Wave Equation, is this problem well posed?
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I am solving $$u_{tt}=4u_{xx}$$ $$u(x,0)=x^2-x$$ $$u_t(x,0)=cos x$$ $$u(0,t)=u(1,t)=0$$
I am trying to solve with the method of separable factors. So I am trying to find a equation like $u(x,t)=X(x)cdot T(t)$. But, I am finding a problem, because if the solution is like this, then $u_t(x,0)=X(x)cdot T'(0) = cos xRightarrow X(x) = frac{cos x}{T'(0)}$.
But if $X(x)$ is this way, then $X(0)=frac{cos 0}{T'(0)}neq 0$.
Am I solving this right or the problem is not well posed?
pde wave-equation
asked Nov 22 at 20:00
Matheus Popst de Campos
504
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2
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I am solving $$u_{tt}=4u_{xx}$$ $$u(x,0)=x^2-x$$ $$u_t(x,0)=cos x$$ $$u(0,t)=u(1,t)=0$$
I am trying to solve with the method of separable factors. So I am trying to find a equation like $u(x,t)=X(x)cdot T(t)$. But, I am finding a problem, because if the solution is like this, then $u_t(x,0)=X(x)cdot T'(0) = cos xRightarrow X(x) = frac{cos x}{T'(0)}$.
But if $X(x)$ is this way, then $X(0)=frac{cos 0}{T'(0)}neq 0$.
Am I solving this right or the problem is not well posed?
pde wave-equation
asked Nov 22 at 20:00
Matheus Popst de Campos
504
3
The problem with your approach is that you only find one solution. If we separate $u(x,t) = X(x)T(t)$ then $T'/T = lambda$ and $4X''/X = lambda$ for some constant $lambda$ that is determined by the boundary conditions (typically this gives $lambda$ is a constant times an integer). The full solution is then a linear combination over all the particular solutions given by different $lambda$-values. $u(x,t) = sum a_lambda X_lambda(x)T_lambda(t)$. It's this solution that equals $x^2-x$. It's the boundary conditions that can be applied on a single solution so $X_lambda(0) = 0$ etc.
– Winther
Nov 22 at 20:11
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am solving $$u_{tt}=4u_{xx}$$ $$u(x,0)=x^2-x$$ $$u_t(x,0)=cos x$$ $$u(0,t)=u(1,t)=0$$
I am trying to solve with the method of separable factors. So I am trying to find a equation like $u(x,t)=X(x)cdot T(t)$. But, I am finding a problem, because if the solution is like this, then $u_t(x,0)=X(x)cdot T'(0) = cos xRightarrow X(x) = frac{cos x}{T'(0)}$.
But if $X(x)$ is this way, then $X(0)=frac{cos 0}{T'(0)}neq 0$.
Am I solving this right or the problem is not well posed?
pde wave-equation
asked Nov 22 at 20:00
Matheus Popst de Campos
504
I am solving $$u_{tt}=4u_{xx}$$ $$u(x,0)=x^2-x$$ $$u_t(x,0)=cos x$$ $$u(0,t)=u(1,t)=0$$
I am trying to solve with the method of separable factors. So I am trying to find a equation like $u(x,t)=X(x)cdot T(t)$. But, I am finding a problem, because if the solution is like this, then $u_t(x,0)=X(x)cdot T'(0) = cos xRightarrow X(x) = frac{cos x}{T'(0)}$.
But if $X(x)$ is this way, then $X(0)=frac{cos 0}{T'(0)}neq 0$.
Am I solving this right or the problem is not well posed?
pde wave-equation
pde wave-equation
asked Nov 22 at 20:00
Matheus Popst de Campos
504
asked Nov 22 at 20:00
Matheus Popst de Campos
504
asked Nov 22 at 20:00
Matheus Popst de Campos
504
asked Nov 22 at 20:00
Matheus Popst de Campos
504
asked Nov 22 at 20:00
Matheus Popst de Campos
504
504
3
The problem with your approach is that you only find one solution. If we separate $u(x,t) = X(x)T(t)$ then $T'/T = lambda$ and $4X''/X = lambda$ for some constant $lambda$ that is determined by the boundary conditions (typically this gives $lambda$ is a constant times an integer). The full solution is then a linear combination over all the particular solutions given by different $lambda$-values. $u(x,t) = sum a_lambda X_lambda(x)T_lambda(t)$. It's this solution that equals $x^2-x$. It's the boundary conditions that can be applied on a single solution so $X_lambda(0) = 0$ etc.
– Winther
Nov 22 at 20:11
add a comment |
3
The problem with your approach is that you only find one solution. If we separate $u(x,t) = X(x)T(t)$ then $T'/T = lambda$ and $4X''/X = lambda$ for some constant $lambda$ that is determined by the boundary conditions (typically this gives $lambda$ is a constant times an integer). The full solution is then a linear combination over all the particular solutions given by different $lambda$-values. $u(x,t) = sum a_lambda X_lambda(x)T_lambda(t)$. It's this solution that equals $x^2-x$. It's the boundary conditions that can be applied on a single solution so $X_lambda(0) = 0$ etc.
