The series is a series of differential equations 3Blue1Brown video notes, the original video can be seen: https://www.bilibili.com/video/av50290975 or https://www.youtube.com/watch?v=p_di4Zn4wz4&list=PLZHQObOWTQDNPOjrT6KVlfJuKtYTftqH6
Due to the limited level of the author, the text will inevitably some shortcomings and wrong, honesty please criticism.
1 Introduction
In 3B1B differential equations Series Notes (b) we introduced by thermal conduction equation understanding of partial differential equations, this chapter will continue to heat conduction equation is based on solving partial differential equations introduced, solving partial differential equations, partial differential equations satisfied Solutions are many, but not all solutions are able to meet our requirements, so we have only partial differential equations is not enough, also need to analyze the boundary conditions, the final solution finally determined by the initial conditions.
2 Partial Differential Equations
First, the specific sine function is very simple equations of heat conduction, because the second derivative of a sine function with itself is proportional, by a different linear combination of the sine function, more complex solutions can be obtained. Fourier series tells us that all functions and can be expressed as a sine curve. All these conditions laid a superiority sine function solution. In fact, the sine function in many places will be very easy to deal with, solving differential equations is just one example.
We start with the solution of partial differential equations by a sinusoidal function.
We assume that the temperature of a metal rod in line with the sine function \ (SiN (X) \) : \
[T (X, 0) = \ SiN (X) \]
By \ (T \) to (X \) \ taking partial derivative and second partial derivatives can be obtained:
\ [\ FRAC {\ partial T} {\ partial X} (X, 0) = \ COS (X) \]
\[ \frac{\partial^{2} T}{\partial x^{2}}(x, 0)=-\sin (x) \]
The one dimensional heat conduction equation:
\ [\ FRAC {\ partial T} {\ partial T} (X, T) = \ Alpha \ CDOT \ FRAC {\ partial ^ {2} T} {\ partial X ^ {2}} (x, t) \]
can be derived to obtain:
\ [\ FRAC {\ partial T} {\ partial T} (X, 0) = \ Alpha \ CDOT \ FRAC {\ partial ^ {2} T} {\ partial X ^ {2}} (x,
0) = - \ alpha \ cdot T (x, 0) \] we find an interesting properties: in the initial state, the rate of temperature change of each point to this point are proportional to the temperature, and the proportion coefficient equal everywhere. After a small time period, will reduce the amplitude of the curve, the curve will continue to reduce the amplitude of the smaller of the next period. After calculation, we found that this property are established at any time. Represents a sine function of the temperature profile is a big advantage, in conjunction with special scales down sinusoidal thereof, i.e., at each time \ (T \) , the amplitude reduction curves are like \ (sin (x) \ ) multiplied by a constant.
Here we have the right side of the heat conduction equation equals roughly describe it, that is to say the partial derivatives with respect to spatial relationships that we've got the general idea, now we need Another to describe the left side of the equal sign, that partial derivatives of the time. Represents a sine function according to the nature of the temperature profile, we know that the temperature is proportional to the time rate of change of temperature with itself. When we see a certain amount of the rate of change is proportional to itself, we will first think of the exponential function:
\ [E KX ^ {} \]
then how should we use exponential function to update the temperature expression (ie (1) ) to reflect the relationship between time of it? For heat conduction equation, the expression on the right represents the sine wave, i.e., the spatial relationship of the deflector, will equal the \ (- \ alpha \) multiplied by a sinusoidal function of temperature per se, i.e. (5). To ensure equal equation, the expression for the temperature to be allowed time to reduce the partial derivative \ (- \ alpha \) times, so that we will only need to \ (sin (x) \) multiplying a base \ (e ^ {- \ alpha t} \) can, so the temperature can be changed to the expression:
\ [T (X, T) = \ SiN (X) E ^ {- \ Alpha T} \]
Next we verify:
It can be seen that the equation \ (T (x, t) = \ sin (x) e ^ {- \ alpha t} \) time \ (T \) seeking a position and a partial derivative \ (X \) the second order partial derivative values are identical demand, which is consistent with our heat conduction equation.
but! If it is really so simple, you do not need boundary conditions what happened!
