Problem Set 9   -   Solutions

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Therefore

YES,   f (x, y ) is harmonic on all of 2.

Release from rest     Þ     q (x) º 0.
The d’Alembert solution

is therefore

A = 1 ,       B = –6 ,       C = 8     Þ     D = 36 - 4(8) = 4 > 0
Therefore

the PDE is hyperbolic.

This is a homogeneous second order linear PDE with constant coefficients, for which the d’Alembert method may be used to find a general solution.

Applying the additional information:



The complete solution is therefore

A = 4 ,       B = 12 ,       C = 9     Þ     D = 144 - 4(4)(9) = 0
Therefore

the PDE is parabolic.

where   h(x, y)   is any non-trivial linear function of x and y that is not a multiple of   y + lx.
A simple choice, such as   h(x, y) = x , will do.
Therefore one form of the complementary function is
     
The right side function is a constant (78), while the left hand side involves only second partial derivatives.   The particular solution must therefore be some function of x and y of second order only, so we try

The second partial derivatives of this trial particular solution are

Substituting into the PDE we obtain

There is only one condition on the three constants.
We therefore have a free choice for two of the constants.
Choose   b = 0   and   c = a     26a = 78     a = 3
The particular solution is therefore  
and the general solution is

In order to find the complete solution, we employ the additional information:



The complete solution to the PDE is therefore

t   is playing the role of   y.
A = 4 ,       B = 0 ,       C = 0     Þ     D = 0 - 4(4)(0) = 0
This is also the PDE for heat diffusion.   Therefore

the PDE is parabolic.

From page 4.28 of the lecture notes (Example 4.5.1), the complete solution of the heat equation (with the boundary conditions of constant temperatures T₁ and T₂ at the ends x = 0 and x = L respectively and an initial temperature profile on [0, L] of u(x, 0) = f (x) ) is

In this case   k = 4, L = 100, T₁ = 0, T₂ = 100   and   f (x) = 2x – (x/10)².

The coefficients in the Fourier sine series become

Integration by parts:

Also   2/L = 1/50.   After some simplification of constants and substitution of (2k – 1) for the index of summation n, the complete solution becomes

The Fourier series for x converges rapidly.
The steady state solution is

Two snapshots of   u (x, t):
   

[The Maple file used to generate these diagrams is available here.]

A = 4 ,       B = 4 ,       C = 1     Þ     D = 16 - 4(4) = 0
Therefore

the PDE is parabolic.

Assume a complementary function of the form
       
where   h(x, y)   is any non-trivial linear function of x and y that is not a multiple of   y + lx.
A simple choice, such as   h(x, y) = x , will do.
Therefore one form of the general solution is

The functional forms of   f   and   g   remain arbitrary in the absence of any further information (such as boundary conditions).   Several alternatives are possible, such as
.

A = 1 ,       B = 0 ,       C = 1     Þ     D = 0 - 4(1) = -4 < 0
Therefore

the PDE is elliptic.

Assuming a complementary function of the form   u_C = f (ax+by) + g(ax+by),

Þ     u_C(x, y)   =   f (y - jx) + g(y+jx)

Many possibilities exist for the particular integral.   Among them is:
Try     u_P(x, y)   =   c (x³ + y³)



Therefore the P.I. is   u_P(x, y)   =   x³ + y³
and the general solution is

u(x, y)   =   f (y - jx)   +   g(y+jx)   +   x3 + y3

The functional forms of   f   and   g   remain arbitrary in the absence of any further information (such as boundary conditions).