Inhomogeneous heat conduction equation homogeneous boundary conditions. Fourier method for the heat equation. Instantaneous point source
Below, we will consider several problems for determining temperature fields for relatively simple geometric and physical conditions that allow analytical solutions that are simple in form and at the same time provide a useful illustration of the characteristic physical processes associated with heat transfer in a solid.
Consider a rod with a thermally insulated side surface (Fig. 38). In this case, heat transfer can take place along the rod. If the rod is aligned with the axis of the Cartesian coordinate system, then the stationary heat equation will have the formAt constant values of the thermal conductivity coefficient of the volumetric power of heat release, the last equation can be integrated twice
(75)
The constants of integration can be found from the boundary conditions. For example, if the temperature at the ends of the rod is , . Then from (75) we have
From here we find the constants of integration and . The solution under the indicated boundary conditions will take the form
It can be seen from the last formula that in the absence of heat sources . The temperature in the rod varies linearly from one limit value to another
Let us now consider another combination of boundary conditions. Let an external source create a heat flux at the left end of the rod. At the right end of the rod, we keep the previous condition, so we have
Expressing these conditions with the help of the general integral (75), we obtain a system with respect to the constants of integration
Having found unknown constants from the resulting system, we obtain a solution in the form
As in the previous example, in the absence of internal sources of heat release, the temperature distribution along the rod will be linear
In this case, the temperature at the left end of the rod, where the external heat source is located, will be equal to .
As the next example, let's find a stationary temperature distribution along the radius in a continuous long circular cylinder (Fig. 39). In this case, the use of a cylindrical coordinate system will significantly simplify the task. In the case of a cylinder with a large length-to-radius ratio and distribution constants
As an internal source of heat release, the temperature far from the ends of the cylinder can be considered independent of the axial coordinate of the cylindrical system . Then the stationary heat equation (71) takes the form
Double integration of the last equation (for a constant ) gives
The symmetry condition for the temperature distribution on the axis of the cylinder () gives
Where do we get
The last condition will be satisfied for . Let the temperature be set on the surface of the cylinder (). Then one can find the second constant of integration from the equation
From here we find and write down the solution in the final form
As a numerical example of the application of the result obtained, we consider the temperature distribution in the plasma of a cylindrical arc discharge with a radius of mm. The discharge channel boundary is formed as a region where ionization processes stop. We saw above that noticeable ionization of the gas during heating stops at K. Therefore, the reduced value can be taken as the boundary K. The volume density of the heat release power in the discharge plasma can be found from the Joule–Lenz law , where σ
is the electrical conductivity of the plasma, E- electric field strength in the discharge channel. Characteristic values for an arc discharge are 1/Ohm m, V/m. The thermal conductivity of the arc plasma is higher than in a neutral gas; at temperatures of the order of 10,000 K, its value can be taken equal to . Thus, the parameter 
. The temperature distribution along the radius is shown in fig. 39. In this case, the temperature on the discharge axis () will be 8000 K.
In the following example, we consider a thermal field with spherical symmetry. Such conditions arise, in particular, if a small heat source is located in a large array, for example, an inter-turn arc fault in the winding of a large electrical machine. In this case, by combining the center of the spherical coordinate system with the source of heat release, we can bring the stationary heat equation (64) to the form:
Integrating this equation twice, we find
Returning to our example, suppose that the arc fault takes place inside a spherical cavity of radius (Fig. 40). Let us take the resistance of the arc discharge equal to Ohm, the discharge current A. Then the power released in the cavity will be . Let us consider the solution outside the scope of the heat source.Then the integral of the heat equation becomes simpler
To calculate the integration constants, we firstly use the condition at points infinitely distant from the discharge site, where C is the ambient temperature. From the last expression we find . To determine the constant, we assume that the thermal energy released in the discharge is uniformly distributed over the surface of a spherical cavity of radius . Therefore, the heat flux at the boundary of the cavity will be
Since , then from the last two equations we have
and the final decision
In this case, the temperature at the boundary of the cavity (mm) at W/mK will be 
K (Fig. 40).
