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Solving $fracpartial^2u(s, t)partialspartialt = -frac1sigma^2f(s,t)$ by direct integration
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I'm told that the PDE
$$fracpartial^2u(s, t)partialspartialt = -frac1sigma^2f(s,t)$$
can be solved by direct integration to get
$$u(s, t) = F(s) + G(t) - phi(s, t)$$
where $F(s)$ and $G(t)$ are arbitrary functions and
$$phi(s, t) = frac1sigma^2 iint f(x(s, t), y(s, t)) ds dt$$
Can someone please demonstrate the steps for how this was done?
The reason I am confused is because I had another integral (see this question), and it was solved using the fundamental theorem of calculus, as follows:
$$dfracdudt = f$$
$$therefore int_0^t fracdudt' dt' = int_0^t f(x(s, t'), y(s, t')) dt'$$
$$therefore u(s, t) = u(s, 0) + F(t), F(t) = int_0^t f(x(s, t'), y(s, t')) dt'$$
However, it doesn't seem like the former integral uses the FTC? I'm confused as to why it's used in one case and not the other?
Thanks to all who kindly take the time to explain this.
real-analysis integration multivariable-calculus pde
asked Aug 6 at 1:05
handler's handle
1508
add a comment |Â
up vote
1
down vote
favorite
I'm told that the PDE
$$fracpartial^2u(s, t)partialspartialt = -frac1sigma^2f(s,t)$$
can be solved by direct integration to get
$$u(s, t) = F(s) + G(t) - phi(s, t)$$
where $F(s)$ and $G(t)$ are arbitrary functions and
$$phi(s, t) = frac1sigma^2 iint f(x(s, t), y(s, t)) ds dt$$
Can someone please demonstrate the steps for how this was done?
The reason I am confused is because I had another integral (see this question), and it was solved using the fundamental theorem of calculus, as follows:
$$dfracdudt = f$$
$$therefore int_0^t fracdudt' dt' = int_0^t f(x(s, t'), y(s, t')) dt'$$
$$therefore u(s, t) = u(s, 0) + F(t), F(t) = int_0^t f(x(s, t'), y(s, t')) dt'$$
However, it doesn't seem like the former integral uses the FTC? I'm confused as to why it's used in one case and not the other?
Thanks to all who kindly take the time to explain this.
real-analysis integration multivariable-calculus pde
asked Aug 6 at 1:05
handler's handle
1508
1
With $k = -1/sigma^2$ beginalign u_st &= kf \ implies u_s &= k int f dt + g(s) \ implies u &= k int int f dt ds + int g(s) ds + h(t) endalign Or, if you would like beginalign u_st &= kf \ implies u_s &= k int_0^t f(s, t') dt' + g(s) \ implies u &= k int_0^s int_0^t f(s', t') dt' ds' + int_0^s g(s') ds' + h(t) endalign Use the FTOC to differentiate both, you'll see they're equal.
– Mattos
Aug 6 at 4:02
@Mattos Your explanation is clear. Thank you very much for this!
– handler's handle
Aug 7 at 10:19
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm told that the PDE
$$fracpartial^2u(s, t)partialspartialt = -frac1sigma^2f(s,t)$$
can be solved by direct integration to get
$$u(s, t) = F(s) + G(t) - phi(s, t)$$
where $F(s)$ and $G(t)$ are arbitrary functions and
$$phi(s, t) = frac1sigma^2 iint f(x(s, t), y(s, t)) ds dt$$
Can someone please demonstrate the steps for how this was done?
The reason I am confused is because I had another integral (see this question), and it was solved using the fundamental theorem of calculus, as follows:
$$dfracdudt = f$$
$$therefore int_0^t fracdudt' dt' = int_0^t f(x(s, t'), y(s, t')) dt'$$
$$therefore u(s, t) = u(s, 0) + F(t), F(t) = int_0^t f(x(s, t'), y(s, t')) dt'$$
However, it doesn't seem like the former integral uses the FTC? I'm confused as to why it's used in one case and not the other?
Thanks to all who kindly take the time to explain this.
real-analysis integration multivariable-calculus pde
asked Aug 6 at 1:05
handler's handle
1508
I'm told that the PDE
$$fracpartial^2u(s, t)partialspartialt = -frac1sigma^2f(s,t)$$
can be solved by direct integration to get
$$u(s, t) = F(s) + G(t) - phi(s, t)$$
where $F(s)$ and $G(t)$ are arbitrary functions and
$$phi(s, t) = frac1sigma^2 iint f(x(s, t), y(s, t)) ds dt$$
Can someone please demonstrate the steps for how this was done?
