One-dimensional heat conduction boundary value problem

Let us consider the one-dimensional heat conduction problem on the interval from $\alpha$ to $\beta$. The heat conduction equation (1.8$\displaystyle \frac{\partial T(M, t)}{\partial t}$ $\displaystyle = a^2 \cdot \Delta T(M, t)$ $\displaystyle + f(M, t),$) in this case takes the form

The Dirichlet boundary conditions (1.13$\displaystyle T(M, t)$ $\displaystyle = \Phi(M, t), \quad M$ $\displaystyle \in S.$) in the one-dimensional case have the form

To get rid of the inhomogeneity, we introduce $T(x,t)$ $= \widehat{T}(x,t)$ $+ U(x,t)$, let us choose the function $U(x,t)$ $= a_1(t)$ $+ a_2(t) \cdot x$, and compute the coefficients, in accordance with (2.2$\displaystyle \widehat{T}(M, t)$ $\displaystyle + U(M, t)$ $\displaystyle = \Phi(M, t), \quad \widehat{T}(M, t)$ $\displaystyle = 0 \;$ $\displaystyle \Rightarrow\; U(M, t)$ $\displaystyle = \Phi(M, t), \quad M$ $\displaystyle \in S.$)

Then equation (2.23$\displaystyle \frac{\partial T(x, t)}{\partial t}$ $\displaystyle = a^2 \cdot \frac{\partial^2 T(x, t)}{\partial x^2}$ $\displaystyle + f(x, t).$) and the boundary conditions (2.24$\displaystyle T(x, 0)$ $\displaystyle = T_0(x),$$\displaystyle T(\alpha, t)$ $\displaystyle = \Phi_\alpha(t),$$\displaystyle T(\beta, t)$ $\displaystyle = \Phi_\beta(t)$) take the form

where $\widehat{f}(x, t)$ $= f(x, t)$ $- \frac{\displaystyle da_1(t)}{\displaystyle dt}$ $- \frac{\displaystyle da_2(t)}{\displaystyle dt} \cdot x$.

The Neumann boundary conditions (1.14$\displaystyle \lambda \cdot \frac{\partial T(M, t)}{\partial \vec{n}}$ $\displaystyle = \Phi(M, t), \quad M$ $\displaystyle \in S,$) in the one-dimensional case have the form

To get rid of the inhomogeneity, we introduce $T(x,t)$ $= \widehat{T}(x,t)$ $+ U(x,t)$, let us choose the function $U(x,t)$ $= a_1(t) \cdot x$ $+ a_2(t) \cdot x^2$, and compute the coefficients, in accordance with (2.3$\displaystyle \frac{\partial \widehat{T}(M, t)}{\partial \vec{n}}$ $\displaystyle + \frac{\partial U(M, t)}{\partial \vec{n}}$ $\displaystyle = \Phi(M, t), \quad \frac{\partial \widehat{T}(M, t)}{\partial \vec{n}}$ $\displaystyle = 0 \;$ $\displaystyle \Rightarrow\; \frac{\partial U(M, t)}{\partial \vec{n}}$ $\displaystyle = \Phi(M, t), \quad M$ $\displaystyle \in S.$)

Then equation (2.23$\displaystyle \frac{\partial T(x, t)}{\partial t}$ $\displaystyle = a^2 \cdot \frac{\partial^2 T(x, t)}{\partial x^2}$ $\displaystyle + f(x, t).$) and the boundary conditions (2.28$\displaystyle T(x, 0)$ $\displaystyle = T_0(x),$$\displaystyle \frac{\displaystyle \partial T(x, t)}{\displaystyle \partial x} \bigg|_{x=\alpha}$ $\displaystyle = \Phi_\alpha(t),$$\displaystyle \frac{\displaystyle \partial T(x, t)}{\displaystyle \partial x} \bigg|_{x=\beta}$ $\displaystyle = \Phi_\beta(t)$) take the form

where $\widehat{f}(x, t)$ $= f(x, t)$ $- \frac{\displaystyle da_1(t)}{\displaystyle dt} \cdot x$ $- \frac{\displaystyle da_2(t)}{\displaystyle dt} \cdot x^2$ $+ 2 \cdot a^2 \cdot a_2(t)$.

