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Heat Equation Calculator

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Example
Created on 2024-06-20Asked by Michael Adams (Solvelet student)
Solve the one-dimensional heat equation ut=α2uxx u_t = \alpha^2 u_{xx} with boundary conditions u(0,t)=0 u(0,t) = 0 , u(L,t)=0 u(L,t) = 0 , and initial condition u(x,0)=f(x) u(x,0) = f(x) .

Solution

To solve the one-dimensional heat equation ut=α2uxx u_t = \alpha^2 u_{xx} with boundary conditions u(0,t)=0 u(0,t) = 0 , u(L,t)=0 u(L,t) = 0 , and initial condition u(x,0)=f(x) u(x,0) = f(x) : We use the method of separation of variables. Assume u(x,t)=X(x)T(t) u(x,t) = X(x)T(t) . Then: ut=X(x)T′(t) u_t = X(x)T'(t) uxx=X′′(x)T(t) u_{xx} = X''(x)T(t) Substitute into the heat equation: X(x)T′(t)=α2X′′(x)T(t) X(x)T'(t) = \alpha^2 X''(x)T(t) Divide by α2X(x)T(t) \alpha^2 X(x)T(t) : T′(t)α2T(t)=X′′(x)X(x)=−λ \frac{T'(t)}{\alpha^2 T(t)} = \frac{X''(x)}{X(x)} = -\lambda This gives two ordinary differential equations: T′(t)+α2λT(t)=0 T'(t) + \alpha^2 \lambda T(t) = 0 X′′(x)+λX(x)=0 X''(x) + \lambda X(x) = 0 Solving T′(t)+α2λT(t)=0 T'(t) + \alpha^2 \lambda T(t) = 0 : T(t)=e−α2λt T(t) = e^{-\alpha^2 \lambda t} Solving X′′(x)+λX(x)=0 X''(x) + \lambda X(x) = 0 with boundary conditions X(0)=0 X(0) = 0 and X(L)=0 X(L) = 0 : X(x)=sin⁡(nπxL),λ=(nπL)2 X(x) = \sin\left(\frac{n\pi x}{L}\right), \quad \lambda = \left(\frac{n\pi}{L}\right)^2 The solution is: u(x,t)=∑n=1∞Bnsin⁡(nπxL)e−α2(nπL)2t u(x,t) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{L}\right) e^{-\alpha^2 \left(\frac{n\pi}{L}\right)^2 t} Using the initial condition u(x,0)=f(x) u(x,0) = f(x) : f(x)=∑n=1∞Bnsin⁡(nπxL) f(x) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{L}\right) The coefficients Bn B_n are determined by the Fourier sine series: Bn=2L∫0Lf(x)sin⁡(nπxL) dx B_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx Therefore, the solution is: u(x,t)=∑n=1∞(2L∫0Lf(x)sin⁡(nπxL) dx)sin⁡(nπxL)e−α2(nπL)2t u(x,t) = \sum_{n=1}^{\infty} \left( \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx \right) \sin\left(\frac{n\pi x}{L}\right) e^{-\alpha^2 \left(\frac{n\pi}{L}\right)^2 t} Solved on Solvelet with Basic AI Model
Some of the related questions asked by Ava Rivera on Solvelet
1. Solve the one-dimensional heat equation ∂u∂t−k∂2u∂x2=0\frac{\partial u}{\partial t} - k\frac{\partial^2 u}{\partial x^2} = 0 subject to the initial condition u(x,0)=f(x)u(x, 0) = f(x) and boundary conditions u(0,t)=u(L,t)=0u(0, t) = u(L, t) = 0,2. Find the temperature distribution in a metal rod of length L=1L = 1 meter with initial temperature distribution u(x,0)=sin⁡(πx)u(x, 0) = \sin(\pi x) and zero temperature at the ends, subject to the heat equation ∂u∂t−∂2u∂x2=0\frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = 0.,
DefinitionThe heat equation is a parabolic partial differential equation that describes the distribution of heat (or variation in temperature) in a good region over time. It is an equation at the heart of the theory of heat conduction, fluid dynamics, and thermodynamics. The standard for of the heat equation is expressed as∂t∂u=α∇2u,where u is the temperature, t is time,α is the thermal diffusivity and ∇2 is the Laplacian operator. Ex: Solving the heat equation for a rod with insulated ends: finding the temperature distribution u(x,t) along the rod with respect to time.
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