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Wave equation - use parallelogram rule to solve the problem

0

1

$begingroup$

Given wave equation: $ u_tt-c^2u_xx=0 $, let u be a solution.

Points A, B, C, D are vertices of parallelogram of two pairs of characteristic lines: $ x-ct=c1, x-ct=c2, x+ct=d1, x+ct=d2 $

Use parallelogram rule to find u that satisfies this:

$ u_tt-u_xx=0, u(-a,a)=a,$ $ $ $ u(a,a)=a^2 $ for $ a>0 $

Guys, how can I use parallelogram rule here? I know that if we draw the characteristic lines we will get picture like that.

pde wave-equation

share|cite|improve this question

asked Mar 27 at 20:29

Alpys RauanAlpys Rauan

111

$endgroup$

add a comment | 

0

1

$begingroup$

Given wave equation: $ u_tt-c^2u_xx=0 $, let u be a solution.

Points A, B, C, D are vertices of parallelogram of two pairs of characteristic lines: $ x-ct=c1, x-ct=c2, x+ct=d1, x+ct=d2 $

Use parallelogram rule to find u that satisfies this:

$ u_tt-u_xx=0, u(-a,a)=a,$ $ $ $ u(a,a)=a^2 $ for $ a>0 $

Guys, how can I use parallelogram rule here? I know that if we draw the characteristic lines we will get picture like that.

pde wave-equation

share|cite|improve this question

asked Mar 27 at 20:29

Alpys RauanAlpys Rauan

111

$endgroup$

add a comment | 

$begingroup$

Given wave equation: $ u_tt-c^2u_xx=0 $, let u be a solution.

Points A, B, C, D are vertices of parallelogram of two pairs of characteristic lines: $ x-ct=c1, x-ct=c2, x+ct=d1, x+ct=d2 $

Use parallelogram rule to find u that satisfies this:

$ u_tt-u_xx=0, u(-a,a)=a,$ $ $ $ u(a,a)=a^2 $ for $ a>0 $

Guys, how can I use parallelogram rule here? I know that if we draw the characteristic lines we will get picture like that.

pde wave-equation

share|cite|improve this question

asked Mar 27 at 20:29

Alpys RauanAlpys Rauan

111

$endgroup$

share|cite|improve this question

asked Mar 27 at 20:29

Alpys RauanAlpys Rauan

111

share|cite|improve this question

asked Mar 27 at 20:29

Alpys RauanAlpys Rauan

111

asked Mar 27 at 20:29

Alpys RauanAlpys Rauan

111

asked Mar 27 at 20:29

Alpys RauanAlpys Rauan

111

1 Answer
1

active

oldest

votes

0

$begingroup$

You are given data on the "angle" $y=-x,, xle 0$ and $y=x,, xge 0$. If you take any point $P(x_0,y_0)$ inside this angle (above the given lines) and you draw characteristic lines parallel to the sides of the angle, the value of $u$ there is the sum of the values carried by the characteristics (this is what the general solution of the wave equation is). The intersections with the sides of the angle are $(fracx_0+y_02, fracx_0+y_02)$ (characteristic with slope $-1$) and $(fracx_0-y_02, fracy_0-x_02)$ (characteristic with slope $+1$). Then,
$$
u(x_0,y_0)=fracy_0-x_02+left(fracx_0+y_02right)^2
$$
The solution is not determined uniquely outside of the angle.

The parallelogram rule just tells you that the sums of the values of the solution at opposite vertices of a parallelogram are equal. In your case the parallelogram is formed by $(x_0,y_0)$, $(0,0)$ and the intersections of the characteristics with the initial angle I mentioned before. You get the same answer of course.

Hope this helps.

share|cite|improve this answer

edited Mar 28 at 2:59

answered Mar 27 at 23:09

GReyesGReyes

2,57315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear GReyes, can you explain how did you find that equation about ?
  $endgroup$
  – Alpys Rauan
  Mar 29 at 12:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The parallelogram rule is a direct consequence of the form of the general solution $u(x,t)=F(x+ct)+G(x-ct)$ of the wave equation $u_tt=c^2u_xx$. If you form a parallelogram in the $(x,t)$-plane with vertices $A,B,C,D$ (say, in that order clockwise) with slopes $+c$ and $-c$, then $u(A)+u(C)=u(B)+u(D)$. This is a straightforward consequence of the fact that $F$ is a constant along two of the opposite sides and $G$ is constant along the other two. You choose your vertices $A(0,0)$, $C(x_0,y_0)$, and $B$ and $D$ are the intersections of the characteristics with your initial angle.
  $endgroup$
  – GReyes
  Mar 30 at 16:21
  

  
  
  
  

add a comment | 

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0

$begingroup$

You are given data on the "angle" $y=-x,, xle 0$ and $y=x,, xge 0$. If you take any point $P(x_0,y_0)$ inside this angle (above the given lines) and you draw characteristic lines parallel to the sides of the angle, the value of $u$ there is the sum of the values carried by the characteristics (this is what the general solution of the wave equation is). The intersections with the sides of the angle are $(fracx_0+y_02, fracx_0+y_02)$ (characteristic with slope $-1$) and $(fracx_0-y_02, fracy_0-x_02)$ (characteristic with slope $+1$). Then,
$$
u(x_0,y_0)=fracy_0-x_02+left(fracx_0+y_02right)^2
$$
The solution is not determined uniquely outside of the angle.