As the first example of this group, consider the thermal field in the cross section of a round wire with a cooling channel (Fig. 41, a). Wires with cooling channels are used in the windings of powerful electrical machines and coils to produce strong magnetic fields. These devices are characterized by a long flow of currents with an amplitude of hundreds and even thousands of amperes. For example, a liquid is pumped, such as water, or a gas (hydrogen, air), which ensures the selection of thermal energy from the inner surface of the channel and the cooling of the wire as a whole. In this case, we are dealing with forced convective cooling of the channel surface, for which we can use the boundary condition of the third kind justified above (67). If we combine the axis of the cylindrical coordinate system with the axis of the wire, then the temperature will depend only on the radial coordinate. The general integral of the stationary heat equation for this case was obtained by us earlier
The volumetric density of the heat release power is found from the Joule-Lenz law: , j- current density, σ - electrical conductivity,
where R- radius of the wire section, a- radius of the cooling channel. The wire is surrounded on the outside by layers of insulation, which, in comparison with the conductor, has a relatively low thermal conductivity. Therefore, in the first approximation, we accept the outer surface of the wire as heat-insulated, i.e., the heat flux on it
On the surface of the cooling channel, the heat flux is determined by the condition of the third kind
where is the heat transfer coefficient, is the temperature of the cooling flow. The minus sign on the right side is taken due to the fact that the normal to the inner surface of the channel is directed in the opposite direction to the axis.
Substituting the expression for the temperature (76) into the first of the written boundary conditions, we obtain
where . The second boundary condition gives
where do we find
However, from (76)
Comparing the last two expressions, we find
After substituting the found constants into the general solution (76) and transformations, we obtain
The temperature at the boundaries of the wire section from the obtained solution will be calculated by the formulas
Temperature distribution along the section radius for a wire with a cooling channel with parameters: A, W/mK, 
1/Ohm m, o C, mm, cm is shown in fig. 41, b.
From fig. 41, b it follows that within the wire cross section, the temperature change is relatively small compared to its average value, which is explained by the high thermal conductivity λ and relatively small cross-sectional dimensions of the wire.
A different situation arises in the temperature distribution along the wire, which consists of separate sections in contact with each other. The deterioration of the quality of the contacts between the connected conductors leads to an increase in heat generation at the junction of the two wires compared to the wire itself. Remote measurement of wire temperature using thermal imagers or pyrometers allows diagnosing the quality of contact connections.Let us calculate the temperature distribution along the wire in the presence of a defective contact. The previous example showed that even under the most severe conditions, the temperature change within the wire section is very small. Therefore, for our calculation, we can, in the first approximation, assume that the temperature distribution within the wire cross section is uniform. The distribution of heat release along the wire depends on the distribution of electrical resistance along the wire, which is uniform far from the contact and increases when approaching it. Let's combine the axis of the Cartesian coordinate system with the axis of the wire, and the origin of coordinates - with the center of the contact area (Fig. 42). As a model for the distribution of resistance along the wire, we take the following distribution of linear resistance
where , is a parameter characterizing the linear size of the contact area . The heat dissipation power per unit length of the wire is 
. Per unit volume, the heat release power is
where S- section of the wire. The wire is cooled by natural convection from its surface. The convective heat flux per unit length of the wire is
where α - heat transfer coefficient, - ambient air temperature, p- the perimeter of the wire section. Heat transfer to the environment per unit volume of the conductor will be
The stationary temperature distribution along the wire will obey the heat conduction equation
For further transformations of the resulting equation, we take the thermal conductivity coefficient constant along the wire, substitute the above expressions for and , and also as the desired function instead of T let's take :
we arrive at a linear inhomogeneous differential equation
We will seek the solution of the resulting equation in the form of the sum of the general solution of the homogeneous equation
and a particular solution in the form of the right side
.
Solving algebraic equations by Newton's method
A fairly popular method for solving equations is tangent method, or Newton's method. In this case, an equation of the form f(x) = 0 is solved as follows. First, the zero approximation is chosen (point x 0). At this point, a tangent to the graph is drawn y = f(x). The point of intersection of this tangent with the x-axis is the next approximation for the root (point x one). At this point, a tangent is again constructed, and so on. Point sequence x 0 , x 1 , x 2 ... should result in the true value of the root. The convergence condition is .
