Solving the wave equation with Neumann boundary conditions












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$begingroup$



Solve the wave equation $u_{tt} = c^2 u_{xx}$ for $0 < x < pi$, with the boundary conditions $u_x(0,t) = u_x(pi,t) = 0$ and initial conditions $u(x,0) = cos(x)$ and $u_t(x,0) = cos^{2}(x)$.




Attempted solution - We know that the general solution for Neumann boundary conditions for the wave equation for $0 < x < l$ is
$$u(x,t) = frac{1}{2}A_0 + frac{1}{2}B_0 t + sum_{n=1}^{infty}left(A_ncos frac{npi c t}{l} + B_n sinfrac{npi c t}{l}right)cosfrac{npi x}{l}$$
Thus in our case we have
$$u(x,t) = frac{1}{2}A_0 + frac{1}{2}B_0 t + sum_{n=1}^{infty}left(A_ncos(nct) + B_n sin(nct)right)cos(nx)$$
Note that
$$A_n = frac{2}{pi}int_{0}^{pi}phi(x)cos(nx)dx$$
and $$B_n = frac{2}{pi}int_{0}^{pi}phi(x)sin(nx)dx$$



Applying the initial conditions we have



$$u(x,0) = frac{1}{2}A_0 + sum_{n=1}^{infty}A_ncos(nx) = cos(x)$$



Thus,



$$A_0 = frac{2}{pi}int_{0}^{pi}cos(x)dx = 0$$
and
$$A_n = frac{2}{pi}int_{0}^{pi}cos(x)cos(nx)dx = ... = -frac{2nsin(npi)}{pi(n^2-1)} text{according to Wolfram}$$



Next,



$$u_t(x,0) = frac{1}{2}B_0 + sum_{n=1}^{infty}B_ntimes ntimes c cos(nx) = cos^{2}(x)$$



From here onward I am not sure how to proceed and I find the answer from wolfram rather odd, if anyone can provide details of what I did wrong or should revise please let me know.



The answer should be $$u(x,t) = frac{1}{2}t + cos(ct)cos(x) + frac{1}{4c}sin(2ct) + cos(2x)$$






pde wave-equation







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  • 1




    $begingroup$
    You find the coefficients by using the initial conditions: what's $phi(x)$ in your case?
    $endgroup$
    – NickD
    Jun 24 '17 at 23:39










  • $begingroup$
    @Nick I edited my post, I know we use the initial conditions to find the coefficients but I am not sure how to proceed from where I left off
    $endgroup$
    – justanewb
    Jun 24 '17 at 23:46










  • $begingroup$
    @Nick I do believe that $A_0 = 0$ in this case
    $endgroup$
    – justanewb
    Jun 24 '17 at 23:47










  • $begingroup$
    If you can identify what $phi(x)$ is in your case, you can subtitute it into the integrals for $A_n$ and $B_n$ and find all the coefficients.
    $endgroup$
    – NickD
    Jun 24 '17 at 23:49










  • $begingroup$
    $phi(x) = cos(x)$ correct?
    $endgroup$
    – justanewb
    Jun 24 '17 at 23:50
















1












$begingroup$



Solve the wave equation $u_{tt} = c^2 u_{xx}$ for $0 < x < pi$, with the boundary conditions $u_x(0,t) = u_x(pi,t) = 0$ and initial conditions $u(x,0) = cos(x)$ and $u_t(x,0) = cos^{2}(x)$.




