Mixing
Partial Differential Equations Question: Find an explicit expression for the solution of the IVP [closed]
$begingroup$
Find an explicit expression for the solution of the IVP
$$
begin{cases}
u_{t}(x,t)+u_{x}(x,t)+u(x,t)=e^{t+2x}\
\
u(0,x)=0,
end{cases}
$$
by using the method of characteristics
pde characteristics
edited Jan 20 at 17:06
Daniele Tampieri
2,3372922
asked Jan 20 at 16:34
JennyJenny
624
$endgroup$
closed as off-topic by max_zorn, Riccardo.Alestra, mrtaurho, metamorphy, José Carlos Santos Jan 21 at 12:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – max_zorn, Riccardo.Alestra, mrtaurho, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Find an explicit expression for the solution of the IVP
$$
begin{cases}
u_{t}(x,t)+u_{x}(x,t)+u(x,t)=e^{t+2x}\
\
u(0,x)=0,
end{cases}
$$
by using the method of characteristics
pde characteristics
edited Jan 20 at 17:06
Daniele Tampieri
2,3372922
asked Jan 20 at 16:34
JennyJenny
624
$endgroup$
closed as off-topic by max_zorn, Riccardo.Alestra, mrtaurho, metamorphy, José Carlos Santos Jan 21 at 12:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – max_zorn, Riccardo.Alestra, mrtaurho, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Hi and welcome to the Math.SE. Questions sounding like "do this for me, please" are poorly received from other members. I you need help, help other users help you for example by explaining what you tried and where did you get stuck.
$endgroup$
– Daniele Tampieri
Jan 20 at 17:08
add a comment |
$begingroup$
Find an explicit expression for the solution of the IVP
$$
begin{cases}
u_{t}(x,t)+u_{x}(x,t)+u(x,t)=e^{t+2x}\
\
u(0,x)=0,
end{cases}
$$
by using the method of characteristics
pde characteristics
edited Jan 20 at 17:06
Daniele Tampieri
2,3372922
asked Jan 20 at 16:34
JennyJenny
624
$endgroup$
Find an explicit expression for the solution of the IVP
$$
begin{cases}
u_{t}(x,t)+u_{x}(x,t)+u(x,t)=e^{t+2x}\
\
u(0,x)=0,
end{cases}
$$
by using the method of characteristics
pde characteristics
pde characteristics
edited Jan 20 at 17:06
Daniele Tampieri
2,3372922
asked Jan 20 at 16:34
JennyJenny
624
edited Jan 20 at 17:06
Daniele Tampieri
2,3372922
asked Jan 20 at 16:34
JennyJenny
624
edited Jan 20 at 17:06
Daniele Tampieri
2,3372922
edited Jan 20 at 17:06
Daniele Tampieri
2,3372922
edited Jan 20 at 17:06
Daniele Tampieri
2,3372922
2,3372922
asked Jan 20 at 16:34
JennyJenny
624
asked Jan 20 at 16:34
JennyJenny
624
asked Jan 20 at 16:34
JennyJenny
624
624
closed as off-topic by max_zorn, Riccardo.Alestra, mrtaurho, metamorphy, José Carlos Santos Jan 21 at 12:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – max_zorn, Riccardo.Alestra, mrtaurho, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by max_zorn, Riccardo.Alestra, mrtaurho, metamorphy, José Carlos Santos Jan 21 at 12:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – max_zorn, Riccardo.Alestra, mrtaurho, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Hi and welcome to the Math.SE. Questions sounding like "do this for me, please" are poorly received from other members. I you need help, help other users help you for example by explaining what you tried and where did you get stuck.
$endgroup$
– Daniele Tampieri
Jan 20 at 17:08
add a comment |
$begingroup$
Hi and welcome to the Math.SE. Questions sounding like "do this for me, please" are poorly received from other members. I you need help, help other users help you for example by explaining what you tried and where did you get stuck.
$endgroup$
– Daniele Tampieri
Jan 20 at 17:08
$begingroup$
Hi and welcome to the Math.SE. Questions sounding like "do this for me, please" are poorly received from other members. I you need help, help other users help you for example by explaining what you tried and where did you get stuck.
$endgroup$
– Daniele Tampieri
Jan 20 at 17:08
$begingroup$
Hi and welcome to the Math.SE. Questions sounding like "do this for me, please" are poorly received from other members. I you need help, help other users help you for example by explaining what you tried and where did you get stuck.
