Verify that $H(x-ct)$ is a weak solution of wave equation $u_tt=c^2u_xx$weak solution of wave equationTwo weak formulations for a time-dependent PDE: are they equivalent?Fundamental solution of wave equation in 3Dreversibility scalar conservation lawweak solution of wave equationIntegration by parts in time derivative of energy integral for wave equationFinite speed of propagation for weak solutions?Entropy solution to scalar conservation lawConvergence of integral (Lebesgue and Bochner spaces)Evans PDE Conservation Law IntegralEnergy equalities and estimates for weak solutions
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Verify that $H(x-ct)$ is a weak solution of wave equation $u_tt=c^2u_xx$
weak solution of wave equationTwo weak formulations for a time-dependent PDE: are they equivalent?Fundamental solution of wave equation in 3Dreversibility scalar conservation lawweak solution of wave equationIntegration by parts in time derivative of energy integral for wave equationFinite speed of propagation for weak solutions?Entropy solution to scalar conservation lawConvergence of integral (Lebesgue and Bochner spaces)Evans PDE Conservation Law IntegralEnergy equalities and estimates for weak solutions
$begingroup$
The problem is in Strauss, Partial Diferential Equations 2nd edition, 12.1 Ex.5.
I want to verify by direct substitution that Heaviside DISTRIBUTION $H(x-ct)$ is a weak solution of wave equation $u_tt=c^2u_xx$.
I found almost identical question at weak solution of wave equation here but I think my problem is slightly different from this, since I considered that $(x,t)inmathbb Rtimes(mathbb R^+cup0)$.
$$int_infty^infty int_0^infty H(x-ct)(phi_tt-c^2phi_xx)dtdx=0$$
should be true for all $phiinmathcal D(mathbb Rtimes(mathbb R^+))$.
Using $phi$ is a $mathcal C^infty$ function with compact support, I found it can be reduced to
$$
beginalign
& int_0^inftyint_0^x/c phi_ttdtdx - c^2int_0^inftyint_ct^infty phi_xxdxdt \
= & int_0^infty (phi_t (x, dfracxc) - phi_t (x, 0))dx + cint_0^inftyphi_x(t,dfractc)dt \
= & int_0^infty cdfracdphids(s,dfracsc)ds-int_0^infty phi_t (x, 0)dx \
= & -cphi(0,0) - int_0^infty phi_t (x, 0)dx.
endalign$$
Since the second term cannot be integrated explicitly, I considered a closed curve on $mathbb Rtimes(mathbb R^+cup0)$ which connects $(0,0),(X,0),(X,T),(0,T)$ by line segments, then send $X$ and $T$ to $+infty$. Since $phi$ and its derivatives vanish over some radius R, the second term would be $-phi(0,0)$, not $-cphi(0,0)$.
Maybe I has made a mistake on integration, but can't find that. Could anyone give me some help?
pde
edited Mar 15 at 18:45
Spencer
8,65012056
asked Mar 15 at 18:13
Jinmu YouJinmu You
339210
$endgroup$
add a comment |
$begingroup$
The problem is in Strauss, Partial Diferential Equations 2nd edition, 12.1 Ex.5.
I want to verify by direct substitution that Heaviside DISTRIBUTION $H(x-ct)$ is a weak solution of wave equation $u_tt=c^2u_xx$.
I found almost identical question at weak solution of wave equation here but I think my problem is slightly different from this, since I considered that $(x,t)inmathbb Rtimes(mathbb R^+cup0)$.
$$int_infty^infty int_0^infty H(x-ct)(phi_tt-c^2phi_xx)dtdx=0$$
should be true for all $phiinmathcal D(mathbb Rtimes(mathbb R^+))$.
