Page 323 - Schaum's Outline of Differential Equations

306 AN INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS [CHAP. 31

Algebraic simplification shows that

6
7 6
because both sides reduce to 720x t — IIQQjc't . Hence, our solution is verified.
31.5. Let u = u(x, y). By integration, find the general solution to u x = 0.
The solution is arrived at by "partial integration", much like the technique employed when solving "exact"
equations (See Chapter 5). Hence, u(x, y) =f(y), where f(y) is any differentiable function of y. We can write this
symbolically as

We note that a "+ C" is not needed because it is "absorbed" into/(;y); that is, f(y) is the most general "constant"
with respect to x.

31.6. Let u = u(x, y, z). By integration, find the general solution to u x = 0.
Here, we see by inspection that our solution can be written as/(;y, z).

31.7. Let u = u(x, y). By integration, find the general solution to u x = 2x.
2
Since, one antiderivative of 2x (with respect to x) is x , the general solution is J 2x dx = x + f(y)', where/(y)
is any differentiable function of y.

31.8. Let u = u(x, y). By integration, find the general solution to u x = 2x, u(0, y) = In y.
2
2
By Problem 31.7, the solution to the PDE is u(x, y) = x +f(y). Letting x = 0 implies u(0, y) = O +f(y) = In y.
Therefore/(;y) = In y, so our solution is u(x, y)=x 2 + In y.
31.9. Let u = u(x, y). By integration, find the general solution to u y = 2x.
Noting that an antiderivative of 2x with respect to y is 2xy, the general solution is given by 2xy + g(x), where
g(x) is any differentiable function of x.

31.10. Let u = u(x, y). By integration, find the general solution to u^ = 2x.
Integrating first with respect to y, we have u x = 2xy +f(x), where f(x) is any differentiable function of x. We
2
now integrate u x with respect to x, we arrive at u(x, y) = x y + g(x) + h(y), where g(x) is an antiderivative of f(x), and
where h(y) is any differentiable function of y.
We note that if the PDE was written as u yx = 2x, our results would be the same.

31.11. Let u(x, t) represent the temperature of a very thin rod of length n, which is placed on the interval
{xlO <x< TT), at position x and time t. The PDE which governs the heat distribution is given by

where u, x, t and k are given in proper units. We further assume that both ends are insulated; that is,
u(0, t) = u(n, t) = 0 are impose "boundary condition" for t > 0. Given an initial temperature distribution
of u(x, 0) = 2 sin 4x - 11 sin 7x, for 0 < x < n, use the technique of separation of variables to find
a (non-trivial) solution, u(x, t).