temperature 0 for all time .
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So, let us look at a different kind of the boundary value problem still heat equation in example 2 were still looking at a heat equation but we have been given slightly different conditions here so, let us see how to interpret your physical interpretation for each element of this boundary value problem .
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So, we have U of T = 1 4th U of X X that is the same heat equation is before Except we been given a specific value of α ² here and that is a constant which measures how susceptible the material E rod is to conducting heat.
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Constant of thermal diffusivity and I switched these around the little that the U of X-0 means we plugged in T = 0 so, that is the initial distribution of heat in rod or the initial temperature in rod of the rod distribution of heat in rod .
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So, this is something that an Experimenter would have to go through and measure the temperature at each place along the rod and tell you the mathematician what the temperature is at each place at each position X and at each time well at time 0 .
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Then, here is the new element for this problem is set of U sub0 is that a 0T we have U of X-0T so, how is that different well what that means is that if U of X of 0T means we plugged in X = 0 so, looking at the left end of the rod and if U of X = 0 that means the rate of change in X direction is 0.
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Which means there is no change in X direction so, this is telling us that there is no change in X direction so, there is no heat flow this is at the left end of the rod because se X = 0 no heat flow across the left end of the rod.
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Which means we somehow prevented any heat from flowing in or out of the left end of the rod which means we must a packet in some kind of insulating material.
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So, the left end is packed in insulation now, I want to emphasize the distinction there before we had U of 0T was = to 0 which with the temperature was forced to be = to 0 now, we had U of X of 0T = 0 that means the temperature flow the rate of change E temperature is forced to be 0.
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So, this is really affect the U of X being = to 0 there is no change in temperature across the left end of the rod before we packed the left end in ice there could be a lot of heat sort of flowing out into the ice bath here we are saying that there is no heat flowing in or out so, we must be packing in some kind of insulating material to prevent any change in temperature across the left end.
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Now, we plugged in X = 1 for the second condition these are both boundary conditions because se they tell us what is happening at the edges of the rod boundary conditions and so, since we plugged in X = 1 that must mean L = 1 must be dealing with a rod of length 1 and since L = 0 that means there is no heat flow across the right ends of the rod .
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So, that right end must also, be packed in insulation so, what were doing here is we are actually solving the problem or not doing any math at this point were just learning how to interpret each of these elements of a boundary value problem that would all be given to us as part of actually solving a heat equation .
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So, the beginning here this was the in this is always the same except that that value of α might change will always be constant and it is a constant reflecting the kind of material you have in your rod.
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Material that conducts heat more easily would have a higher value of α ² material that is less conductive of heat would have a lower value of α ² .
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Now, the difference from this problem to the previous 1 from example 1 is E previous 1 that was adjusting U in here we have a U of X what that represents is that at the left end of the rod X = 0 and the right E rod X = 1 we got no change in temperature across the boundary.
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Which means that there is no heat flowing in or out across the boundary which must mean that we have artificially preventing heat from flowing in or out which means we must have packed it in some kind of insulating material.
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That is how we ended up concluding from these boundary conditions of the left end and the right end must be packed in insulation .
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Finally we have this initial condition here which says that U of X of X, 0 = some function of X that is telling you the initial distribution of heat throughout the rod so, that would be something that the Experimenter has to go and measure before presenting all of these conditions together as a package called a boundary value problem. 2100 So, example 3 were also, going to give a physical interpretation for each element of the boundary value problem associated with the wave equation so, let us see how to interpret each 1 of these elements of the wave equation.
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U TT = C ² U X X that is just the partial differential equation for the wave equation and the only thing that ever changes there is that C ² is always a constant but it might be different constant depending on what kind of material you have in your string.
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So, remember the wave equation measures the displacement of a vibrating string at time T and it position X so, X but varies between 0 and L and T goes from 0 to infinity because se we want to watch this string sort of oscillate infinitely.
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Now, that C ² is the constant of elasticity it tells you how elastic the material of your string is in other words if if you are dealing with for example when the rod instead of a string that is not going to be elastic at all that is not going to vibrate up and down at all.