Attempted solution - We know that the general solution for Neumann boundary conditions for the wave equation for $0 < x < l$ is
$$u(x,t) = frac{1}{2}A_0 + frac{1}{2}B_0 t + sum_{n=1}^{infty}left(A_ncos frac{npi c t}{l} + B_n sinfrac{npi c t}{l}right)cosfrac{npi x}{l}$$
Thus in our case we have
$$u(x,t) = frac{1}{2}A_0 + frac{1}{2}B_0 t + sum_{n=1}^{infty}left(A_ncos(nct) + B_n sin(nct)right)cos(nx)$$
Note that
$$A_n = frac{2}{pi}int_{0}^{pi}phi(x)cos(nx)dx$$
and $$B_n = frac{2}{pi}int_{0}^{pi}phi(x)sin(nx)dx$$



Applying the initial conditions we have



$$u(x,0) = frac{1}{2}A_0 + sum_{n=1}^{infty}A_ncos(nx) = cos(x)$$



Thus,



$$A_0 = frac{2}{pi}int_{0}^{pi}cos(x)dx = 0$$
and
$$A_n = frac{2}{pi}int_{0}^{pi}cos(x)cos(nx)dx = ... = -frac{2nsin(npi)}{pi(n^2-1)} text{according to Wolfram}$$



Next,



$$u_t(x,0) = frac{1}{2}B_0 + sum_{n=1}^{infty}B_ntimes ntimes c cos(nx) = cos^{2}(x)$$



From here onward I am not sure how to proceed and I find the answer from wolfram rather odd, if anyone can provide details of what I did wrong or should revise please let me know.



The answer should be $$u(x,t) = frac{1}{2}t + cos(ct)cos(x) + frac{1}{4c}sin(2ct) + cos(2x)$$






pde wave-equation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You find the coefficients by using the initial conditions: what's $phi(x)$ in your case?
    $endgroup$
    – NickD
    Jun 24 '17 at 23:39










  • $begingroup$
    @Nick I edited my post, I know we use the initial conditions to find the coefficients but I am not sure how to proceed from where I left off
    $endgroup$
    – justanewb
    Jun 24 '17 at 23:46










  • $begingroup$
    @Nick I do believe that $A_0 = 0$ in this case
    $endgroup$
    – justanewb
    Jun 24 '17 at 23:47










  • $begingroup$
    If you can identify what $phi(x)$ is in your case, you can subtitute it into the integrals for $A_n$ and $B_n$ and find all the coefficients.
    $endgroup$
    – NickD
    Jun 24 '17 at 23:49










  • $begingroup$
    $phi(x) = cos(x)$ correct?
    $endgroup$
    – justanewb
    Jun 24 '17 at 23:50














1












1








1





$begingroup$



Solve the wave equation $u_{tt} = c^2 u_{xx}$ for $0 < x < pi$, with the boundary conditions $u_x(0,t) = u_x(pi,t) = 0$ and initial conditions $u(x,0) = cos(x)$ and $u_t(x,0) = cos^{2}(x)$.




Attempted solution - We know that the general solution for Neumann boundary conditions for the wave equation for $0 < x < l$ is
$$u(x,t) = frac{1}{2}A_0 + frac{1}{2}B_0 t + sum_{n=1}^{infty}left(A_ncos frac{npi c t}{l} + B_n sinfrac{npi c t}{l}right)cosfrac{npi x}{l}$$
Thus in our case we have
$$u(x,t) = frac{1}{2}A_0 + frac{1}{2}B_0 t + sum_{n=1}^{infty}left(A_ncos(nct) + B_n sin(nct)right)cos(nx)$$
Note that
$$A_n = frac{2}{pi}int_{0}^{pi}phi(x)cos(nx)dx$$
and $$B_n = frac{2}{pi}int_{0}^{pi}phi(x)sin(nx)dx$$



Applying the initial conditions we have



$$u(x,0) = frac{1}{2}A_0 + sum_{n=1}^{infty}A_ncos(nx) = cos(x)$$



Thus,



$$A_0 = frac{2}{pi}int_{0}^{pi}cos(x)dx = 0$$
and
$$A_n = frac{2}{pi}int_{0}^{pi}cos(x)cos(nx)dx = ... = -frac{2nsin(npi)}{pi(n^2-1)} text{according to Wolfram}$$



Next,



$$u_t(x,0) = frac{1}{2}B_0 + sum_{n=1}^{infty}B_ntimes ntimes c cos(nx) = cos^{2}(x)$$



From here onward I am not sure how to proceed and I find the answer from wolfram rather odd, if anyone can provide details of what I did wrong or should revise please let me know.