Since the equation of a straight line passing through a point x 0 , f(x 0) (and this is the tangent) is written as
and as the next approximation x 1 for the root of the original equation, the point of intersection of this line with the abscissa axis is taken, then it should be put at this point y = 0:
from which immediately follows the equation for finding the next approximation in terms of the previous one:
On Fig. Figure 3 shows the implementation of Newton's method using Excel. In cell B3, enter the initial approximation ( x 0 = -3), and then all intermediate values are calculated in the remaining cells of the column up to the calculation x one . To perform the second step, the value from cell B10 is entered into cell C3 and the calculation process is repeated in column C. Then, having selected cells C2:C10, you can extend it to columns D:F by dragging the marker in the lower right corner of the selected area. As a result, the value 0 is received in cell F6, i.e. the value in cell F3 is the root of the equation.
The same result can be obtained using cyclic calculations. Then after filling the first column and getting the first value x 1, enter the formula =H10 in cell H3. In this case, the computational process will be looped and in order for it to be executed, in the menu Service | Parameters tab Computing checkbox must be checked Iterations and indicate the limit number of steps of the iterative process and the relative error (the default value of 0.001 is clearly not enough in many cases), upon reaching which the computational process will stop.
As is known, such physical processes as heat transfer, mass transfer in the process of diffusion obey Fick's law
where l- coefficient of thermal conductivity (diffusion), and T is the temperature (concentration), and is the flow of the corresponding value. It is known from mathematics that the divergence of the flow is equal to the bulk density of the source Q this value, i.e.
or, for the two-dimensional case, when the temperature distribution in one plane is investigated, this equation can be written as:
The solution of this equation is analytically possible only for areas of a simple shape: a rectangle, a circle, a ring. In other situations, the exact solution of this equation is impossible, i.e. it is also impossible to determine the distribution of temperature (or concentration of a substance) in complex cases. Then one has to use approximate methods for solving such equations.
Approximate solution of equation (4) in the region of complex shape consists of several stages: 1) grid construction; 2) construction of a difference scheme; 3) solution of a system of algebraic equations. Let's consider successively each of the stages and their implementation using the Excel package.
Building a grid. Let the region have the shape shown in Fig. 4. With this form, the exact analytical solution of equation (4), for example, by the method of separation of variables, is impossible. Therefore, we will look for an approximate solution of this equation at individual points. Apply to the region a uniform grid consisting of squares with sides h. Now, instead of looking for a continuous solution of Eq. (4) defined at each point of the region, we will look for an approximate solution defined only at the nodal points of the grid applied to the region, i.e. at the corners of the squares.
Construction of a difference scheme. To construct a difference scheme, consider an arbitrary internal mesh node C (central) (Fig. 5). Four nodes are adjacent to it: B (upper), H (lower), L (left) and P (right). Recall that the distance between nodes in the grid is h. Then, using expression (2) for an approximate notation of the second derivatives in equation (4), we can approximately write:
whence it is easy to obtain an expression relating the temperature value at the central point with its values at neighboring points:
Expression (5) allows us, knowing the temperature values at neighboring points, to calculate its value at the central point. Such a scheme, in which derivatives are replaced by finite differences, and only the values at the nearest neighboring points are used to search for values at a grid point, is called the central-difference scheme, and the method itself is called the finite difference method.
It should be understood that we obtain an equation similar to (5) FOR EVERY point of the grid, which, thus, turn out to be connected with each other. That is, we have a system of algebraic equations in which the number of equations is equal to the number of grid nodes. Such a system of equations can be solved by various methods.
Solution of a system of algebraic equations. Iteration method. Let the temperature at the boundary nodes be given and equal to 20, and the heat source power equal to 100. The dimensions of our area are given and equal to 6 vertically and 8 horizontally, so the side of the grid square (step) h= 1. Then expression (5) for calculating the temperature at internal points takes the form
Let's assign a cell on an Excel sheet to each NODE. In the cells corresponding to the boundary points, enter the number 20 (in Fig. 6 they are highlighted in gray). In the remaining cells, we write the formula (6). For example, in cell F2 it would look like this: =(F1 + F3 + E2 + G2)/4 + 100*(1^2)/4. By writing this formula in cell F2, you can copy it and paste it into the remaining cells of the area corresponding to the internal nodes. In this case, Excel will report the impossibility of performing calculations due to the looping of the results:
Click "Cancel" and go to the window Service|Options|Calculations, where check the box in the "Iterations" section, specifying 0.00001 as the relative error, and 10000 as the maximum number of iterations:
Such values will provide us with a small COUNTING error and guarantee that the iterative process will reach the specified error.