– Winther
Nov 22 at 20:11
3
3
The problem with your approach is that you only find one solution. If we separate $u(x,t) = X(x)T(t)$ then $T'/T = lambda$ and $4X''/X = lambda$ for some constant $lambda$ that is determined by the boundary conditions (typically this gives $lambda$ is a constant times an integer). The full solution is then a linear combination over all the particular solutions given by different $lambda$-values. $u(x,t) = sum a_lambda X_lambda(x)T_lambda(t)$. It's this solution that equals $x^2-x$. It's the boundary conditions that can be applied on a single solution so $X_lambda(0) = 0$ etc.
– Winther
Nov 22 at 20:11
The problem with your approach is that you only find one solution. If we separate $u(x,t) = X(x)T(t)$ then $T'/T = lambda$ and $4X''/X = lambda$ for some constant $lambda$ that is determined by the boundary conditions (typically this gives $lambda$ is a constant times an integer). The full solution is then a linear combination over all the particular solutions given by different $lambda$-values. $u(x,t) = sum a_lambda X_lambda(x)T_lambda(t)$. It's this solution that equals $x^2-x$. It's the boundary conditions that can be applied on a single solution so $X_lambda(0) = 0$ etc.
– Winther
Nov 22 at 20:11
add a comment |
1 Answer
1
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The separated solutions of $u_{tt}=4u_{xx}$ have the form $u(x,t)=X(x)T(t)$ where
$$
frac{T''}{T}=lambda = frac{X''}{X}
$$
leading to
$$
T(t) = Asin(sqrt{lambda}t)+Bcos(sqrt{lambda}t) \
X(x) = Csin(sqrt{lambda}x)+Dcos(sqrt{lambda}x).
$$
The condition $X(0)=X(1)=0$ requires $sqrt{lambda}=npi$ for $n=1,2,3,cdots$, and
$$
X_n(t)=C_nsin(npi x) \
T_n(t)=A_ncos(npi t)+B_nsin(npi t).
$$
The general solution (after combining constants) is
$$
u(x,t)=sum_{n=1}^{infty}(A_ncos(npi t)+B_nsin(npi t))sin(npi x)
$$
The constants are determined by the initial conditions:
$$
x^2-x = u(x,0) = sum_{n=1}^{infty}A_nsin(npi x) \
cos(x) = u_t(x,0) = sum_{n=1}^{infty}B_n npisin(npi x).
$$
So,
$$
A_n = frac{int_{0}^{1}(x^2-x)sin(npi x)dx}{int_{0}^{1}sin^2(npi x)dx}, \
B_n = frac{int_{0}^{1}cos(x)sin(npi x)dx}{npiint_{0}^{1}sin^2(npi x)dx}.
$$
answered Nov 23 at 4:31
DisintegratingByParts
58k42477
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The separated solutions of $u_{tt}=4u_{xx}$ have the form $u(x,t)=X(x)T(t)$ where
$$
frac{T''}{T}=lambda = frac{X''}{X}
$$
leading to
$$
T(t) = Asin(sqrt{lambda}t)+Bcos(sqrt{lambda}t) \
X(x) = Csin(sqrt{lambda}x)+Dcos(sqrt{lambda}x).
$$
The condition $X(0)=X(1)=0$ requires $sqrt{lambda}=npi$ for $n=1,2,3,cdots$, and
$$
X_n(t)=C_nsin(npi x) \
T_n(t)=A_ncos(npi t)+B_nsin(npi t).
$$
The general solution (after combining constants) is
$$
u(x,t)=sum_{n=1}^{infty}(A_ncos(npi t)+B_nsin(npi t))sin(npi x)
$$
The constants are determined by the initial conditions:
$$
x^2-x = u(x,0) = sum_{n=1}^{infty}A_nsin(npi x) \
cos(x) = u_t(x,0) = sum_{n=1}^{infty}B_n npisin(npi x).
$$
So,
$$
A_n = frac{int_{0}^{1}(x^2-x)sin(npi x)dx}{int_{0}^{1}sin^2(npi x)dx}, \
B_n = frac{int_{0}^{1}cos(x)sin(npi x)dx}{npiint_{0}^{1}sin^2(npi x)dx}.
$$
answered Nov 23 at 4:31
DisintegratingByParts
58k42477
add a comment |
up vote
0
down vote
accepted
The separated solutions of $u_{tt}=4u_{xx}$ have the form $u(x,t)=X(x)T(t)$ where
$$
frac{T''}{T}=lambda = frac{X''}{X}
$$
leading to
$$
T(t) = Asin(sqrt{lambda}t)+Bcos(sqrt{lambda}t) \
X(x) = Csin(sqrt{lambda}x)+Dcos(sqrt{lambda}x).