The answer should be $$u(x,t) = frac{1}{2}t + cos(ct)cos(x) + frac{1}{4c}sin(2ct) + cos(2x)$$






pde wave-equation







share|cite|improve this question











$endgroup$





Solve the wave equation $u_{tt} = c^2 u_{xx}$ for $0 < x < pi$, with the boundary conditions $u_x(0,t) = u_x(pi,t) = 0$ and initial conditions $u(x,0) = cos(x)$ and $u_t(x,0) = cos^{2}(x)$.




Attempted solution - We know that the general solution for Neumann boundary conditions for the wave equation for $0 < x < l$ is
$$u(x,t) = frac{1}{2}A_0 + frac{1}{2}B_0 t + sum_{n=1}^{infty}left(A_ncos frac{npi c t}{l} + B_n sinfrac{npi c t}{l}right)cosfrac{npi x}{l}$$
Thus in our case we have
$$u(x,t) = frac{1}{2}A_0 + frac{1}{2}B_0 t + sum_{n=1}^{infty}left(A_ncos(nct) + B_n sin(nct)right)cos(nx)$$
Note that
$$A_n = frac{2}{pi}int_{0}^{pi}phi(x)cos(nx)dx$$
and $$B_n = frac{2}{pi}int_{0}^{pi}phi(x)sin(nx)dx$$



Applying the initial conditions we have



$$u(x,0) = frac{1}{2}A_0 + sum_{n=1}^{infty}A_ncos(nx) = cos(x)$$



Thus,



$$A_0 = frac{2}{pi}int_{0}^{pi}cos(x)dx = 0$$
and
$$A_n = frac{2}{pi}int_{0}^{pi}cos(x)cos(nx)dx = ... = -frac{2nsin(npi)}{pi(n^2-1)} text{according to Wolfram}$$



Next,



$$u_t(x,0) = frac{1}{2}B_0 + sum_{n=1}^{infty}B_ntimes ntimes c cos(nx) = cos^{2}(x)$$



From here onward I am not sure how to proceed and I find the answer from wolfram rather odd, if anyone can provide details of what I did wrong or should revise please let me know.



The answer should be $$u(x,t) = frac{1}{2}t + cos(ct)cos(x) + frac{1}{4c}sin(2ct) + cos(2x)$$






pde wave-equation




pde wave-equation






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share|cite|improve this question













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edited Jun 25 '17 at 0:17







justanewb

















asked Jun 24 '17 at 23:31









justanewbjustanewb

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  • 1




    $begingroup$
    You find the coefficients by using the initial conditions: what's $phi(x)$ in your case?
    $endgroup$
    – NickD
    Jun 24 '17 at 23:39










  • $begingroup$
    @Nick I edited my post, I know we use the initial conditions to find the coefficients but I am not sure how to proceed from where I left off
    $endgroup$
    – justanewb
    Jun 24 '17 at 23:46










  • $begingroup$
    @Nick I do believe that $A_0 = 0$ in this case
    $endgroup$
    – justanewb
    Jun 24 '17 at 23:47










  • $begingroup$
    If you can identify what $phi(x)$ is in your case, you can subtitute it into the integrals for $A_n$ and $B_n$ and find all the coefficients.
    $endgroup$
    – NickD
    Jun 24 '17 at 23:49










  • $begingroup$
    $phi(x) = cos(x)$ correct?
    $endgroup$
    – justanewb
    Jun 24 '17 at 23:50














  • 1




    $begingroup$
    You find the coefficients by using the initial conditions: what's $phi(x)$ in your case?
    $endgroup$
    – NickD
    Jun 24 '17 at 23:39










  • $begingroup$
    @Nick I edited my post, I know we use the initial conditions to find the coefficients but I am not sure how to proceed from where I left off
    $endgroup$
    – justanewb
    Jun 24 '17 at 23:46