Using $phi$ is a $mathcal C^infty$ function with compact support, I found it can be reduced to
$$
beginalign
& int_0^inftyint_0^x/c phi_ttdtdx - c^2int_0^inftyint_ct^infty phi_xxdxdt \
= & int_0^infty (phi_t (x, dfracxc) - phi_t (x, 0))dx + cint_0^inftyphi_x(t,dfractc)dt \
= & int_0^infty cdfracdphids(s,dfracsc)ds-int_0^infty phi_t (x, 0)dx \
= & -cphi(0,0) - int_0^infty phi_t (x, 0)dx.
endalign$$
Since the second term cannot be integrated explicitly, I considered a closed curve on $mathbb Rtimes(mathbb R^+cup0)$ which connects $(0,0),(X,0),(X,T),(0,T)$ by line segments, then send $X$ and $T$ to $+infty$. Since $phi$ and its derivatives vanish over some radius R, the second term would be $-phi(0,0)$, not $-cphi(0,0)$.
Maybe I has made a mistake on integration, but can't find that. Could anyone give me some help?
pde
edited Mar 15 at 18:45
Spencer
8,65012056
asked Mar 15 at 18:13
Jinmu YouJinmu You
339210
$endgroup$
$begingroup$
If $phi in mathcal D(mathbb R_x times mathbb R^+_t)$, does this mean that the support of $phi$ in time is bounded away from 0?
$endgroup$
– Calvin Khor
Mar 15 at 18:56
$begingroup$
@CalvinKhor I meant that $phi$ is a test function in distribution space en.wikipedia.org/wiki/Distribution_(mathematics). So $phi$ and its derivatives will be identically zero outside some bounded interval.
$endgroup$
– Jinmu You
Mar 16 at 2:23
1
$begingroup$
I know what a distribution and what a test function is, I'm asking you to check that you mean $t=0$ is included in the support of $phi$. In this case, i think the definition of a weak solution should also have some terms with $t=0$ from the integration by parts
$endgroup$
– Calvin Khor
Mar 16 at 8:31
$begingroup$
@CalvinKhor Sorry for misunderstanding. I think I cannot be sure that $t=0$ is included in the support of $phi$. Maybe $phi(0,0)$ can get some nonzero value. My expectation was that the last integral term would be $-cphi(0,0)$ and make the integration zero, regardless of the $phi$. But I can't justify that.
$endgroup$
– Jinmu You
Mar 16 at 14:26
add a comment |
$begingroup$
The problem is in Strauss, Partial Diferential Equations 2nd edition, 12.1 Ex.5.
I want to verify by direct substitution that Heaviside DISTRIBUTION $H(x-ct)$ is a weak solution of wave equation $u_tt=c^2u_xx$.
I found almost identical question at weak solution of wave equation here but I think my problem is slightly different from this, since I considered that $(x,t)inmathbb Rtimes(mathbb R^+cup0)$.
$$int_infty^infty int_0^infty H(x-ct)(phi_tt-c^2phi_xx)dtdx=0$$
should be true for all $phiinmathcal D(mathbb Rtimes(mathbb R^+))$.
Using $phi$ is a $mathcal C^infty$ function with compact support, I found it can be reduced to
$$
beginalign
& int_0^inftyint_0^x/c phi_ttdtdx - c^2int_0^inftyint_ct^infty phi_xxdxdt \
= & int_0^infty (phi_t (x, dfracxc) - phi_t (x, 0))dx + cint_0^inftyphi_x(t,dfractc)dt \
= & int_0^infty cdfracdphids(s,dfracsc)ds-int_0^infty phi_t (x, 0)dx \
= & -cphi(0,0) - int_0^infty phi_t (x, 0)dx.
endalign$$
Since the second term cannot be integrated explicitly, I considered a closed curve on $mathbb Rtimes(mathbb R^+cup0)$ which connects $(0,0),(X,0),(X,T),(0,T)$ by line segments, then send $X$ and $T$ to $+infty$. Since $phi$ and its derivatives vanish over some radius R, the second term would be $-phi(0,0)$, not $-cphi(0,0)$.
Maybe I has made a mistake on integration, but can't find that. Could anyone give me some help?
pde
edited Mar 15 at 18:45
Spencer
8,65012056
asked Mar 15 at 18:13
Jinmu YouJinmu You
339210
$endgroup$
The problem is in Strauss, Partial Diferential Equations 2nd edition, 12.1 Ex.5.
