Mixing
regular Cauchy data for wave equation
$begingroup$
From Salsa, PDEs in Action, chapter 5.
Let $w_phi$ be the solution of the Cauchy problem
begin{array}{l}
w_{tt}-c^2 Delta w=0 quad xinmathbb{R}^3,; t>0 \
w(x,0) = 0 quad xinmathbb{R}^3 \
w_t(x,0)=phi(x) quad xinmathbb{R}^3
end{array}
such that $win C^3(mathbb{R}^3times[0,+infty))$.
Then $vequiv partial_t w_phi$ is the solution of the Cauchy problem
begin{array}{l}
v_{tt}-c^2 Delta v=0 quad xinmathbb{R}^3,; t>0 \
v(x,0) = phi(x) quad xinmathbb{R}^3 \
v_t(x,0)=0 quad xinmathbb{R}^3
end{array}
This lemma, combined with the superposition principle, simplifies the solution of the Cauchy problem with both Cauchy data.
I cannot figure out how to verify that $ v_t(x,0) = 0$.
Here are the steps I have made:
begin{equation*}
v_t(x,0) = partial_{tt} w_phi(x,0)
= lim_{tto 0^+} partial_{tt} w_phi(x,t)
= lim_{tto 0^+} c^2Delta w_phi(x,t)
= c^2Delta w_phi(x,0)
end{equation*}
where the limits hold due to the regularity of $w_phi$.
Why is the last term equal to zero?
In other words, why is $w_phi(cdot,0)$ harmonic?
Thanks in advance.
pde wave-equation initial-value-problems
asked Sep 3 '18 at 15:34
Giovanni MarianiGiovanni Mariani
335
$endgroup$
add a comment |
$begingroup$
From Salsa, PDEs in Action, chapter 5.
Let $w_phi$ be the solution of the Cauchy problem
begin{array}{l}
w_{tt}-c^2 Delta w=0 quad xinmathbb{R}^3,; t>0 \
w(x,0) = 0 quad xinmathbb{R}^3 \
w_t(x,0)=phi(x) quad xinmathbb{R}^3
end{array}
such that $win C^3(mathbb{R}^3times[0,+infty))$.
Then $vequiv partial_t w_phi$ is the solution of the Cauchy problem
begin{array}{l}
v_{tt}-c^2 Delta v=0 quad xinmathbb{R}^3,; t>0 \
v(x,0) = phi(x) quad xinmathbb{R}^3 \
v_t(x,0)=0 quad xinmathbb{R}^3
end{array}
This lemma, combined with the superposition principle, simplifies the solution of the Cauchy problem with both Cauchy data.
I cannot figure out how to verify that $ v_t(x,0) = 0$.
Here are the steps I have made:
begin{equation*}
v_t(x,0) = partial_{tt} w_phi(x,0)
= lim_{tto 0^+} partial_{tt} w_phi(x,t)
= lim_{tto 0^+} c^2Delta w_phi(x,t)
= c^2Delta w_phi(x,0)
end{equation*}
where the limits hold due to the regularity of $w_phi$.
Why is the last term equal to zero?
In other words, why is $w_phi(cdot,0)$ harmonic?
Thanks in advance.
pde wave-equation initial-value-problems
asked Sep 3 '18 at 15:34
Giovanni MarianiGiovanni Mariani
335
$endgroup$
1
$begingroup$
You have specified that $w(x, 0) = 0$. Doesn't this imply $Delta w_phi(x, 0) = 0$, since $w_phi = w$ when $t = 0$?
$endgroup$
– Robert Lewis
Sep 3 '18 at 15:45
1
$begingroup$
Then why bother using the wave equation in the first place? Since $v(x,0)=phi(x)$, then $v_t(x,0)=0$ because $phi$ does not depend on $t$. I have always thought that you should apply the operator first and then calculate the value of the function on the specified point.
$endgroup$
– Giovanni Mariani
Sep 3 '18 at 15:51
$begingroup$
Those are two different concepts: the difference between $lim_{x rightarrow 0} frac{d}{dx}(w(x))$ and $frac{d}{dx}(lim_{x rightarrow 0} w(x))$
$endgroup$
– DaveNine
Sep 3 '18 at 18:03
$begingroup$
@DaveNine and that's exactly my point--and my issue: $Delta w_phi (x,0) =lim_{tto 0^+} Delta w_phi (x,t) neq Delta lim_{tto 0^+} w_phi (x,t) =Delta w_phi (x,0) =Delta 0 = 0$
$endgroup$
– Giovanni Mariani
Sep 10 '18 at 18:52
add a comment |
$begingroup$
From Salsa, PDEs in Action, chapter 5.
