2 Method of separation of variables - general approach

In Section 25.2 we showed that

1. $u\left(x,y\right)=sinxcoshy$

   is a solution of the two-dimensional Laplace equation

2. $u\left(x,t\right)=e^{-2{\pi }^{2}t}sin\pi x$

   is a solution of the one-dimensional heat conduction equation

3. $u\left(x,t\right)=u_{0}sin\left(\frac{\pi x}{\ell }\right)cos\left(\frac{\pi ct}{\ell }\right)$

   is a solution of the one-dimensional wave equation.

All three solutions here have a specific form: in 1. $u\left(x,y\right)$ is a product of a function of $x$ alone, $sinx$ , and a function of $y$ alone, $coshy$ . Similarly, in both 2. and 3., $u\left(x,t\right)$ 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.

$u\left(x,t\right)=X\left(x\right)T\left(t\right)\left(\text{or}u\left(x,y\right)=X\left(x\right)Y\left(y\right)\right)$

where $X\left(x\right)$ is a function of $x$ only, $T\left(t\right)$ is a function of $t$ only and $Y\left(y\right)$ 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

$u\left(x,y\right)=x^2-y^2$

(which is not of the form $u\left(x,y\right)=X\left(x\right)Y\left(y\right)$ ) 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.

2.1 Heat conduction equation

$\frac{{\partial }^{2}u}{\partial x^2}=\frac{1}{k}\frac{\partial u}{\partial t}k>0$ (1)

Assuming that

$$ (2)

then

Substituting into the original PDE (1)

$X^{″}T=\frac{1}{k}X{T}^{\prime }$

which can be re-arranged as

$\frac{X^{″}}{X}=\frac{1}{k}\frac{T^{\prime }}{T}$ (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 $T\left(t\right)$ is a function such that

$\frac{1}{k}\frac{T^{\prime }}{T}=K$ (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.

$\frac{X^{″}}{X}=K$ (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

$\frac{{\partial }^{2}u}{\partial x^2}=\frac{1}{c^2}\frac{{\partial }^{2}u}{\partial t^2}$ (6)

Task!

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

$u\left(x,t\right)=X\left(x\right)T\left(t\right)$

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

First obtain $\frac{{\partial }^{2}u}{\partial x^2}$ and $\frac{{\partial }^{2}u}{\partial t^2}$ :

Answer

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:

Answer

Finally, write down the required ordinary differential equations:

Answer

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

Laplace’s equation

$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$ (9)

Task!

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

Answer