heat equation separation of variables

Our building-block solutions are, \[u_n(x,t)=X_n(x)T_n(t)= \sin \left( \frac{n \pi}{L}x \right) e^{\frac{-n^2 \pi^2}{L^2}kt}. The plot of $u(x,t)$ confirms this intuition. Now suppose the ends of the wire are insulated. Let us recall that a partial differential equation or PDE is an equation containing the partial derivatives with respect to several independent variables. If $t>0$, then these coefficients go to zero faster than any $\frac{1}{n^P}$ for any power $p$. First order partial differential equations, method of characteristics. It is relatively easy to see that the maximum temperature will always be at $x=0.5$, in the middle of the wire. Each of our examples will illustrate behavior that is typical for the whole class. Figure 4.14: Initial distribution of temperature in the wire. The method of separation of variables is to try to find solutions that are sums or products of functions of one variable. We are looking for nontrivial solutions $X$ of the eigenvalue problem $X''+ \lambda X=0,$ $X'(0)=0,$ $X'(L)=0,$. “x”) appear on one side of the equation, while all terms containing the other variable (e.g. Figure 4.17: Plot of the temperature of the insulated wire at position $x$ at time $t$. Verify the principle of superposition for the heat equation. ��w��HY��2��)��@�VQ# �M��v,ȷ��p�)/��S�fa��|�8��R�Θh7#ОќH��2� AX_ ��A\��WD��߁ :�n��c�m��}��;�rYe��Nؑ�C��z. Solution of the heat equation: separation of variables To illustrate the method we consider the heat equation (2.48) with the boundary conditions (2.49) for all time and the initial condition, at , is (2.50) where is a given function of . This gives us our third separation constant, which we call n2. That the desired solution we are looking for is of this form is too much to hope for. Thus even if the function $f(x)$ has jumps and corners, then for a fixed $t>0$, the solution $u(x,t)$ as a function of $x$ is as smooth as we want it to be. We are solving the following PDE problem, \[u_t=0.003u_{xx}, \\ u_x(0,t)= u_x(1,t)=0, \\ u(x,0)= 50x(1-x) ~~~~ {\rm{for~}} 0 0 proportional to the temperature gradient. Hence, the solution to the PDE problem, plotted in Figure 4.17, is given by the series, \[ u(x,t)=\frac{25}{3}+\sum^{\infty}_{\underset{n~ {\rm{even}} }{n=2}} \left( \frac{-200}{\pi^2 n^2} \right) \cos(n \pi x) e^{-n^2 \pi^2 0.003t}.\]. That is, the change in heat at a specific point is proportional to the second derivative of the heat along the wire. Finally, let us answer the question about the maximum temperature. That is. We solve, \[ 6.25=\frac{400}{\pi^3}e^{-\pi^2 0.003t}.\], \[ t=\frac{\ln{\frac{6.25 \pi^3}{400}}}{-\pi^2 0.003} \approx 24.5.\]. Heat equation. The equation governing this setup is the so-called one-dimensional heat equation: \[\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2}, \]. It is one of the oldest and most common methods for solving PDEs. \], If, on the other hand, the ends are also insulated we get the conditions, \[ u_x(0,t)=0 ~~~~~ {\rm{and}} ~~~~~ u_x(L,t)=0. We begin by looking for functions of the form v(x, t) = X(x)T(t) that are not identically zero and satisfy Up: Heat equation. Separation of Variables. Assume that the sides of the rod are insulated so that heat energy neither enters nor leaves the rod through its sides. 1. Browse other questions tagged partial-differential-equations heat-equation or ask your own question. 2 2. Finally, plugging in $t=0$, we notice that $T_n(0)=1$ and so, \[ u(x,0)= \sum^{\infty}_{n=1}b_n u_n (x,0)= \sum^{\infty}_{n=1}b_n \sin(\frac{n \pi}{L}x)=f(x).\]. