Mixing
Confused about Laplace Wave Equation Transformation
Clash Royale CLAN TAG#URR8PPP
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$$fracpartial^2partial t^2 u(x,t) -fracpartial^2partial x^2 u(x,t) =f(x), quad 0<x<1, quad t>0\ u(x,0)=0, quad fracpartialpartial t u(x,0)=0\ u(0,t)=0, quad u(1,t)=0$$
I am supposed to take the Laplace transform of the wave equation that yields a non-homogeneous ordinary differential equation in terms of $mathcalL f(x)$ and $mathcalL u(x,t)$ and its $x$-derivatives.
Can someone please explain the relationship between the two functions, and how it's $x$-derivatives changes will be reflected in the function? This question was posed as a challenge question, and I am seeking some guidance.
How does one setup this Laplace transformation? Why is there $t$ and $x$ in the same system?
differential-equations pde laplace-transform
edited 5 hours ago
Rócherz
2,1811417
asked 6 hours ago
FireMeUP
406
 |Â
show 1 more comment
up vote
5
down vote
favorite
$$fracpartial^2partial t^2 u(x,t) -fracpartial^2partial x^2 u(x,t) =f(x), quad 0<x<1, quad t>0\ u(x,0)=0, quad fracpartialpartial t u(x,0)=0\ u(0,t)=0, quad u(1,t)=0$$
I am supposed to take the Laplace transform of the wave equation that yields a non-homogeneous ordinary differential equation in terms of $mathcalL f(x)$ and $mathcalL u(x,t)$ and its $x$-derivatives.
Can someone please explain the relationship between the two functions, and how it's $x$-derivatives changes will be reflected in the function? This question was posed as a challenge question, and I am seeking some guidance.
How does one setup this Laplace transformation? Why is there $t$ and $x$ in the same system?
differential-equations pde laplace-transform
edited 5 hours ago
Rócherz
2,1811417
asked 6 hours ago
FireMeUP
406
how can we singularly transform the t variable when we have x'' as well.
– FireMeUP
6 hours ago
Likewise, this is my first time seeing IVP in a coordinate fashion. What does u(x,0) = 0 mean?
– FireMeUP
6 hours ago
Just hold it constant. Should I be treating the u(x,t) terms as constants?
– FireMeUP
6 hours ago
If you transform $t$ then I would keep $x$ but treat it as constant regarding the Laplace transformation.
– mvw
6 hours ago
Based on your suggestions, then do I simply just do L(t'')u(x't) = f'(x)? Do I completely ignore the x^2 term?
– FireMeUP
6 hours ago
 |Â
show 1 more comment
up vote
5
down vote
favorite
up vote
5
down vote
favorite
$$fracpartial^2partial t^2 u(x,t) -fracpartial^2partial x^2 u(x,t) =f(x), quad 0<x<1, quad t>0\ u(x,0)=0, quad fracpartialpartial t u(x,0)=0\ u(0,t)=0, quad u(1,t)=0$$
I am supposed to take the Laplace transform of the wave equation that yields a non-homogeneous ordinary differential equation in terms of $mathcalL f(x)$ and $mathcalL u(x,t)$ and its $x$-derivatives.
Can someone please explain the relationship between the two functions, and how it's $x$-derivatives changes will be reflected in the function? This question was posed as a challenge question, and I am seeking some guidance.
How does one setup this Laplace transformation? Why is there $t$ and $x$ in the same system?
differential-equations pde laplace-transform
edited 5 hours ago
Rócherz
2,1811417
asked 6 hours ago
FireMeUP
406
$$fracpartial^2partial t^2 u(x,t) -fracpartial^2partial x^2 u(x,t) =f(x), quad 0<x<1, quad t>0\ u(x,0)=0, quad fracpartialpartial t u(x,0)=0\ u(0,t)=0, quad u(1,t)=0$$
I am supposed to take the Laplace transform of the wave equation that yields a non-homogeneous ordinary differential equation in terms of $mathcalL f(x)$ and $mathcalL u(x,t)$ and its $x$-derivatives.
Can someone please explain the relationship between the two functions, and how it's $x$-derivatives changes will be reflected in the function? This question was posed as a challenge question, and I am seeking some guidance.
How does one setup this Laplace transformation? Why is there $t$ and $x$ in the same system?
differential-equations pde laplace-transform
edited 5 hours ago
Rócherz
2,1811417
asked 6 hours ago
FireMeUP
406
edited 5 hours ago
Rócherz
2,1811417
edited 5 hours ago
Rócherz
2,1811417
edited 5 hours ago
Rócherz
2,1811417
2,1811417
asked 6 hours ago
FireMeUP
406
asked 6 hours ago
FireMeUP
406
asked 6 hours ago
FireMeUP
406
406
how can we singularly transform the t variable when we have x'' as well.