Equation (2.8$\displaystyle \Delta \Psi(M)$ $\displaystyle + \gamma^2 \cdot \Psi(M)$ $\displaystyle = 0.$) takes the form

The general solution of this equation has the form

where $c_1$ and $c_2$ are coefficients determined from the boundary conditions, and the weight ensuring the orthogonality of the eigenfunctions is $\rho(x)$ $= 1$.

Let us substitute the general solution into the boundary conditions (2.27$\displaystyle \frac{\displaystyle \partial \widehat{T}(x, t)}{\displaystyle \partial t}$ $\displaystyle = a^2 \cdot \frac{\displaystyle \partial^2 \widehat{T}(x, t)}{\displaystyle \partial x^2}$ $\displaystyle + \widehat{f}(x, t),$$\displaystyle \widehat{T}(x, 0)$ $\displaystyle = T_0(x)$ $\displaystyle - a_1(0)$ $\displaystyle - a_2(0) \cdot x,$$\displaystyle \widehat{T}(\alpha, t)$ $\displaystyle = 0, \widehat{T}(\beta, t)$ $\displaystyle = 0,$) and obtain

Clearly, from the first equation $c_1$ $= 0$ and the solution makes sense only when $c_2$ $\neq 0$, which is possible only when $\sin[\gamma \cdot (\beta - \alpha)]$ $= 0$

Zero and negative values of $n$ are no good: for $n$ $= 0$ we have $\gamma$ $= 0$, and $\Psi(x)$ $= c_2 \cdot \sin[0 \cdot (x - \alpha)]$ $\equiv 0$ — identically zero, which is not an eigenfunction; to negative $n$ corresponds the same eigenvalue $\gamma^2$ (since $\gamma_{-n}$ $= -\gamma_n$), and $\sin[-\gamma_n \cdot (x - \alpha)]$ $= -\sin[\gamma_n \cdot (x - \alpha)]$ is linearly dependent with the solution for positive $n$ and gives no new contribution. Therefore $n$ $\in (1..\infty)$.

Thus, the eigenvalues and eigenfunctions have the form

where $c_{2n}$ is an arbitrary constant factor: the eigenfunction is defined up to it. Let us set $c_{2n}$ $= 1$ — this factor is absorbed into the expansion coefficients anyway $T_{0n}$ and $f_n$, as in the general solution.

To expand functions into a Fourier series in $\Psi_n(x)$ we need to compute the norm $\|\Psi_n(x)\|^2$; the weight for Cartesian coordinates is $\rho(x)$ $= 1$.

Now let us substitute the solution into the boundary conditions (2.31$\displaystyle \frac{\displaystyle \partial \widehat{T}(x, t)}{\displaystyle \partial t}$ $\displaystyle = a^2 \cdot \frac{\displaystyle \partial^2 \widehat{T}(x, t)}{\displaystyle \partial x^2}$ $\displaystyle + \widehat{f}(x, t),$$\displaystyle \widehat{T}(x, 0)$ $\displaystyle = T_0(x)$ $\displaystyle - a_1(0) \cdot x$ $\displaystyle - a_2(0) \cdot x^2,$$\displaystyle \frac{\displaystyle \partial \widehat{T}(x, t)}{\displaystyle \partial x} \bigg|_{x=\alpha}$ $\displaystyle = 0, \frac{\displaystyle \partial \widehat{T}(x, t)}{\displaystyle \partial x} \bigg|_{x=\beta}$ $\displaystyle = 0,$) and obtain

Clearly, from the first equation $c_2$ $= 0$ and the solution makes sense only when $\sin[\gamma_n \cdot (\beta - \alpha)]$ $= 0$

Here, unlike the Dirichlet problem, the mode $n$ $= 0$ survives: for $\gamma_0$ $= 0$ the eigenfunction $\Psi_0(x)$ $= c_{10} \cdot \cos[0 \cdot (x - \alpha)]$ $= c_{10}$ — a nonzero constant, that is, a fully fledged eigenfunction (the constant mode). Hence the indexing starts from zero.

