Title: Default for 25.3 Solution Using Separation of Variables
Course: HELM Workbooks 2023-2024

Solution of important PDEs

We shall just consider two analytic solution techniques for PDEs:

  1. Direct integration
  2. Separation of variables

The method of direct integration is a straightforward extension of solving very simple ODEs by integration, and will be considered first. The method of separation of variables is more important and we will study it in detail shortly.
You should note that many practical problems involving PDEs have to be solved by numerical methods but that is another story (introduced in  32 and  33).

Exercise

Solve the ODE

\(\qquad\displaystyle \dfrac{d^2y}{dx^2}=x^2+2\)

given that \(y=[value]\) when \(x=0\) and \(\dfrac{dy}{dx}=[value]\) when \(x=0.\)

1. First find \(\dfrac{dy}{dx}\) by integrating once, not forgetting the arbitrary constant of integration, \(A\):
2. Now find \(y\) by integrating again, not forgetting to include another arbitrary constant, \(B\):
3. Now find \(A\) and \(B\) by inserting the two given initial conditions and so find the solution:
Feedback: \(y(0) = 1\) gives \(B = 1\) | \(y'(0) = 2\) gives \(A = 2\) so the required solution is \(y =\dfrac{x^4}{12}+x^2+2x+1\)

Consider now a similar type of PDE i.e. one that can also be solved by direct integration.

Suppose we require the general solution of

\(\qquad\displaystyle \dfrac{\partial^2u}{\partial x^2}=2xe^t\)

where \(u\) is a function of \(x\) and \(t\).

Integrating with respect to \(x\) gives us

\(\qquad\displaystyle \dfrac{\partial u}{\partial x}=x^2e^t+f(t)\)

where the arbitrary function \(f(t)\) replaces the normal “arbitrary constant” of ordinary integration. This function of \(t\) only is needed because we are integrating “partially” with respect to \(x\) i.e. we are reversing a partial differentiation with respect to \(x\) at constant \(t\).

Integrating again with respect to \(x\) gives the general solution:

\(\qquad\displaystyle u=\dfrac{x^3}{3}e^t+x\ f(t)+g(t)\)

where \(g(t)\) is a second arbitrary function. We have now obtained the general solution of the given PDE but to find the arbitrary function we must know two initial conditions.

Suppose, for the sake of example, that these conditions are \(\qquad\displaystyle u (0,t)=t\ ,\qquad \dfrac{\partial u}{\partial x}(0,t)=e^t\)

Inserting the first of these conditions into the general solution gives\(\ \ \displaystyle g(t)=t\).

Inserting the second condition into the general solution gives \(\ \ \displaystyle f(t)=e^t\).

So the final solution is\(\ \ \displaystyle u=\dfrac{x^3}{3}e^t+xe^t+t.\)

Exercise

Solve the PDE

\(\qquad\displaystyle \dfrac{\partial^2u}{\partial x\partial y}=\sin x\ \cos y\)

subject to the conditions

\(\qquad\displaystyle \dfrac{\partial u}{\partial x}=2x\ \ \text{at}\ \ y=\dfrac{\pi}{2},\ \ u=2\sin y\ \ \text{at}\ \ x=\pi.\)

1. First integrate the PDE with respect to \(y\): (it is equally valid to integrate first with respect to \(x\)). Don’t forget the arbitrary function \(f(x)\).

2. Since one of the given conditions is on \(\dfrac{\partial u}{\partial x}\), impose this condition to determine the arbitrary function \(f(x)\):

