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Wave equation with Neumann BC on semi-infinite domain

2

$begingroup$

This problem is from our recitation which I do not have solutions for, and I'm stuck on the very last part where I need to satisfy the $u_t(x,0)$ initial condition.

The problem is:

$$
left{
beginsplit
u_tt &= c^2u_xx, quadqquad x>0,t>0\
u(&0,t) = 0 qquadqquadqquad t>0\
u(&x,0) = sin(x) = f(x);quad x>0\
u_t&(x,0) = e^-x = g(x)quadquad x>0
endsplit
right.
$$
Since we have Neumann conditions we use an even extension. Let
$$
u(x,0) =
begincases
f(x) &textif x > 0 \
f(-x) &textif x < 0
endcases
$$
and
$$
u_t(x,0) =
begincases g(x) &textif x > 0 \
g(-x) &textif x < 0
endcases
$$
Then D'Alembert's formula gives:
$$
beginsplit
u(x,t) &= frac12big(f(x-ct)+f(x+ct)big) + frac12cint_x-ct^x+ct g(s),ds\
&= frac12big(f(x)+f(x)big) + frac12cint_x^x g(s),ds \
&= f(x)=sin(x)
endsplit
$$
thus the First initial condition is satisfied. For the second one we have
$$
u_t(x,t) = frac12big(-cf(x-ct)+cf(x+ct)big) + fracddtbigg [ frac12c int_x-ct^x+ct g(s),ds bigg ]
$$

My problem is I'm not sure what to do with the
$$
fracddtbigg [ frac12c int_x-ct^x+ct g(s),ds bigg ]
$$
term. Should I differentiate the integral and then plug in $t = 0$ for the whole expression? How do I get the term $e^-x$ from this?

pde boundary-value-problem wave-equation

share|cite|improve this question

edited 2 days ago

Daniele Tampieri

2,68721022

asked Apr 8 at 0:44

mathmath

392

$endgroup$

add a comment | 

2

$begingroup$

This problem is from our recitation which I do not have solutions for, and I'm stuck on the very last part where I need to satisfy the $u_t(x,0)$ initial condition.

The problem is:

$$
left{
beginsplit
u_tt &= c^2u_xx, quadqquad x>0,t>0\
u(&0,t) = 0 qquadqquadqquad t>0\
u(&x,0) = sin(x) = f(x);quad x>0\
u_t&(x,0) = e^-x = g(x)quadquad x>0
endsplit
right.
$$
Since we have Neumann conditions we use an even extension. Let
$$
u(x,0) =
begincases
f(x) &textif x > 0 \
f(-x) &textif x < 0
endcases
$$
and
$$
u_t(x,0) =
begincases g(x) &textif x > 0 \
g(-x) &textif x < 0
endcases
$$
Then D'Alembert's formula gives:
$$
beginsplit
u(x,t) &= frac12big(f(x-ct)+f(x+ct)big) + frac12cint_x-ct^x+ct g(s),ds\
&= frac12big(f(x)+f(x)big) + frac12cint_x^x g(s),ds \
&= f(x)=sin(x)
endsplit
$$
thus the First initial condition is satisfied. For the second one we have
$$
u_t(x,t) = frac12big(-cf(x-ct)+cf(x+ct)big) + fracddtbigg [ frac12c int_x-ct^x+ct g(s),ds bigg ]
$$

My problem is I'm not sure what to do with the
$$
fracddtbigg [ frac12c int_x-ct^x+ct g(s),ds bigg ]
$$
term. Should I differentiate the integral and then plug in $t = 0$ for the whole expression? How do I get the term $e^-x$ from this?

pde boundary-value-problem wave-equation

share|cite|improve this question

edited 2 days ago

Daniele Tampieri

2,68721022

asked Apr 8 at 0:44

mathmath

392

$endgroup$

add a comment | 

$begingroup$

This problem is from our recitation which I do not have solutions for, and I'm stuck on the very last part where I need to satisfy the $u_t(x,0)$ initial condition.

