Calc1231W3T3 - The Wave Equation

Saturday, 4 August 2018

9:23 PM

Untitled picture.png Machine generated alternative text:
Given a function, it is relatively easy to find the partial derivatives. For example suppose u(x, t) 
Dt 
We say this function u satisfies, or is a solution to, the differential equation 
sin(x + ct) then 
Du 
Dt 
Du 
Similarly if t'(x, t) = sin@ — ct) then 
Dv 
Dt 
So this function v satisfies the differential equation 
Dv D-v 
Dt DC 

Untitled picture.png Machine generated alternative text:
Suppose u@, t) = sin@ + ct) where c is a constant. There are four possible derivatives of order two. These are 
32 u 
Dt2 
D2u 
DtDx 
D2u 
32 u 
Dx2 
So this function satisfies a very famous partial differential equation, which originated from the study of a vibrating string on a musical instrument, called the one dimensional wave equation. 
32 u c2 D2u 
Dt2 
Dx2 
The function v(x, t) = sin@ — ct) is also a solution to this equatiom The functions u(x, t), v(x, t) and u@,t) + v(x, t) are plotted in red, orange and purple in the picture below with 
c=l: 
Assuming c > O then the red function sin@ + ct) is a wave that moves to the lett 
O over time; while the orange function sin(x 
ct) is a wave that moves to the 
right 
O over time The sum ofthe left and right moving waves is the standing wave sin@ + ct) + sin@ — ct) in purple.
Untitled picture.png Machine generated alternative text:
Suppose u@, t) = sin@ + ct) where c is a constant. There are four possible derivatives of order two. These are 
32 u 
Dt2 
D2u 
DtDx 
D2u 
32 u 
Dx2 
So this function satisfies a very famous partial differential equation, which originated from the study of a vibrating string on a musical instrument, called the one dimensional wave equation. 
32 u c2 D2u 
Dt2 
Dx2 
The function v(x, t) = sin@ — ct) is also a solution to this equatiom The functions u(x, t), v(x, t) and u@,t) + v(x, t) are plotted in red, orange and purple in the picture below with 
c=l: 
Assuming c > O then the red function sin@ + ct) is a wave that moves to the lett 
O over time; while the orange function sin(x 
ct) is a wave that moves to the 
right 
O over time The sum ofthe left and right moving waves is the standing wave sin@ + ct) + sin@ — ct) in purple. 

Untitled picture.png Machine generated alternative text:
The function u@, t) 
ei(x+ct) 
is another solution to the wave equation 
D2u 
Dt2 
c2 D2u 
Let's investiage the case when c = 2 + 4i _ Then 
• The real part of u is 
• The imaginary part of u is 
o fit 
So the function oscillates in the complex plane; but these oscillations are dampened by the exponential term. As time goes on 
lim u(x,t) 




















Untitled picture.png Machine generated alternative text:
D2u 
Dt2 
D2u 
-cos(x+ct) 
Dx2 
02 v 
Dt2 
02 v 
-cos(x-ct) 
Dx2 
These functions are both solutions to the one-dimensional wave equation because 
32 u 
Dt2 
and 
D2t' 
Dt2 
Consider u(x, t) = cos(x + ct) and v(x, t) 
—o 
—1 
cos@ 
ct.). Then 
c2 D2u 
DC 2 
c2 D2v 
Dx2 
The functions u@,t),t'(x, t) and t) + t'(x, t) are plotted in red, orange and purple below with c 
311/2 
Ink Drawings
Ink Drawings
Ink Drawings











Untitled picture.png Machine generated alternative text:
D2u 
Dt2 
D2u 
-cos(x+ct) 
Dx2 
02 v 
Dt2 
02 v 
-cos(x-ct) 
Dx2 
These functions are both solutions to the one-dimensional wave equation because 
32 u 
Dt2 
and 
D2t' 
Dt2 
Consider u(x, t) = cos(x + ct) and v(x, t) 
—o 
—1 
cos@ 
ct.). Then 
c2 D2u 
DC 2 
c2 D2v 
Dx2 
The functions u@,t),t'(x, t) and t) + t'(x, t) are plotted in red, orange and purple below with c 
311/2 

Untitled picture.png Machine generated alternative text:
We have seen that the one-dimensional wave equation has lots of possible solutions In fact there is a whole vector subspace of solutions (you may like to prove this). We can narrow down the 
range of possible solutions by adding extra conditions that the function must satisfy. These are usually called initial conditions. 
For example, suppose that u@, t) satisfies the one-dimensional wave equation: 
D 2 u c2 D2u 
Dt2 
Dx2 
In addition, suppose u@, t) satisfies the initial conditions: 
Du 
O) = and 
In 1746; Jean-Baptiste le Rond d'Alembert discovered that there is only one possible solution which satisfies these conditions: 
u(x, t) — 
1 
1 
2 
g(s)ds. 
x—ct 
Jean-Baptise le Rond d'Alembert (1717 -1783) 
For example, if we have the wave equation with c 
and initial conditions f@) = sin(x) and g@) — cos@) then 
x —ct 
So by d'Alembert's formula, the solution is 
g(s)ds 
02 u 
Dt2 
cos(s)ds 
02 u 
Dx2 
sin(x+t) - sin(x-t)
Untitled picture.png Machine generated alternative text:
We have seen that the one-dimensional wave equation has lots of possible solutions In fact there is a whole vector subspace of solutions (you may like to prove this). We can narrow down the 
range of possible solutions by adding extra conditions that the function must satisfy. These are usually called initial conditions. 
For example, suppose that u@, t) satisfies the one-dimensional wave equation: 
D 2 u c2 D2u 
Dt2 
Dx2 
In addition, suppose u@, t) satisfies the initial conditions: 
Du 
O) = and 
In 1746; Jean-Baptiste le Rond d'Alembert discovered that there is only one possible solution which satisfies these conditions: 
u(x, t) — 
1 
1 
2 
g(s)ds. 
x—ct 
Jean-Baptise le Rond d'Alembert (1717 -1783) 
For example, if we have the wave equation with c 
and initial conditions f@) = sin(x) and g@) — cos@) then 
x —ct 
So by d'Alembert's formula, the solution is 
g(s)ds 
02 u 
Dt2 
cos(s)ds 
02 u 
Dx2 
sin(x+t) - sin(x-t)

 

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