However, these values DO NOT PROVIDE a small error of the method itself, since the latter depends on the error when the second derivatives are replaced by finite differences. Obviously, this error is the smaller, the smaller the grid step, i.e. the size of the square on which our difference scheme is built. This means that the exactly CALCULATED temperature value at the grid nodes, shown in Fig. 6 may actually turn out to be completely untrue. There is only one way to check the found solution: find it on a finer grid and compare it with the previous one. If these solutions differ little, then we can assume that the found temperature distribution corresponds to reality.
Let's cut the step in half. Instead of 1, it will become equal to ½. The number of nodes will change accordingly. Vertically, instead of 7 nodes (there were 6 steps, i.e. 7 nodes), it will become 13 (12 squares, i.e. 13 nodes), and horizontally, instead of 9, it will become 17. At the same time, one should not forget that the step size has halved and now in formula (6) instead of 1 2 you need to substitute (1/2) 2 on the right side. As a control point at which we will compare the found solutions, we take the point with the maximum temperature, marked in Fig. 6 in yellow. The result of calculations is shown in fig. nine:
It can be seen that a decrease in the step led to a significant change in the temperature value at the control point: by 4%. To improve the accuracy of the found solution, the grid step should be further reduced. For h= ¼ we get 199.9 at the control point, and for h = 1/8 the corresponding value is 200.6. You can plot the dependence of the found value on the step size:
From the figure, we can conclude that a further decrease in the step will not lead to a significant change in the temperature at the control point, and the accuracy of the solution found can be considered satisfactory.
Using the capabilities of the Excel package, you can build a temperature surface that visually represents its distribution in the study area.
with initial conditions
and boundary conditions
We will seek the solution of this problem in the form of a Fourier series with respect to the system of eigenfunctions (94)
those. in the form of decomposition
while considering t parameter.
Let the functions f(x, t) is continuous and has a piecewise continuous derivative of the 1st order with respect to X and for all t>0 conditions are met
Let us now assume that the functions f(x,
t)
and 
can be expanded into a Fourier series in terms of sines
, (117)
(118)
, (119)
. (120)
We substitute (116) into equation (113) and, taking into account (117), we obtain
.
This equality holds when
, (121)
or if 
, then this equation (121) can be written as
. (122)
Using the initial condition (114), taking into account (116), (117), and (119), we obtain
. (123)
Thus, to find the desired function 
we arrive at the Cauchy problem (122), (123) for an ordinary inhomogeneous differential equation of the first order. Using the Euler formula, we can write the general solution of equation (122)
,
and taking into account (123), the solution of the Cauchy problem
.
Therefore, when we substitute the value of this function into expression (116), we will eventually obtain a solution to the original problem
(124)
where functions f(x,
t)
and 
are defined by formulas (118) and (120).
Example 14 Find a solution to an inhomogeneous equation of parabolic type
under the initial condition
(14.2)
and boundary conditions
. (14.3)
▲ Let us first choose such a function to satisfy the boundary conditions (14.3). Let, for example, = xt 2. Then
Therefore, the function defined as
satisfies the equation
(14.5)
homogeneous boundary conditions
and zero initial conditions
. (14.7)
Applying the Fourier method to solve the homogeneous equation
under conditions (14.6), (14.7), we set
.
We arrive at the following Sturm-Liouville problem:
,
.
