1 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 HELM booklet  32 and HELM booklet  33).

Task!

Solve the ODE

d 2 y d x 2 = x 2 + 2

given that y = 1 when x = 0 and d y d x = 2 when x = 0.

First find d y d x by integrating once, not forgetting the arbitrary constant of integration:

d y d x = x 3 3 + 2 x + A

Now find y by integrating again, not forgetting to include another arbitrary constant:

y = x 4 12 + x 2 + A x + B

Now find A and B by inserting the two given initial conditions and so find the solution:

y ( 0 ) = 1 gives B = 1 y ′ ( 0 ) = 2 gives A = 2

so the required solution is

y = x 4 12 + x 2 + 2 x + 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

∂ 2 u ∂ x 2 = 2 x e t

where u is a function of x and t .

Integrating with respect to x gives us

∂ u ∂ x = x 2 e 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:

u = 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

u ( 0 , t ) = t , ∂ u ∂ x ( 0 , t ) = e t

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

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

So the final solution is u = x 3 3 e t + x e t + t .

Task!

Solve the PDE

∂ 2 u ∂ x ∂ y = sin x cos y

subject to the conditions

∂ u ∂ x = 2 x at y = π 2 , u = 2 sin y at x = π .

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

Recall that ∂ 2 u ∂ x ∂ y = ∂ ∂ y ∂ u ∂ x

Hence integration with respect to y gives ∂ u ∂ x = sin x sin y + f ( x )

Since one of the given conditions is on ∂ u ∂ x , impose this condition to determine the arbitrary function f ( x ) :

At y = π ∕ 2 the condition gives sin x sin π ∕ 2 + f ( x ) = 2 x i.e. f ( x ) = 2 x − sin x

So ∂ u ∂ x = sin x sin y + 2 x − sin x

Now integrate again to determine u :

Integrating now with respect to x gives u = − cos x sin y + x 2 + cos x + g ( y )

Next, obtain the arbitrary function g ( y ) :

The condition u ( π , y ) = 2 sin y gives − cos π sin y + π 2 + cos π + g ( y ) = 2 sin y

∴ sin y + π 2 − 1 + g ( y ) = 2 sin y

∴ g ( y ) = sin y + 1 − π 2

Now write down the final answer for u ( x , y ) :

u ( x , y ) = x 2 + cos x ( 1 − sin y ) + sin y + 1 − π 2