Mixing
inhomogenous wave equation with initial value; PDE
$begingroup$
Find the general solution of the inhomogeneous wave equation
$$u_{tt} − c^2u_{xx} = sin t$$
What is the solution to the initial-value problem for this equation with $u(x, 0) = 0$ and $u_t(x, 0) = 1$?
I know that the general solution to the homogenous equation is $u(x,t) = f(x+ct) + g(x-ct)$ and I know how to solve the initial value problem from there. It's the inhomogenous part of this equation that is confusing me.
Thanks in advance!
pde wave-equation
edited Jan 30 at 9:26
Dylan
14.2k31127
asked Jan 29 at 21:33
meff11meff11
466
$endgroup$
add a comment |
$begingroup$
Find the general solution of the inhomogeneous wave equation
$$u_{tt} − c^2u_{xx} = sin t$$
What is the solution to the initial-value problem for this equation with $u(x, 0) = 0$ and $u_t(x, 0) = 1$?
I know that the general solution to the homogenous equation is $u(x,t) = f(x+ct) + g(x-ct)$ and I know how to solve the initial value problem from there. It's the inhomogenous part of this equation that is confusing me.
Thanks in advance!
pde wave-equation
edited Jan 30 at 9:26
Dylan
14.2k31127
asked Jan 29 at 21:33
meff11meff11
466
$endgroup$
$begingroup$
Consider $u(x,t) = v(x,t) + 2 t - sin(t)$ to obtain a much simpler initial value problem for $v$.
$endgroup$
– Christoph
Jan 30 at 4:15
add a comment |
$begingroup$
Find the general solution of the inhomogeneous wave equation
$$u_{tt} − c^2u_{xx} = sin t$$
What is the solution to the initial-value problem for this equation with $u(x, 0) = 0$ and $u_t(x, 0) = 1$?
I know that the general solution to the homogenous equation is $u(x,t) = f(x+ct) + g(x-ct)$ and I know how to solve the initial value problem from there. It's the inhomogenous part of this equation that is confusing me.
Thanks in advance!
pde wave-equation
edited Jan 30 at 9:26
Dylan
14.2k31127
asked Jan 29 at 21:33
meff11meff11
466
$endgroup$
Find the general solution of the inhomogeneous wave equation
$$u_{tt} − c^2u_{xx} = sin t$$
What is the solution to the initial-value problem for this equation with $u(x, 0) = 0$ and $u_t(x, 0) = 1$?
I know that the general solution to the homogenous equation is $u(x,t) = f(x+ct) + g(x-ct)$ and I know how to solve the initial value problem from there. It's the inhomogenous part of this equation that is confusing me.
Thanks in advance!
pde wave-equation
pde wave-equation
edited Jan 30 at 9:26
Dylan
14.2k31127
asked Jan 29 at 21:33
meff11meff11
466
edited Jan 30 at 9:26
Dylan
14.2k31127
asked Jan 29 at 21:33
meff11meff11
466
edited Jan 30 at 9:26
Dylan
14.2k31127
edited Jan 30 at 9:26
Dylan
14.2k31127
edited Jan 30 at 9:26
Dylan
14.2k31127
14.2k31127
asked Jan 29 at 21:33
meff11meff11
466
asked Jan 29 at 21:33
meff11meff11
466
asked Jan 29 at 21:33
meff11meff11
466
466
$begingroup$
Consider $u(x,t) = v(x,t) + 2 t - sin(t)$ to obtain a much simpler initial value problem for $v$.
$endgroup$
– Christoph
Jan 30 at 4:15
add a comment |
$begingroup$
Consider $u(x,t) = v(x,t) + 2 t - sin(t)$ to obtain a much simpler initial value problem for $v$.
$endgroup$
– Christoph
Jan 30 at 4:15
$begingroup$
Consider $u(x,t) = v(x,t) + 2 t - sin(t)$ to obtain a much simpler initial value problem for $v$.
$endgroup$
– Christoph
Jan 30 at 4:15
$begingroup$
Consider $u(x,t) = v(x,t) + 2 t - sin(t)$ to obtain a much simpler initial value problem for $v$.
$endgroup$
– Christoph
Jan 30 at 4:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First, observe that a particular solution is $u_p(t) = -sin t$.