$endgroup$
– Daniele Tampieri
Jan 20 at 17:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is a linear PDE so
$$
u = u^h + u^p
$$
with
$$
u_t^h+u_x^h=-u^h\
u_t^p+u_x^p+u^p=e^{t+2x}
$$
Using the Characteristics method with $u^h$ we have
$$
begin{cases}
frac{dt}{dtau} = 1 & t(0) = 0 & t(tau) = tau\
frac{dx}{dtau} = 1 & x(0) = s & x(tau) = s+tau\
frac{du^h}{dtau} = -u^h & u^h(0) = phi(s) & -ln u^h(tau) = phi(s)+tau
end{cases}
$$
or
$$
u^h(t,x) = e^{-t}psi(x-t)
$$
regarding the particular we have that $u^p(t,x) = frac 14e^{t+2x}$ verifies the particular equation then
$$
u(t,x) = e^{-t}psi(x-t)+frac 14e^{t+2x}
$$
and for $t=0$ we have
$$
psi(x) + frac 14e^{2x}=0
$$
then $psi(x) = -frac 14 e^{2x}$ and finally
$$
u(t,x) = frac 14left(-e^{-t}e^{2(x-t)}+e^{t+2x}right)
$$
answered Jan 20 at 17:19
CesareoCesareo
9,2013517
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add a comment |
$begingroup$
Yes, by using the method of characteristics, one can find the explicit expression for the solution :
$$u(t,x)=frac14left(e^{t+2x}-e^{-3t+2x} right)$$
The question is concise. The answer as well.
answered Jan 20 at 18:16
JJacquelinJJacquelin
44.3k21854
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a linear PDE so
$$
u = u^h + u^p
$$
with
$$
u_t^h+u_x^h=-u^h\
u_t^p+u_x^p+u^p=e^{t+2x}
$$
Using the Characteristics method with $u^h$ we have
$$
begin{cases}
frac{dt}{dtau} = 1 & t(0) = 0 & t(tau) = tau\
frac{dx}{dtau} = 1 & x(0) = s & x(tau) = s+tau\
frac{du^h}{dtau} = -u^h & u^h(0) = phi(s) & -ln u^h(tau) = phi(s)+tau
end{cases}
$$
or
$$
u^h(t,x) = e^{-t}psi(x-t)
$$
regarding the particular we have that $u^p(t,x) = frac 14e^{t+2x}$ verifies the particular equation then
$$
u(t,x) = e^{-t}psi(x-t)+frac 14e^{t+2x}
$$
and for $t=0$ we have
$$
psi(x) + frac 14e^{2x}=0
$$
then $psi(x) = -frac 14 e^{2x}$ and finally
$$
u(t,x) = frac 14left(-e^{-t}e^{2(x-t)}+e^{t+2x}right)
$$
answered Jan 20 at 17:19
CesareoCesareo
9,2013517
$endgroup$
add a comment |
$begingroup$
This is a linear PDE so
$$
u = u^h + u^p
$$
with
$$
u_t^h+u_x^h=-u^h\
u_t^p+u_x^p+u^p=e^{t+2x}
$$
Using the Characteristics method with $u^h$ we have
$$
begin{cases}
frac{dt}{dtau} = 1 & t(0) = 0 & t(tau) = tau\
frac{dx}{dtau} = 1 & x(0) = s & x(tau) = s+tau\
frac{du^h}{dtau} = -u^h & u^h(0) = phi(s) & -ln u^h(tau) = phi(s)+tau
end{cases}
$$
or
$$
u^h(t,x) = e^{-t}psi(x-t)
$$
regarding the particular we have that $u^p(t,x) = frac 14e^{t+2x}$ verifies the particular equation then
$$
u(t,x) = e^{-t}psi(x-t)+frac 14e^{t+2x}
$$
and for $t=0$ we have
$$
psi(x) + frac 14e^{2x}=0
$$
then $psi(x) = -frac 14 e^{2x}$ and finally
$$
u(t,x) = frac 14left(-e^{-t}e^{2(x-t)}+e^{t+2x}right)
$$
answered Jan 20 at 17:19
CesareoCesareo
9,2013517
$endgroup$
add a comment |
$begingroup$
This is a linear PDE so
$$
u = u^h + u^p
$$
with
$$
u_t^h+u_x^h=-u^h\
u_t^p+u_x^p+u^p=e^{t+2x}
$$
Using the Characteristics method with $u^h$ we have
$$
begin{cases}
frac{dt}{dtau} = 1 & t(0) = 0 & t(tau) = tau\
frac{dx}{dtau} = 1 & x(0) = s & x(tau) = s+tau\