Let $w_phi$ be the solution of the Cauchy problem
begin{array}{l}
w_{tt}-c^2 Delta w=0 quad xinmathbb{R}^3,; t>0 \
w(x,0) = 0 quad xinmathbb{R}^3 \
w_t(x,0)=phi(x) quad xinmathbb{R}^3
end{array}
such that $win C^3(mathbb{R}^3times[0,+infty))$.
Then $vequiv partial_t w_phi$ is the solution of the Cauchy problem
begin{array}{l}
v_{tt}-c^2 Delta v=0 quad xinmathbb{R}^3,; t>0 \
v(x,0) = phi(x) quad xinmathbb{R}^3 \
v_t(x,0)=0 quad xinmathbb{R}^3
end{array}
This lemma, combined with the superposition principle, simplifies the solution of the Cauchy problem with both Cauchy data.
I cannot figure out how to verify that $ v_t(x,0) = 0$.
Here are the steps I have made:
begin{equation*}
v_t(x,0) = partial_{tt} w_phi(x,0)
= lim_{tto 0^+} partial_{tt} w_phi(x,t)
= lim_{tto 0^+} c^2Delta w_phi(x,t)
= c^2Delta w_phi(x,0)
end{equation*}
where the limits hold due to the regularity of $w_phi$.
Why is the last term equal to zero?
In other words, why is $w_phi(cdot,0)$ harmonic?
Thanks in advance.
pde wave-equation initial-value-problems
asked Sep 3 '18 at 15:34
Giovanni MarianiGiovanni Mariani
335
$endgroup$
From Salsa, PDEs in Action, chapter 5.
Let $w_phi$ be the solution of the Cauchy problem
begin{array}{l}
w_{tt}-c^2 Delta w=0 quad xinmathbb{R}^3,; t>0 \
w(x,0) = 0 quad xinmathbb{R}^3 \
w_t(x,0)=phi(x) quad xinmathbb{R}^3
end{array}
such that $win C^3(mathbb{R}^3times[0,+infty))$.
Then $vequiv partial_t w_phi$ is the solution of the Cauchy problem
begin{array}{l}
v_{tt}-c^2 Delta v=0 quad xinmathbb{R}^3,; t>0 \
v(x,0) = phi(x) quad xinmathbb{R}^3 \
v_t(x,0)=0 quad xinmathbb{R}^3
end{array}
This lemma, combined with the superposition principle, simplifies the solution of the Cauchy problem with both Cauchy data.
I cannot figure out how to verify that $ v_t(x,0) = 0$.
Here are the steps I have made:
begin{equation*}
v_t(x,0) = partial_{tt} w_phi(x,0)
= lim_{tto 0^+} partial_{tt} w_phi(x,t)
= lim_{tto 0^+} c^2Delta w_phi(x,t)
= c^2Delta w_phi(x,0)
end{equation*}
where the limits hold due to the regularity of $w_phi$.
Why is the last term equal to zero?
In other words, why is $w_phi(cdot,0)$ harmonic?
Thanks in advance.
pde wave-equation initial-value-problems
pde wave-equation initial-value-problems
asked Sep 3 '18 at 15:34
Giovanni MarianiGiovanni Mariani
335
asked Sep 3 '18 at 15:34
Giovanni MarianiGiovanni Mariani
335
edited Dec 24 '18 at 12:52
Giovanni Mariani
asked Sep 3 '18 at 15:34
Giovanni MarianiGiovanni Mariani
335
asked Sep 3 '18 at 15:34
Giovanni MarianiGiovanni Mariani
335
asked Sep 3 '18 at 15:34
Giovanni MarianiGiovanni Mariani
335
335
1
$begingroup$
You have specified that $w(x, 0) = 0$. Doesn't this imply $Delta w_phi(x, 0) = 0$, since $w_phi = w$ when $t = 0$?
$endgroup$
– Robert Lewis
Sep 3 '18 at 15:45
1
$begingroup$
Then why bother using the wave equation in the first place? Since $v(x,0)=phi(x)$, then $v_t(x,0)=0$ because $phi$ does not depend on $t$. I have always thought that you should apply the operator first and then calculate the value of the function on the specified point.
$endgroup$
– Giovanni Mariani
Sep 3 '18 at 15:51
$begingroup$
Those are two different concepts: the difference between $lim_{x rightarrow 0} frac{d}{dx}(w(x))$ and $frac{d}{dx}(lim_{x rightarrow 0} w(x))$
$endgroup$
– DaveNine
Sep 3 '18 at 18:03
$begingroup$
@DaveNine and that's exactly my point--and my issue: $Delta w_phi (x,0) =lim_{tto 0^+} Delta w_phi (x,t) neq Delta lim_{tto 0^+} w_phi (x,t) =Delta w_phi (x,0) =Delta 0 = 0$
$endgroup$
– Giovanni Mariani
Sep 10 '18 at 18:52
add a comment |
1
$begingroup$
You have specified that $w(x, 0) = 0$. Doesn't this imply $Delta w_phi(x, 0) = 0$, since $w_phi = w$ when $t = 0$?