Then, find the remaining homogeneous solution $v(x,t)$ such that $u = u_p + v$. This gives the BVP
begin{cases}
v_{tt} - c^2 v_{xx} = 0 \
v(x,0) = u(x,0) - u_p(0) = 0 \
v_t(x,0) = u_t(x,0) - u_p'(0) = 2
end{cases}
d'Alembert's formula gives
$$v(x,t) = frac{1}{2c}int_{x-ct}^{x+ct} 2 ds = frac{1}{c}big[(x+ct) - (x-ct) big] = frac{2ct}{c} = 2t$$
answered Jan 30 at 9:29
DylanDylan
14.2k31127
$endgroup$
$begingroup$
That particular solution does not agree with the initial condition of du/dt(x,0) = 1. Your particular solution gives -1
$endgroup$
– meff11
Feb 1 at 16:39
$begingroup$
@meff11 It doesn't have to! The particular solution only needs to satisfy the inhomogeneous equation. The initial condition is corrected by the remaining homogeneous solution. See my notes on $v(x,t)$
$endgroup$
– Dylan
Feb 1 at 16:56
$begingroup$
makes sense thanks! also, when I do d'Alembert's formula I keep getting t not 2t. Am I missing something? or could you show your calculations? thanks
$endgroup$
– meff11
Feb 1 at 17:10
$begingroup$
@meff11 No problem. I added the calculation.
$endgroup$
– Dylan
Feb 1 at 17:45
add a comment |
$begingroup$
$$u_{tt} − c^2u_{xx} = sin t$$
You know that the general solution to the homogenous equation is $u(x,t) = f(x+ct) + g(x-ct)$
Then you have to find a particular solution of the inhomogeneous equation (any one). Only by inspection it is easy to see that $u_p=-sin(t)$ is a solution.
If you don't see it, try any fonction of $t$ only, that is on the form $u_p(t)$. So, you solve the ODE $u_p''(t)=sin(t)$ . No need for all solutions. Any one among them is sufficient, for example $u_p=-sin(t)$ .
Thus the general solution of the PDE is :
$$u(x,t)=f(x+ct)+g(x-ct)-sin(t)$$
$u_t=cf'(x+ct)-cg'(x-ct)-cos(t)$
Conditions :
$begin{cases}
u(x,0)=f(x)+g(x)=0quadimpliesquad g(x)=-f(x) \
u-t(x,0)=cf'(x)-cg'(x)-1=1 quadimpliesquad 2cf'(x)=2
end{cases}$
Integrating the second equation leads to $quad f(x)=frac{x}{c}+C quad$ and $quad g(x)=-frac{x}{c}-C$ .
Putting them into the above general solution gives :
$$u(x,t)=(frac{x+ct}{c}+C)+(-frac{x-ct}{c}-C)-sin(t)$$
Finally, after simplification the solution is :
$$u(x,t)=2t-sin(t)$$
answered Feb 2 at 15:53
JJacquelinJJacquelin
45.3k21856
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
First, observe that a particular solution is $u_p(t) = -sin t$.
Then, find the remaining homogeneous solution $v(x,t)$ such that $u = u_p + v$. This gives the BVP
begin{cases}
v_{tt} - c^2 v_{xx} = 0 \
v(x,0) = u(x,0) - u_p(0) = 0 \
v_t(x,0) = u_t(x,0) - u_p'(0) = 2
end{cases}
d'Alembert's formula gives
$$v(x,t) = frac{1}{2c}int_{x-ct}^{x+ct} 2 ds = frac{1}{c}big[(x+ct) - (x-ct) big] = frac{2ct}{c} = 2t$$
answered Jan 30 at 9:29
DylanDylan
14.2k31127
$endgroup$
$begingroup$
That particular solution does not agree with the initial condition of du/dt(x,0) = 1. Your particular solution gives -1
$endgroup$
– meff11
Feb 1 at 16:39
$begingroup$
@meff11 It doesn't have to! The particular solution only needs to satisfy the inhomogeneous equation. The initial condition is corrected by the remaining homogeneous solution. See my notes on $v(x,t)$
$endgroup$
– Dylan
Feb 1 at 16:56
$begingroup$
makes sense thanks! also, when I do d'Alembert's formula I keep getting t not 2t. Am I missing something? or could you show your calculations? thanks
$endgroup$
– meff11
Feb 1 at 17:10
$begingroup$
@meff11 No problem. I added the calculation.