frac{du^h}{dtau} = -u^h & u^h(0) = phi(s) & -ln u^h(tau) = phi(s)+tau
end{cases}
$$
or
$$
u^h(t,x) = e^{-t}psi(x-t)
$$
regarding the particular we have that $u^p(t,x) = frac 14e^{t+2x}$ verifies the particular equation then
$$
u(t,x) = e^{-t}psi(x-t)+frac 14e^{t+2x}
$$
and for $t=0$ we have
$$
psi(x) + frac 14e^{2x}=0
$$
then $psi(x) = -frac 14 e^{2x}$ and finally
$$
u(t,x) = frac 14left(-e^{-t}e^{2(x-t)}+e^{t+2x}right)
$$
answered Jan 20 at 17:19
CesareoCesareo
9,2013517
$endgroup$
This is a linear PDE so
$$
u = u^h + u^p
$$
with
$$
u_t^h+u_x^h=-u^h\
u_t^p+u_x^p+u^p=e^{t+2x}
$$
Using the Characteristics method with $u^h$ we have
$$
begin{cases}
frac{dt}{dtau} = 1 & t(0) = 0 & t(tau) = tau\
frac{dx}{dtau} = 1 & x(0) = s & x(tau) = s+tau\
frac{du^h}{dtau} = -u^h & u^h(0) = phi(s) & -ln u^h(tau) = phi(s)+tau
end{cases}
$$
or
$$
u^h(t,x) = e^{-t}psi(x-t)
$$
regarding the particular we have that $u^p(t,x) = frac 14e^{t+2x}$ verifies the particular equation then
$$
u(t,x) = e^{-t}psi(x-t)+frac 14e^{t+2x}
$$
and for $t=0$ we have
$$
psi(x) + frac 14e^{2x}=0
$$
then $psi(x) = -frac 14 e^{2x}$ and finally
$$
u(t,x) = frac 14left(-e^{-t}e^{2(x-t)}+e^{t+2x}right)
$$
answered Jan 20 at 17:19
CesareoCesareo
9,2013517
edited Jan 20 at 18:41
answered Jan 20 at 17:19
CesareoCesareo
9,2013517
answered Jan 20 at 17:19
CesareoCesareo
9,2013517
answered Jan 20 at 17:19
CesareoCesareo
9,2013517
9,2013517
add a comment |
add a comment |
$begingroup$
Yes, by using the method of characteristics, one can find the explicit expression for the solution :
$$u(t,x)=frac14left(e^{t+2x}-e^{-3t+2x} right)$$
The question is concise. The answer as well.
answered Jan 20 at 18:16
JJacquelinJJacquelin
44.3k21854
$endgroup$
add a comment |
$begingroup$
Yes, by using the method of characteristics, one can find the explicit expression for the solution :
$$u(t,x)=frac14left(e^{t+2x}-e^{-3t+2x} right)$$
The question is concise. The answer as well.
answered Jan 20 at 18:16
JJacquelinJJacquelin
44.3k21854
$endgroup$
add a comment |
$begingroup$
Yes, by using the method of characteristics, one can find the explicit expression for the solution :
$$u(t,x)=frac14left(e^{t+2x}-e^{-3t+2x} right)$$
The question is concise. The answer as well.
answered Jan 20 at 18:16
JJacquelinJJacquelin
44.3k21854
$endgroup$
Yes, by using the method of characteristics, one can find the explicit expression for the solution :
$$u(t,x)=frac14left(e^{t+2x}-e^{-3t+2x} right)$$
The question is concise. The answer as well.
answered Jan 20 at 18:16
JJacquelinJJacquelin
44.3k21854
answered Jan 20 at 18:16
JJacquelinJJacquelin
44.3k21854
answered Jan 20 at 18:16
JJacquelinJJacquelin
44.3k21854
answered Jan 20 at 18:16
JJacquelinJJacquelin
44.3k21854
44.3k21854
add a comment |
add a comment |
$begingroup$
Hi and welcome to the Math.SE. Questions sounding like "do this for me, please" are poorly received from other members. I you need help, help other users help you for example by explaining what you tried and where did you get stuck.
$endgroup$
– Daniele Tampieri
Jan 20 at 17:08