Thus, the eigenvalues and eigenfunctions have the form

where $c_{1n}$ is an arbitrary constant factor: the eigenfunction is defined up to it. Let us set $c_{1n}$ $= 1$ — this factor is absorbed into the expansion coefficients anyway $T_{0n}$ and $f_n$, as in the general solution.

We compute the norm of the eigenfunctions in the same way as in the sine case, reducing the power of the cosine by the half-angle formula. For $n$ $\ge 1$

since the integral of the cosine over an integer number of half-periods equals zero: $\sin \left[ 2 \cdot \gamma_n \cdot (\beta - \alpha) \right]$ $= \sin(2 \cdot \pi \cdot n)$ $= 0$. The special case $n$ $= 0$ (for $\gamma_0$ $= 0$ the eigenfunction $\Psi_0(x)$ $= 1$) we compute separately

The final solution of the one-dimensional Dirichlet boundary value problem with inhomogeneous boundary conditions takes the form

In the Neumann problem, unlike the Dirichlet one, the spectrum contains the zero mode ($\gamma_0$ $= 0$, $\Psi_0(x)$ $= 1$) — the case $\gamma_n$ $= 0$, which in the general solution (2.18$\displaystyle T(M)$ $\displaystyle = \sum_{n=1}^{\infty} \frac{\displaystyle \Psi_n(M)}{\displaystyle \|\Psi_n(M)\|^2} \cdot \frac{\displaystyle 1}{\displaystyle a^2 \cdot \gamma_n^2} \cdot \iiint_G \rho(M) \cdot f(M) \cdot \Psi_n(M) \,dG, \quad \gamma_n$ $\displaystyle \neq 0.$) was set aside for separate consideration; it is precisely what gives the first (non-decaying) term. The final solution of the one-dimensional Neumann boundary value problem with inhomogeneous boundary conditions takes the form

To obtain the stationary solutions, we let time tend to infinity in the solutions found ($t$ $\to \infty$), as was done for the general case in (2.18$\displaystyle T(M)$ $\displaystyle = \sum_{n=1}^{\infty} \frac{\displaystyle \Psi_n(M)}{\displaystyle \|\Psi_n(M)\|^2} \cdot \frac{\displaystyle 1}{\displaystyle a^2 \cdot \gamma_n^2} \cdot \iiint_G \rho(M) \cdot f(M) \cdot \Psi_n(M) \,dG, \quad \gamma_n$ $\displaystyle \neq 0.$). The term with the initial condition vanishes, since $e^{- a^2 \cdot \gamma_n^2 \cdot t}$ $\to 0$; the sources and boundary conditions stop depending on time, and the time integral for $\gamma_n$ $\neq 0$ gives the factor $\frac{1}{a^2 \cdot \gamma_n^2}$:

The stationary solution of the one-dimensional Dirichlet boundary value problem takes the form

The stationary solution of the one-dimensional Neumann boundary value problem is not so simple. Unlike the Dirichlet problem, the purely Neumann stationary problem is not always solvable. The reduced function $\widehat{T}$ satisfies the homogeneous Neumann conditions $\widehat{T}'(\alpha)$ $= \widehat{T}'(\beta)$ $= 0$ and the stationary equation $a^2 \cdot \widehat{T}''(x)$ $+ \widehat{f}(x)$ $= 0$. Let us integrate it over the interval $[\alpha, \beta]$:

The boundary term vanished due to the homogeneous Neumann conditions, and what remains is the solvability condition (2.22$\displaystyle \iiint_G f(M) \,dM$ $\displaystyle = - a^2 \cdot \iint_S \Phi(M) \,dS,$): $\int_\alpha^\beta \widehat{f}(\omega) \,d\omega$ $= 0$ — the total reduced source must vanish (otherwise heat accumulates, the mean grows without bound and there is no steady state). When this condition holds, the solution is defined only up to an arbitrary additive constant $C$: it corresponds to the zero mode ($\gamma_0$ $= 0$, $\Psi_0(x)$ $= 1$), whose amplitude is not fixed by the stationary equation. With this caveat the solution takes the form