– FireMeUP
6 hours ago
Likewise, this is my first time seeing IVP in a coordinate fashion. What does u(x,0) = 0 mean?
– FireMeUP
6 hours ago
Just hold it constant. Should I be treating the u(x,t) terms as constants?
– FireMeUP
6 hours ago
If you transform $t$ then I would keep $x$ but treat it as constant regarding the Laplace transformation.
– mvw
6 hours ago
Based on your suggestions, then do I simply just do L(t'')u(x't) = f'(x)? Do I completely ignore the x^2 term?
– FireMeUP
6 hours ago
 |Â
show 1 more comment
how can we singularly transform the t variable when we have x'' as well.
– FireMeUP
6 hours ago
Likewise, this is my first time seeing IVP in a coordinate fashion. What does u(x,0) = 0 mean?
– FireMeUP
6 hours ago
Just hold it constant. Should I be treating the u(x,t) terms as constants?
– FireMeUP
6 hours ago
If you transform $t$ then I would keep $x$ but treat it as constant regarding the Laplace transformation.
– mvw
6 hours ago
Based on your suggestions, then do I simply just do L(t'')u(x't) = f'(x)? Do I completely ignore the x^2 term?
– FireMeUP
6 hours ago
how can we singularly transform the t variable when we have x'' as well.
– FireMeUP
6 hours ago
how can we singularly transform the t variable when we have x'' as well.
– FireMeUP
6 hours ago
Likewise, this is my first time seeing IVP in a coordinate fashion. What does u(x,0) = 0 mean?
– FireMeUP
6 hours ago
Likewise, this is my first time seeing IVP in a coordinate fashion. What does u(x,0) = 0 mean?
– FireMeUP
6 hours ago
Just hold it constant. Should I be treating the u(x,t) terms as constants?
– FireMeUP
6 hours ago
Just hold it constant. Should I be treating the u(x,t) terms as constants?
– FireMeUP
6 hours ago
If you transform $t$ then I would keep $x$ but treat it as constant regarding the Laplace transformation.
– mvw
6 hours ago
If you transform $t$ then I would keep $x$ but treat it as constant regarding the Laplace transformation.
– mvw
6 hours ago
Based on your suggestions, then do I simply just do L(t'')u(x't) = f'(x)? Do I completely ignore the x^2 term?
– FireMeUP
6 hours ago
Based on your suggestions, then do I simply just do L(t'')u(x't) = f'(x)? Do I completely ignore the x^2 term?
– FireMeUP
6 hours ago
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
I've not used Laplace transform a lot but I think that it would act similarly as the Fourier transform (the only difference is that Fourier transform acts on position variable, so $x,y,z,cdots$ and Laplace transform acts on the time variable). With that in mind if you take the Laplace transform on both side of the equation you would get: $$mathcalLpartial^2_ttu-mathcalLpartial^2_xxu = mathcalLf(x)\ colorredmathcalLpartial^2_ttu-underbracepartial^2_xxmathcalLu = f(x)colorbluemathcalL1_textthe transform acts only\ texton time so it's independent\ textof the x\ colorreds^2U(x,s)-su(x,0^+)-partial_tu(x,t) - partial^2_xxU(x,s)=fracf(x)colorblues$$
The red one follows from the property of derivation for the Laplace transform and the blue from the Laplace transform of $1$.
As you can see you began with an ode with two partial derivation and you arrived at a second order ode in the function $U(x,s)$. This means that one you've found the solution $U(x,s)$ you can directly find $u(x,t)$ by anti-transforming! If this isn't the most beautiful thing you've ever seen, I don't know what it would be!
answered 6 hours ago
Davide Morgante
1,566118
1
Let me try on a pice of paper! As I said I'm always working with Fourier transform and convolutions come up a lot, probably even in this case can help
– Davide Morgante
6 hours ago
1
From the properties of the Laplace transform of the derivative $$mathcalLf^(n)(t) = s^(n)F(s) -sum_k=1^n s^n-kf^(k-1)(0^+)$$
– Davide Morgante
5 hours ago
1
Well, you have to be careful $fracd^2dx^2$ it isn't a number, it's an operator you cannot divide by a derivative! Once arrived to the transformed equation you just have to solve a second order linear partial differential equation in the variable x with one of the myriad of methods that exists to solve them!
