Partial differential equation 1

[mona123]

hi can someone help me please to do b) of this problem i didn't know how to skech the region:
Problem . Consider transport equation ut + 3_ux = 0, −∞ < x < ∞, t > 0.
a) Find solution to this problem satisfying initial condition u(x, 0) = e^{−x^2}
for −∞ < x < ∞.
b) Sketch a region on (x, t) plane where u(x, t) > 1/e.
thanks

 

[Ishuda]

hi can someone help me please to do b) of this problem i didn't know how to skech the region:
Problem . Consider transport equation ut + 3_ux = 0, −∞ < x < ∞, t > 0.
a) Find solution to this problem satisfying initial condition u(x, 0) = e^{−x^2}
for −∞ < x < ∞.
b) Sketch a region on (x, t) plane where u(x, t) > 1/e.
thanks

Have you done part (a). If so, what was your answer? If not, try the method of characteristics for a solution. Possibly this link would help
http://en.wikipedia.org/wiki/Method_of_characteristics

 

[mona123]

equation differentielle 1

hi Ishuda the solution is u(x,t)=e^{−(x-3t)^2 and i used to find to find it} the method of characteristics
but i didn't know how to do b)
can you help me please.thanks

 

[Ishuda]

hi Ishuda the solution is u(x,t)=e^{−(x-3t)^2 and i used to find to find it} the method of characteristics
but i didn't know how to do b)
can you help me please.thanks

(b) Sketch a region on (x, t) plane where u(x, t) > 1/e.
So we want
u(t,x) = e^{-(x-3t)^2} > 1/ e
or
e = e¹ > e^{(x-3t)^2}
Since the exponential function is monotonic increasing, this means we want
1 > (x-3t)²
This happens if
1 > |x - 3 t|
which is equivalent to either
1 > x - 3t
or
1 > -x + 3t

The general way to solve these type of equations is to look at the place where they are equal since that is the dividing line which separates the two sides of the inequality. So sketch two lines:
x = 3t - 1
and
x = 3t + 1.
That will divide the plane into three regions. For which region can the inequality be true.



Hint: Dos the point (0,0) satisfy the inequality?