3. Now integrate again to determine \(u\):

4. Next, obtain the arbitrary function \(g(y)\):

5. Now write down the final answer for \(u(x,y)\):
Feedback: 1. Recall that \(\displaystyle \dfrac{\partial^2u}{\partial x\partial y}\)=\(\dfrac{\partial }{\partial y}\)(\(\dfrac{\partial u}{\partial x}\)) Hence integration with respect to \(y\) gives \(\displaystyle \dfrac{\partial u}{\partial x}= sin(x)sin(y) + f(x)\) 2. At \(y = \pi/2\) the condition gives \(sin(x)sin(\pi/2) + f(x) = 2x\) i.e. \(f(x) = 2x − sin(x)\) So \(\dfrac{\partial u}{\partial x}= sin(x)sin(y) + 2x − sin(x)\) 3. Integrating now with respect to \(x\) gives \(u = − cos(x)sin(y) + x^2 + cos(x) + g(y)\) 4. The condition \(u(\pi, y) = 2 sin(y)\) gives \(− cos(\pi) sin(y) + \pi^ 2 + cos(\pi) + g(y) = 2 sin(y)\) Therefore \(sin(y) + \pi^2 − 1 + g(y) = 2 sin(y)\) Hence, \(g(y) = sin(y) + 1 − \pi^2\) 5. A correct answer is \(u(x, y) = x^2 + cos(x)(1 − sin(y)) + sin(y) + 1 −\pi^2\)

Method of separation of variables - general approach

In Section 25.2 we showed that

  1. \(u(x,y)=\sin x \cosh y\)
is a solution of the two-dimensional Laplace equation
  2. \(u(x,t)=e^{-2\pi^2t}\sin \pi x\)
is a solution of the one-dimensional heat conduction equation
  3. \(u(x,t)=u_0\sin \left(\dfrac{\pi x}{\ell}\right)\cos \left(\dfrac{\pi c t}{\ell}\right)\)
is a solution of the one-dimensional wave equation.

All three solutions here have a specific form: in (a) \(u(x,y)\) is a product of a function of \(x\) alone, \(\sin x\), and a function of \(y\) alone, \(\cosh y\). Similarly, in both (b) and (c), \(u(x,t)\) is a product of a function of \(x\) alone and a function of \(t\) alone.

The method of separation of variables involves finding solutions of PDEs which are of this product form. In the method we assume that a solution to a PDE has the form.

\(\qquad\displaystyle u(x,t)=X(x) T(t)\) | \(\text{or}\qquad u(x,y)=X(x) Y(y))\)

where \(X(x)\) is a function of \(x\) only, \(T(t)\) is a function of \(t\) only and \(Y(y)\) is a function \(y\) only. You should note that not all solutions to PDEs are of this type; for example, it is easy to verify that

\(\qquad\displaystyle u(x,y)=x^2-y^2\)

(which is not of the form \(u(x,y)=X(x) Y(y)\)) is a solution of the Laplace equation.
However, many interesting and useful solutions of PDEs are obtainable which are of the product form. We shall firstly consider the types of solution obtainable for our three basic PDEs using trial solutions of the product form.

Heat conduction equation

\(\dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{k}\dfrac{\partial u}{\partial t}\) | \(k>0\) | \[(1)\]

Assuming that

\(\qquad u=X(x)T(t)\) | \[(2)\]

then \(=XT \qquad \) for short

\(\qquad \dfrac{\partial u}{\partial x}=\dfrac{dX}{dx}T=X'T\qquad\) for short

\(\qquad \dfrac{\partial^2u}{\partial x^2}=\dfrac{d^2X}{dx^2}T=X''T\qquad\) for short

\(\qquad \dfrac{\partial u}{\partial t}=X\dfrac{dT}{dt}=XT'\qquad\) for short
Substituting into the original PDE (1)

\(\qquad\displaystyle X''T=\dfrac{1}{k}XT'\)
which can be re-arranged as

\(\qquad\displaystyle \dfrac{X''}{X}=\dfrac{1}{k}\dfrac{T'}{T}\) |  | \[(3)\]

Now the left-hand side of \((3)\) involves functions of \(x\) only and the right-hand side expression contain functions of \(t\) only. Thus altering the value of \(t\) cannot change the left-hand side of \((3)\) i.e. it stays constant. Hence so must the right-hand side be constant. We conclude that \(T(t)\) is a function such that

\(\qquad\displaystyle \dfrac{1}{k}\dfrac{T'}{T}=K\) |  | \((4)\)

where \(K\) is a constant whose sign is yet to be determined.
By a similar argument, altering the value of \(x\) cannot change the right-hand side of \((3)\) and consequently the left-hand side must be a constant, i.e.