– Davide Morgante
5 hours ago
1
Nope, be carful $$fracd^2 Udx^2 = U''$$ sort of, if you want to look it that way
– Davide Morgante
5 hours ago
1
Sure it is! Happy to have been useful!
– Davide Morgante
5 hours ago
 |Â
show 15 more comments
up vote
2
down vote
Assuming $mathcalL$ transforms the $t$ variable into the $s$ variable.
E.g. $u(x, t)$ into $U(x, s)$.
I added the $t$ subscript to the operator to indicate it acts on $t$.
The subscripts $xx$ mean second order partial derivative regarding $x$,
$tt$ mean second order partial derivative regarding $t$.
My guess would be:
$$
mathcalL_t u_tt(x, t) - u_xx(x, t) = mathcalL_t f(x) iff \
s^2 , U(x, s) - s , u(x, 0+) - u_t(x, 0+) - U_xx(x, s) = f(x) mathcalL_t 1
$$
The idea is to apply the initial conditions here, then get an ordinary differential equation for $U$, solve for $U$ and then use the inverse Laplace transform to get a solution $u$.
answered 6 hours ago
mvw
30.1k22150
Thank you for you response. Can you show how you got big U from s^2U?
– FireMeUP
6 hours ago
So solving for U, then I would set up the equation as f(x)L1 + su(x,0+) + ut(x,0+) = U(x,s) (s^2 - 1)? Then divide by s^2-1?
– FireMeUP
5 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I've not used Laplace transform a lot but I think that it would act similarly as the Fourier transform (the only difference is that Fourier transform acts on position variable, so $x,y,z,cdots$ and Laplace transform acts on the time variable). With that in mind if you take the Laplace transform on both side of the equation you would get: $$mathcalLpartial^2_ttu-mathcalLpartial^2_xxu = mathcalLf(x)\ colorredmathcalLpartial^2_ttu-underbracepartial^2_xxmathcalLu = f(x)colorbluemathcalL1_textthe transform acts only\ texton time so it's independent\ textof the x\ colorreds^2U(x,s)-su(x,0^+)-partial_tu(x,t) - partial^2_xxU(x,s)=fracf(x)colorblues$$
The red one follows from the property of derivation for the Laplace transform and the blue from the Laplace transform of $1$.
As you can see you began with an ode with two partial derivation and you arrived at a second order ode in the function $U(x,s)$. This means that one you've found the solution $U(x,s)$ you can directly find $u(x,t)$ by anti-transforming! If this isn't the most beautiful thing you've ever seen, I don't know what it would be!
answered 6 hours ago
Davide Morgante
1,566118
1
Let me try on a pice of paper! As I said I'm always working with Fourier transform and convolutions come up a lot, probably even in this case can help
– Davide Morgante
6 hours ago
1
From the properties of the Laplace transform of the derivative $$mathcalLf^(n)(t) = s^(n)F(s) -sum_k=1^n s^n-kf^(k-1)(0^+)$$
– Davide Morgante
5 hours ago
1
Well, you have to be careful $fracd^2dx^2$ it isn't a number, it's an operator you cannot divide by a derivative! Once arrived to the transformed equation you just have to solve a second order linear partial differential equation in the variable x with one of the myriad of methods that exists to solve them!
– Davide Morgante
5 hours ago
1
Nope, be carful $$fracd^2 Udx^2 = U''$$ sort of, if you want to look it that way
– Davide Morgante
5 hours ago
1
Sure it is! Happy to have been useful!
– Davide Morgante
5 hours ago
 |Â
show 15 more comments
up vote
2
down vote
accepted
I've not used Laplace transform a lot but I think that it would act similarly as the Fourier transform (the only difference is that Fourier transform acts on position variable, so $x,y,z,cdots$ and Laplace transform acts on the time variable). With that in mind if you take the Laplace transform on both side of the equation you would get: $$mathcalLpartial^2_ttu-mathcalLpartial^2_xxu = mathcalLf(x)\ colorredmathcalLpartial^2_ttu-underbracepartial^2_xxmathcalLu = f(x)colorbluemathcalL1_textthe transform acts only\ texton time so it's independent\ textof the x\ colorreds^2U(x,s)-su(x,0^+)-partial_tu(x,t) - partial^2_xxU(x,s)=fracf(x)colorblues$$
The red one follows from the property of derivation for the Laplace transform and the blue from the Laplace transform of $1$.