$endgroup$
– Robert Lewis
Sep 3 '18 at 15:45
1
$begingroup$
Then why bother using the wave equation in the first place? Since $v(x,0)=phi(x)$, then $v_t(x,0)=0$ because $phi$ does not depend on $t$. I have always thought that you should apply the operator first and then calculate the value of the function on the specified point.
$endgroup$
– Giovanni Mariani
Sep 3 '18 at 15:51
$begingroup$
Those are two different concepts: the difference between $lim_{x rightarrow 0} frac{d}{dx}(w(x))$ and $frac{d}{dx}(lim_{x rightarrow 0} w(x))$
$endgroup$
– DaveNine
Sep 3 '18 at 18:03
$begingroup$
@DaveNine and that's exactly my point--and my issue: $Delta w_phi (x,0) =lim_{tto 0^+} Delta w_phi (x,t) neq Delta lim_{tto 0^+} w_phi (x,t) =Delta w_phi (x,0) =Delta 0 = 0$
$endgroup$
– Giovanni Mariani
Sep 10 '18 at 18:52
1
1
$begingroup$
You have specified that $w(x, 0) = 0$. Doesn't this imply $Delta w_phi(x, 0) = 0$, since $w_phi = w$ when $t = 0$?
$endgroup$
– Robert Lewis
Sep 3 '18 at 15:45
$begingroup$
You have specified that $w(x, 0) = 0$. Doesn't this imply $Delta w_phi(x, 0) = 0$, since $w_phi = w$ when $t = 0$?
$endgroup$
– Robert Lewis
Sep 3 '18 at 15:45
1
1
$begingroup$
Then why bother using the wave equation in the first place? Since $v(x,0)=phi(x)$, then $v_t(x,0)=0$ because $phi$ does not depend on $t$. I have always thought that you should apply the operator first and then calculate the value of the function on the specified point.
$endgroup$
– Giovanni Mariani
Sep 3 '18 at 15:51
$begingroup$
Then why bother using the wave equation in the first place? Since $v(x,0)=phi(x)$, then $v_t(x,0)=0$ because $phi$ does not depend on $t$. I have always thought that you should apply the operator first and then calculate the value of the function on the specified point.
$endgroup$
– Giovanni Mariani
Sep 3 '18 at 15:51
$begingroup$
Those are two different concepts: the difference between $lim_{x rightarrow 0} frac{d}{dx}(w(x))$ and $frac{d}{dx}(lim_{x rightarrow 0} w(x))$
$endgroup$
– DaveNine
Sep 3 '18 at 18:03
$begingroup$
Those are two different concepts: the difference between $lim_{x rightarrow 0} frac{d}{dx}(w(x))$ and $frac{d}{dx}(lim_{x rightarrow 0} w(x))$
$endgroup$
– DaveNine
Sep 3 '18 at 18:03
$begingroup$
@DaveNine and that's exactly my point--and my issue: $Delta w_phi (x,0) =lim_{tto 0^+} Delta w_phi (x,t) neq Delta lim_{tto 0^+} w_phi (x,t) =Delta w_phi (x,0) =Delta 0 = 0$
$endgroup$
– Giovanni Mariani
Sep 10 '18 at 18:52
$begingroup$
@DaveNine and that's exactly my point--and my issue: $Delta w_phi (x,0) =lim_{tto 0^+} Delta w_phi (x,t) neq Delta lim_{tto 0^+} w_phi (x,t) =Delta w_phi (x,0) =Delta 0 = 0$
$endgroup$
– Giovanni Mariani
Sep 10 '18 at 18:52
add a comment |
1 Answer
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oldest
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$begingroup$
I have found a reason why $c^2 Delta w_phi(x,0)=0$.
Since the Laplacian acts only on the space variables $x$ and leaves the time variable $t$ unchanged, we have
$$ (Delta w_phi)(x,0) = Delta (w_phi(x,0)) = 0 quadtext{since}quad w_phi(x,0)=0 $$
In other words, taking the Laplacian and then letting $t=0$ or vice-versa yield the same result.
Incidentally, this is why one must use the wave equation via the limits and cannot simply stop at $partial_{tt} w_phi$: actually
$$ (partial_{tt} w_phi)(x,0) neq partial_{tt} (w_phi(x,0))$$
My comment above on this point was misleading.
answered Jan 6 at 22:47
Giovanni MarianiGiovanni Mariani
335
$endgroup$
add a comment |
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$begingroup$
I have found a reason why $c^2 Delta w_phi(x,0)=0$.