The parallelogram rule just tells you that the sums of the values of the solution at opposite vertices of a parallelogram are equal. In your case the parallelogram is formed by $(x_0,y_0)$, $(0,0)$ and the intersections of the characteristics with the initial angle I mentioned before. You get the same answer of course.

Hope this helps.

share|cite|improve this answer

edited Mar 28 at 2:59

answered Mar 27 at 23:09

GReyesGReyes

2,57315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear GReyes, can you explain how did you find that equation about ?
  $endgroup$
  – Alpys Rauan
  Mar 29 at 12:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The parallelogram rule is a direct consequence of the form of the general solution $u(x,t)=F(x+ct)+G(x-ct)$ of the wave equation $u_tt=c^2u_xx$. If you form a parallelogram in the $(x,t)$-plane with vertices $A,B,C,D$ (say, in that order clockwise) with slopes $+c$ and $-c$, then $u(A)+u(C)=u(B)+u(D)$. This is a straightforward consequence of the fact that $F$ is a constant along two of the opposite sides and $G$ is constant along the other two. You choose your vertices $A(0,0)$, $C(x_0,y_0)$, and $B$ and $D$ are the intersections of the characteristics with your initial angle.
  $endgroup$
  – GReyes
  Mar 30 at 16:21
  

  
  
  
  

add a comment | 

0

$begingroup$

You are given data on the "angle" $y=-x,, xle 0$ and $y=x,, xge 0$. If you take any point $P(x_0,y_0)$ inside this angle (above the given lines) and you draw characteristic lines parallel to the sides of the angle, the value of $u$ there is the sum of the values carried by the characteristics (this is what the general solution of the wave equation is). The intersections with the sides of the angle are $(fracx_0+y_02, fracx_0+y_02)$ (characteristic with slope $-1$) and $(fracx_0-y_02, fracy_0-x_02)$ (characteristic with slope $+1$). Then,
$$
u(x_0,y_0)=fracy_0-x_02+left(fracx_0+y_02right)^2
$$
The solution is not determined uniquely outside of the angle.

The parallelogram rule just tells you that the sums of the values of the solution at opposite vertices of a parallelogram are equal. In your case the parallelogram is formed by $(x_0,y_0)$, $(0,0)$ and the intersections of the characteristics with the initial angle I mentioned before. You get the same answer of course.

Hope this helps.

share|cite|improve this answer

edited Mar 28 at 2:59

answered Mar 27 at 23:09

GReyesGReyes

2,57315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear GReyes, can you explain how did you find that equation about ?
  $endgroup$
  – Alpys Rauan
  Mar 29 at 12:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The parallelogram rule is a direct consequence of the form of the general solution $u(x,t)=F(x+ct)+G(x-ct)$ of the wave equation $u_tt=c^2u_xx$. If you form a parallelogram in the $(x,t)$-plane with vertices $A,B,C,D$ (say, in that order clockwise) with slopes $+c$ and $-c$, then $u(A)+u(C)=u(B)+u(D)$. This is a straightforward consequence of the fact that $F$ is a constant along two of the opposite sides and $G$ is constant along the other two. You choose your vertices $A(0,0)$, $C(x_0,y_0)$, and $B$ and $D$ are the intersections of the characteristics with your initial angle.
  $endgroup$
  – GReyes
  Mar 30 at 16:21
  

  
  
  
  

add a comment | 

$begingroup$

You are given data on the "angle" $y=-x,, xle 0$ and $y=x,, xge 0$. If you take any point $P(x_0,y_0)$ inside this angle (above the given lines) and you draw characteristic lines parallel to the sides of the angle, the value of $u$ there is the sum of the values carried by the characteristics (this is what the general solution of the wave equation is). The intersections with the sides of the angle are $(fracx_0+y_02, fracx_0+y_02)$ (characteristic with slope $-1$) and $(fracx_0-y_02, fracy_0-x_02)$ (characteristic with slope $+1$). Then,
$$
u(x_0,y_0)=fracy_0-x_02+left(fracx_0+y_02right)^2
$$
The solution is not determined uniquely outside of the angle.

The parallelogram rule just tells you that the sums of the values of the solution at opposite vertices of a parallelogram are equal. In your case the parallelogram is formed by $(x_0,y_0)$, $(0,0)$ and the intersections of the characteristics with the initial angle I mentioned before. You get the same answer of course.

Hope this helps.

share|cite|improve this answer

edited Mar 28 at 2:59

answered Mar 27 at 23:09

GReyesGReyes

2,57315

$endgroup$

share|cite|improve this answer

edited Mar 28 at 2:59

answered Mar 27 at 23:09

GReyesGReyes

2,57315

answered Mar 27 at 23:09

GReyesGReyes

2,57315

answered Mar 27 at 23:09

GReyesGReyes

2,57315