As you can see you began with an ode with two partial derivation and you arrived at a second order ode in the function $U(x,s)$. This means that one you've found the solution $U(x,s)$ you can directly find $u(x,t)$ by anti-transforming! If this isn't the most beautiful thing you've ever seen, I don't know what it would be!
answered 6 hours ago
Davide Morgante
1,566118
1
Let me try on a pice of paper! As I said I'm always working with Fourier transform and convolutions come up a lot, probably even in this case can help
– Davide Morgante
6 hours ago
1
From the properties of the Laplace transform of the derivative $$mathcalLf^(n)(t) = s^(n)F(s) -sum_k=1^n s^n-kf^(k-1)(0^+)$$
– Davide Morgante
5 hours ago
1
Well, you have to be careful $fracd^2dx^2$ it isn't a number, it's an operator you cannot divide by a derivative! Once arrived to the transformed equation you just have to solve a second order linear partial differential equation in the variable x with one of the myriad of methods that exists to solve them!
– Davide Morgante
5 hours ago
1
Nope, be carful $$fracd^2 Udx^2 = U''$$ sort of, if you want to look it that way
– Davide Morgante
5 hours ago
1
Sure it is! Happy to have been useful!
– Davide Morgante
5 hours ago
 |Â
show 15 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I've not used Laplace transform a lot but I think that it would act similarly as the Fourier transform (the only difference is that Fourier transform acts on position variable, so $x,y,z,cdots$ and Laplace transform acts on the time variable). With that in mind if you take the Laplace transform on both side of the equation you would get: $$mathcalLpartial^2_ttu-mathcalLpartial^2_xxu = mathcalLf(x)\ colorredmathcalLpartial^2_ttu-underbracepartial^2_xxmathcalLu = f(x)colorbluemathcalL1_textthe transform acts only\ texton time so it's independent\ textof the x\ colorreds^2U(x,s)-su(x,0^+)-partial_tu(x,t) - partial^2_xxU(x,s)=fracf(x)colorblues$$
The red one follows from the property of derivation for the Laplace transform and the blue from the Laplace transform of $1$.
As you can see you began with an ode with two partial derivation and you arrived at a second order ode in the function $U(x,s)$. This means that one you've found the solution $U(x,s)$ you can directly find $u(x,t)$ by anti-transforming! If this isn't the most beautiful thing you've ever seen, I don't know what it would be!
answered 6 hours ago
Davide Morgante
1,566118
I've not used Laplace transform a lot but I think that it would act similarly as the Fourier transform (the only difference is that Fourier transform acts on position variable, so $x,y,z,cdots$ and Laplace transform acts on the time variable). With that in mind if you take the Laplace transform on both side of the equation you would get: $$mathcalLpartial^2_ttu-mathcalLpartial^2_xxu = mathcalLf(x)\ colorredmathcalLpartial^2_ttu-underbracepartial^2_xxmathcalLu = f(x)colorbluemathcalL1_textthe transform acts only\ texton time so it's independent\ textof the x\ colorreds^2U(x,s)-su(x,0^+)-partial_tu(x,t) - partial^2_xxU(x,s)=fracf(x)colorblues$$
The red one follows from the property of derivation for the Laplace transform and the blue from the Laplace transform of $1$.
As you can see you began with an ode with two partial derivation and you arrived at a second order ode in the function $U(x,s)$. This means that one you've found the solution $U(x,s)$ you can directly find $u(x,t)$ by anti-transforming! If this isn't the most beautiful thing you've ever seen, I don't know what it would be!
answered 6 hours ago
Davide Morgante
1,566118
edited 6 hours ago
answered 6 hours ago
Davide Morgante
1,566118
answered 6 hours ago
Davide Morgante
1,566118
answered 6 hours ago
Davide Morgante
1,566118
1,566118
1
Let me try on a pice of paper! As I said I'm always working with Fourier transform and convolutions come up a lot, probably even in this case can help
– Davide Morgante
6 hours ago
1
From the properties of the Laplace transform of the derivative $$mathcalLf^(n)(t) = s^(n)F(s) -sum_k=1^n s^n-kf^(k-1)(0^+)$$
– Davide Morgante
5 hours ago
1
Well, you have to be careful $fracd^2dx^2$ it isn't a number, it's an operator you cannot divide by a derivative! Once arrived to the transformed equation you just have to solve a second order linear partial differential equation in the variable x with one of the myriad of methods that exists to solve them!