I want to verify by direct substitution that Heaviside DISTRIBUTION $H(x-ct)$ is a weak solution of wave equation $u_tt=c^2u_xx$.
I found almost identical question at weak solution of wave equation here but I think my problem is slightly different from this, since I considered that $(x,t)inmathbb Rtimes(mathbb R^+cup0)$.
$$int_infty^infty int_0^infty H(x-ct)(phi_tt-c^2phi_xx)dtdx=0$$
should be true for all $phiinmathcal D(mathbb Rtimes(mathbb R^+))$.
Using $phi$ is a $mathcal C^infty$ function with compact support, I found it can be reduced to
$$
beginalign
& int_0^inftyint_0^x/c phi_ttdtdx - c^2int_0^inftyint_ct^infty phi_xxdxdt \
= & int_0^infty (phi_t (x, dfracxc) - phi_t (x, 0))dx + cint_0^inftyphi_x(t,dfractc)dt \
= & int_0^infty cdfracdphids(s,dfracsc)ds-int_0^infty phi_t (x, 0)dx \
= & -cphi(0,0) - int_0^infty phi_t (x, 0)dx.
endalign$$
Since the second term cannot be integrated explicitly, I considered a closed curve on $mathbb Rtimes(mathbb R^+cup0)$ which connects $(0,0),(X,0),(X,T),(0,T)$ by line segments, then send $X$ and $T$ to $+infty$. Since $phi$ and its derivatives vanish over some radius R, the second term would be $-phi(0,0)$, not $-cphi(0,0)$.
Maybe I has made a mistake on integration, but can't find that. Could anyone give me some help?
pde
pde
edited Mar 15 at 18:45
Spencer
8,65012056
asked Mar 15 at 18:13
Jinmu YouJinmu You
339210
edited Mar 15 at 18:45
Spencer
8,65012056
asked Mar 15 at 18:13
Jinmu YouJinmu You
339210
edited Mar 15 at 18:45
Spencer
8,65012056
edited Mar 15 at 18:45
Spencer
8,65012056
edited Mar 15 at 18:45
Spencer
8,65012056
8,65012056
asked Mar 15 at 18:13
Jinmu YouJinmu You
339210
asked Mar 15 at 18:13
Jinmu YouJinmu You
339210
asked Mar 15 at 18:13
Jinmu YouJinmu You
339210
339210
$begingroup$
If $phi in mathcal D(mathbb R_x times mathbb R^+_t)$, does this mean that the support of $phi$ in time is bounded away from 0?
$endgroup$
– Calvin Khor
Mar 15 at 18:56
$begingroup$
@CalvinKhor I meant that $phi$ is a test function in distribution space en.wikipedia.org/wiki/Distribution_(mathematics). So $phi$ and its derivatives will be identically zero outside some bounded interval.
$endgroup$
– Jinmu You
Mar 16 at 2:23
1
$begingroup$
I know what a distribution and what a test function is, I'm asking you to check that you mean $t=0$ is included in the support of $phi$. In this case, i think the definition of a weak solution should also have some terms with $t=0$ from the integration by parts
$endgroup$
– Calvin Khor
Mar 16 at 8:31
$begingroup$
@CalvinKhor Sorry for misunderstanding. I think I cannot be sure that $t=0$ is included in the support of $phi$. Maybe $phi(0,0)$ can get some nonzero value. My expectation was that the last integral term would be $-cphi(0,0)$ and make the integration zero, regardless of the $phi$. But I can't justify that.
$endgroup$
– Jinmu You
Mar 16 at 14:26
add a comment |
$begingroup$
If $phi in mathcal D(mathbb R_x times mathbb R^+_t)$, does this mean that the support of $phi$ in time is bounded away from 0?
$endgroup$
– Calvin Khor
Mar 15 at 18:56
$begingroup$
@CalvinKhor I meant that $phi$ is a test function in distribution space en.wikipedia.org/wiki/Distribution_(mathematics). So $phi$ and its derivatives will be identically zero outside some bounded interval.