\(\qquad\displaystyle \dfrac{X''}{X}=K\) |  | \((5)\)

We see that the effect of assuming a product trial solution of the form \((2)\) converts the PDE \((1)\) into the two ODEs \((4)\) and \((5)\).
Both these ODEs are types whose solution we revised at the beginning of this Workbook but we shall not attempt to solve them yet. In particular the solution of \((5)\) depends on whether the constant \(K\) is positive or negative.

Wave equation

\(\qquad\displaystyle \dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{c^2}\dfrac{\partial^2u}{\partial t^2}\) | \((6)\)

Example

By following a similar procedure to the above, assume a product solution

\(\qquad\displaystyle u(x,t)=X(x)T(t)\)

for the wave equation and find the two ODEs satisfied by \(X(x)\) and \(T(t)\).

1. First obtain \(\dfrac{\partial^2u}{\partial x^2}\) and \(\dfrac{\partial^2u}{\partial t^2}\):
2. Now substitute these results into \((6)\) and transpose so the variables are separated i.e. all functions of \(x\) are on the left-hand side, all funtions of \(t\) on the right-hand side:
3. Finally, write down the required ordinary differential equations:

Solution

1. \( u=X(x)T(t)\) gives \(\dfrac{\partial^2u}{\partial x^2}=X''T\) and \(\dfrac{\partial^2u}{\partial t^2}=XT''\)
2. We get \(X''T= \dfrac{1}{c^2}XT''\) and, transposing, \(\dfrac{X''}{X}=\dfrac{1}{c^2} \dfrac{T''}{T}\)
3. Equating both sides to the same constant \(K\) gives
\(\qquad \dfrac{X''}{X}=K\) or \(\dfrac{d^2X}{dx^2}-KX=0\)
and
\(\qquad \dfrac{1}{c^2} \dfrac{T''}{T}=K \qquad\) or \(\qquad \dfrac{d^2 T}{dt^2}-Kc^2T=0\)

The solution of the ODEs \((7)\) and \((8)\) has been obtained earlier, and will depend on the sign of \(K\).

Laplace’s equation

\(\qquad\displaystyle \dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2}=0\) | \((9)\)

Example

Separating the variables for Laplace’s equation follows similar lines to the previous Example. Obtain the ODEs satisfied by \(X(x)\) and \(Y(y)\).

Solution

Assuming \(u(x, y) = X(x)Y(y)\) leads to: \(\displaystyle \dfrac{\partial^2u}{\partial x^2}=X''Y\) \(\qquad\displaystyle \dfrac{\partial^2u}{\partial y^2}=XY''\) so

\(\qquad X''Y + XY''=0\) \(\qquad\) or \(\qquad\) \(\dfrac{X''}{X}=-\dfrac{Y''}{Y}\)
Equating each side to a constant \(K\)

\(\qquad \dfrac{X''}{X}=K\)\(\qquad\) or\(\qquad\) \(\displaystyle \frac{\mathrm{d^2}X}{\mathrm{d}x^2}-KX=0\) | \((10a)\)
\(\qquad \dfrac{Y''}{Y}=-K\)\(\qquad\) or\(\qquad\)\(\displaystyle \frac{\mathrm{d^2}Y}{\mathrm{d}y^2}+KY=0\) | \((10b)\)

(Note carefully the different signs in the two ODEs. Yet again the sign of the “separation constant” \(K\) will determine the solutions.)

Method of separation of variables - specific solutions

We shall now study some specific problems which can be fully solved by the separation of variables method.