– Davide Morgante
5 hours ago
1
Nope, be carful $$fracd^2 Udx^2 = U''$$ sort of, if you want to look it that way
– Davide Morgante
5 hours ago
1
Sure it is! Happy to have been useful!
– Davide Morgante
5 hours ago
 |Â
show 15 more comments
1
Let me try on a pice of paper! As I said I'm always working with Fourier transform and convolutions come up a lot, probably even in this case can help
– Davide Morgante
6 hours ago
1
From the properties of the Laplace transform of the derivative $$mathcalLf^(n)(t) = s^(n)F(s) -sum_k=1^n s^n-kf^(k-1)(0^+)$$
– Davide Morgante
5 hours ago
1
Well, you have to be careful $fracd^2dx^2$ it isn't a number, it's an operator you cannot divide by a derivative! Once arrived to the transformed equation you just have to solve a second order linear partial differential equation in the variable x with one of the myriad of methods that exists to solve them!
– Davide Morgante
5 hours ago
1
Nope, be carful $$fracd^2 Udx^2 = U''$$ sort of, if you want to look it that way
– Davide Morgante
5 hours ago
1
Sure it is! Happy to have been useful!
– Davide Morgante
5 hours ago
1
1
Let me try on a pice of paper! As I said I'm always working with Fourier transform and convolutions come up a lot, probably even in this case can help
– Davide Morgante
6 hours ago
Let me try on a pice of paper! As I said I'm always working with Fourier transform and convolutions come up a lot, probably even in this case can help
– Davide Morgante
6 hours ago
1
1
From the properties of the Laplace transform of the derivative $$mathcalLf^(n)(t) = s^(n)F(s) -sum_k=1^n s^n-kf^(k-1)(0^+)$$
– Davide Morgante
5 hours ago
From the properties of the Laplace transform of the derivative $$mathcalLf^(n)(t) = s^(n)F(s) -sum_k=1^n s^n-kf^(k-1)(0^+)$$
– Davide Morgante
5 hours ago
1
1
Well, you have to be careful $fracd^2dx^2$ it isn't a number, it's an operator you cannot divide by a derivative! Once arrived to the transformed equation you just have to solve a second order linear partial differential equation in the variable x with one of the myriad of methods that exists to solve them!
– Davide Morgante
5 hours ago
Well, you have to be careful $fracd^2dx^2$ it isn't a number, it's an operator you cannot divide by a derivative! Once arrived to the transformed equation you just have to solve a second order linear partial differential equation in the variable x with one of the myriad of methods that exists to solve them!
– Davide Morgante
5 hours ago
1
1
Nope, be carful $$fracd^2 Udx^2 = U''$$ sort of, if you want to look it that way
– Davide Morgante
5 hours ago
Nope, be carful $$fracd^2 Udx^2 = U''$$ sort of, if you want to look it that way
– Davide Morgante
5 hours ago
1
1
Sure it is! Happy to have been useful!
– Davide Morgante
5 hours ago
Sure it is! Happy to have been useful!
– Davide Morgante
5 hours ago
 |Â
show 15 more comments
up vote
2
down vote
Assuming $mathcalL$ transforms the $t$ variable into the $s$ variable.
E.g. $u(x, t)$ into $U(x, s)$.
I added the $t$ subscript to the operator to indicate it acts on $t$.
The subscripts $xx$ mean second order partial derivative regarding $x$,
$tt$ mean second order partial derivative regarding $t$.
My guess would be:
$$
mathcalL_t u_tt(x, t) - u_xx(x, t) = mathcalL_t f(x) iff \
s^2 , U(x, s) - s , u(x, 0+) - u_t(x, 0+) - U_xx(x, s) = f(x) mathcalL_t 1
$$
The idea is to apply the initial conditions here, then get an ordinary differential equation for $U$, solve for $U$ and then use the inverse Laplace transform to get a solution $u$.
answered 6 hours ago
mvw
30.1k22150
Thank you for you response. Can you show how you got big U from s^2U?
– FireMeUP
6 hours ago
So solving for U, then I would set up the equation as f(x)L1 + su(x,0+) + ut(x,0+) = U(x,s) (s^2 - 1)? Then divide by s^2-1?
– FireMeUP
5 hours ago
add a comment |Â
up vote
2
down vote
Assuming $mathcalL$ transforms the $t$ variable into the $s$ variable.
E.g. $u(x, t)$ into $U(x, s)$.
I added the $t$ subscript to the operator to indicate it acts on $t$.
The subscripts $xx$ mean second order partial derivative regarding $x$,
$tt$ mean second order partial derivative regarding $t$.
