91Ó°ÊÓ

Heat conduction in a semi-infinite rod with initial temperature \(g(x)\) leads to the equations $$ \left\\{\begin{array}{l} u_{t}=u_{x x} \quad \text { for } x>0, t>0 \\ u(x, 0)=g(x) \quad \text { for } x>0 \end{array}\right. $$ Assume that \(g\) is continuous and bounded for \(x \geq 0\). (a) If \(g(0)=0\) and the rod has its end maintained at zero temperature. then we must include the boundary condition \(u(0, t)=0\) for \(t>0\). Find a formula for the solution \(u(x, t)\). (b) If the rod has its end insulated so that there is no heat flow at \(x=0\), then we must include the boundary condition \(u_{x}(0, t)=0\) for \(t>0\). Find a formula for the solution \(u(x, t)\). Do you need to require \(g^{\prime}(0)=0\) ?

Step by step solution

01

Apply Boundary Conditions

For case (a), we apply the boundary condition \(u(0, t)=0\), that's \(u(x, t)=T(t)X(x)\), \(T(0)X(x)=0\). As \(X(x)\) is not zero, therefore this implies \(T(0)=0\). For case (b), we apply the boundary condition \(u_x(0, t)=0\), that's \(T(t)X'(0)=0\). As \(T(t)\) is not zero, this implies \(X'(0)=0\).

02

Separate the Variables

Separate the variables to solve the PDE, assume that the solution \(u(x, t)\) can be expressed as product of functions of \(x\) and \(t\) separately, as \(u(x, t)=T(t)X(x)\). Substitute \(u(x, t)=T(t)X(x)\) into \(u_t=u_{xx}\) to have equation \(T'(t)X(x)=T(t)X''(x)\). Divide both sides by \(T(t)X(x)\) to get \(\frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}\), since the left side is function of \(t\) only and the right side is function of \(x\) only and they are equal, this implies they equal to a constant. Let it be \(-\lambda\), this will give \(\frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=-\lambda\), it gives rise to two ordinary differential equations as \(T'(t)+\lambda T(t)=0\) and \(X''(x)+\lambda X(x)=0\).

03

Solve the Ordinary Differential Equations

Solving the ODE for \(T(t)\), use the method of integrating factor. For Case (a): Integrate \(T'(t)+\lambda T(t)=0\) to get \(T(t) = c_1 \exp(-\lambda t)\). For Case (b): Integrate \(X''(x)-\lambda X(x)=0\) to get \(X(x) = c_2 \sin(\sqrt{\lambda} x) + c_3 \cos(\sqrt{\lambda} x)\), but as \(X'(0) = 0\), it turns out that \(c_2 = 0\). So, \(X(x) = c_3 \cos(\sqrt{\lambda} x)\).

04

Compute Integral Transforms

For case (a): The solution is given by \(u(x, t) = \int_0^\infty \exp(-\lambda t)g(x) \, d\lambda\). For this we need the fourthier sine transform of \(g(x)\) and then we can find \(u(x, t)\). For case (b): The solution is given by \(u(x, t) = \int_0^\infty \exp(-\lambda t)g(x) \, d\lambda\). For this we need the fourthier cosine transform of \(g(x)\) and then we can find \(u(x, t)\).

05

Combine the Solutions

Finally combine the solutions of \(T(t)\) and \(X(x)\) to get the solution of \(u(x, t)\). Case (a): \(u(x, t) = \int_0^\infty \exp(-\lambda t) \sin(\sqrt{\lambda} x) g(\sqrt{\lambda}) \, d\lambda\). Case (b): \(u(x, t) = \int_0^\infty \exp(-\lambda t) \cos(\sqrt{\lambda} x) g(\sqrt{\lambda}) \, d\lambda\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Temperature Function

The initial temperature function, denoted by g(x), represents the temperature distribution along the rod at the beginning, at time t = 0. It is critical in determining the future states of the system since the heat conduction equation involves time derivatives, meaning it describes how the temperature evolves over time from this initial state.

The property of continuity and boundedness for g(x), as stated in the problem, ensures that the temperature distribution does not have abrupt, infinite, or undefined values, which are essential conditions for the mathematical solutions to be physically meaningful and calculable.

Separation of Variables

Separation of variables is a method used to solve partial differential equations (PDEs), like the heat conduction equation. By assuming that the solution can be written as a product of functions where each function depends on only one variable, we turn the PDE into simpler, ordinary differential equations (ODEs).

This works because if the product of two functions equals a constant, each function must individually equal a constant or its negative. This powerful technique leverages the fact that if the original PDE holds true, then the simpler ODEs must also hold true.

Boundary Conditions

Boundary conditions specify the values that a solution to a differential equation must take on at the boundaries of the domain. For heat conduction, this could mean setting the temperature at the rod's end (as in u(0, t) = 0 for a rod held at zero temperature), or it might define the heat flux (as in u_x(0, t) = 0 for an insulated rod end).

These conditions are crucial as they ensure the solution to the PDE is unique. Without them, we could have infinitely many solutions, rendering the problem unsolvable in a practical sense. They ground the mathematical model in the physical world by introducing real-world constraints.

Ordinary Differential Equations

Ordinary differential equations (ODEs) involve functions of a single variable and their derivatives. In the context of heat conduction, once we separate variables, we reduce the original PDE into two ODEs – one for time, T(t), and one for space, X(x).

Solving these ODEs is often more straightforward than solving a PDE and can typically be done using established techniques such as integrating factors, characteristic equations, or comparison to known forms.

Integral Transforms

Integral transforms, like Fourier transforms, are used to solve PDEs by changing the domain from the time or space domain to the frequency domain, where many problems become algebraically simpler.

In the case of the heat conduction problem, after separation of variables and applying boundary conditions, we are left with integrals that in principle resemble a transform. Specifically, the sine or cosine transform is used here to handle the initial temperature function, converting it from the spatial domain into the frequency domain, which simplifies the integration process to solve for u(x, t).