Find the solution to the heat equation ∂u/∂t = α²∂²u/∂x² with boundary conditions u(0,t) = u(L,t) = 0 and initial condition u(x,0) = sin(πx/L).
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Step-by-Step Solution
Step 1: Assume a Separable Solution
We begin by assuming a separable solution of the form u(x,t)=X(x)T(t), where X(x) is a function of x only and T(t) is a function of t only. This method is commonly used for solving partial differential equations like the heat equation.
Step 2: Substitute into the Heat Equation
Substituting u(x,t)=X(x)T(t) into the heat equation ∂t∂u=α2∂x2∂2u and separating variables leads to two ordinary differential equations. Since one side depends only on t and the other only on x, both must be equal to a constant, which we denote as −λ2 for convenience in solving the resulting equations.
Step 3: Solve the Spatial Equation
The spatial equation X′′(x)+λ2X(x)=0 with boundary conditions X(0)=0 and X(L)=0 yields a general solution involving sines and cosines. Applying the boundary conditions forces C1=0 and λ to be quantized as Lnπ, leading to eigenfunctions Xn(x)=Bnsin(Lnπx).
Step 4: Solve the Temporal Equation
The temporal equation T′(t)+α2λ2T(t)=0 is a first-order linear differential equation. Its solution is an exponential decay function, where λ takes the values determined from the spatial problem.
Step 5: Form the General Solution
Combining the spatial and temporal solutions, and using the principle of superposition, we form the general solution as an infinite series. Each term in the series corresponds to a particular value of n.
Step 6: Apply Initial Condition
To find the coefficients Bn, we apply the initial condition u(x,0)=sin(Lπx). This involves using the orthogonality of sine functions. For the given initial condition, only the n=1 term will be non-zero.
Step 7: Determine Coefficients and Final Solution
Evaluating the integral for Bn, we find that B1=1 and all other Bn=0. Substituting these coefficients back into the general solution gives the particular solution that satisfies all given conditions.