The reason I am confused is because I had another integral (see this question), and it was solved using the fundamental theorem of calculus, as follows:
$$dfracdudt = f$$
$$therefore int_0^t fracdudt' dt' = int_0^t f(x(s, t'), y(s, t')) dt'$$
$$therefore u(s, t) = u(s, 0) + F(t), F(t) = int_0^t f(x(s, t'), y(s, t')) dt'$$
However, it doesn't seem like the former integral uses the FTC? I'm confused as to why it's used in one case and not the other?
Thanks to all who kindly take the time to explain this.
real-analysis integration multivariable-calculus pde
asked Aug 6 at 1:05
handler's handle
1508
edited Aug 6 at 2:03
asked Aug 6 at 1:05
handler's handle
1508
asked Aug 6 at 1:05
handler's handle
1508
asked Aug 6 at 1:05
handler's handle
1508
1508
1
With $k = -1/sigma^2$ beginalign u_st &= kf \ implies u_s &= k int f dt + g(s) \ implies u &= k int int f dt ds + int g(s) ds + h(t) endalign Or, if you would like beginalign u_st &= kf \ implies u_s &= k int_0^t f(s, t') dt' + g(s) \ implies u &= k int_0^s int_0^t f(s', t') dt' ds' + int_0^s g(s') ds' + h(t) endalign Use the FTOC to differentiate both, you'll see they're equal.
– Mattos
Aug 6 at 4:02
@Mattos Your explanation is clear. Thank you very much for this!
– handler's handle
Aug 7 at 10:19
add a comment |Â
1
With $k = -1/sigma^2$ beginalign u_st &= kf \ implies u_s &= k int f dt + g(s) \ implies u &= k int int f dt ds + int g(s) ds + h(t) endalign Or, if you would like beginalign u_st &= kf \ implies u_s &= k int_0^t f(s, t') dt' + g(s) \ implies u &= k int_0^s int_0^t f(s', t') dt' ds' + int_0^s g(s') ds' + h(t) endalign Use the FTOC to differentiate both, you'll see they're equal.
– Mattos
Aug 6 at 4:02
@Mattos Your explanation is clear. Thank you very much for this!
– handler's handle
Aug 7 at 10:19
1
1
With $k = -1/sigma^2$ beginalign u_st &= kf \ implies u_s &= k int f dt + g(s) \ implies u &= k int int f dt ds + int g(s) ds + h(t) endalign Or, if you would like beginalign u_st &= kf \ implies u_s &= k int_0^t f(s, t') dt' + g(s) \ implies u &= k int_0^s int_0^t f(s', t') dt' ds' + int_0^s g(s') ds' + h(t) endalign Use the FTOC to differentiate both, you'll see they're equal.
– Mattos
Aug 6 at 4:02
With $k = -1/sigma^2$ beginalign u_st &= kf \ implies u_s &= k int f dt + g(s) \ implies u &= k int int f dt ds + int g(s) ds + h(t) endalign Or, if you would like beginalign u_st &= kf \ implies u_s &= k int_0^t f(s, t') dt' + g(s) \ implies u &= k int_0^s int_0^t f(s', t') dt' ds' + int_0^s g(s') ds' + h(t) endalign Use the FTOC to differentiate both, you'll see they're equal.
– Mattos
Aug 6 at 4:02
@Mattos Your explanation is clear. Thank you very much for this!
– handler's handle
Aug 7 at 10:19
@Mattos Your explanation is clear. Thank you very much for this!
– handler's handle
Aug 7 at 10:19
add a comment |Â
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With $k = -1/sigma^2$ beginalign u_st &= kf \ implies u_s &= k int f dt + g(s) \ implies u &= k int int f dt ds + int g(s) ds + h(t) endalign Or, if you would like beginalign u_st &= kf \ implies u_s &= k int_0^t f(s, t') dt' + g(s) \ implies u &= k int_0^s int_0^t f(s', t') dt' ds' + int_0^s g(s') ds' + h(t) endalign Use the FTOC to differentiate both, you'll see they're equal.
– Mattos
Aug 6 at 4:02
@Mattos Your explanation is clear. Thank you very much for this!
– handler's handle
Aug 7 at 10:19