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Next, we will study thewave equation, which is an example of a hyperbolic PDE. But the left hand side does not depend on $x$ and the right hand side does not depend on $t$. The first term in the series is already a very good approximation of the function. The heat equation “smoothes” out the function $f(x)$ as $t$ grows. In particular, if $u_1$ and $u_2$ are solutions that satisfy $u(0,t)=0$ and $u_(L,t)=0$, and $c_1,c_2$ are constants, then $ u= c_1u_1+c_2u_2$ is still a solution that satisfies $u(0,t)=0$ and$u_(L,t)=0$. Hence, let us pick solutions, \[X_n(x)= \cos(\frac{n \pi}{L}x)~~~~ {\rm{and}}~~~~ X_0(x)=1.\], \[T'_n(t)+ \frac{n^2 \pi^2}{L^2}kT_n(t)=0.\], \[T_n(t)= e^{\frac{-n^2 \pi^2}{L^2}kt}.\], For $n=0$, we have $T'_0(t)=0$ and hence $T_0(t)=1$. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. This makes sense; if at a fixed $t$ the graph of the heat distribution has a maximum (the graph is concave down), then heat flows away from the maximum. Boundary and Initial Conditions u(0,t)=u(L,t)=0. Suppose that we have an insulated wire of length $1$, such that the ends of the wire are embedded in ice (temperature 0). Then suppose that initial heat distribution is $u(x,0)=50x(1-x)$. See Figure 4.13. Using the Principle of Superposition we’ll find a solution to the problem and then apply the final boundary condition to determine the value of the constant(s) that are left in the problem. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Let us plot the function $0.5,t$, the temperature at the midpoint of the wire at time $t$, in Figure 4.16. Solving heat equation on a circle. As a first example, we will assume that the perfectly insulated rod is of finite length Land has its ends maintained at zero temperature. So the maximum temperature drops to half at about $t=24.5$. Inhomogeneous heat equation Neumann boundary conditions with f(x,t)=cos(2x). That is, we find the Fourier series of the even periodic extension of $f(x)$. In illustrating its use with the Heat Equation it will become evident how PDEs … Let us write $f(x)$ as the sine series, \[ f(x)= \sum_{n=1}^{\infty} b_n \sin \left( \frac{n \pi}{L}x \right).\], That is, we find the Fourier series of the odd periodic extension of $f(x)$. We have previously found that the only eigenvalues are $\lambda_n=\frac{n^2 \pi^2}{L^2}$, for integers $ n \geq 0$, where eigenfunctions are $\cos(\frac{n \pi}{L})X$ (we include the constant eigenfunction). Hence, let us pick the solutions, \[ X_n(x)= \sin \left( \frac{n \pi}{L}x \right).\], The corresponding $T_n$ must satisfy the equation, \[ T'_n(t) + \frac{n^2 \pi^2}{L^2}kT_n(t)=0.\], By the method of integrating factor, the solution of this problem is, It will be useful to note that $T_n(0)=1$. Featured on Meta Feature Preview: Table Support First, we will study the heat equation, which is an example of a parabolic PDE. �/pb�@�z�×fCrV��' _ �ר+8��|z[%U�_�3j��O*w�2E�Δv�&�d@kq��J�� &��K�J�R_^!��RQ�y+J們��$o�x$? We always have two conditions along the $x$ axis as there are two derivatives in the $x$ direction. The Problem Let u(x,t) denote the temperature at position x and time t in a long, thin rod of length ℓ that runs from x = 0 to x = ℓ. The goal is to rewrite the differential equation so that all terms containing one variable (e.g. Separation of Variables is a special method to solve some Differential Equations. Let us first study the heat equation. We will write $u_t$ instead of $ \frac{\partial u}{\partial t}$, and we will write $u_{xx}$ instead of $\frac{\partial^2 u}{\partial x^2} $. Hence, each side must be a constant. The equation … Suppose that we have a wire (or a thin metal rod) of length $L$ that is insulated except at the endpoints. While solving any partial differentiation equation using