  • $begingroup$
    @Nick I do believe that $A_0 = 0$ in this case
    $endgroup$
    – justanewb
    Jun 24 '17 at 23:47










  • $begingroup$
    If you can identify what $phi(x)$ is in your case, you can subtitute it into the integrals for $A_n$ and $B_n$ and find all the coefficients.
    $endgroup$
    – NickD
    Jun 24 '17 at 23:49










  • $begingroup$
    $phi(x) = cos(x)$ correct?
    $endgroup$
    – justanewb
    Jun 24 '17 at 23:50








1




1




$begingroup$
You find the coefficients by using the initial conditions: what's $phi(x)$ in your case?
$endgroup$
– NickD
Jun 24 '17 at 23:39




$begingroup$
You find the coefficients by using the initial conditions: what's $phi(x)$ in your case?
$endgroup$
– NickD
Jun 24 '17 at 23:39












$begingroup$
@Nick I edited my post, I know we use the initial conditions to find the coefficients but I am not sure how to proceed from where I left off
$endgroup$
– justanewb
Jun 24 '17 at 23:46




$begingroup$
@Nick I edited my post, I know we use the initial conditions to find the coefficients but I am not sure how to proceed from where I left off
$endgroup$
– justanewb
Jun 24 '17 at 23:46












$begingroup$
@Nick I do believe that $A_0 = 0$ in this case
$endgroup$
– justanewb
Jun 24 '17 at 23:47




$begingroup$
@Nick I do believe that $A_0 = 0$ in this case
$endgroup$
– justanewb
Jun 24 '17 at 23:47












$begingroup$
If you can identify what $phi(x)$ is in your case, you can subtitute it into the integrals for $A_n$ and $B_n$ and find all the coefficients.
$endgroup$
– NickD
Jun 24 '17 at 23:49




$begingroup$
If you can identify what $phi(x)$ is in your case, you can subtitute it into the integrals for $A_n$ and $B_n$ and find all the coefficients.
$endgroup$
– NickD
Jun 24 '17 at 23:49












$begingroup$
$phi(x) = cos(x)$ correct?
$endgroup$
– justanewb
Jun 24 '17 at 23:50




$begingroup$
$phi(x) = cos(x)$ correct?
$endgroup$
– justanewb
Jun 24 '17 at 23:50










1 Answer
1






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oldest

votes


















1












$begingroup$

Assuming $u(t,x)=T(t)X(x)$, the separated equations are
$$
frac{T''}{c^2 T} = lambda, ;; lambda = frac{X''}{X}, ; X'(0)=X'(pi)=0.
$$
The $X$ solutions dictate the values of $lambda$ to be $-n^2$ for $n=1,2,3,cdots$, and the corresponding eigenfunctions are unique up to multiplicative constants, and are given by
$$
X_n(x) = cos(n x),;;; n=0,1,2,3,cdots.
$$
The general solution is
$$
u(x,t) = (A_0+B_0t)+sum_{n=1}^{infty}left(A_ncos(nc t)+B_nsin(nc t)right)cos(n x).
$$
The constants $A_n,B_n$ are determined by the initial conditions:
$$
cos(x) = u(x,0) = A_0+sum_{n=1}^{infty}A_ncos(n x), \
cos^2(x) = u_{t}(x,0) = B_0+sum_{n=1}^{infty}nc B_ncos(n x).
$$
The mutual orthogonality of the functions ${ cos(npi x) }_{n=0}^{infty}$ in $L^2[0,pi]$ is used to determine the coefficients $A_n$ and $B_n$ in the usual manner of Fourier, which is simplified after applying the identity
$$
cos^2(x) = frac{1+cos(2x)}{2}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you give more detail on how to find $A_0,A_n$ and $B_n$ with the identity you mention at the bottom. I am still stuck...
    $endgroup$
    – justanewb
    Jun 29 '17 at 19:14