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If you are dealing with something more flexible like a rubber band that something is going to be very elastic and so, would have a much higher constant of elasticity. 2248 Now, U of X-0 that means we plugged in T = 0 so, that means were looking at an initial condition here in initial condition and so, that would measure the initial displacement of the string .
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Maybe if you taken the string and you stretched it out beautifully plucked it and you are about to let it go that F of X would tell you the position of the string at any given point at time = 0 .
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That U of X-0 as U of 0T means we plugged in X = 0 which means were looking at the left end of the string and if that= 0 that means it is not displaced at all for all time which means you pin down the left hand and that the string.
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So, left end you pin down the left end of the string of string is pinned down and then the same kind of thing here we plugged in X = L into U of LT and we been told that U LT = 0 so, that means the right hand and of the string is also, pinned down for all time.
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So, what all these equations mean the first 1 is just the PDE which tells you the general behaviour of a vibrating string and the C ² is a constant of elasticity which depends on the material of the string .
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Is it actually string is it guitar wire is its rubber band is it something completely rigid like oh a wooden bar so, you get different constants of elasticity limit with the context of any 1 problem that number will always be constant so, it will stay fixed it C ².
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Now, the other conditions here are essentially keeping track of everything U would need to know, about in order to do is determine and predict the behaviour of that string U of X-0 tells U the position of the string when you initially pluck it or when you measure it at time 0.
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Such a stretch it up and then let it go when U stretch up measure the initial position of that string and that would B F of X that so, that is the initial condition U 0T and U of LT means U plugged in X = 0 and X = L and what it is telling you since their 0 is that your keeping the left hand and E right hand then pinned down for all time .
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So, let us go ahead and look at some possible solutions to differential equations so, were to look at the solution to a possible solutions to the wave equation U TT = 9 U X X .
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So, what were going to do here is look at each 1 of several different types of functions and I may calculate U TT for each 1 of them may calculate U X X , for each 1 and were to see if they work out this post workout being told to check that each 1 works out.
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So, we will see for each 1 of these functions the first 1 is U of XT = cosine X sin of 3T so, let us figure out U of TT and then we will figure out U of X X for each 1 of these functions.
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Now, cosine X x sin of 3T we figure out U of TT that means that this cosine of X want to think of that is being constant and so, we will just have cosine of X coming down now, sin of 3T were going to take 2 derivatives of that so, the first derivative of sin is cosine a 3T and then we multiply on a 3 by the chain rule.
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That is the first relative and E second derivative of the negative sin of 3T and we multiply out another 3 so, we end up getting -9 there and so, U of TT would be -9 cosine X x sin of 3T .
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Now, U of XX works just the other way around now, we think of sin of 3T is a constant so, sin of 3T is just going to come down the constant but cosine X me to take the second derivative of that while the first revenue cosine X is negative sin X the second derivative of cosine X is negative cosine X.
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So, U of X X is negative cosine X x sin of 3T and if you compare these to each other what we see is the U of TT is exactly 9 x U of XX and so, it satisfies the differential equation so, we just checked that our first function there satisfies the partial differ differential equation.
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So, it is going try that with the second function here U X at U so, that is T = cosine X U of cosine a 3T so, U of TT is cosine X because se we think of that is just being a constant for taking the derivative with respect to T cosine a 3T only take 2 derivatives of that .
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The first relative is negative sin of 3T multiplied by 3 by the general and the second derivative through the sin is cosine a 3T but we saw that negative and now, we get another 3 popping out so, -9 cosine 3T so, -9 cosine of X cosine of 3T down the other side we were looking at U of X X so, that cosine a 3T would just come down as a constant since you think of T is being a constant .
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Now, through a cosine on the second cosine this negative sin derivative of that is negative cosine her second derivative is negative cosine X x cosine a 3T again if you compare those 2 to each other on the left we have negative 9 x what we have on the right.
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So, U of TT is indeed = to 9 x U of XX let us try that again with the third function here U XT = E T x E X over 3 U of TT means we think of X is a constant we take the second derivative of ET which is just E T and then E X over 3 is just comes down as a constant .
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Each of the X over 3 U of X X means the E T comes down the constant or take the second derivative of E X over 3 will the first relative is E X over 3×1/3 .