Solving this problem, we find the eigenvalues
and their corresponding eigenfunctions
. (14.8)
The solution of problem (14.5)-(14.7) is sought in the form of a series
, (14.9)
(14.10)
Substituting 
from (14.9) to (14.5) we obtain
. (14.11)
To find a function T n (t) expand the function (1- X) in a Fourier series in the system of functions (14.8) on the interval (0,1):
. (14.12)
,
and from (14.11) and (14.12) we obtain the equation
, (14.13)
which is an ordinary inhomogeneous linear differential equation of the first order. We find its general solution using the Euler formula
and taking into account condition (14.10), we find the solution of the Cauchy problem
. (14.14)
From (14.4), (14.9) and (14.14) we find the solution of the original problem (14.1)-(14.3)
Tasks for independent work
Solve initial boundary value problems
3.4. Cauchy problem for the heat equation
First of all, let's consider Cauchy problem for homogeneous heat equation.
satisfying
Let's start by changing the variables x
and t on 
and introduce the function 
. Then the functions 
will satisfy the equations
where 
- Green's function, defined by the formula
, (127)
and having properties
; (130)
. (131)
Multiplying the first equation by G* , and the second to and and then adding the results obtained, we obtain the equality
. (132)
After integration by parts of equality (132) over 
ranging from -∞ to +∞ and 
ranging from 0 to t, we get
If we assume that the function 
and its derivative 
limited at 
, then, due to properties (131), the integral on the right-hand side of (133) is equal to zero. Therefore, one can write
Replacing in this equality with 
, a 
on 
, we get the ratio
.
Hence, using formula (127), we finally obtain
. (135)
Formula (135) is called Poisson's formula and determines the solution of the Cauchy problem (125), (126) for a homogeneous heat equation with a non-homogeneous initial condition.
The solution the Cauchy problem for the inhomogeneous heat equation
satisfying inhomogeneous initial condition
is the sum of the solutions:
where is the solution of the Cauchy problem for the homogeneous heat equation . , satisfying an inhomogeneous initial condition, and is a solution satisfying a homogeneous initial condition. Thus, the solution of the Cauchy problem (136), (137) is determined by the formula
Example 15 Find a solution to the equation
(15.1)
for the following rod temperature distribution:
▲ The rod is infinite, so the solution can be written using the formula (135)
.
Because 
in the interval 
equal to constant temperature 
, and outside this interval the temperature is equal to zero, then the solution takes the form
. (15.3)
Assuming in (15.3) 
, we get
.
Because the
is the integral of probabilities, then the final solution of the original problem (13.1), (13.2) can be expressed by the formula
.▲
Heat equation for non-stationary case
non-stationary, if the temperature of the body depends both on the position of the point and on time.
Denote by and = and(M, t) temperature at the point M homogeneous body bounded by a surface S, at the moment of time t. It is known that the amount of heat dQ absorbed over time dt, is expressed by the equality
where dS− surface element, k− coefficient of internal thermal conductivity, − derivative of the function and in the direction of the outer normal to the surface S. Since it propagates in the direction of decreasing temperature, then dQ> 0 if > 0, and dQ < 0, если < 0.
From equality (1) it follows
Now let's find Q another way. Select an element dV volume V, bounded by the surface S. Quantity of heat dQ received by the element dV during dt, proportional to the temperature increase in this element and the mass of the element itself, i.e.
where is the density of the substance, the coefficient of proportionality, called the heat capacity of the substance.
From equality (2) it follows
Thus,
where . Considering that = , , we get
Replacing the right side of the equality with the help of the Ostrogradsky-Green formula, we obtain
for any volume V. From this we obtain the differential equation
which is called the heat equation for the non-stationary case.
If the body is a rod directed along the axis Oh, then the heat equation has the form
Consider the Cauchy problem for the following cases.
1. The case of an unbounded rod. Find a solution to equation (3) ( t> 0, ) satisfying the initial condition . Using the Fourier method, we obtain a solution in the form
− Poisson integral.
2. Stud Case, limited on one side. The solution of equation (3), which satisfies the initial condition and the boundary condition , is expressed by the formula
3. Stud Case, bounded on both sides. Cauchy's problem is to X= 0 and X = l find a solution to equation (3) that satisfies the initial condition and two boundary conditions, for example, or .
In this case, a particular solution is sought in the form of a series
for boundary conditions ,
and in the form of a series
for boundary conditions .
Example. Find a solution to the equation
satisfying the initial conditions
and boundary conditions.