$endgroup$
– Dylan
Feb 1 at 17:45
add a comment |
$begingroup$
First, observe that a particular solution is $u_p(t) = -sin t$.
Then, find the remaining homogeneous solution $v(x,t)$ such that $u = u_p + v$. This gives the BVP
begin{cases}
v_{tt} - c^2 v_{xx} = 0 \
v(x,0) = u(x,0) - u_p(0) = 0 \
v_t(x,0) = u_t(x,0) - u_p'(0) = 2
end{cases}
d'Alembert's formula gives
$$v(x,t) = frac{1}{2c}int_{x-ct}^{x+ct} 2 ds = frac{1}{c}big[(x+ct) - (x-ct) big] = frac{2ct}{c} = 2t$$
answered Jan 30 at 9:29
DylanDylan
14.2k31127
$endgroup$
$begingroup$
That particular solution does not agree with the initial condition of du/dt(x,0) = 1. Your particular solution gives -1
$endgroup$
– meff11
Feb 1 at 16:39
$begingroup$
@meff11 It doesn't have to! The particular solution only needs to satisfy the inhomogeneous equation. The initial condition is corrected by the remaining homogeneous solution. See my notes on $v(x,t)$
$endgroup$
– Dylan
Feb 1 at 16:56
$begingroup$
makes sense thanks! also, when I do d'Alembert's formula I keep getting t not 2t. Am I missing something? or could you show your calculations? thanks
$endgroup$
– meff11
Feb 1 at 17:10
$begingroup$
@meff11 No problem. I added the calculation.
$endgroup$
– Dylan
Feb 1 at 17:45
add a comment |
$begingroup$
First, observe that a particular solution is $u_p(t) = -sin t$.
Then, find the remaining homogeneous solution $v(x,t)$ such that $u = u_p + v$. This gives the BVP
begin{cases}
v_{tt} - c^2 v_{xx} = 0 \
v(x,0) = u(x,0) - u_p(0) = 0 \
v_t(x,0) = u_t(x,0) - u_p'(0) = 2
end{cases}
d'Alembert's formula gives
$$v(x,t) = frac{1}{2c}int_{x-ct}^{x+ct} 2 ds = frac{1}{c}big[(x+ct) - (x-ct) big] = frac{2ct}{c} = 2t$$
answered Jan 30 at 9:29
DylanDylan
14.2k31127
$endgroup$
First, observe that a particular solution is $u_p(t) = -sin t$.
Then, find the remaining homogeneous solution $v(x,t)$ such that $u = u_p + v$. This gives the BVP
begin{cases}
v_{tt} - c^2 v_{xx} = 0 \
v(x,0) = u(x,0) - u_p(0) = 0 \
v_t(x,0) = u_t(x,0) - u_p'(0) = 2
end{cases}
d'Alembert's formula gives
$$v(x,t) = frac{1}{2c}int_{x-ct}^{x+ct} 2 ds = frac{1}{c}big[(x+ct) - (x-ct) big] = frac{2ct}{c} = 2t$$
answered Jan 30 at 9:29
DylanDylan
14.2k31127
edited Feb 1 at 17:28
answered Jan 30 at 9:29
DylanDylan
14.2k31127
answered Jan 30 at 9:29
DylanDylan
14.2k31127
answered Jan 30 at 9:29
DylanDylan
14.2k31127
14.2k31127
$begingroup$
That particular solution does not agree with the initial condition of du/dt(x,0) = 1. Your particular solution gives -1
$endgroup$
– meff11
Feb 1 at 16:39
$begingroup$
@meff11 It doesn't have to! The particular solution only needs to satisfy the inhomogeneous equation. The initial condition is corrected by the remaining homogeneous solution. See my notes on $v(x,t)$
$endgroup$
– Dylan
Feb 1 at 16:56
$begingroup$
makes sense thanks! also, when I do d'Alembert's formula I keep getting t not 2t. Am I missing something? or could you show your calculations? thanks
$endgroup$
– meff11
Feb 1 at 17:10
$begingroup$
@meff11 No problem. I added the calculation.