$endgroup$
– Jinmu You
Mar 16 at 2:23
1
$begingroup$
I know what a distribution and what a test function is, I'm asking you to check that you mean $t=0$ is included in the support of $phi$. In this case, i think the definition of a weak solution should also have some terms with $t=0$ from the integration by parts
$endgroup$
– Calvin Khor
Mar 16 at 8:31
$begingroup$
@CalvinKhor Sorry for misunderstanding. I think I cannot be sure that $t=0$ is included in the support of $phi$. Maybe $phi(0,0)$ can get some nonzero value. My expectation was that the last integral term would be $-cphi(0,0)$ and make the integration zero, regardless of the $phi$. But I can't justify that.
$endgroup$
– Jinmu You
Mar 16 at 14:26
$begingroup$
If $phi in mathcal D(mathbb R_x times mathbb R^+_t)$, does this mean that the support of $phi$ in time is bounded away from 0?
$endgroup$
– Calvin Khor
Mar 15 at 18:56
$begingroup$
If $phi in mathcal D(mathbb R_x times mathbb R^+_t)$, does this mean that the support of $phi$ in time is bounded away from 0?
$endgroup$
– Calvin Khor
Mar 15 at 18:56
$begingroup$
@CalvinKhor I meant that $phi$ is a test function in distribution space en.wikipedia.org/wiki/Distribution_(mathematics). So $phi$ and its derivatives will be identically zero outside some bounded interval.
$endgroup$
– Jinmu You
Mar 16 at 2:23
$begingroup$
@CalvinKhor I meant that $phi$ is a test function in distribution space en.wikipedia.org/wiki/Distribution_(mathematics). So $phi$ and its derivatives will be identically zero outside some bounded interval.
$endgroup$
– Jinmu You
Mar 16 at 2:23
1
1
$begingroup$
I know what a distribution and what a test function is, I'm asking you to check that you mean $t=0$ is included in the support of $phi$. In this case, i think the definition of a weak solution should also have some terms with $t=0$ from the integration by parts
$endgroup$
– Calvin Khor
Mar 16 at 8:31
$begingroup$
I know what a distribution and what a test function is, I'm asking you to check that you mean $t=0$ is included in the support of $phi$. In this case, i think the definition of a weak solution should also have some terms with $t=0$ from the integration by parts
$endgroup$
– Calvin Khor
Mar 16 at 8:31
$begingroup$
@CalvinKhor Sorry for misunderstanding. I think I cannot be sure that $t=0$ is included in the support of $phi$. Maybe $phi(0,0)$ can get some nonzero value. My expectation was that the last integral term would be $-cphi(0,0)$ and make the integration zero, regardless of the $phi$. But I can't justify that.
$endgroup$
– Jinmu You
Mar 16 at 14:26
$begingroup$
@CalvinKhor Sorry for misunderstanding. I think I cannot be sure that $t=0$ is included in the support of $phi$. Maybe $phi(0,0)$ can get some nonzero value. My expectation was that the last integral term would be $-cphi(0,0)$ and make the integration zero, regardless of the $phi$. But I can't justify that.
$endgroup$
– Jinmu You
Mar 16 at 14:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you want your test functions to have support containing $t=0$, your definition of a weak solution should be adjusted since if $uin C^2$,
$$ int_mathbb Rint_0^infty u_tt(tau) phi(tau) dtau dx = int_mathbb R left( -u_t(0)phi(0) +u(0)phi_t(0) +int_0^infty u(tau)phi_tt(tau) dtau right)dx$$
So for test functions $mathfrak D:=mathcal D(mathbb R times [0,infty))$, we should say that $uinmathfrak D'$ is a weak solution of the IVP for $u_0,u_1inmathcal D(mathbb R)'$,
$$ u_tt = c^2 u_xx\ u(0)=u_0\
u_t(0)=u_1$$
if
$$ -int_mathbb R u_1(x)phi(x,0)dx + int_mathbb R u_0(x)phi_t(x,0)dx +int_mathbb R int_0^infty u(x,t) (phi_tt(x,t)-c^2phi_xx(x,t)) dxdt = 0$$
When $u(x,t)=H(x-ct)$, then $u(x,0)= H(x), u_t(x,0) = -cdelta_0(x),$ so we should use
$$u_0(x) = H(x),\ u_1(x) = -cdelta_0(x). $$
This causes the two extra terms you found to exactly cancel.
