Fourier Series and Partial Differential Equations Lecture Notes
Fourier Series and Partial Differential Equations Lecture Notes
Fourier Series and Partial Differential Equations Lecture Notes
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Chapter 5. Laplace’s equation in the plane 57<br />
Definition A problem is said to be well-posed (well-set) if the following three conditions<br />
are satisfied:<br />
1. EXISTENCE—there is a solution;<br />
2. UNIQUENESS—there is no more than one solution;<br />
3. CONTINUOUS DEPENDENCE—the solution depends continuously on the data.<br />
Example 5.8 As an illustration consider the IVP<br />
∂2y ∂t2 = c2∂2 y<br />
∂x2 −∞ < x < ∞, t > 0, (5.90)<br />
y(x,0) = f(x),<br />
∂y<br />
(x,0) = g(x), −∞ < x < ∞,<br />
∂t<br />
(5.91)<br />
where f <strong>and</strong> g are the initial data. We know that there is exactly one solution, given by<br />
D’Alembert’s formula:<br />
y(x,t) = 1<br />
x+ct 1<br />
[f(x−ct)+f(x+ct)]+ g(s)ds.<br />
2 2c<br />
(5.92)<br />
Thus 1. <strong>and</strong> 2. hold.<br />
Suppose we are interested in making predictions in the time interval 0 < t < T for<br />
time T. Consider a similar problem<br />
∂2y ∂t2 = c2∂2 y<br />
∂x2 −∞ < x < ∞, t > 0, (5.93)<br />
y(x,0) = F(x),<br />
∂y<br />
(x,0) = G(x), −∞ < x < ∞,<br />
∂t<br />
(5.94)<br />
whereF <strong>and</strong>Garedifferent initial data. Again, weknow that thereis exactly onesolution:<br />
Y(x,t) = 1<br />
x+ct 1<br />
[F(x−ct)+F(x+ct)]+ G(s)ds,<br />
2 2c<br />
(5.95)<br />
<strong>and</strong><br />
Y(x,t)−y(x,t) = 1<br />
[(F(x−ct)−f(x−ct))+(F(x+ct)−f(x+ct))] (5.96)<br />
2<br />
+ 1<br />
x+ct<br />
[G(s)−g(s)] ds. (5.97)<br />
2c<br />
x−ct<br />
Now let ǫ > 0 be arbitrary <strong>and</strong> suppose that<br />
Then<br />
x−ct<br />
x−ct<br />
|F(x)−f(x)| < δ <strong>and</strong> |G(x)−g(x)| < δ for −∞ < x < ∞. (5.98)<br />
|Y(x,t)−y(x,t)| ≤ 1<br />
2 |F(x−ct)−f(x−ct)|<br />
+ 1<br />
2 |F(x+ct)−f(x+ct)|<br />
+ 1<br />
2c<br />
x+ct<br />
|G(s)−g(s)|ds, (5.99)<br />
x−ct<br />
x+ct<br />
< 1 1 1<br />
δ + δ + δds,<br />
2 2 2c x−ct<br />
(5.100)<br />
= 1 1 1<br />
δ + δ + ·2ctδ,<br />
2 2 2c<br />
(5.101)<br />
= (1+t)δ < (1+T)δ. (5.102)