  • $begingroup$
    @Wolfgang : You can see from the equation $cos(x)=u(x,0)=A_0+sum_{n=1}^{infty}A_ncos(nx)$ that $A_0$ is the coefficient of $1$ in the orthogonal set ${1,cos(x),cos(2x),cdots}$. That means $A_0=0$ and $A_1=1$ and $A_n=0$ for $n > 1$. Then $frac{1+cos(2x)}{2}=B_0+cB_1cos(x)+2cB_2cos(2x)+3cB_3cos(3x)$ gives $B_0=1/2$, $B_1=0$, $B_2=1/4c$, and $B_n=0$ for $n ge 3$.
    $endgroup$
    – DisintegratingByParts
    Jun 29 '17 at 19:21











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1 Answer
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1 Answer
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active

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active

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active

oldest

votes









1












$begingroup$

Assuming $u(t,x)=T(t)X(x)$, the separated equations are
$$
frac{T''}{c^2 T} = lambda, ;; lambda = frac{X''}{X}, ; X'(0)=X'(pi)=0.
$$
The $X$ solutions dictate the values of $lambda$ to be $-n^2$ for $n=1,2,3,cdots$, and the corresponding eigenfunctions are unique up to multiplicative constants, and are given by
$$
X_n(x) = cos(n x),;;; n=0,1,2,3,cdots.
$$
The general solution is
$$
u(x,t) = (A_0+B_0t)+sum_{n=1}^{infty}left(A_ncos(nc t)+B_nsin(nc t)right)cos(n x).
$$
The constants $A_n,B_n$ are determined by the initial conditions:
$$
cos(x) = u(x,0) = A_0+sum_{n=1}^{infty}A_ncos(n x), \
cos^2(x) = u_{t}(x,0) = B_0+sum_{n=1}^{infty}nc B_ncos(n x).
$$
The mutual orthogonality of the functions ${ cos(npi x) }_{n=0}^{infty}$ in $L^2[0,pi]$ is used to determine the coefficients $A_n$ and $B_n$ in the usual manner of Fourier, which is simplified after applying the identity
$$
cos^2(x) = frac{1+cos(2x)}{2}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you give more detail on how to find $A_0,A_n$ and $B_n$ with the identity you mention at the bottom. I am still stuck...
    $endgroup$
    – justanewb
    Jun 29 '17 at 19:14










  • $begingroup$
    @Wolfgang : You can see from the equation $cos(x)=u(x,0)=A_0+sum_{n=1}^{infty}A_ncos(nx)$ that $A_0$ is the coefficient of $1$ in the orthogonal set ${1,cos(x),cos(2x),cdots}$. That means $A_0=0$ and $A_1=1$ and $A_n=0$ for $n > 1$. Then $frac{1+cos(2x)}{2}=B_0+cB_1cos(x)+2cB_2cos(2x)+3cB_3cos(3x)$ gives $B_0=1/2$, $B_1=0$, $B_2=1/4c$, and $B_n=0$ for $n ge 3$.
    $endgroup$
    – DisintegratingByParts
    Jun 29 '17 at 19:21
















1












$begingroup$

Assuming $u(t,x)=T(t)X(x)$, the separated equations are
$$
frac{T''}{c^2 T} = lambda, ;; lambda = frac{X''}{X}, ; X'(0)=X'(pi)=0.
$$
The $X$ solutions dictate the values of $lambda$ to be $-n^2$ for $n=1,2,3,cdots$, and the corresponding eigenfunctions are unique up to multiplicative constants, and are given by
$$
X_n(x) = cos(n x),;;; n=0,1,2,3,cdots.
$$
The general solution is
$$
u(x,t) = (A_0+B_0t)+sum_{n=1}^{infty}left(A_ncos(nc t)+B_nsin(nc t)right)cos(n x).
$$
The constants $A_n,B_n$ are determined by the initial conditions:
$$
cos(x) = u(x,0) = A_0+sum_{n=1}^{infty}A_ncos(n x), \
cos^2(x) = u_{t}(x,0) = B_0+sum_{n=1}^{infty}nc B_ncos(n x).
$$
The mutual orthogonality of the functions ${ cos(npi x) }_{n=0}^{infty}$ in $L^2[0,pi]$ is used to determine the coefficients $A_n$ and $B_n$ in the usual manner of Fourier, which is simplified after applying the identity
$$
cos^2(x) = frac{1+cos(2x)}{2}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you give more detail on how to find $A_0,A_n$ and $B_n$ with the identity you mention at the bottom. I am still stuck...
    $endgroup$
    – justanewb
    Jun 29 '17 at 19:14