For the definition of a weak solution you suggested, you should use the fact that functions in $mathcal D(mathbb R times (0,infty))$ vanish at $t=0$.
answered Mar 16 at 20:29
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
$begingroup$
Thanks for perfect explanations. Aside from this problem, does the conclusion $ int_0^infty phi_t(x,0)dx=-phi(0,0) $ when $phiinmathcal D$ makes sense? I got the idea from contour integral in complex analysis, but I feel unsure of using this.
$endgroup$
– Jinmu You
Mar 17 at 12:53
1
$begingroup$
@JinmuYou if I understand correctly, the Cauchy's theorem that is required to say integrals on loops are 0 is unavailable in this setting.
$endgroup$
– Calvin Khor
Mar 17 at 13:16
add a comment |
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$begingroup$
If you want your test functions to have support containing $t=0$, your definition of a weak solution should be adjusted since if $uin C^2$,
$$ int_mathbb Rint_0^infty u_tt(tau) phi(tau) dtau dx = int_mathbb R left( -u_t(0)phi(0) +u(0)phi_t(0) +int_0^infty u(tau)phi_tt(tau) dtau right)dx$$
So for test functions $mathfrak D:=mathcal D(mathbb R times [0,infty))$, we should say that $uinmathfrak D'$ is a weak solution of the IVP for $u_0,u_1inmathcal D(mathbb R)'$,
$$ u_tt = c^2 u_xx\ u(0)=u_0\
u_t(0)=u_1$$
if
$$ -int_mathbb R u_1(x)phi(x,0)dx + int_mathbb R u_0(x)phi_t(x,0)dx +int_mathbb R int_0^infty u(x,t) (phi_tt(x,t)-c^2phi_xx(x,t)) dxdt = 0$$
When $u(x,t)=H(x-ct)$, then $u(x,0)= H(x), u_t(x,0) = -cdelta_0(x),$ so we should use
$$u_0(x) = H(x),\ u_1(x) = -cdelta_0(x). $$
This causes the two extra terms you found to exactly cancel.
For the definition of a weak solution you suggested, you should use the fact that functions in $mathcal D(mathbb R times (0,infty))$ vanish at $t=0$.
answered Mar 16 at 20:29
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
$begingroup$
Thanks for perfect explanations. Aside from this problem, does the conclusion $ int_0^infty phi_t(x,0)dx=-phi(0,0) $ when $phiinmathcal D$ makes sense? I got the idea from contour integral in complex analysis, but I feel unsure of using this.
$endgroup$
– Jinmu You
Mar 17 at 12:53
1
$begingroup$
@JinmuYou if I understand correctly, the Cauchy's theorem that is required to say integrals on loops are 0 is unavailable in this setting.
$endgroup$
– Calvin Khor
Mar 17 at 13:16
add a comment |
$begingroup$
If you want your test functions to have support containing $t=0$, your definition of a weak solution should be adjusted since if $uin C^2$,
$$ int_mathbb Rint_0^infty u_tt(tau) phi(tau) dtau dx = int_mathbb R left( -u_t(0)phi(0) +u(0)phi_t(0) +int_0^infty u(tau)phi_tt(tau) dtau right)dx$$
So for test functions $mathfrak D:=mathcal D(mathbb R times [0,infty))$, we should say that $uinmathfrak D'$ is a weak solution of the IVP for $u_0,u_1inmathcal D(mathbb R)'$,
$$ u_tt = c^2 u_xx\ u(0)=u_0\
u_t(0)=u_1$$
if
$$ -int_mathbb R u_1(x)phi(x,0)dx + int_mathbb R u_0(x)phi_t(x,0)dx +int_mathbb R int_0^infty u(x,t) (phi_tt(x,t)-c^2phi_xx(x,t)) dxdt = 0$$
When $u(x,t)=H(x-ct)$, then $u(x,0)= H(x), u_t(x,0) = -cdelta_0(x),$ so we should use
$$u_0(x) = H(x),\ u_1(x) = -cdelta_0(x). $$
This causes the two extra terms you found to exactly cancel.