Example

Solve the heat conduction equation

\(\qquad\displaystyle \dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{2}\dfrac{\partial u}{\partial t}\)

over    \(0<x<3,\ \ \ t>0\)    for the boundary conditions

\(\qquad\displaystyle u(0,t)=u(3,t)=0\)

and the initial condition

\(\qquad\displaystyle u(x,0)=5\sin 4\pi x.\)

Solution

Assuming \(u(x, t) = X(x)T(t)\) gives rise to the differential equations \((4)\) and \((5)\) with the parameter \(k = 2\):

\(\qquad \displaystyle \frac{\mathrm{d}T}{\mathrm{d}t}=2KT\) \(\qquad\) \(\displaystyle \frac{\mathrm{d^2}X}{\mathrm{d}x^2}=KX\)

The \(T\) equation has general solution

\(\qquad\displaystyle T=Ae^{2Kt}\)

which will increase exponentially with increasing \(t\) if \(K\) is positive and decrease with \(t\) if \(K\) is negative. In any physical problem the latter is the meaningful situation. To emphasise that \(K\) is being taken as negative we put

\(\qquad\displaystyle K=-\lambda^2\)

so

\(\qquad\displaystyle T=Ae^{-2\lambda^2t}.\)

The \(X\) equation then becomes

\(\qquad\displaystyle \dfrac{d^2X}{dx^2}=-\lambda^2X\)

which has solution

\(\qquad\displaystyle X(x)=B\cos \lambda x+C\sin \lambda x.\)

Hence

\(\qquad \displaystyle u(x,t)=X(x)T(t)=(D\cos\lambda x+E\sin \lambda x)e^{-2\lambda^2t}\) | \[(11)\]

where \(D=AB\) and \(E=AC\).

(You should always try to keep the number of arbitrary constants down to an absolute minimum by multiplying them together in this way.)

We now insert the initial and boundary conditions to obtain the constant \(D\) and \(E\) and also the separation constant \(\lambda\).

The initial condition \(u(0,t)=0\) gives

\(\qquad\displaystyle (D\cos 0+E\sin 0)e^{-2\lambda^2t}=0\)\(\qquad\)for all \(t\).

Since \(\sin 0=0\) and \(\cos 0=1\) this must imply that \(D=0\).

The other initial condition \(u(3,t)=0\) then gives

\(\qquad\displaystyle E\sin (3\lambda) e^{-2\lambda^2t}=0\qquad\)for all \(t.\)

We cannot deduce that the constant \(E\) has to be zero because then the solution \((11)\) would be the trivial solution \(u\equiv 0\). The only sensible deduction is that

\(\qquad\sin 3\lambda =0\) i.e. \(3\lambda=n\pi \qquad\) (where \(n\) is some integer).

Hence solutions of the form \((11)\) satisfying the 2 boundary conditions have the form

\(\qquad\displaystyle u(x,t)=E_n\sin\left(\dfrac{n\pi x}{3}\right)e^{-\frac{2n^2\pi^2t}{9}}\)

where we have written \(E_n\) for \(E\) to allow for the possibility of a different value for the constant for each different value of \(n\).
We obtain the value of \(n\) by using the initial condition \(u(x,0)=5\sin 4 \pi x\) and forcing this solution to agree with it. That is,

\(\qquad\displaystyle u(x,0)=E_n\sin\left(\dfrac{n\pi x}{3}\right)=5\sin 4\pi x\)

so we must choose \(n=12\) with \(E_{12}=5\).

Hence, finally,

\(\qquad\displaystyle u(x,t)=5\sin\left(\dfrac{12\pi x}{3}\right)e^{-\frac{2}{9}(12)^2\pi^2t}=5\sin (4\pi x)\ e^{-32\pi^2t}.\)

Example

Solve the \(1\)-dimensional wave equation \(\quad\displaystyle \dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{16}\dfrac{\partial^2u}{\partial t^2}\)   for \(0<x<2,\ \ \ t>0\)

The boundary conditions are

\(\qquad\displaystyle u(0,t)=u(2,t)=0\)

The initial conditions are

(i)  \(u(x,0)=6\sin\pi x-3\sin 4\pi x\) \(\qquad\qquad\)(ii)   \(\dfrac{\partial u}{\partial t}(x,0)=0\)