$endgroup$
– Dylan
Feb 1 at 17:45
add a comment |
$begingroup$
That particular solution does not agree with the initial condition of du/dt(x,0) = 1. Your particular solution gives -1
$endgroup$
– meff11
Feb 1 at 16:39
$begingroup$
@meff11 It doesn't have to! The particular solution only needs to satisfy the inhomogeneous equation. The initial condition is corrected by the remaining homogeneous solution. See my notes on $v(x,t)$
$endgroup$
– Dylan
Feb 1 at 16:56
$begingroup$
makes sense thanks! also, when I do d'Alembert's formula I keep getting t not 2t. Am I missing something? or could you show your calculations? thanks
$endgroup$
– meff11
Feb 1 at 17:10
$begingroup$
@meff11 No problem. I added the calculation.
$endgroup$
– Dylan
Feb 1 at 17:45
$begingroup$
That particular solution does not agree with the initial condition of du/dt(x,0) = 1. Your particular solution gives -1
$endgroup$
– meff11
Feb 1 at 16:39
$begingroup$
That particular solution does not agree with the initial condition of du/dt(x,0) = 1. Your particular solution gives -1
$endgroup$
– meff11
Feb 1 at 16:39
$begingroup$
@meff11 It doesn't have to! The particular solution only needs to satisfy the inhomogeneous equation. The initial condition is corrected by the remaining homogeneous solution. See my notes on $v(x,t)$
$endgroup$
– Dylan
Feb 1 at 16:56
$begingroup$
@meff11 It doesn't have to! The particular solution only needs to satisfy the inhomogeneous equation. The initial condition is corrected by the remaining homogeneous solution. See my notes on $v(x,t)$
$endgroup$
– Dylan
Feb 1 at 16:56
$begingroup$
makes sense thanks! also, when I do d'Alembert's formula I keep getting t not 2t. Am I missing something? or could you show your calculations? thanks
$endgroup$
– meff11
Feb 1 at 17:10
$begingroup$
makes sense thanks! also, when I do d'Alembert's formula I keep getting t not 2t. Am I missing something? or could you show your calculations? thanks
$endgroup$
– meff11
Feb 1 at 17:10
$begingroup$
@meff11 No problem. I added the calculation.
$endgroup$
– Dylan
Feb 1 at 17:45
$begingroup$
@meff11 No problem. I added the calculation.
$endgroup$
– Dylan
Feb 1 at 17:45
add a comment |
$begingroup$
$$u_{tt} − c^2u_{xx} = sin t$$
You know that the general solution to the homogenous equation is $u(x,t) = f(x+ct) + g(x-ct)$
Then you have to find a particular solution of the inhomogeneous equation (any one). Only by inspection it is easy to see that $u_p=-sin(t)$ is a solution.
If you don't see it, try any fonction of $t$ only, that is on the form $u_p(t)$. So, you solve the ODE $u_p''(t)=sin(t)$ . No need for all solutions. Any one among them is sufficient, for example $u_p=-sin(t)$ .
Thus the general solution of the PDE is :
$$u(x,t)=f(x+ct)+g(x-ct)-sin(t)$$
$u_t=cf'(x+ct)-cg'(x-ct)-cos(t)$
Conditions :
$begin{cases}
u(x,0)=f(x)+g(x)=0quadimpliesquad g(x)=-f(x) \
u-t(x,0)=cf'(x)-cg'(x)-1=1 quadimpliesquad 2cf'(x)=2
end{cases}$
Integrating the second equation leads to $quad f(x)=frac{x}{c}+C quad$ and $quad g(x)=-frac{x}{c}-C$ .
Putting them into the above general solution gives :
$$u(x,t)=(frac{x+ct}{c}+C)+(-frac{x-ct}{c}-C)-sin(t)$$
Finally, after simplification the solution is :
$$u(x,t)=2t-sin(t)$$
answered Feb 2 at 15:53
JJacquelinJJacquelin
45.3k21856
$endgroup$
add a comment |
$begingroup$
$$u_{tt} − c^2u_{xx} = sin t$$
You know that the general solution to the homogenous equation is $u(x,t) = f(x+ct) + g(x-ct)$
Then you have to find a particular solution of the inhomogeneous equation (any one). Only by inspection it is easy to see that $u_p=-sin(t)$ is a solution.
If you don't see it, try any fonction of $t$ only, that is on the form $u_p(t)$. So, you solve the ODE $u_p''(t)=sin(t)$ . No need for all solutions. Any one among them is sufficient, for example $u_p=-sin(t)$ .