Since the Laplacian acts only on the space variables $x$ and leaves the time variable $t$ unchanged, we have
$$ (Delta w_phi)(x,0) = Delta (w_phi(x,0)) = 0 quadtext{since}quad w_phi(x,0)=0 $$
In other words, taking the Laplacian and then letting $t=0$ or vice-versa yield the same result.
Incidentally, this is why one must use the wave equation via the limits and cannot simply stop at $partial_{tt} w_phi$: actually
$$ (partial_{tt} w_phi)(x,0) neq partial_{tt} (w_phi(x,0))$$
My comment above on this point was misleading.
answered Jan 6 at 22:47
Giovanni MarianiGiovanni Mariani
335
$endgroup$
add a comment |
$begingroup$
I have found a reason why $c^2 Delta w_phi(x,0)=0$.
Since the Laplacian acts only on the space variables $x$ and leaves the time variable $t$ unchanged, we have
$$ (Delta w_phi)(x,0) = Delta (w_phi(x,0)) = 0 quadtext{since}quad w_phi(x,0)=0 $$
In other words, taking the Laplacian and then letting $t=0$ or vice-versa yield the same result.
Incidentally, this is why one must use the wave equation via the limits and cannot simply stop at $partial_{tt} w_phi$: actually
$$ (partial_{tt} w_phi)(x,0) neq partial_{tt} (w_phi(x,0))$$
My comment above on this point was misleading.
answered Jan 6 at 22:47
Giovanni MarianiGiovanni Mariani
335
$endgroup$
add a comment |
$begingroup$
I have found a reason why $c^2 Delta w_phi(x,0)=0$.
Since the Laplacian acts only on the space variables $x$ and leaves the time variable $t$ unchanged, we have
$$ (Delta w_phi)(x,0) = Delta (w_phi(x,0)) = 0 quadtext{since}quad w_phi(x,0)=0 $$
In other words, taking the Laplacian and then letting $t=0$ or vice-versa yield the same result.
Incidentally, this is why one must use the wave equation via the limits and cannot simply stop at $partial_{tt} w_phi$: actually
$$ (partial_{tt} w_phi)(x,0) neq partial_{tt} (w_phi(x,0))$$
My comment above on this point was misleading.
answered Jan 6 at 22:47
Giovanni MarianiGiovanni Mariani
335
$endgroup$
I have found a reason why $c^2 Delta w_phi(x,0)=0$.
Since the Laplacian acts only on the space variables $x$ and leaves the time variable $t$ unchanged, we have
$$ (Delta w_phi)(x,0) = Delta (w_phi(x,0)) = 0 quadtext{since}quad w_phi(x,0)=0 $$
In other words, taking the Laplacian and then letting $t=0$ or vice-versa yield the same result.
Incidentally, this is why one must use the wave equation via the limits and cannot simply stop at $partial_{tt} w_phi$: actually
$$ (partial_{tt} w_phi)(x,0) neq partial_{tt} (w_phi(x,0))$$
My comment above on this point was misleading.
answered Jan 6 at 22:47
Giovanni MarianiGiovanni Mariani
335
edited Jan 6 at 22:53
answered Jan 6 at 22:47
Giovanni MarianiGiovanni Mariani
335
answered Jan 6 at 22:47
Giovanni MarianiGiovanni Mariani
335
answered Jan 6 at 22:47
Giovanni MarianiGiovanni Mariani
335
335
add a comment |
add a comment |
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$begingroup$
You have specified that $w(x, 0) = 0$. Doesn't this imply $Delta w_phi(x, 0) = 0$, since $w_phi = w$ when $t = 0$?
$endgroup$
– Robert Lewis
Sep 3 '18 at 15:45
1
$begingroup$
Then why bother using the wave equation in the first place? Since $v(x,0)=phi(x)$, then $v_t(x,0)=0$ because $phi$ does not depend on $t$. I have always thought that you should apply the operator first and then calculate the value of the function on the specified point.
$endgroup$
– Giovanni Mariani
Sep 3 '18 at 15:51
$begingroup$
Those are two different concepts: the difference between $lim_{x rightarrow 0} frac{d}{dx}(w(x))$ and $frac{d}{dx}(lim_{x rightarrow 0} w(x))$
$endgroup$
– DaveNine
Sep 3 '18 at 18:03
$begingroup$
@DaveNine and that's exactly my point--and my issue: $Delta w_phi (x,0) =lim_{tto 0^+} Delta w_phi (x,t) neq Delta lim_{tto 0^+} w_phi (x,t) =Delta w_phi (x,0) =Delta 0 = 0$
$endgroup$
– Giovanni Mariani
Sep 10 '18 at 18:52