1. Firstly, either using \((7)\) and \((8)\) or by working from first principles assuming the product solution
\(\qquad u(x,t)=X(x)T(t),\)
write down the ODEs satisfied by \(X(x)\) and \(T(t)\):
2. Now decide on the appropriate sign for \(K\) and then write down the solution to these equations:
3. Now obtain the general solution \(u(x,t)\) by multiplying \(X(x)\) by \(T(t)\) and insert the two boundary conditions to obtain information about two of the constants:
4. Now insert the initial condition
\(\qquad\displaystyle \dfrac{\partial u}{\partial t}(x,0)=0\ \ \text{for all}\ x\qquad 0<x<2.\)
and deduce the value of \(F\):
5. Finally using the other the initial condition  \(u(x,0)=6\sin(\pi x)-3\sin (4\pi x)\) deduce the form of \(u(x,t)\):

Solution

1. \(\dfrac{X''}{X}=K \qquad\qquad \dfrac{T''}{16T}=K\)
2. Choosing \(K\) as negative (say \(K = −\lambda^2)\) will produce Sinusoidal solutions for \(X\) and \(T\) which are appropriate in the context of the wave equation where oscillatory solutions can be expected.
Then \(X'' = −\lambda^2 X\) gives
\(\qquad X = A\cos\lambda x + B\sin\lambda x\)
Similarly \(T'' = −16\lambda^2 T\) gives
\(\qquad T = C\cos4\lambda t + D\sin4\lambda t\)
3. \(u(x, t) = (A\cos\lambda x + B\sin\lambda x)(C\cos4\lambda t + D\sin4\lambda t)\)
\(u(0, t) = 0\) for all \(t\) gives
\(\qquad A(C\cos4\lambda t + D\sin4\lambda t) = 0\)
which implies that \(A = 0\).
\(u(2, t) = 0\) for all \(t\) gives
\(\qquad B\sin2\lambda (C\cos4\lambda t + D\sin4\lambda t) = 0\)
so, for a non-trivial solution,
\(\qquad \sin2\lambda = 0 \qquad \) i.e. \(\qquad \lambda = \dfrac{n\pi}{2}\) for some integer \(n\).
At this stage we write the solution as
\(\qquad u(x, t) = \sin\left(\dfrac{n\pi x}{2} \right) (E\cos2n\pi t + F\sin2n\pi t)\)
where we have multiplied constants and put \(E = BC\) and \(F = BD\).
4. Differentiating partially with respect to \(t\)
\(\qquad \dfrac{\partial u}{\partial t} = \sin\left(\dfrac{n\pi x}{2}\right)(−2n\pi E \sin 2n\pi t + 2n\pi F\cos 2n\pi t)\)
so at \(t = 0\)
\(\qquad \dfrac{\partial u}{\partial t}(x,0) = \sin\left(\dfrac{n\pi x}{2}\right)2n\pi F = 0\)
from which we must have that \(F = 0\).
5. At this stage the solution reads

\(\qquad u(x, t) = E\sin\left(\dfrac{n\pi x}{2}\right)\cos(2n\pi t)\) | \((12)\)

We now have to insert the last condition i.e. the initial condition

\(\qquad u(x, 0) = 6\sin \pi x − 3 \sin 4\pi x\) | \((13)\)

This seems strange because, putting \(t = 0\) in our solution \((12)\) suggests
\(\qquad u(x, 0) = E \sin\left(\dfrac{n\pi x}{2}\right)\)
At this point we seem to have incompatability because no single value of \(n\) will enable us to satisfy \((13)\). However, in the solution \((12)\), any positive integer value of \(n\) is acceptable and we can in fact, superpose solutions of the form (\(12)\) and still have a valid solution to the PDE Hence we first write, instead of \((12)\)

\(\qquad u(x, t) = \sum_{n=1}^\infty E_n\sin\left(\dfrac{n\pi x}{2}\right)\cos(2n\pi t)\) | \[(14)\]

from which

\(\qquad u(x, 0) = \sum_{n=1}^\infty E_n\sin\left(\dfrac{n\pi x}{2}\right)\) | \[(15)\]

(which looks very much like, and indeed is, a Fourier series.)

To make the solution \((15)\) fit the initial condition \((13)\) we do not require all the terms in the infinite Fourier series. We need only the terms with \(n = 2\) with coefficient \(E_2 = 6\) and the term for which \(n = 8\) with \(E_8 = −3\). All the other coefficients \(E_n\) have to be chosen as zero.
Using these results in \((14)\) we obtain the solution
\(\qquad u(x, t) = 6 \sin \pi x \cos 4\pi t − 3 \sin 4\pi x \cos 16\pi t\)

The above solution perhaps seems rather involved but there is a definite sequence of logical steps which can be readily applied to other similar problems.

