Show that ##f(x,y)=u(x+cy)+v(x-cy)## is a solution of the given PDE

[chwala]

Homework Statement

See attached.

Relevant Equations

wave equation.

Looks pretty straightforward, i approached it as follows,

$f_x = u(x+cy) + v(x-cy)$

$f_{xx}=u(x+cy) + v(x-cy)$

$f_y= cu(x+cy) -cv(x-cy)$

$f_{yy}=c^2u(x+cy)+c^2v(x-cy)$

Therefore,

$f_{xx} -\dfrac{1}{c^2} f_{yy} = u(x+cy) + v(x-cy) - \dfrac{1}{c^2}⋅ c^2 \left[u(x+cy)+v(x-cy) \right]$

$f_{xx} -\dfrac{1}{c^2} f_{yy} = u(x+cy) + v(x-cy) - u(x+cy) - v(x-cy) =0$

thus shown.

Unless there is a different way to look at it...cheers.

Amendment in post $5$.

 

[Orodruin]

Looks ok, but as always you can of course also look at it in a different way.

Consider the wave front coordinates:
$$
\xi = x - vt,\qquad \eta = x + vt
$$
and use the chain rule to rewrite the wave equation in terms of those coordinates. The resulting PDE is very straightforward to integrate.

 

[PeroK]

Looks pretty straightforward, i approached it as follows,

$f_x = u(x+cy) + v(x-cy)$

Shouldn't the derivatives of $u$ and $v$ appear somewhere in that equation?

 

[Steve4Physics]

Homework Statement: See attached.
Relevant Equations: wave equation.

View attachment 336265

Looks pretty straightforward, i approached it as follows,

$f_x = u(x+cy) + v(x-cy)$

Are you sure the above is correct? It seems you are saying that $f = f_x$ (as both would be equal to $u(x+cy) + v(x-cy)$.

Note that $u$ and $v$ can be any twice differntiable functions of a single variable. For example $u(z) = z^2$ and $v(z)=z^3$. These would give $f(x,y) = (x+cy)^2 + (x-cy)^3$.

It should be clear that $f_x$ and $f$ would then be different functions of $x$ and $y$.

 

[chwala]

Shouldn't the derivatives of $u$ and $v$ appear somewhere in that equation?

I see...i ought to make use of chain rule differentiation ...let me amend and repost. A minute.

...should be,

$f_x = u^{'}(x+cy) + v^{'}(x-cy)$

$f_{xx}=u^{''}(x+cy) + v^{''}(x-cy)$

$f_y= cu^{'}(x+cy) -cv^{'}(x-cy)$

$f_{yy}=c^2u^{''}(x+cy)+c^2v^{''}(x-cy)$

Therefore,

$f_{xx} -\dfrac{1}{c^2} f_{yy} = u^{''}(x+cy) + v^{''}(x-cy) - \dfrac{1}{c^2}⋅ c^2 \left[u^{''}(x+cy)+v^{''}(x-cy) \right]$

$f_{xx} -\dfrac{1}{c^2} f_{yy} = u^{''}(x+cy) + v^{''}(x-cy) - u^{''}(x+cy) - v^{''}(x-cy) =0$

 

[Orodruin]

Shouldn't the derivatives of $u$ and $v$ appear somewhere in that equation?

I guess I am so used to this argument that my mind added the derivatives by itself …

 

[Steve4Physics]

...should be,

$f_x = u^{'}(x+cy) + v^{'}(x-cy)$

$f_{xx}=u^{''}(x+cy) + v^{''}(x-cy)$

$f_y= cu^{'}(x+cy) -cv^{'}(x-cy)$

$f_{yy}=c^2u^{''}(x+cy)+c^2v^{''}(x-cy)$

Therefore,

$f_{xx} -\dfrac{1}{c^2} f_{yy} = u^{''}(x+cy) + v^{''}(x-cy) - \dfrac{1}{c^2}⋅ c^2 \left[u^{''}(x+cy)+v^{''}(x-cy) \right]$

$f_{xx} -\dfrac{1}{c^2} f_{yy} = u^{''}(x+cy) + v^{''}(x-cy) - u^{''}(x+cy) - v^{''}(x-cy) =0$

Edit: My reply was b*ll*cks so I've struck it through.

Be careful because your notation is ambiguous.

For example, does $u^{’}$ mean $\frac {\partial u}{\partial x}$ or $\frac {\partial u}{\partial y}$? You have used $u^{’}$ to mean both! Similarly for $u^{''}$.

 

[Orodruin]

Be careful because your notation is ambiguous.

For example, does $u^{’}$ mean $\frac {\partial u}{\partial x}$ or $\frac {\partial u}{\partial y}$? You have used $u^{’}$ to mean both! Similarly for $u^{''}$.

Neither, it means $u'$. The function $u$ is a function of one parameter only and the derivative is with respect to that parameter. There is nothing ambiguous about this notation. That you then put a composite function as the argument is another issue.

 

[Steve4Physics]

Neither, it means $u'$. The function $u$ is a function of one parameter only and the derivative is with respect to that parameter. There is nothing ambiguous about this notation. That you then put a composite function as the argument is another issue.

Aagh. Yes, I was being daft.

 

[chwala]

Looks ok, but as always you can of course also look at it in a different way.

Consider the wave front coordinates:
$$
\xi = x - vt,\qquad \eta = x + vt
$$
and use the chain rule to rewrite the wave equation in terms of those coordinates. The resulting PDE is very straightforward to integrate.

I will look at this and get back. Looks like characteristic equation ...