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Then the second derivative is E X over 3 multiplied by another thirds we get 1/9 so, 1/9 U of T x E X over 3 and again we can see that the left-hand side = 9 x the right-hand side so, if we compare those to each other they definitely do satisfy the partial differential equation E T x E X over 3 is 9×1/9 E T x E X over 3.
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Try again with U of XT U of XT = E 3T x E X so, U of TT means the X can be a constant E 3T second derivative of that, first derivative is 3 E 3T second related 9 E 3T.
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So, we get 9 E 3T x E X over on U of X X that means E 3T is a constant and the second derivative of E X is just E X and so, I wrote a 3 here it should have be 9.
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So, what we see comparing the left-hand side the right-hand side the left-hand side really is 9 x as big as the right-hand side so, looking back up our differential equation U of TT = 9 U of XX.
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Finally in our last 1 we have U of XT is = 3T + X we calculate that 1 a little more carefully U of T = the derivative sin x cosine of 3T + X x 3 the derivative of the inside and then U of TT is a derivative of cosine is negative sin of 3T + X .
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So, 3 x that first 3 would be 9 now, let us look at U of X U of X is cosine of 3T + X and U of X X is just negative sin of 3T + X.
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You do not have to worry about the chain rule here because the derivative of the inside stuff with respect to X is just 1 .
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My U of TT is -9 sin of 3T + X is copying that over from left there my U of X X is negative sin of 3T + X now, if you compare these 2 to each other you see again that the left-hand side is 9 x as big as the right-hand side and so, again we have a solution to the partial differential equation.
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So, in each case we have U of TT = 9 x U of X X so, all of these functions work as solutions to the wave equation U TT = 9 U of XX .
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So, let me remind you how we did that in each 1 of these functions we worked out U of T and U of TT and U of X and U of X X and then I just kind of listed what we got on each of those for each of those second partial derivatives over on the right here.
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You are pointing to remember when you are taking this these partial derivatives is that whatever variable you are differentiating with respect to you want to hold the other variable constant.
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So, for example when I took the derivative with respect to T I hold the X part constant so, if even if I have cosine a 3X or something like that it just comes down as 1 big constant so, what I did was take the second derivative with respect to T secondary with respect to X I did that for each 1 of these functions holding the other variable constant each time .
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I checked on each 1 that what I got on the left for U of TT really was 9 x when I got on the right for U of XX and it worked out in every1 and so, I can say that each 1 of these functions really is a solution to that wave equation .
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Example 5 where the check for each 1 of the following functions as a solution to the heat equation U T is = 4 U X X were going to also, determine which 1 also, satisfies the boundary condition U of 0T = 0 .
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So, let me first just work out UT and U X X for each 1 of these functions so, let us look at U of T and U of X X for each 1 of these functions so, for the first 1 U T would be E X of a constant x E 4T x the derivative of 4T so, that would be E X E 4T U of X X just means you are taking the second partial derivative with respect to X which the second partial derivative of E X is just E X and then either 4T comes down as a constant .
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And if you compare both sides here it does look like the left-hand side is 4 x as big as the right-hand side so, we do have a solution to our heat equation therefore partial differential equation.
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In our second function here got E X over 2 on the left and E T on the right so, U of T is just the derivative of E T because se E X over to as a constant so, E T derivative is just itself some nothing really happened there.
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U of X X well take 1 derivative here we get 1 half E X over 2 if you take a second derivative 1 half x 1 half so, 1 4th E X over 2 then E T just comes down as a constant so, we get 1 4th E X over 2x E T .
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If you compare those to each other you do see that the left-hand side is 4 x as big as the right-hand side so, again we have a solution for partial differential equation for part C if we do U of T well looks like the sin X comes down as a big constant and the derivative of E negative 4T is -4 E negative 4T .
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Then if we take the second derivative with respect to X the E the 4 negative 4T just can be a constant but sin X the first derivative is cosine X the second derivative is negative sin X derivative of cosine is negative sin X and so, we get negative sin X E negative 4T .
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Now, we notice if we look at the bulk that both sides there the left-hand side is exactly 4 x the right-hand side and so, again our U of T is 4 x U of X X so, we do have a solution to the partial differential equation.
