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Suppose an ice core are perfectly insulated and heated at one end at a rate $\alpha$ and cooled from the other end at a rate $\beta$. Consider the following boundary value problem $$u_t-u_{xx}=0, \ u_x(0)=\beta, \ u_x(l)=\alpha,$$ where $u$ is temperature, $t$ is time and $l$ is the length of the core.

What should $\beta$ be such that the ice cores temperature remains stable ($u_t=0$)?

My attempt:

If $u_t=0$, then this implies $u_{xx}=0$. if we integrate both sides with respect to $x$, we get $$u_x=A$$ for some $A\in\mathbb{R}$. So, $$u_x(0)=\beta\implies A=\beta,\\u_x(l)=\alpha\implies A=\alpha.$$ Hence, we conclude that $\beta=\alpha\in\mathbb{R}$.

For part (b), assuming $\alpha=1, l=10$ and the average temperature of the core is $-15$, what is the solution for $u$ in the stable case.

Finally, If the cooling mechanism were to fail ($\beta=0$) how long would it take before the ice core started to melt (i.e. when would $u$ rise above $0$ at any point)?

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$\def\d{\mathrm{d}}\def\peq{\mathrm{\phantom{=}}{}}$For the first question, your solution so far is correct, but it would be better to complete it through further verification.

If $β = α$, then $u(x, t) = αx + c$, where $c$ is a constant, are solutions and the uniqueness of solutions to PDE of the second type boundary condition (up to a constant) implies that these are all the solutions. It is easy to verify that $u_t = 0$.

For the second question, since $u(x, t) = x + c$ and the average temperature is$$ -15 = \frac{1}{l} \int_0^l u(x, t) \,\d x = \frac{1}{10} \int_0^{10} (x + c) \,\d x = c + 5, $$ then $c = -20$ and $u(x, t) = x - 20$.

For the third question, suppose $u(x, t) = v(x, t) + \dfrac{αx^2}{2l}$, thus$$ \begin{cases} v_t - v_{xx} = \dfrac{α}{l}\\ v(x, 0) = -\dfrac{αx^2}{2l} + αx + c\\ v_x(0) = v_x(l) = 0 \end{cases}. \tag{1} $$ Using separation of variables, plugging in $v(x, t) = X(x) T(t)$ yields$$ \begin{cases} \dfrac{X''}{X} = \dfrac{T'}{T} = -λ\\ X'(0) = X'(l) = 0 \end{cases}. $$ For $λ = 0$, $X_0(x) = 1$. For $λ > 0$,$$ \begin{cases} X'' + λX = 0\\ X'(0) = X'(l) = 0 \end{cases} \Longrightarrow \begin{cases} λ_k = \dfrac{k^2 π^2}{l^2}\\ X_k(x) = \cos(\sqrt{λ_k} x) \end{cases}. $$ Thus the solution to (1) can be written as$$ v(x, t) = \sum_{k = 0}^∞ X_k(x) T_k(t). $$ Because\begin{gather*} \frac{α}{l} = \sum_{k = 0}^∞ \frac{\displaystyle \int_0^l \frac{α}{l} · X_k(x) \,\d x}{\displaystyle \int_0^l X_k^2(x) \,\d x} X_k(x) = \frac{α}{l} X_0(x), \quad c = cX_0(x),\\ x = \sum_{k = 0}^∞ \frac{\displaystyle \int_0^l x · X_k(x) \,\d x}{\displaystyle \int_0^l X_k^2(x) \,\d x} X_k(x) = \frac{l}{2} X_0(x) + \sum_{k = 1}^∞ \frac{2l}{k^2 π^2} ((-1)^k - 1) X_k(x),\\ x^2 = \sum_{k = 0}^∞ \frac{\displaystyle \int_0^l x · X_k(x) \,\d x}{\displaystyle \int_0^l X_k^2(x) \,\d x} X_k(x) = \frac{l^2}{3} X_0(x) + \sum_{k = 1}^∞ \frac{4l^2}{k^2 π^2} (-1)^k X_k(x), \end{gather*}\begin{align*} v_t - v_{xx} &= \sum_{k = 0}^∞ c_k X_k(x) T_k'(t) - \sum_{k = 0}^∞ X_k''(x) T_k(t)\\ &= \sum_{k = 0}^∞ X_k(x) T_k'(t) - \sum_{k = 0}^∞ (-λ_k X_k(x)) T_k(t)\\ &= \sum_{k = 0}^∞ X_k(x) (T_k'(t) + λ_k T_k(t)), \end{align*} then comparing coefficients of $X_k$'s yields$$ \begin{cases} T_0'(t) = \dfrac{α}{l}\\ T_0(0) = \dfrac{1}{3} αl + c \end{cases}, \quad \begin{cases} T_k'(t) + λ_k T_k(t) = 0\\ T_k(0) = -\dfrac{2αl}{k^2 π^2} \end{cases}\ (k \geqslant 1), $$ which implies$$ T_0(t) = \frac{αt}{l} + \frac{1}{3} αl + c, \quad T_k(t) = -\frac{2αl}{k^2 π^2} \exp(-λ_k t)\ (k \geqslant 1). $$ Therefore,\begin{align*} &\peq u(x, t) = v(x, t) + \frac{αx^2}{2l}\\ &= \frac{αx^2}{2l} + \frac{αt}{l} + \frac{1}{3} αl + c - \sum_{k = 1}^∞ \frac{2αl}{k^2 π^2} \cos\left( \frac{kπx}{l} \right) \exp\left( -\frac{k^2 π^2 t}{l^2} \right). \end{align*}

$u(x, t)$ is plotted below.

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  • $\begingroup$ Does this confirm the result? $\endgroup$ – Bell Oct 11 '18 at 6:15
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    $\begingroup$ @Bell It confirms the boundary conditions only, but $u_t-u_{xx}=0$ is way easier to confirm directly. Also, it seems that $T≈116$, so the series terms can be omitted when numerically finding the zero of $u$. $\endgroup$ – Saad Oct 11 '18 at 6:18

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