Engineering Example

Heat conduction through a furnace wall

Introduction

Conduction is a mode of heat transfer through molecular collision inside a material without any motion of the material as a whole. If one end of a solid material is at a higher temperature, then heat will be transferred towards the colder end because of the relative movement of the particles. They will collide with the each other with a net transfer of energy.

Energy flows through heat conductive materials by a thermal process generally known as ’gradient heat transport’. Gradient heat transport depends on three quantities: the heat conductivity of the material, the cross-sectional area of the material which is available for heat transfer and the spatial gradient of temperature (driving force for the process). The larger the conductivity, the gradient, and the cross section, the faster the heat flows.

The temperature profile within a body depends upon the rate of heat transfer to the atmosphere, its capacity to store some of this heat, and its rate of thermal conduction to its boundaries (where the heat is transferred to the surrounding environment). Mathematically this is stated by the heat equation

\(\qquad\displaystyle \dfrac{\partial}{\partial x}\left( k \dfrac{\partial T}{\partial x} \right)= \rho c \dfrac {\partial T}{\partial x}\) | \((1)\)

The thermal diffusivity \(\alpha\) is related to the thermal conductivity \(k\), the specific heat \(c\), and the density of solid material \(\rho\), by

\(\qquad\displaystyle \alpha = \dfrac{k}{\rho c}\)

Problem in Words

The wall (thickness \(L\)) of a furnace, with inside temperature \(800^{\circ}\) C, is comprised of brick material [thermal conductivity \(= 0.02\) W m\(^{-1}\) K\(^{-1}\))]. Given that the wall thickness is \(12\) cm, the atmospheric temperature is \(0^{\circ}\) C, the density and heat capacity of the brick material are \(1.9\) \(^{-3}\) and \(6.0\ \text{J}\ \text{kg}^{-1}\ \text{K}^{-1}\) respectively, estimate the temperature profile within the brick wall after \(2\) hours.

Mathematical statement of problem

Solve the partial differential equation

\(\qquad\displaystyle \dfrac{\partial}{\partial x}\left( k \dfrac{\partial T}{\partial x} \right)= \rho c \dfrac {\partial T}{\partial t}\) | \((2)\)

subject to the initial condition

\(\qquad\displaystyle T(x,0) = 800 \sin \dfrac{\pi x}{2 L}\) | \((3)\)

and boundary conditions at the inner (\(x = L\)) and outer (\(x = 0\)) walls of

\(\qquad\displaystyle T = 0 \hspace{2.0cm} \text{at} \hspace{1.0cm} x = 0\) | \((4)\)

and

\(\qquad\displaystyle \frac{\partial T}{\partial x} = 0 \hspace{1.5cm} \text{at} \hspace{1.0cm} x = L\) | \((4b)\)

Find the temperature profile at \(T = 7200 \hspace{0.1cm} \text{seconds} = 2 \hspace{0.1cm} \text{hours}.\)

Mathematical analysis

Using separation of variables

\(\qquad\displaystyle T(x,t) = X(x) \times Y(t)\) | \((5)\)

so Equation \((2)\) becomes

\(\qquad\displaystyle \dfrac{Y'}{Y} = \alpha \dfrac{X''}{X} = K\) | \((6)\)

Using values of \(K\) which are zero or positive does not allow a solution which satisfies the initial and boundary conditions. Thus, \(K\) is assumed to be negative i.e. \(K = -\lambda^2\). Equation \((6)\) separates into the two ordinary differential equations

\(\qquad \dfrac{dY}{dt} = -\lambda^2 Y \)

\(\qquad \dfrac{d^2X}{dx^2} = -\lambda^2 \alpha X\)

with solutions

\(\qquad Y = C e^{-\lambda^2 t}\)

\(\qquad X = A^* \cos \dfrac{\lambda}{\sqrt{\alpha}}x + B^* \sin \dfrac{\lambda}{\sqrt{\alpha}}x\)

and

\(\qquad\displaystyle T = X \times Y \qquad = e^{-\lambda^2 t} \left\{ A \cos \dfrac{\lambda}{\sqrt{\alpha}}x + B \sin \dfrac{\lambda}{\sqrt{\alpha}}x \right\}\) | \[(8)\]

where \(A = A^* \times C\) and \(B = B^*\times C\).