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Let us get look at U of XT = to cosine of X over 2 E negative T so, our U of T just has cosine of X over to coming down as a constant E negative T the derivative of that is negative E negative TU of X X the E negative T is going to come down as a constant the derivative of cosine X over 2 is negative sin of X over 2.
00:38:49.000 --> 00:39:25.000
Then we have to multiply by 1 half by the chain rule and derivative of sin is cosine of X over 2 multiply by 1 half again by the generals we get -1/4 get -1/4 cosine of X over 2 x E negative T and again if you compare the left-hand side of the right-hand side each the left-hand side is exactly 4 x the right-hand side .
00:39:25.000 --> 00:39:51.000
So, again we have a solution to the partial differential equation there so, let me recap what happened here each 1 of these functions will we did was we tried it out in partial differential equation when we tried out was just by calculating the T derivative holding X constant and then the second X derivative holding T constant .
00:39:51.000 --> 00:40:11.000
So, here all the derivatives that we calculated on each side each time and then for each 1 we checked that the T derivative the U of T was exactly 4 x the U of X X and it did not work out every time and so, each 1 of those really was a solution to the partial differential equation that we were given .
00:40:11.000 --> 00:40:33.000
Now, our X also, to determine which 1 also, satisfies the boundary condition U 0T = 0 so, we do that on the next slide so, next slide is where we’re going to try to satisfy the boundary condition U 0T = 0 .
00:40:33.000 --> 00:41:24.000
Let me remind you of the 4 functions that we had, here are first 1 was U of XT = E X E 4T our second 1 was U of XT = E the X over 2 x E T third 1 is U of XT = sin X x E negative 4T and our last 1 was U of XT is cosine of X over 2 x E negative T now, for each 1 of those were in a plug-in X = 0.
00:41:24.000 --> 00:41:51.000
So, let us try plugging in a single 0 for each 1 of those who would get E 0 x E 4T which is just in 0s once was E 4T here again we get in 0 x E T which is just E T here we get sin of 0 x E negative 4T.
00:41:51.000 --> 00:42:16.000
Now, sin 0 0 so, this just cancels a way to 0 and finally here we get cosine of 0 x E negative T now, cosine of 0 is 1 so, this is even negative T and what were looking for is 1 where U 0T comes out 30 so, clearly the first 1 fails the second 1 fails the third 1 works and the 4th 1 fails .
00:42:16.000 --> 00:42:34.000
So, that is our solution right there that is the 1 that satisfies both the partial differential equation and the boundary condition X 0T = 0 so, just to remind you what we did for each function there.
00:42:34.000 --> 00:42:59.000
We are plenty of X = 0 so, each 1 of these were finding U of 0T plug-in X = 0 and simplifying it down U 0T U is 0T U are starting to look like Y for me I clean up those U there .
00:42:59.000 --> 00:43:16.000
For 1 of those the function simplify down to be = to 0 which is what we are looking for and so, were to say that that 1 is the actual solution to both the differential equation and to the boundary condition at U 0T = 0 .
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So, that wraps of this lecture on the partial differential equations and in particular the heat equation we remind you that we did not actually learn how to solve the heat equation in this lecture we will learn that in the next lecture .
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We relearned about Fourier series so, and actually that I think it is 2 lectures for now, word to learn about Fourier series next lectures can be on separation of variables.
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So, let me just remind what we did in this lecture will learn how interpret all the different conditions that come package with a partial differential equation.
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We learn what each 1 of them means physically been actually learn how to solve a partial differential equation but I hope now, that you are little more comfortable with interpreting each 1 of those initial conditions and boundary conditions that you would receive as part of a boundary value problem along with a partial differential equation.
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So, hopefully you can interpret each 1 of those you know, what each 1 of them means and in the next few lectures were going to learn about separation of variables in order to learn about Fourier series.
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We will actually learn how to solve all that together and solve the heat equation.
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So, in the meantime you have been watching educator.com , these are the differential equations lectures and my name is Will Murray. We will look forward to talking to you in the next lectures. In the meantime, thanks for watching, bye bye.