For the definition of a weak solution you suggested, you should use the fact that functions in $mathcal D(mathbb R times (0,infty))$ vanish at $t=0$.
answered Mar 16 at 20:29
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
$begingroup$
Thanks for perfect explanations. Aside from this problem, does the conclusion $ int_0^infty phi_t(x,0)dx=-phi(0,0) $ when $phiinmathcal D$ makes sense? I got the idea from contour integral in complex analysis, but I feel unsure of using this.
$endgroup$
– Jinmu You
Mar 17 at 12:53
1
$begingroup$
@JinmuYou if I understand correctly, the Cauchy's theorem that is required to say integrals on loops are 0 is unavailable in this setting.
$endgroup$
– Calvin Khor
Mar 17 at 13:16
add a comment |
$begingroup$
If you want your test functions to have support containing $t=0$, your definition of a weak solution should be adjusted since if $uin C^2$,
$$ int_mathbb Rint_0^infty u_tt(tau) phi(tau) dtau dx = int_mathbb R left( -u_t(0)phi(0) +u(0)phi_t(0) +int_0^infty u(tau)phi_tt(tau) dtau right)dx$$
So for test functions $mathfrak D:=mathcal D(mathbb R times [0,infty))$, we should say that $uinmathfrak D'$ is a weak solution of the IVP for $u_0,u_1inmathcal D(mathbb R)'$,
$$ u_tt = c^2 u_xx\ u(0)=u_0\
u_t(0)=u_1$$
if
$$ -int_mathbb R u_1(x)phi(x,0)dx + int_mathbb R u_0(x)phi_t(x,0)dx +int_mathbb R int_0^infty u(x,t) (phi_tt(x,t)-c^2phi_xx(x,t)) dxdt = 0$$
When $u(x,t)=H(x-ct)$, then $u(x,0)= H(x), u_t(x,0) = -cdelta_0(x),$ so we should use
$$u_0(x) = H(x),\ u_1(x) = -cdelta_0(x). $$
This causes the two extra terms you found to exactly cancel.
For the definition of a weak solution you suggested, you should use the fact that functions in $mathcal D(mathbb R times (0,infty))$ vanish at $t=0$.
answered Mar 16 at 20:29
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
If you want your test functions to have support containing $t=0$, your definition of a weak solution should be adjusted since if $uin C^2$,
$$ int_mathbb Rint_0^infty u_tt(tau) phi(tau) dtau dx = int_mathbb R left( -u_t(0)phi(0) +u(0)phi_t(0) +int_0^infty u(tau)phi_tt(tau) dtau right)dx$$
So for test functions $mathfrak D:=mathcal D(mathbb R times [0,infty))$, we should say that $uinmathfrak D'$ is a weak solution of the IVP for $u_0,u_1inmathcal D(mathbb R)'$,
$$ u_tt = c^2 u_xx\ u(0)=u_0\
u_t(0)=u_1$$
if
$$ -int_mathbb R u_1(x)phi(x,0)dx + int_mathbb R u_0(x)phi_t(x,0)dx +int_mathbb R int_0^infty u(x,t) (phi_tt(x,t)-c^2phi_xx(x,t)) dxdt = 0$$
When $u(x,t)=H(x-ct)$, then $u(x,0)= H(x), u_t(x,0) = -cdelta_0(x),$ so we should use
$$u_0(x) = H(x),\ u_1(x) = -cdelta_0(x). $$
This causes the two extra terms you found to exactly cancel.