□ We will seek the solution of the Cauchy problem in the form
Thus,
Heat equation for the stationary case
The distribution of heat in a body is called stationary if the body temperature and depends on the position of the point M(X, at, z), but does not depend on time t, i.e.
and = and(M) = and(X, at, z).
In this case, 0 and the heat equation for the stationary case turns into Laplace equation
which is often written as .
To temperature and in the body was determined unambiguously from this equation, you need to know the temperature on the surface S body. Thus, for equation (1), the boundary value problem is formulated as follows.
Find a feature and, satisfying equation (1) inside the volume V and receiving at each point M surfaces S setpoints
This task is called Dirichlet problem or first boundary value problem for equation (1).
If the temperature on the surface of the body is unknown, and the heat flux at each point of the surface is known, which is proportional to , then on the surface S instead of the boundary condition (2) we will have the condition
The problem of finding a solution to equation (1) that satisfies the boundary condition (3) is called Neumann problem or second boundary value problem.
For plane figures, the Laplace equation is written as
The same form has the Laplace equation for space, if and does not depend on the coordinate z, i.e. and(M) remains constant as the point moves M along a straight line parallel to the axis Oz.
By replacing , equation (4) can be converted to polar coordinates
The concept of a harmonic function is related to the Laplace equation. The function is called harmonic in area D, if in this region it is continuous together with its derivatives up to the second order inclusive and satisfies the Laplace equation.
Example. Find the stationary temperature distribution in a thin rod with a thermally insulated lateral surface, if at the ends of the rod , .
□ We have a one-dimensional case. Need to find a function and, satisfying the equation and boundary conditions , . The general equation of this equation has the form . Taking into account the boundary conditions, we obtain
Thus, the temperature distribution in a thin rod with a thermally insulated side surface is linear. ■
Dirichlet problem for a circle
Let a circle of radius be given R centered at the pole O polar coordinate system. We need to find a function that is harmonic in the circle and satisfies the condition on its circle, where is a given function that is continuous on the circle. The desired function must satisfy the Laplace equation in the circle
Using the Fourier method, one can get
− Poisson integral.
Example. Find the stationary temperature distribution on a homogeneous thin round plate of radius R, the upper half is kept at , and the lower half at .
□ If , then , and if , then . The temperature distribution is expressed by the integral
Let the location point be in the upper semicircle, i.e. ; then changes from to , and this length interval does not contain points . Therefore, we introduce the substitution , whence , . Then we get
Since the right side is negative, then and for satisfies the inequalities . For this case, we get the solution
If the point is located in the lower semicircle, i.e. , then the interval of change contains the point , but does not contain 0, and we can make a substitution , whence , . Then for these values we have
After similar transformations, we find
Since the right side is now positive, then . ■
Finite difference method for solving the heat equation
Let it be required to find a solution to the equation
satisfying:
initial condition
and boundary conditions
So, it is required to find a solution to equation (1) that satisfies conditions (2), (3), (4), i.e. it is required to find a solution in a rectangle bounded by straight lines , , , , if the values of the desired function are given on its three sides , , .
We construct a rectangular grid formed by lines
− step along the axis Oh;
− step along the axis From.
Let us introduce the notation:
From the concept of finite differences, we can write
likewise
Taking into account formulas (6), (7) and the introduced notation, we write equation (1) in the form
From here we get the calculation formula
From (8) it follows that if three values of k are known k th layer of the grid: , , , then you can determine the value in ( k+ 1)th layer.