  • $begingroup$
    @Wolfgang : You can see from the equation $cos(x)=u(x,0)=A_0+sum_{n=1}^{infty}A_ncos(nx)$ that $A_0$ is the coefficient of $1$ in the orthogonal set ${1,cos(x),cos(2x),cdots}$. That means $A_0=0$ and $A_1=1$ and $A_n=0$ for $n > 1$. Then $frac{1+cos(2x)}{2}=B_0+cB_1cos(x)+2cB_2cos(2x)+3cB_3cos(3x)$ gives $B_0=1/2$, $B_1=0$, $B_2=1/4c$, and $B_n=0$ for $n ge 3$.
    $endgroup$
    – DisintegratingByParts
    Jun 29 '17 at 19:21














1












1








1





$begingroup$

Assuming $u(t,x)=T(t)X(x)$, the separated equations are
$$
frac{T''}{c^2 T} = lambda, ;; lambda = frac{X''}{X}, ; X'(0)=X'(pi)=0.
$$
The $X$ solutions dictate the values of $lambda$ to be $-n^2$ for $n=1,2,3,cdots$, and the corresponding eigenfunctions are unique up to multiplicative constants, and are given by
$$
X_n(x) = cos(n x),;;; n=0,1,2,3,cdots.
$$
The general solution is
$$
u(x,t) = (A_0+B_0t)+sum_{n=1}^{infty}left(A_ncos(nc t)+B_nsin(nc t)right)cos(n x).
$$
The constants $A_n,B_n$ are determined by the initial conditions:
$$
cos(x) = u(x,0) = A_0+sum_{n=1}^{infty}A_ncos(n x), \
cos^2(x) = u_{t}(x,0) = B_0+sum_{n=1}^{infty}nc B_ncos(n x).
$$
The mutual orthogonality of the functions ${ cos(npi x) }_{n=0}^{infty}$ in $L^2[0,pi]$ is used to determine the coefficients $A_n$ and $B_n$ in the usual manner of Fourier, which is simplified after applying the identity
$$
cos^2(x) = frac{1+cos(2x)}{2}.
$$






share|cite|improve this answer









$endgroup$



Assuming $u(t,x)=T(t)X(x)$, the separated equations are
$$
frac{T''}{c^2 T} = lambda, ;; lambda = frac{X''}{X}, ; X'(0)=X'(pi)=0.
$$
The $X$ solutions dictate the values of $lambda$ to be $-n^2$ for $n=1,2,3,cdots$, and the corresponding eigenfunctions are unique up to multiplicative constants, and are given by
$$
X_n(x) = cos(n x),;;; n=0,1,2,3,cdots.
$$
The general solution is
$$
u(x,t) = (A_0+B_0t)+sum_{n=1}^{infty}left(A_ncos(nc t)+B_nsin(nc t)right)cos(n x).
$$
The constants $A_n,B_n$ are determined by the initial conditions:
$$
cos(x) = u(x,0) = A_0+sum_{n=1}^{infty}A_ncos(n x), \
cos^2(x) = u_{t}(x,0) = B_0+sum_{n=1}^{infty}nc B_ncos(n x).
$$
The mutual orthogonality of the functions ${ cos(npi x) }_{n=0}^{infty}$ in $L^2[0,pi]$ is used to determine the coefficients $A_n$ and $B_n$ in the usual manner of Fourier, which is simplified after applying the identity
$$
cos^2(x) = frac{1+cos(2x)}{2}.
$$