a variable separable method which is of order 1 or 2, we use the formula of Fourier series to find the coefficients at last. In this section we go through the complete separation of variables process, including solving the two ordinary differential equations the process generates. Second order partial differential equations: wave equation, heat equation, Laplace's equation, separation of variables. For a fixed $t$, the solution is a Fourier series with coefficients $b_n e^{\frac{-n^2 \pi^2}{L^2}kt}$. speciﬁc heat of the material and ‰ its density (mass per unit volume). If $u_1$ and $u_2$ are solutions and $c_1,c_2$ are constants, then $ u= c_1u_1+c_2u_2$ is also a solution. Let us call this constant $- \lambda$ (the minus sign is for convenience later). The method of separation of variables relies upon the assumption that a function of the form, u(x,t) = φ(x)G(t) (1) (1) u (x, t) = φ (x) G (t) will be a solution to a linear homogeneous partial differential equation in x x and t t. D��\j��s�D�:�4&P7��l� 9�-�$��M�#;;1ϛA�r�(?��87EW��X�{�߽߮ �5pP�ޒ�THU��7��ǉ��ԕ�A,�I��۫��×��>�avR�.�>��ZS$��h��/��0o��|�Vl�ґ��ՙ�&F+k��OVh7��$VjH�(�x�6D�$(��T��k� �Y�+�2��U�i��@�@n�'l��+t��>)dF´��#1�� The heat equation is linear as $u$ and its derivatives do not appear to any powers or in any functions. The figure also plots the approximation by the first term. We will apply separation of variables to each problem and find a product solution that will satisfy the differential equation and the three homogeneous boundary conditions. Suppose that we have a wire (or a thin metal rod)... 4.6.2 Separation of variables. (12) Because each side only depends on one independent variable, both sides of this equation must be constant. Have questions or comments? What is perfectly reasonable to ask, however, is to find enough “building-block” solutions of the form $ u(x,t)=X(x)T(t)$ using this procedure so that the desired solution to the PDE is somehow constructed from these building blocks by the use of superposition. In other words, the Fourier series has infinitely many derivatives everywhere. Finally, we will study the Laplace equation, which is an example of an elliptic PDE. Normalizing as for the 1D case, x κ x˜ = , t˜ = t, l l2 Eq. Let us recall that a partial differential equation or PDE is an equation containing the partial derivatives with respect to several independent variables. and consequently the heat equation (2,3,1) implies that 2.3.2 Separation ofVariables where ¢(x) is only a function of x and G(t) only a function of t, Equation (2,3.4) must satisfy the linear homogeneous partial differential equation (2.3,1) and bound ary conditions (2,3,2), but … 2 2D and 3D Wave equation The 1D wave equation can be generalized to a 2D or 3D wave equation, in scaled coordinates, u 2= Our method of solving this problem is called separation of variables (not to be confused with method of separation of variables used in Section 2.2 for solving ordinary differential equations). That is, $f(x)= \sum^{\infty}_{n=1}b_n \sin(n \pi x)$, where, \[ b_n= 2 \int^1_0 50x(1-x) \sin(n \pi x)dx = \frac{200}{\pi^3 n^3}-\frac{200(-1)^n}{\pi^3 n^3}= \left\{ \begin{array}{cc} 0 & {\rm{if~}} n {\rm{~even,}} \\ \frac{400}{\pi^3 n^3} & {\rm{if~}} n {\rm{~odd.