Thus the general solution of the PDE is :
$$u(x,t)=f(x+ct)+g(x-ct)-sin(t)$$
$u_t=cf'(x+ct)-cg'(x-ct)-cos(t)$
Conditions :
$begin{cases}
u(x,0)=f(x)+g(x)=0quadimpliesquad g(x)=-f(x) \
u-t(x,0)=cf'(x)-cg'(x)-1=1 quadimpliesquad 2cf'(x)=2
end{cases}$
Integrating the second equation leads to $quad f(x)=frac{x}{c}+C quad$ and $quad g(x)=-frac{x}{c}-C$ .
Putting them into the above general solution gives :
$$u(x,t)=(frac{x+ct}{c}+C)+(-frac{x-ct}{c}-C)-sin(t)$$
Finally, after simplification the solution is :
$$u(x,t)=2t-sin(t)$$
answered Feb 2 at 15:53
JJacquelinJJacquelin
45.3k21856
$endgroup$
add a comment |
$begingroup$
$$u_{tt} − c^2u_{xx} = sin t$$
You know that the general solution to the homogenous equation is $u(x,t) = f(x+ct) + g(x-ct)$
Then you have to find a particular solution of the inhomogeneous equation (any one). Only by inspection it is easy to see that $u_p=-sin(t)$ is a solution.
If you don't see it, try any fonction of $t$ only, that is on the form $u_p(t)$. So, you solve the ODE $u_p''(t)=sin(t)$ . No need for all solutions. Any one among them is sufficient, for example $u_p=-sin(t)$ .
Thus the general solution of the PDE is :
$$u(x,t)=f(x+ct)+g(x-ct)-sin(t)$$
$u_t=cf'(x+ct)-cg'(x-ct)-cos(t)$
Conditions :
$begin{cases}
u(x,0)=f(x)+g(x)=0quadimpliesquad g(x)=-f(x) \
u-t(x,0)=cf'(x)-cg'(x)-1=1 quadimpliesquad 2cf'(x)=2
end{cases}$
Integrating the second equation leads to $quad f(x)=frac{x}{c}+C quad$ and $quad g(x)=-frac{x}{c}-C$ .
Putting them into the above general solution gives :
$$u(x,t)=(frac{x+ct}{c}+C)+(-frac{x-ct}{c}-C)-sin(t)$$
Finally, after simplification the solution is :
$$u(x,t)=2t-sin(t)$$
answered Feb 2 at 15:53
JJacquelinJJacquelin
45.3k21856
$endgroup$
$$u_{tt} − c^2u_{xx} = sin t$$
You know that the general solution to the homogenous equation is $u(x,t) = f(x+ct) + g(x-ct)$
Then you have to find a particular solution of the inhomogeneous equation (any one). Only by inspection it is easy to see that $u_p=-sin(t)$ is a solution.
If you don't see it, try any fonction of $t$ only, that is on the form $u_p(t)$. So, you solve the ODE $u_p''(t)=sin(t)$ . No need for all solutions. Any one among them is sufficient, for example $u_p=-sin(t)$ .
Thus the general solution of the PDE is :
$$u(x,t)=f(x+ct)+g(x-ct)-sin(t)$$
$u_t=cf'(x+ct)-cg'(x-ct)-cos(t)$
Conditions :
$begin{cases}
u(x,0)=f(x)+g(x)=0quadimpliesquad g(x)=-f(x) \
u-t(x,0)=cf'(x)-cg'(x)-1=1 quadimpliesquad 2cf'(x)=2
end{cases}$
Integrating the second equation leads to $quad f(x)=frac{x}{c}+C quad$ and $quad g(x)=-frac{x}{c}-C$ .
Putting them into the above general solution gives :
$$u(x,t)=(frac{x+ct}{c}+C)+(-frac{x-ct}{c}-C)-sin(t)$$
Finally, after simplification the solution is :
$$u(x,t)=2t-sin(t)$$
answered Feb 2 at 15:53
JJacquelinJJacquelin
45.3k21856
answered Feb 2 at 15:53
JJacquelinJJacquelin
45.3k21856
answered Feb 2 at 15:53
JJacquelinJJacquelin
45.3k21856
answered Feb 2 at 15:53
JJacquelinJJacquelin
45.3k21856
45.3k21856
add a comment |
add a comment |
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$begingroup$
Consider $u(x,t) = v(x,t) + 2 t - sin(t)$ to obtain a much simpler initial value problem for $v$.
$endgroup$
– Christoph
Jan 30 at 4:15