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answered Jun 25 '17 at 1:18









DisintegratingByPartsDisintegratingByParts

59.3k42580




59.3k42580












  • $begingroup$
    Can you give more detail on how to find $A_0,A_n$ and $B_n$ with the identity you mention at the bottom. I am still stuck...
    $endgroup$
    – justanewb
    Jun 29 '17 at 19:14










  • $begingroup$
    @Wolfgang : You can see from the equation $cos(x)=u(x,0)=A_0+sum_{n=1}^{infty}A_ncos(nx)$ that $A_0$ is the coefficient of $1$ in the orthogonal set ${1,cos(x),cos(2x),cdots}$. That means $A_0=0$ and $A_1=1$ and $A_n=0$ for $n > 1$. Then $frac{1+cos(2x)}{2}=B_0+cB_1cos(x)+2cB_2cos(2x)+3cB_3cos(3x)$ gives $B_0=1/2$, $B_1=0$, $B_2=1/4c$, and $B_n=0$ for $n ge 3$.
    $endgroup$
    – DisintegratingByParts
    Jun 29 '17 at 19:21


















  • $begingroup$
    Can you give more detail on how to find $A_0,A_n$ and $B_n$ with the identity you mention at the bottom. I am still stuck...
    $endgroup$
    – justanewb
    Jun 29 '17 at 19:14










  • $begingroup$
    @Wolfgang : You can see from the equation $cos(x)=u(x,0)=A_0+sum_{n=1}^{infty}A_ncos(nx)$ that $A_0$ is the coefficient of $1$ in the orthogonal set ${1,cos(x),cos(2x),cdots}$. That means $A_0=0$ and $A_1=1$ and $A_n=0$ for $n > 1$. Then $frac{1+cos(2x)}{2}=B_0+cB_1cos(x)+2cB_2cos(2x)+3cB_3cos(3x)$ gives $B_0=1/2$, $B_1=0$, $B_2=1/4c$, and $B_n=0$ for $n ge 3$.
    $endgroup$
    – DisintegratingByParts
    Jun 29 '17 at 19:21
















$begingroup$
Can you give more detail on how to find $A_0,A_n$ and $B_n$ with the identity you mention at the bottom. I am still stuck...
$endgroup$
– justanewb
Jun 29 '17 at 19:14




$begingroup$
Can you give more detail on how to find $A_0,A_n$ and $B_n$ with the identity you mention at the bottom. I am still stuck...
$endgroup$
– justanewb
Jun 29 '17 at 19:14












$begingroup$
@Wolfgang : You can see from the equation $cos(x)=u(x,0)=A_0+sum_{n=1}^{infty}A_ncos(nx)$ that $A_0$ is the coefficient of $1$ in the orthogonal set ${1,cos(x),cos(2x),cdots}$. That means $A_0=0$ and $A_1=1$ and $A_n=0$ for $n > 1$. Then $frac{1+cos(2x)}{2}=B_0+cB_1cos(x)+2cB_2cos(2x)+3cB_3cos(3x)$ gives $B_0=1/2$, $B_1=0$, $B_2=1/4c$, and $B_n=0$ for $n ge 3$.
$endgroup$
– DisintegratingByParts
Jun 29 '17 at 19:21




$begingroup$
@Wolfgang : You can see from the equation $cos(x)=u(x,0)=A_0+sum_{n=1}^{infty}A_ncos(nx)$ that $A_0$ is the coefficient of $1$ in the orthogonal set ${1,cos(x),cos(2x),cdots}$. That means $A_0=0$ and $A_1=1$ and $A_n=0$ for $n > 1$. Then $frac{1+cos(2x)}{2}=B_0+cB_1cos(x)+2cB_2cos(2x)+3cB_3cos(3x)$ gives $B_0=1/2$, $B_1=0$, $B_2=1/4c$, and $B_n=0$ for $n ge 3$.
$endgroup$
– DisintegratingByParts
Jun 29 '17 at 19:21


















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