}} 3. We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. It seems to be very random and I can't find a way to do the next problem once looking at old problems? Method of characteristics for second order hyperbolic partial differential equations. $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$, 4.6: PDEs, separation of variables, and the heat equation, [ "article:topic", "targettag:lower", "authortag:lebl", "authorname:lebl", "showtoc:no" ], $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$. The temperature, , is assumed seperable in and and we write 3. A PDE is said to be linear if the dependent variable and its derivatives appear at most to the first power and in no functions. Our building-block solutions will be, \[u_n(x,t)=X_n(x)T_n(t)= \cos \left( \frac{n \pi}{L} x \right) e^{\frac{-n^2 \pi^2}{L^2}kt},\], We note that $u_n(x,0) =\cos \left( \frac{n \pi}{L} x \right)$. Separation of Variables is a standard method of solving differential equations. These side conditions are called homogeneous (that is, $u$ or a derivative of $u$ is set to zero). Note: 2 lectures, §9.5 in , §10.5 in . Separation of Variables The first technique to solve the PDE above is by Separation of Variables. That is, when is the temperature at the midpoint $12.5/2=6.25$. Similarly for the side conditions $u_x(0,t)=0$ and $u_x(L,t)=0$. If one can re-arrange an ordinary diﬀerential equation into the follow-ing standard form: dy dx = f(x)g(y), then the solution may be found by the technique of SEPARATION OF VARIABLES: Z dy g(y) = Z f(x)dx. “y”) appear on the opposite side. K c x u c t u. 6.1. First note that it is a solution to the heat equation by superposition. It satisfies $u(0,t)=0$ and $u(L,t)=0$ , because $x=0$ or $x=L$ makes all the sines vanish. Furthermore, suppose that we know the initial temperature distribution at time $t=0$. Solution of heat equation. That If you are interested in behavior for large enough $t$, only the first one or two terms may be necessary. Question: (a) Solve The Heat Equation By The Method Of Separation Of Variables ди A²u A 0 At @x2 V U(0,1)= U(4,1)= 0 And U(x,0) = X-4, 0 This question hasn't been answered yet Ask an expert 2. Missed the LibreFest? At this point we are ready to now resume our work on solving the three main equations: the heat equation, Laplace’s equation and the wave equa- tion using the method of separation of variables. Solve heat equation using separation of variables. In this case, we are solving the equation, \[ u_t=ku_{xx}~~~~ {\rm{with}}~~~u_x(0,t)=0,~~~u_x(L,t)=0,~~~{\rm{and}}~~~u(x,0)=f(x).\], Yet again we try a solution of the form $u(x,t)=X(x)T(t)$. Chapter 6. Fourier method - separation of variables. Let us get back to the question of when is the maximum temperature one half of the initial maximum temperature. This result is obtained by dividing the standard form by g(y), and then integrating both sides with respect to x. Toc JJ II J I Back We obtain the two equations, \[ \frac{T'(t)}{kT(t)}= - \lambda = \frac{X''(x)}{X(x)}.\], \[ X''(x) + \lambda X(x)=0, \\ T'(t) + \lambda k T(t)=0.\], The boundary condition $u(0,t)=0$ implies $ X(0)T(t)=0$. Thus the principle of superposition still applies for the heat equation (without side conditions). Solving the heat equation using the separation of variables. vRO�� w�'��ģ�#��n�:.յ�l��f�\l��y>1�]�Ѱ��^�)��akL��G��L��z?y#�B�8�=1\�� -�[��38��8�l��#1�a��PچC�vP]��f\� � �� ɧ��Z�m�-q�wg��M��(�w where $k>0$ is a constant (the thermal conductivity of the material). Finally, we use superposition to write the solution as, \[ u(x,t)= \sum^{\infty}_{n=1}b_n u_n (x,t)= \sum^{\infty}_{n=1}b_n \sin(\frac{n \pi}{L}x)e^{\frac{-n^2 \pi^2}{L^2}kt}.\], Why does this solution work? Similarly, $u_x(L,t)=0$ implies $X'(L)=0$. We are solving the following PDE problem: \[u_t=0.003u_{xx}, \\ u(0,t)= u(1,t)=0, \\ u(x,0)= 50x(1-x) ~~~~ {\rm{for~}} 00 (4.1) subject to the initial and boundary conditions Solution of the HeatEquation by Separation of Variables. The approximation gets better and better as $t$ gets larger as the other terms decay much faster. 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