The initial condition (2) allows finding all values on the line ; boundary conditions (3), (4) allow us to find the values on the lines and . According to formula (8), we find the values at all internal points of the next layer, i.e. for k= 1. The values of the desired function at the extreme points are known from the boundary conditions (3), (4). Passing from one grid layer to another, we determine the values of the desired solution at all grid nodes. ;
Let's solve the first mixed problem for the heat equation: find a solution u(x, t) of the equation satisfying the initial condition and boundary conditions Let's start with the simplest problem: find the solution u(x, t) of the homogeneous equation satisfying the initial condition and zero (homogeneous) boundary conditions Method Fourier for the heat equation We will look for non-trivial solutions of Eq. (4) that satisfy the boundary conditions (6), in the form *) of the form (7) satisfying the boundary conditions (6), it is necessary to find non-trivial solutions of equation (10) that satisfy the boundary conditions for which there are non-trivial solutions to the problem This problem was considered in the previous chapter. It was shown there that nontrivial solutions exist only for . For A = An the general solution of Eq. (9) has the form satisfy Eq. (4) and boundary conditions (6). We form a formal series Requiring that the function u(x) t) defined by formula (12) satisfies the initial condition, we obtain Series (13) is the expansion of a given function in a Fourier series in terms of sines in the interval (O, I). The coefficients an of the expansion are determined by the well-known formulas Fourier method for the heat equation Let us assume that Then series (13) with coefficients determined by formulas (14) converges absolutely and uniformly to the function. Since for then the series for also converges absolutely and uniformly. Therefore, the function u(x, t) - the sum of the series (12) - is continuous in the region and satisfies the initial and boundary conditions. It remains to show that the function u(x, t) satisfies Eq. (4) in the domain 0. To do this, it suffices to show that the series obtained from (12) by term-by-term differentiation with respect to t once and by term-by-term differentiation with respect to x twice are also absolutely and converge uniformly at. But this follows from the fact that for any t > 0 if n is large enough. The uniqueness of the solution of problem (4)-(6) and the continuous dependence of the solution on the initial function have already been established earlier. Thus, for t > 0 problem (4)-(6) is well-posed; on the contrary, for negative t this problem is incorrect. Comment. In contrast to the House equation, the equation is non-metrical about time t: if we replace t with -t, then we get an equation of a different kind that describes irreversible processes: m was this and for the time t before the moment in question. This distinction between prediction and prehistory is typical of the parabolic equation and does not occur, for example, in the wave equation; in the case of the latter, it is as easy to look into the past as into the future. Example. Find the temperature distribution in a homogeneous rod of length w if the initial temperature of the rod and zero temperature is maintained at the ends of the rod. 4 The problem is reduced to solving the equation under the initial condition and boundary conditions Using the Fourier method, we are looking for non-trivial solutions of equation (15) that satisfy boundary conditions (17), in the form Substituting u(x,t) in form (18) into equation (15) and separating the variables, we get from where the Eigenvalues of the problem. eigenfunctions Xn(x) = mn nx. For A = An, the general solution of Eq. (19) has the form Tn(t) = ape a n\, so that the solution of problem (15)–(17) is sought in the form of a series. Therefore, the solution of the original problem will be the function 2. Let us now consider the following problem: find a solution rx(x, t) of the inhomogeneous equation _ satisfying the initial condition and homogeneous boundary conditions. Suppose that the function / is continuous, has a continuous derivative, and for all t > 0 the condition is met. The solution of problem (1)-(3) will be sought in the form )-(7) in the form of a series in terms of eigenfunctions (of the boundary value problem. Substituting t) in the form in equation (5), we obtain functions /(x, t) in a Fourier series, we get! Using the initial condition for v(x, t), Fourier method for the heat equation, we find that Solutions of equations (15) under initial conditions (16) have the form: Substituting the found expressions for Tn(t) into series (11), we obtain the solution Function will be a solution to the original problem (1)-(3). 3. Let us consider the problem: to find a solution to the equation in the domain under the initial condition and inhomogeneous boundary conditions. The Fourier method is not applicable directly due to the inhomogeneity of conditions (20). We introduce a new unknown function v(x, t) by setting where Then the solution of problem (18)–(20) reduces to the solution of problem (1)–(3) considered in Section 2 for the function v(x, J). Exercises 1. An infinite homogeneous rod is given. Show that if the initial temperature is then at any moment the temperature of the rod is 2. The ends of the rod of length x are maintained at a temperature equal to zero. The initial temperature is determined by the formula Determine the temperature of the rod for any time t > 0. 3. The ends of a rod of length I are maintained at a temperature equal to zero. The initial temperature of the rod is determined by the formula Determine the temperature of the rod for any time t > 0. 4. The ends of the rod of length I are maintained at a temperature equal to zero. Initial temperature distribution Determine the temperature of the rod for any time t > 0. Answers