$$
The condition $X(0)=X(1)=0$ requires $sqrt{lambda}=npi$ for $n=1,2,3,cdots$, and
$$
X_n(t)=C_nsin(npi x) \
T_n(t)=A_ncos(npi t)+B_nsin(npi t).
$$
The general solution (after combining constants) is
$$
u(x,t)=sum_{n=1}^{infty}(A_ncos(npi t)+B_nsin(npi t))sin(npi x)
$$
The constants are determined by the initial conditions:
$$
x^2-x = u(x,0) = sum_{n=1}^{infty}A_nsin(npi x) \
cos(x) = u_t(x,0) = sum_{n=1}^{infty}B_n npisin(npi x).
$$
So,
$$
A_n = frac{int_{0}^{1}(x^2-x)sin(npi x)dx}{int_{0}^{1}sin^2(npi x)dx}, \
B_n = frac{int_{0}^{1}cos(x)sin(npi x)dx}{npiint_{0}^{1}sin^2(npi x)dx}.
$$
answered Nov 23 at 4:31
DisintegratingByParts
58k42477
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The separated solutions of $u_{tt}=4u_{xx}$ have the form $u(x,t)=X(x)T(t)$ where
$$
frac{T''}{T}=lambda = frac{X''}{X}
$$
leading to
$$
T(t) = Asin(sqrt{lambda}t)+Bcos(sqrt{lambda}t) \
X(x) = Csin(sqrt{lambda}x)+Dcos(sqrt{lambda}x).
$$
The condition $X(0)=X(1)=0$ requires $sqrt{lambda}=npi$ for $n=1,2,3,cdots$, and
$$
X_n(t)=C_nsin(npi x) \
T_n(t)=A_ncos(npi t)+B_nsin(npi t).
$$
The general solution (after combining constants) is
$$
u(x,t)=sum_{n=1}^{infty}(A_ncos(npi t)+B_nsin(npi t))sin(npi x)
$$
The constants are determined by the initial conditions:
$$
x^2-x = u(x,0) = sum_{n=1}^{infty}A_nsin(npi x) \
cos(x) = u_t(x,0) = sum_{n=1}^{infty}B_n npisin(npi x).
$$
So,
$$
A_n = frac{int_{0}^{1}(x^2-x)sin(npi x)dx}{int_{0}^{1}sin^2(npi x)dx}, \
B_n = frac{int_{0}^{1}cos(x)sin(npi x)dx}{npiint_{0}^{1}sin^2(npi x)dx}.
$$
answered Nov 23 at 4:31
DisintegratingByParts
58k42477
The separated solutions of $u_{tt}=4u_{xx}$ have the form $u(x,t)=X(x)T(t)$ where
$$
frac{T''}{T}=lambda = frac{X''}{X}
$$
leading to
$$
T(t) = Asin(sqrt{lambda}t)+Bcos(sqrt{lambda}t) \
X(x) = Csin(sqrt{lambda}x)+Dcos(sqrt{lambda}x).
$$
The condition $X(0)=X(1)=0$ requires $sqrt{lambda}=npi$ for $n=1,2,3,cdots$, and
$$
X_n(t)=C_nsin(npi x) \
T_n(t)=A_ncos(npi t)+B_nsin(npi t).
$$
The general solution (after combining constants) is
$$
u(x,t)=sum_{n=1}^{infty}(A_ncos(npi t)+B_nsin(npi t))sin(npi x)
$$
The constants are determined by the initial conditions:
$$
x^2-x = u(x,0) = sum_{n=1}^{infty}A_nsin(npi x) \
cos(x) = u_t(x,0) = sum_{n=1}^{infty}B_n npisin(npi x).
$$
So,
$$
A_n = frac{int_{0}^{1}(x^2-x)sin(npi x)dx}{int_{0}^{1}sin^2(npi x)dx}, \
B_n = frac{int_{0}^{1}cos(x)sin(npi x)dx}{npiint_{0}^{1}sin^2(npi x)dx}.
$$
answered Nov 23 at 4:31
DisintegratingByParts
58k42477
answered Nov 23 at 4:31
DisintegratingByParts
58k42477
answered Nov 23 at 4:31
DisintegratingByParts
58k42477
answered Nov 23 at 4:31
DisintegratingByParts
58k42477
58k42477
add a comment |
add a comment |
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The problem with your approach is that you only find one solution. If we separate $u(x,t) = X(x)T(t)$ then $T'/T = lambda$ and $4X''/X = lambda$ for some constant $lambda$ that is determined by the boundary conditions (typically this gives $lambda$ is a constant times an integer). The full solution is then a linear combination over all the particular solutions given by different $lambda$-values. $u(x,t) = sum a_lambda X_lambda(x)T_lambda(t)$. It's this solution that equals $x^2-x$. It's the boundary conditions that can be applied on a single solution so $X_lambda(0) = 0$ etc.
– Winther
Nov 22 at 20:11