3 boundary conditions
In fact, even if the temperature on the pole happens to be this perfect sine curve, as the temperature does not change as the index, just as in the analysis of which, two border points for the pole, we assume that the temperature remains constant. But we think carefully about it, if the heat transfer pole and the outside world does not exist, the temperature boundary points will change in an instant will be the beginning of temperature change to that point and their closely adjacent equal! That system begins at the moment of the border points of the first derivative of the closed beta will always be equal to zero.
In 3B1B differential equations Series notes (b) , we refer to \ (x \) to understand intuitively the second derivative is the value of the United States will tend to be a bit of the average of two adjacent, but at the border point, there are no side adjacent points. Then the value of the boundary points tend to be a bit of neighboring values measured rate of change is proportional to that difference. This also led after the system starts, the slope of the border will always be equal to zero.
Apparently, just the sine function obviously can not satisfy this condition. In other words, find a function in line with the heat equation itself is not enough, it must meet the \ (t> 0 \) , the border must be horizontal, mathematical expression meaning that
\ [\ frac {\ partial T } { \ partial x} (0, t
) = \ frac {\ partial T} {\ partial x} (L, t) = 0 \] where \ (L \) is the length of the rod, and \ (t> 0 \) .
This is an example of boundary conditions, when we actually want to pick up partial differential equations, boundary conditions often will appear, and partial differential equations itself and needs our attention.
After adding boundary conditions, we need to further modify the expression of temperature, can be close to the true solution. Here we just need to make some adjustments, so you can function at the boundary level. Here is not difficult to think of, we can replace the sine function by cosine function. A cosine function are able to meet any of the \ (x = 0 \) the first derivative at 0, but may not be able to meet in \ (x = L \) at the first derivative is also 0, so we need to adjust the cosine function cycle, by \ (X \) before multiplied by a factor \ (\ Omega \) , and \ (\ Omega \) larger sinusoidal vibration means the faster. However, according to the chain rule, there will be a second order function of the new coefficients front guide \ (\ Omega ^ 2 \) : \
[\ COS ({\ Omega} \ CDOT X) \ stackrel {\ FRAC {\ partial} {\ partial x}} {\ rightarrow} - \ omega \ sin (\ omega \ cdot x) \ stackrel {\ frac {\ partial} {\ partial x}} {\ rightarrow} - \ omega ^ {2} \ cos (\ omega \ cdot x) \]
in order to ensure approximately equal to the equation, we need to make a derivative of the exponential term is also multiplied by \ (\ Omega ^ 2 \) , i.e.:
\ [T (X, T) = \ COS (\ omega \ cdot x) e ^ {
- \ alpha \ omega ^ 2 t} \] just mentioned, \ (\ Omega \)Sinusoidal vibration means the larger the faster, that is to say \ (\ Omega \) larger, a sine and cosine wave each derivative zero curvature greater, a function of temperature for a larger curvature, it will cool down faster, and is the square of times faster. The new exponential function perfectly illustrated this point, intuition tells us that the right direction.
Next, let us constraints \ (\ Omega \) values, since the lever length \ (L \) , then the minimum frequency of the cosine function satisfies the boundary conditions is \ (\ PI / L \) , by \ (n-(\ pi / L) \) instead of \ (\ Omega \) , we can temperature function written as:
\ [T (X, T) = \ COS (n-(\ PI / L) X) E ^ {- \ Alpha ( n \ pi / L) ^ {
2} t} \] in this way, we get satisfied partial differential equation and boundary conditions a function of temperature.
Note that the order number and a number corresponding to the video, (c) on the first mentioned here, the remaining part will continue to expand in solving (IV) in the.