Question

Suppose that current flowing through a laterally insulated rod generates heat at a constant rate; then Problem 19 yields the equation $$ \frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}}+C $$ Assume the boundary conditions $u(0, t)=u(L, t)=0$ and $u(x, 0)=f(x) .$ (a) Find the steady-state temperature $u_{\mathrm{ss}}(x)$ determined by $$ 0=k \frac{d^{2} u_{\mathrm{ss}}}{d x^{2}}+C, \quad u_{\mathrm{ss}}(0)=u_{\mathrm{ss}}(L)=0 $$ (b) Show that the transient temperature $$ u_{\mathrm{tr}}(x, t)=u(x, t)-u_{\mathrm{ss}}(x) $$ satisfies the boundary value problem $$ \begin{aligned} \frac{\partial u_{\mathrm{tr}}}{\partial t} &=k \frac{\partial^{2} u_{\mathrm{tr}}}{\partial x^{2}} \\ u_{\mathrm{tr}}(0, t) &=u_{\mathrm{tr}}(L, t)=0 \\ u_{\mathrm{tr}}(x, 0) &=g(x)=f(x)-u_{\mathrm{ss}}(x) \end{aligned} $$ Hence conclude from the formulas in (34) and (35) that $$ u(x, t)=u_{\mathrm{ss}}(x)+\sum_{n=1}^{\infty} c_{n} \exp \left(-n^{2} \pi^{2} k t / L^{2}\right) \sin \frac{n \pi x}{L} $$ where $$ c_{n}=\frac{2}{L} \int_{0}^{L}\left[f(x)-u_{\mathrm{ss}}(x)\right] \sin \frac{n \pi x}{L} d x . $$

          Suppose that current flowing through a laterally insulated rod generates heat at a constant rate; then Problem 19 yields the equation
$$
\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}}+C
$$
Assume the boundary conditions $u(0, t)=u(L, t)=0$ and $u(x, 0)=f(x) .$ (a) Find the steady-state temperature $u_{\mathrm{ss}}(x)$ determined by
$$
0=k \frac{d^{2} u_{\mathrm{ss}}}{d x^{2}}+C, \quad u_{\mathrm{ss}}(0)=u_{\mathrm{ss}}(L)=0
$$
(b) Show that the transient temperature
$$
u_{\mathrm{tr}}(x, t)=u(x, t)-u_{\mathrm{ss}}(x)
$$
satisfies the boundary value problem
$$
\begin{aligned}
\frac{\partial u_{\mathrm{tr}}}{\partial t} &=k \frac{\partial^{2} u_{\mathrm{tr}}}{\partial x^{2}} \\
u_{\mathrm{tr}}(0, t) &=u_{\mathrm{tr}}(L, t)=0 \\
u_{\mathrm{tr}}(x, 0) &=g(x)=f(x)-u_{\mathrm{ss}}(x)
\end{aligned}
$$
Hence conclude from the formulas in (34) and (35) that
$$
u(x, t)=u_{\mathrm{ss}}(x)+\sum_{n=1}^{\infty} c_{n} \exp \left(-n^{2} \pi^{2} k t / L^{2}\right) \sin \frac{n \pi x}{L}
$$
where
$$
c_{n}=\frac{2}{L} \int_{0}^{L}\left[f(x)-u_{\mathrm{ss}}(x)\right] \sin \frac{n \pi x}{L} d x .
$$

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Added by Vanessa S.

Differential Equations and Linear Algebra

Stephen W. Goode, Scott A. Annin 4th Edition

Instant Answer

Solved by Expert Shu-Ting Huang

01/06/2022

Step 1

The general solution is $$ u_{ss}(x) = Ax^2 + Bx + D $$ where A, B, and D are constants. The boundary conditions $u_{ss}(0)=u_{ss}(L)=0$ give us $$ D = 0, \quad AL^2 + BL = 0 $$ which implies that $A = -B/L$. Substituting this into the differential equation Show more…

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Suppose that current flowing through a laterally insulated rod generates heat at a constant rate; then Problem 19 yields the equation $$ \frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}}+C $$ Assume the boundary conditions $u(0, t)=u(L, t)=0$ and $u(x, 0)=f(x) .$ (a) Find the steady-state temperature $u_{\mathrm{ss}}(x)$ determined by $$ 0=k \frac{d^{2} u_{\mathrm{ss}}}{d x^{2}}+C, \quad u_{\mathrm{ss}}(0)=u_{\mathrm{ss}}(L)=0 $$ (b) Show that the transient temperature $$ u_{\mathrm{tr}}(x, t)=u(x, t)-u_{\mathrm{ss}}(x) $$ satisfies the boundary value problem $$ \begin{aligned} \frac{\partial u_{\mathrm{tr}}}{\partial t} &=k \frac{\partial^{2} u_{\mathrm{tr}}}{\partial x^{2}} \\ u_{\mathrm{tr}}(0, t) &=u_{\mathrm{tr}}(L, t)=0 \\ u_{\mathrm{tr}}(x, 0) &=g(x)=f(x)-u_{\mathrm{ss}}(x) \end{aligned} $$ Hence conclude from the formulas in (34) and (35) that $$ u(x, t)=u_{\mathrm{ss}}(x)+\sum_{n=1}^{\infty} c_{n} \exp \left(-n^{2} \pi^{2} k t / L^{2}\right) \sin \frac{n \pi x}{L} $$ where $$ c_{n}=\frac{2}{L} \int_{0}^{L}\left[f(x)-u_{\mathrm{ss}}(x)\right] \sin \frac{n \pi x}{L} d x . $$

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00:01 In this question we are considering the partial differential equation d u over d t equal k d squared u over d x squared plus c with boundary condition u of 0 t u of l t both 0 and initial condition is u of x 0 equal f of x in part a, we are considering the steady -state solution that does not depend on t.

00:51 So we are going to find this u .s.

00:55 As such that 0 equal k -u -s -s double prime of x plus c with the boundary.

01:18 Condition us s of 0, uss of l both of 0.

01:30 We note that this is the same as u .s .s.

01:34 Double prime equal negative c over k.

01:43 Since the second derivative is a constant, and we know that this has two root.

01:54 So it is a second derivative is a constant we can write it as a degree two polynomial.

02:06 And we also know the two roots.

02:08 So it is xx minus l.

02:12 And this coefficient is negative c over 2k.

02:22 Now we come to part b.

02:23 In part b, we consider the chenuan solution, so this is u -tr -x -t equal u of xt minus the steady -state solution uss of x.

02:51 Then if we differentiate it with respect to t, we will get it is a steady -state solution, uss of x.

03:07 U over d t because this does not depend on t so it is now we go to the original equation k d square u over d x square plus c and this u can be written as ut plus u s so it is k d square utr dx square plus k d square u s over dx square plus k d square u s over dx square plus c and we know that this is the first time we just copy it and then the second and third term, because kus as double prime plus c is 0, so it's 0 here...

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