Clarification of Answer by
mathtalk-ga
on
04 Dec 2002 07:12 PST
Hi, awl-ga:
I'm not sure if you are asking how to verify in (b) that U_t = 0 and
U_x = 1, or how I discovered that this is so. These are often quite
different processes in post-secondary mathematics.
Given that U(x,t) = x, it is easy to see that the partial derivatives
are as stated. I mentioned these facts in _verifying_ that:
(b) U_t + U U_x = x = U and U(x,0) = x
How did I discover that this was the solution (and thus that the
partial derivatives were as stated)? The usual phrase is "by
inspection", meaning (in part) a lucky guess.
It is not too difficult to recount why I made this guess. We know
that U(x,0) = x, and I naturally wondered what would be the
consequence if one simply extended U(x,t) = x throughout time.
Fortunately it turns out to be the solution of the PDE.
I'm not sure what your wish to be clarified about the solution U(x,t)
= t to:
(a') U_t + U U_x = 1 and U(x,0) = 0
The way I verified the solution is by taking partial derivatives:
U_t = 1 and U_x = 0
and substituting them into the PDE. The way I discovered this
solution was that after finding the solution to (b) I began to wonder
if the solution to (a) should not be equally simple. I therefore
tried the obvious candidate, i.e. t rather than x. I found it
satisfies the PDE of (a) but not the initial condition U(x,0) = x.
However it would satisfy U(x,0) = 0, which made me suspect as
recounted earlier, that the problem had simply been misstated
someplace along the way. After all, the (a) part of an exercise is
typically easier than the (b) part.
In any case the solution to your original problem:
(a) U_t + U U_x = 1 and U(x,0) = x
is:
U(x,t) = x/(t+1) + (1/2)[ t + 1 - 1/(t+1) ]
Note that although this definition involves division by (t+1), it is
valid for all time t >= 0.
To verify that this is a solution of the PDE, one proceeds to compute
the partial derivatives:
U_t = -x/(t+1)^2 + (1/2)[ 1 + 1/(t+1)^2 ]
U_x = 1/(t+1)
and plug these into the equation:
U_t + U U_x
= -x/(t+1)^2 + (1/2)[ 1 + 1/(t+1)^2 ]
+ (x/(t+1) + (1/2)[ t + 1 - 1/(t+1) ])(1/t+1)
= -x/(t+1)^2 + x/(t+1)^2
+ (1/2)[ 1 + 1/(t+1)^2 ]
+ (1/2)[ 1 - 1/(t+1)^2 ]
= (1/2)[ 1 + 1 ]
= 1
Finally we need to verify the initial condition:
U(x,0) = x/(0+1) + (1/2)[ 0 + 1 - 1/(0+1) ]
= x/1 + (1/2)[ 1 - 1 ]
= x
For this reconfirmed problem (a) the solution was hardly found "by
inspection". Rather I used the "method of characteristics" which will
be explained more fully in my response to your other question. Here
is the outline of the solution however.
Mathematicians have observed, in connection with Burger's equation,
that if one gives x = x(t) as a function of time t, then the "total"
derivative of U with respect to time t (as opposed to the partial
derivative U_t) is given by:
dU(x(t),t)/dt = U_t + U_x (dx/dt)
Comparing this expression with the left hand side of the differential
equation, we see that they will coincide provided:
dx/dt = U
Therefore along a path (x(t),t) parameterized by time, we have:
dU/dt = U_t + U U_x = 1
so that along such a "characteristic curve" the function U(x(t),t)
must take the form:
U(x(t),t) = t + b
in order for the total derivative with respect to time to be 1, as
above.
From this and the above prescription that dx/dt is U, we find:
x(t) = t^2/2 + bt + c
for some constants b and c, which may be different depending on the
particular characteristic curve.
The idea is to tie these curves back to the initial condition at t=0
and thus determine the appropriate values of b and c.
Note that at time t=0:
x(0) = c and U(x,0) = dx/dt = b
Since the initial condition is for U(x,0) = x, we see that both:
b = x(0) = c
Therefore upon using these values for b,c:
x(t) = t^2/2 + x(0)t + x(0)
is the equation for the characteristic curve that meets the x-axis at
(x(0),0). Given that:
U(x(t),t) = t + b = t + x(0)
along such a curve, it remains only to determine for a given point
(x,t) which initial point (x(0),0) will give rise to the
characteristic curve passing through that point:
x = t^2/2 + (t+1) x(0)
x(0) = [ x - t^2/2 ]/(t+1)
Note that since we are concerned with nonnegative time t, the divisor
t+1 here causes no difficulty. Substituting this value for x(0) into
the solution:
U(x,t) = t + x(0)
= t + [ x - t^2/2 ]/(t+1)
= x/(t+1) + [ t(t+1) - t^2/2 ]/(t+1)
= x/(t+1) + [ t^2/2 + t ]/(t+1)
= x/(t+1) + (1/2)[ t + 1 - 1/(t+1) ]
Recall that we have already verified above that this function indeed
satisfies both the PDE and the initial condition prescribed by part
(a) of your original question.
regards, mathtalk-ga
,
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