My guess would be:
$$
mathcalL_t u_tt(x, t) - u_xx(x, t) = mathcalL_t f(x) iff \
s^2 , U(x, s) - s , u(x, 0+) - u_t(x, 0+) - U_xx(x, s) = f(x) mathcalL_t 1
$$
The idea is to apply the initial conditions here, then get an ordinary differential equation for $U$, solve for $U$ and then use the inverse Laplace transform to get a solution $u$.
answered 6 hours ago
mvw
30.1k22150
Thank you for you response. Can you show how you got big U from s^2U?
– FireMeUP
6 hours ago
So solving for U, then I would set up the equation as f(x)L1 + su(x,0+) + ut(x,0+) = U(x,s) (s^2 - 1)? Then divide by s^2-1?
– FireMeUP
5 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Assuming $mathcalL$ transforms the $t$ variable into the $s$ variable.
E.g. $u(x, t)$ into $U(x, s)$.
I added the $t$ subscript to the operator to indicate it acts on $t$.
The subscripts $xx$ mean second order partial derivative regarding $x$,
$tt$ mean second order partial derivative regarding $t$.
My guess would be:
$$
mathcalL_t u_tt(x, t) - u_xx(x, t) = mathcalL_t f(x) iff \
s^2 , U(x, s) - s , u(x, 0+) - u_t(x, 0+) - U_xx(x, s) = f(x) mathcalL_t 1
$$
The idea is to apply the initial conditions here, then get an ordinary differential equation for $U$, solve for $U$ and then use the inverse Laplace transform to get a solution $u$.
answered 6 hours ago
mvw
30.1k22150
Assuming $mathcalL$ transforms the $t$ variable into the $s$ variable.
E.g. $u(x, t)$ into $U(x, s)$.
I added the $t$ subscript to the operator to indicate it acts on $t$.
The subscripts $xx$ mean second order partial derivative regarding $x$,
$tt$ mean second order partial derivative regarding $t$.
My guess would be:
$$
mathcalL_t u_tt(x, t) - u_xx(x, t) = mathcalL_t f(x) iff \
s^2 , U(x, s) - s , u(x, 0+) - u_t(x, 0+) - U_xx(x, s) = f(x) mathcalL_t 1
$$
The idea is to apply the initial conditions here, then get an ordinary differential equation for $U$, solve for $U$ and then use the inverse Laplace transform to get a solution $u$.
answered 6 hours ago
mvw
30.1k22150
edited 6 hours ago
answered 6 hours ago
mvw
30.1k22150
answered 6 hours ago
mvw
30.1k22150
answered 6 hours ago
mvw
30.1k22150
30.1k22150
Thank you for you response. Can you show how you got big U from s^2U?
– FireMeUP
6 hours ago
So solving for U, then I would set up the equation as f(x)L1 + su(x,0+) + ut(x,0+) = U(x,s) (s^2 - 1)? Then divide by s^2-1?
– FireMeUP
5 hours ago
add a comment |Â
Thank you for you response. Can you show how you got big U from s^2U?
– FireMeUP
6 hours ago
So solving for U, then I would set up the equation as f(x)L1 + su(x,0+) + ut(x,0+) = U(x,s) (s^2 - 1)? Then divide by s^2-1?
– FireMeUP
5 hours ago
Thank you for you response. Can you show how you got big U from s^2U?
– FireMeUP
6 hours ago
Thank you for you response. Can you show how you got big U from s^2U?
– FireMeUP
6 hours ago
So solving for U, then I would set up the equation as f(x)L1 + su(x,0+) + ut(x,0+) = U(x,s) (s^2 - 1)? Then divide by s^2-1?
– FireMeUP
5 hours ago
So solving for U, then I would set up the equation as f(x)L1 + su(x,0+) + ut(x,0+) = U(x,s) (s^2 - 1)? Then divide by s^2-1?
– FireMeUP
5 hours ago
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how can we singularly transform the t variable when we have x'' as well.
– FireMeUP
6 hours ago
Likewise, this is my first time seeing IVP in a coordinate fashion. What does u(x,0) = 0 mean?
– FireMeUP
6 hours ago
Just hold it constant. Should I be treating the u(x,t) terms as constants?
– FireMeUP
6 hours ago
If you transform $t$ then I would keep $x$ but treat it as constant regarding the Laplace transformation.
– mvw
6 hours ago
Based on your suggestions, then do I simply just do L(t'')u(x't) = f'(x)? Do I completely ignore the x^2 term?
– FireMeUP
6 hours ago