Setting \(T=0\) where \(x=0\) (Equation \((4a)\)) gives \(A = 0\) i.e.

\(\qquad\displaystyle T = B e^{-\lambda^2 t} \sin \dfrac{\lambda}{\sqrt{\alpha}}x\) | \((9)\)

and hence

\(\qquad\displaystyle \dfrac{dT}{dx} = B \dfrac{\lambda}{\sqrt{\alpha}} e^{-\lambda^2 t} \cos \dfrac{\lambda}{\sqrt{\alpha}}x\) | \((10)\)

Setting \(\dfrac{dT}{dx}=0\) where \(x=L\) (and for all \(t\)), Equation \((4b)\) gives one of the conclusions,

\(\qquad B = 0\)

\(\qquad \lambda = 0\)

\(\qquad \cos \dfrac{\lambda}{\sqrt{\alpha}}L = 0\)

The first two possibilities (\(B=0\) and \(\lambda=0\)) can be discounted as they leave \(T = 0\) for all \(x\) and \(t\) and it is not possible to satisfy the initial condition \((3)\). Hence \(\cos \dfrac{\lambda}{\sqrt{\alpha}}L = 0\) so \(\dfrac{\lambda}{\sqrt{\alpha}}L = (n + \frac{1}{2})\pi\) and we deduce that

\(\qquad\displaystyle \lambda = \dfrac{\sqrt{\alpha}}{L}(n + \frac{1}{2})\pi\) | \((11)\)

and so the temperature \(T\) satisfies

\(\qquad\displaystyle T = B e^{-\dfrac{\alpha}{L^2}(n+\frac{1}{2})^2 t}\sin\left\{(n+\frac{1}{2})\dfrac{\pi x}{L}\right\}\) | \((12)\)

However, this must also satisfy Equation \((2)\) i.e.

\(\qquad\displaystyle 800 \sin \dfrac{\pi x}{2 L} = B \sin \left\{(n+\frac{1}{2})\dfrac{\pi x}{L}\right\}\) | \((13)\)

Equating the arguments of the sine terms

\(\qquad\displaystyle \dfrac{\pi x}{2 L} = (n+\frac{1}{2})\dfrac{\pi x}{L} \hspace{1.0cm} \text{so} \hspace{0.5cm} n = 0\)

Equating the coefficients of the sine terms

\(\qquad\displaystyle 800 = B\)

So the temperature profile is

\(\qquad\displaystyle T = 800 e^{-\frac{\alpha t}{2 L^2}} \sin \dfrac{\pi x}{2 L}\) | \((14)\)

where \(\alpha = \dfrac{k}{\rho c} = \dfrac{0.02}{1900 \times 6} = 1.764 \times 10^{-6}m^2s^{-1}\).

After two hours, \(t = 7200\) so \(-\dfrac{\alpha t}{2L^2} = -0.438\) so

\(\qquad\displaystyle T = 800 \times e^{-0.438} \sin \dfrac{\pi x}{2 L} = 516 \sin \dfrac{\pi x}{2 L}\) | \[(15)\]

so the inner wall of the furnace has cooled from \(800 ^{\circ}\)C to \(516 ^{\circ}\)C.

Interpretation

The boundary conditions \((2)\) and \((3)\) represent approximations to the true boundary conditions, approximations made to enable solution by separation of variables. More realistic conditions would be

\(\qquad -k \dfrac{\partial T(0,t)}{\partial x} = h_{outside} \left\{T_{\infty} - T(0,t) \right\}\)

\(\qquad -k \dfrac{\partial T(L,t)}{\partial x} = h_{inside} \left\{T(0,t) - T_{s} \right\}\)