For the definition of a weak solution you suggested, you should use the fact that functions in $mathcal D(mathbb R times (0,infty))$ vanish at $t=0$.
answered Mar 16 at 20:29
Calvin KhorCalvin Khor
12.4k21439
answered Mar 16 at 20:29
Calvin KhorCalvin Khor
12.4k21439
answered Mar 16 at 20:29
Calvin KhorCalvin Khor
12.4k21439
answered Mar 16 at 20:29
Calvin KhorCalvin Khor
12.4k21439
12.4k21439
$begingroup$
Thanks for perfect explanations. Aside from this problem, does the conclusion $ int_0^infty phi_t(x,0)dx=-phi(0,0) $ when $phiinmathcal D$ makes sense? I got the idea from contour integral in complex analysis, but I feel unsure of using this.
$endgroup$
– Jinmu You
Mar 17 at 12:53
1
$begingroup$
@JinmuYou if I understand correctly, the Cauchy's theorem that is required to say integrals on loops are 0 is unavailable in this setting.
$endgroup$
– Calvin Khor
Mar 17 at 13:16
add a comment |
$begingroup$
Thanks for perfect explanations. Aside from this problem, does the conclusion $ int_0^infty phi_t(x,0)dx=-phi(0,0) $ when $phiinmathcal D$ makes sense? I got the idea from contour integral in complex analysis, but I feel unsure of using this.
$endgroup$
– Jinmu You
Mar 17 at 12:53
1
$begingroup$
@JinmuYou if I understand correctly, the Cauchy's theorem that is required to say integrals on loops are 0 is unavailable in this setting.
$endgroup$
– Calvin Khor
Mar 17 at 13:16
$begingroup$
Thanks for perfect explanations. Aside from this problem, does the conclusion $ int_0^infty phi_t(x,0)dx=-phi(0,0) $ when $phiinmathcal D$ makes sense? I got the idea from contour integral in complex analysis, but I feel unsure of using this.
$endgroup$
– Jinmu You
Mar 17 at 12:53
$begingroup$
Thanks for perfect explanations. Aside from this problem, does the conclusion $ int_0^infty phi_t(x,0)dx=-phi(0,0) $ when $phiinmathcal D$ makes sense? I got the idea from contour integral in complex analysis, but I feel unsure of using this.
$endgroup$
– Jinmu You
Mar 17 at 12:53
1
1
$begingroup$
@JinmuYou if I understand correctly, the Cauchy's theorem that is required to say integrals on loops are 0 is unavailable in this setting.
$endgroup$
– Calvin Khor
Mar 17 at 13:16
$begingroup$
@JinmuYou if I understand correctly, the Cauchy's theorem that is required to say integrals on loops are 0 is unavailable in this setting.
$endgroup$
– Calvin Khor
Mar 17 at 13:16
add a comment |
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$begingroup$
If $phi in mathcal D(mathbb R_x times mathbb R^+_t)$, does this mean that the support of $phi$ in time is bounded away from 0?
$endgroup$
– Calvin Khor
Mar 15 at 18:56
$begingroup$
@CalvinKhor I meant that $phi$ is a test function in distribution space en.wikipedia.org/wiki/Distribution_(mathematics). So $phi$ and its derivatives will be identically zero outside some bounded interval.
$endgroup$
– Jinmu You
Mar 16 at 2:23
1
$begingroup$
I know what a distribution and what a test function is, I'm asking you to check that you mean $t=0$ is included in the support of $phi$. In this case, i think the definition of a weak solution should also have some terms with $t=0$ from the integration by parts
$endgroup$
– Calvin Khor
Mar 16 at 8:31
$begingroup$
@CalvinKhor Sorry for misunderstanding. I think I cannot be sure that $t=0$ is included in the support of $phi$. Maybe $phi(0,0)$ can get some nonzero value. My expectation was that the last integral term would be $-cphi(0,0)$ and make the integration zero, regardless of the $phi$. But I can't justify that.
$endgroup$
– Jinmu You
Mar 16 at 14:26