The problem is:

$$
left{
beginsplit
u_tt &= c^2u_xx, quadqquad x>0,t>0\
u(&0,t) = 0 qquadqquadqquad t>0\
u(&x,0) = sin(x) = f(x);quad x>0\
u_t&(x,0) = e^-x = g(x)quadquad x>0
endsplit
right.
$$
Since we have Neumann conditions we use an even extension. Let
$$
u(x,0) =
begincases
f(x) &textif x > 0 \
f(-x) &textif x < 0
endcases
$$
and
$$
u_t(x,0) =
begincases g(x) &textif x > 0 \
g(-x) &textif x < 0
endcases
$$
Then D'Alembert's formula gives:
$$
beginsplit
u(x,t) &= frac12big(f(x-ct)+f(x+ct)big) + frac12cint_x-ct^x+ct g(s),ds\
&= frac12big(f(x)+f(x)big) + frac12cint_x^x g(s),ds \
&= f(x)=sin(x)
endsplit
$$
thus the First initial condition is satisfied. For the second one we have
$$
u_t(x,t) = frac12big(-cf(x-ct)+cf(x+ct)big) + fracddtbigg [ frac12c int_x-ct^x+ct g(s),ds bigg ]
$$

My problem is I'm not sure what to do with the
$$
fracddtbigg [ frac12c int_x-ct^x+ct g(s),ds bigg ]
$$
term. Should I differentiate the integral and then plug in $t = 0$ for the whole expression? How do I get the term $e^-x$ from this?

pde boundary-value-problem wave-equation

share|cite|improve this question

edited 2 days ago

Daniele Tampieri

2,68721022

asked Apr 8 at 0:44

mathmath

392

$endgroup$

share|cite|improve this question

edited 2 days ago

Daniele Tampieri

2,68721022

asked Apr 8 at 0:44

mathmath

392

share|cite|improve this question

edited 2 days ago

Daniele Tampieri

2,68721022

asked Apr 8 at 0:44

mathmath

392

edited 2 days ago

Daniele Tampieri

2,68721022

edited 2 days ago

Daniele Tampieri

2,68721022

asked Apr 8 at 0:44

mathmath

392

asked Apr 8 at 0:44

mathmath

392

1 Answer
1

active

oldest

votes

1

$begingroup$

FTC says $$fracddtint_a(t)^b(t)f(s)ds = f(b(t))b'(t)-f(a(t))a'(t).$$ Substituting your functions, we get $$fracddtfrac12cint_x-ct^x+ctg(s)ds = frac12c(g(x)c+g(x)c) = g(x),$$
as desired.

share|cite|improve this answer

answered Apr 8 at 0:51

whpowell96whpowell96

53319

$endgroup$

add a comment | 

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1

$begingroup$

FTC says $$fracddtint_a(t)^b(t)f(s)ds = f(b(t))b'(t)-f(a(t))a'(t).$$ Substituting your functions, we get $$fracddtfrac12cint_x-ct^x+ctg(s)ds = frac12c(g(x)c+g(x)c) = g(x),$$
as desired.

share|cite|improve this answer

answered Apr 8 at 0:51

whpowell96whpowell96

53319

$endgroup$

add a comment | 

1

$begingroup$

FTC says $$fracddtint_a(t)^b(t)f(s)ds = f(b(t))b'(t)-f(a(t))a'(t).$$ Substituting your functions, we get $$fracddtfrac12cint_x-ct^x+ctg(s)ds = frac12c(g(x)c+g(x)c) = g(x),$$
as desired.

share|cite|improve this answer

answered Apr 8 at 0:51

whpowell96whpowell96

53319

$endgroup$

add a comment | 

$begingroup$

FTC says $$fracddtint_a(t)^b(t)f(s)ds = f(b(t))b'(t)-f(a(t))a'(t).$$ Substituting your functions, we get $$fracddtfrac12cint_x-ct^x+ctg(s)ds = frac12c(g(x)c+g(x)c) = g(x),$$
as desired.

share|cite|improve this answer

answered Apr 8 at 0:51

whpowell96whpowell96

53319

$endgroup$

share|cite|improve this answer

answered Apr 8 at 0:51

whpowell96whpowell96

53319

answered Apr 8 at 0:51

whpowell96whpowell96

53319

answered Apr 8 at 0:51

whpowell96whpowell96

53319