Question

For 0 < x < 5 and t > 0 , solve the one-dimensional heat flow equation ∂u/ ∂t=4d^2u/dx^2 the conditions u (t, 0)= u( t ,5)= 0 ,u(0,x) = x .

Answer 1

The general one dimensional heat flow equation is given by $\dfrac{\partial u}{\partial t} = c^{2}\dfrac{\partial^{2} u}{\partial x^{2}}.$ Then from the given equation, we have $c^{2} = 4.$

Let $u(t,x)=T(t)X(x)$ be the solution of the one dimensional heat flow equation. Then,

$\dfrac{\partial u}{\partial t}=T'X\\ \dfrac{\partial^{2} u}{\partial x^{2}}=TX''\\$ . The given equation becomes,

$T'X=4TX''\\ \text{By separating variables,}\\ \dfrac{X''}{X}=\dfrac{1}{4}\dfrac{T'}{T}=k(say).$

Then $X''-kX=0~ \text{and}~ \dfrac{T'}{T}=4k$ .We have three cases depending on the value of k.

$\text{Case i: Let $k$ be positive, say $k=p^{2}$}.\\ \text{Then, $X''-p^{2}X=0$ and $T' - 4p^{2}T=0$}.$

The auxiliary equations are,

$m^{2}-4=0~\text{and}~m-4p^{2}=0 \\ m=\pm 2~\text{and}~m=4p^{2}\\ \text{Therefore,} X = c_{1}e^{2x}+c_{2}e^{-2x}~ \text{and}~ T = c_{3}e^{4p^{2}t}\\ \text{Using this, we have a solution}\\ u(t,x)=c_{3}e^{4p^{2}t}(c_{1}e^{2x}+c_{2}e^{-2x})\hspace{0.5in}-(A)$

$\text{Case ii: Let $k$ be negative, say $k=-p^{2}$}.\\ \text{Then, $X''+p^{2}X=0$ and $T' + 4p^{2}T=0$}.$

The auxiliary equations are,

$m^{2}+4=0~\text{and}~m+4p^{2}=0 \\ m=\pm 2i~\text{and}~m=-4p^{2}\\ \text{Therefore,} X = (c_{4}\cos2x + c_{5}\sin2x)~ \text{and}~ T = c_{6}e^{-4p^{2}t}\\ \text{Using this, we have a solution}\\ u(t,x)=c_{6}e^{-4p^{2}t}(c_{4}\cos2x + c_{5}\sin2x)\hspace{0.3in}-(B)$

$\text{Case iii: Let $k = 0$}\\ \text{Then, $X''= 0$ and $T' =0$}.\\ X = c_{7}x+c_{8}~\text{and}~ T = c_{9}\\ \text{Using this we have a solution}\\ u(t,x)=c_{9}(c_{7}x+c_{8})\hspace{1.3in}-(C)$

Of these three solutions, we have to choose the solution which is consistent with the physical nature of the problem and the boundary conditions. Hence the suitable solution of the heat equation is given by,

$u(t,x)=(A\cos px+B\sin px)e^{-4p^{2}t}~\text{(since $c^2=4$)}\hspace{0.3in}-(1)\\$

The given initial and boundary conditions are,

$u(t,0)=0\hspace{1in}-(i)\\ u(t,5)=0\hspace{1in}-(ii)\\ u(0,x)=x\hspace{0.956in}-(iii)\\$

Using the condition (i) in equation (1),

$u(t,0)=Ae^{-4p^{2}t}\\ 0=Ae^{-4p^{2}t}\\ A=0~~(\text{since $e^{-4p^{2}t} \ne 0$}).$

Equation (1) becomes,

$u(t,x)=Be^{-4p^{2}t}\sin px \hspace{1in}-(2)$

Using the condition (ii) in equation (2),

$u(t,5)=Be^{-4p^{2}t}\sin5p\\ 0=Be^{-4p^{2}t}\sin5p\\ \sin 5p=0~~(\text{since $B \neq 0$, $e^{-4p^{2}t}\ne0$})\\ \sin 5p = \sin n\pi\\ p=\dfrac{n\pi}{5}$

Using the value of p in equation (2),

$u(t,x)=Be^{\frac{-4n^{2}\pi^{2}t}{25}}\sin(\frac{n\pi x}{5})\hspace{0.8in}-(3)$

The most general solution is obtained by a linear combination of solution of the form (3). Hence

$u(t,x) =\displaystyle \sum_{n=1}^{\infty}B_{n}\sin\left(\frac{n\pi x}{5}\right)e^{-\frac{4n^{2}\pi^{2}t}{25}}\hspace{0.4in}-(4)$

Using the condition (iii) in (4),

$u(0,x) =x= \displaystyle\sum_{n=1}^{\infty}B_{n}\sin\left(\frac{n\pi x}{5}\right)$ , which is the half range sine series representing the function $x$ in $(0,5)$ .

Therefore,

$B_{n} = \dfrac{2}{5}\displaystyle\int_{0}^{5}x \sin\left(\frac{n\pi x}{5}\right)dx\\ =\dfrac{2}{5}\left.\left\{\biggl(x \left(\frac{-\cos(\frac{n\pi x}{5})}{(\frac{n\pi}{5})}\right) + \frac{\sin(\frac{n\pi x}{5})}{(\frac{n\pi}{5})^2} \biggr)\right|_{0}^{5}\right\}\\ =\dfrac{2}{5}\left\{\frac{-25\cos n\pi}{n\pi}+ \frac{25\sin n\pi}{n^{2}\pi^{2}}-0\right\}\\ =\dfrac{-10(-1)^{n}}{n\pi}~~(\text{since $\cos n\pi= (-1)^n$, $\sin n\pi = 0$ for $n = 1,2,3,\cdots$})\\ B_{n}=\dfrac{10(-1)^{n+1}}{n\pi}$

Using the value of $B_{n}$ in equation (4), the general solution is,

$u(t,x) =\displaystyle \sum_{n=1}^{\infty}\frac{10(-1)^{n+1}}{n\pi}\sin\left(\frac{n\pi x}{5}\right)e^{-\frac{4n^{2}\pi^{2}t}{25}}$