Linear algebra and differential equations (Math 54): Lecture 24 Vivek Shende April 23, 2019
Hello and welcome to class!
Hello and welcome to class! For some time now
Hello and welcome to class! For some time now We have been studying linear ordinary differential equations.
Hello and welcome to class! For some time now We have been studying linear ordinary differential equations. This time
Hello and welcome to class! For some time now We have been studying linear ordinary differential equations. This time We turn to linear partial differential equations.
Hello and welcome to class! For some time now We have been studying linear ordinary differential equations. This time We turn to linear partial differential equations. These are equations in which the function for which we are solving may depend on many variables, and we may take derivatives with respect to all of them.
Hello and welcome to class! For some time now We have been studying linear ordinary differential equations. This time We turn to linear partial differential equations. These are equations in which the function for which we are solving may depend on many variables, and we may take derivatives with respect to all of them. Rather than develop a systematic theory, we will mostly focus on a few examples of physical importance.
The heat equation
The heat equation Let us try and understand how heat flows in a substance.
The heat equation Let us try and understand how heat flows in a substance. We will first discuss the one dimensional case.
The heat equation Let us try and understand how heat flows in a substance. We will first discuss the one dimensional case. This means either you should imagine the substance is one dimensional,
The heat equation Let us try and understand how heat flows in a substance. We will first discuss the one dimensional case. This means either you should imagine the substance is one dimensional, or close to it, e.g. a very thin wire,
The heat equation Let us try and understand how heat flows in a substance. We will first discuss the one dimensional case. This means either you should imagine the substance is one dimensional, or close to it, e.g. a very thin wire, or just that the temperature is constant in two of the dimensions, and only varies along the third.
The heat equation Let us try and understand how heat flows in a substance. We will first discuss the one dimensional case. This means either you should imagine the substance is one dimensional, or close to it, e.g. a very thin wire, or just that the temperature is constant in two of the dimensions, and only varies along the third. We will moreover imagine that the material in question is uniform.
Temperature facts Imagine that we add energy (heat) to the system.
Temperature facts Imagine that we add energy (heat) to the system. The temperature will change proportionally.
Temperature facts Imagine that we add energy (heat) to the system. The temperature will change proportionally. That is, per unit time: change in temperature ∼ change in heat amount of stuff
Temperature facts Imagine that we add energy (heat) to the system. The temperature will change proportionally. That is, per unit time: change in temperature ∼ change in heat amount of stuff Also, as it turns out, having a temperature difference causes heat to flow, proportionally to the difference
Temperature facts Imagine that we add energy (heat) to the system. The temperature will change proportionally. That is, per unit time: change in temperature ∼ change in heat amount of stuff Also, as it turns out, having a temperature difference causes heat to flow, proportionally to the difference per unit time ∼ change in temperature heat flow distance to flow
The heat equation Let’s rewrite things in terms of symbols.
The heat equation Let’s rewrite things in terms of symbols.
The heat equation Let’s rewrite things in terms of symbols. We’ll write u ( x , t ) for the function giving the temperature at location x and time t , and H ( x , t ) for the heat flowing.
The heat equation Let’s rewrite things in terms of symbols. We’ll write u ( x , t ) for the function giving the temperature at location x and time t , and H ( x , t ) for the heat flowing. Then the first equation change in temperature change in heat ∼ per unit time (amount of stuff)(per unit time) says δ u ( x , t ) ∼ δ H δ t δ x δ t
The heat equation The second equation per unit time ∼ change in temperature heat flow distance to flow says δ H δ t ∼ δ u δ x
The heat equation The second equation per unit time ∼ change in temperature heat flow distance to flow says δ H δ t ∼ δ u δ x Combining this with the previous equation δ u δ H δ x δ t , δ t ∼
The heat equation The second equation per unit time ∼ change in temperature heat flow distance to flow says δ H δ t ∼ δ u δ x Combining this with the previous equation δ u δ H δ x δ t , and δ t ∼ replacing difference with derivative,
The heat equation The second equation per unit time ∼ change in temperature heat flow distance to flow says δ H δ t ∼ δ u δ x Combining this with the previous equation δ u δ H δ x δ t , and δ t ∼ replacing difference with derivative, we get the heat equation ∂ t ∼ ∂ 2 u ∂ u ∂ x 2
The heat equation We restore now the proportionality constant. It will depend on various properties of the substance; the heat capacity, the density, the thermal conductivity; but is a positive constant. ∂ t = β ∂ 2 u ∂ u ∂ x 2
The heat equation We restore now the proportionality constant. It will depend on various properties of the substance; the heat capacity, the density, the thermal conductivity; but is a positive constant. ∂ t = β ∂ 2 u ∂ u ∂ x 2 The analogue of the “initial value problem” in this setting is the following.
The heat equation We restore now the proportionality constant. It will depend on various properties of the substance; the heat capacity, the density, the thermal conductivity; but is a positive constant. ∂ t = β ∂ 2 u ∂ u ∂ x 2 The analogue of the “initial value problem” in this setting is the following. One should specify the temperature distribution in the wire at some initial time t 0 .
The heat equation We restore now the proportionality constant. It will depend on various properties of the substance; the heat capacity, the density, the thermal conductivity; but is a positive constant. ∂ t = β ∂ 2 u ∂ u ∂ x 2 The analogue of the “initial value problem” in this setting is the following. One should specify the temperature distribution in the wire at some initial time t 0 . This is the data of a function in one variable, u ( x , t 0 ).
The heat equation We restore now the proportionality constant. It will depend on various properties of the substance; the heat capacity, the density, the thermal conductivity; but is a positive constant. ∂ t = β ∂ 2 u ∂ u ∂ x 2 The analogue of the “initial value problem” in this setting is the following. One should specify the temperature distribution in the wire at some initial time t 0 . This is the data of a function in one variable, u ( x , t 0 ). One should also say how the ends of the wire will behave during the whole evolution.
Separation of variables
Separation of variables Recall that we understand linear vector ODE, i.e., equations like d dt u = A u
Separation of variables Recall that we understand linear vector ODE, i.e., equations like d dt u = A u Given a function of two variables like u ( x , t ), one can specify one of the variables to get a function in the other. I.e., u ( x , 0) and u ( x , 1) are just functions in x .
Separation of variables Recall that we understand linear vector ODE, i.e., equations like d dt u = A u Given a function of two variables like u ( x , t ), one can specify one of the variables to get a function in the other. I.e., u ( x , 0) and u ( x , 1) are just functions in x . So, one way to think of a function of two variables like u ( t , x )
Separation of variables Recall that we understand linear vector ODE, i.e., equations like d dt u = A u Given a function of two variables like u ( x , t ), one can specify one of the variables to get a function in the other. I.e., u ( x , 0) and u ( x , 1) are just functions in x . So, one way to think of a function of two variables like u ( t , x ) is as a function of t valued in the vector space of functions in x .
Separation of variables Now our equation dt u ( x , t ) = β d 2 d dx 2 u ( x , t ) looks a lot like d dt u ( t ) = A u ( t )
Separation of variables Now our equation dt u ( x , t ) = β d 2 d dx 2 u ( x , t ) looks a lot like d dt u ( t ) = A u ( t ) where now the vector space V in which u ( t ) takes values is replaced by the vector space of functions in x ,
Separation of variables Now our equation dt u ( x , t ) = β d 2 d dx 2 u ( x , t ) looks a lot like d dt u ( t ) = A u ( t ) where now the vector space V in which u ( t ) takes values is replaced by the vector space of functions in x , and the linear transformation A on V is replaced by the linear d 2 transformation dx 2 on the space of functions of x .
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