Solving $frac{partial}{partial s} (u(x+as, t+s)) = f(x+as, t+s)$
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I am doing a course on PDEs and I am struggling to understand how to solve this:
$$frac{partial}{partial s} (u(x+as, t+s) = f(x+as, t+s)$$
The notes say to integrate from $s = -t$ to $s= 0$ and they obtain $$u(x, t) = u(x-at, 0) + int_{-t}^0 f(x+as, t+s) ds$$ but I don't understand how they have arrived here. This is probably fairly simple, but I'm looking for someone to help me fill in the details.
I am particularly struggling with the arguments on $u$, so how it has gone from $u(x+as, t+s)$ to $u(x, t)$ and how we have obtained the $u(x-at, 0)$ term
pde
asked Jan 15 at 17:35
PhysicsMathsLovePhysicsMathsLove
1,315516
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add a comment |
$begingroup$
I am doing a course on PDEs and I am struggling to understand how to solve this:
$$frac{partial}{partial s} (u(x+as, t+s) = f(x+as, t+s)$$
The notes say to integrate from $s = -t$ to $s= 0$ and they obtain $$u(x, t) = u(x-at, 0) + int_{-t}^0 f(x+as, t+s) ds$$ but I don't understand how they have arrived here. This is probably fairly simple, but I'm looking for someone to help me fill in the details.
I am particularly struggling with the arguments on $u$, so how it has gone from $u(x+as, t+s)$ to $u(x, t)$ and how we have obtained the $u(x-at, 0)$ term
pde
asked Jan 15 at 17:35
PhysicsMathsLovePhysicsMathsLove
1,315516
$endgroup$
add a comment |
$begingroup$
I am doing a course on PDEs and I am struggling to understand how to solve this:
$$frac{partial}{partial s} (u(x+as, t+s) = f(x+as, t+s)$$
The notes say to integrate from $s = -t$ to $s= 0$ and they obtain $$u(x, t) = u(x-at, 0) + int_{-t}^0 f(x+as, t+s) ds$$ but I don't understand how they have arrived here. This is probably fairly simple, but I'm looking for someone to help me fill in the details.
I am particularly struggling with the arguments on $u$, so how it has gone from $u(x+as, t+s)$ to $u(x, t)$ and how we have obtained the $u(x-at, 0)$ term
pde
asked Jan 15 at 17:35
PhysicsMathsLovePhysicsMathsLove
1,315516
$endgroup$
I am doing a course on PDEs and I am struggling to understand how to solve this:
$$frac{partial}{partial s} (u(x+as, t+s) = f(x+as, t+s)$$
The notes say to integrate from $s = -t$ to $s= 0$ and they obtain $$u(x, t) = u(x-at, 0) + int_{-t}^0 f(x+as, t+s) ds$$ but I don't understand how they have arrived here. This is probably fairly simple, but I'm looking for someone to help me fill in the details.
I am particularly struggling with the arguments on $u$, so how it has gone from $u(x+as, t+s)$ to $u(x, t)$ and how we have obtained the $u(x-at, 0)$ term
pde
pde
asked Jan 15 at 17:35
PhysicsMathsLovePhysicsMathsLove
1,315516
asked Jan 15 at 17:35
PhysicsMathsLovePhysicsMathsLove
1,315516
asked Jan 15 at 17:35
PhysicsMathsLovePhysicsMathsLove
1,315516
asked Jan 15 at 17:35
PhysicsMathsLovePhysicsMathsLove
1,315516
asked Jan 15 at 17:35
PhysicsMathsLovePhysicsMathsLove
1,315516
1,315516
add a comment |
add a comment |
1 Answer
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It is just the fundamental theorem of analysis. We have
$$int_a^b f(s),ds = F(b) - F(a) tag{1}$$
with $F'(s) =f(s)$.
Now introduce the functions $$F_L(s) = u(x+as, t+s)$$ for the left-hand side and $$f_R(s) = f(x+as, t+s)$$
for the right-hand side.
From $f_L(s)= f_R(s)$, we obtain (from (1) with $a=-t$ and $b=0$)
$$ F_L(0) - F_L(-t) = int_{-t}^0f_R(s),ds = int_{-t}^0 f(x+as, t+s),ds $$
or equivalently
$$ u(x,t) - u(x- a t, 0) = int_{-t}^0 f(x+as, t+s),ds,.$$
answered Jan 15 at 17:44
FabianFabian
20.2k3774
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1 Answer
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1 Answer
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$begingroup$
It is just the fundamental theorem of analysis. We have
$$int_a^b f(s),ds = F(b) - F(a) tag{1}$$
with $F'(s) =f(s)$.
Now introduce the functions $$F_L(s) = u(x+as, t+s)$$ for the left-hand side and $$f_R(s) = f(x+as, t+s)$$
for the right-hand side.
From $f_L(s)= f_R(s)$, we obtain (from (1) with $a=-t$ and $b=0$)
$$ F_L(0) - F_L(-t) = int_{-t}^0f_R(s),ds = int_{-t}^0 f(x+as, t+s),ds $$
or equivalently
$$ u(x,t) - u(x- a t, 0) = int_{-t}^0 f(x+as, t+s),ds,.$$
answered Jan 15 at 17:44
FabianFabian
20.2k3774
$endgroup$
add a comment |
$begingroup$
It is just the fundamental theorem of analysis. We have
$$int_a^b f(s),ds = F(b) - F(a) tag{1}$$
with $F'(s) =f(s)$.
Now introduce the functions $$F_L(s) = u(x+as, t+s)$$ for the left-hand side and $$f_R(s) = f(x+as, t+s)$$
for the right-hand side.
From $f_L(s)= f_R(s)$, we obtain (from (1) with $a=-t$ and $b=0$)
$$ F_L(0) - F_L(-t) = int_{-t}^0f_R(s),ds = int_{-t}^0 f(x+as, t+s),ds $$
or equivalently
$$ u(x,t) - u(x- a t, 0) = int_{-t}^0 f(x+as, t+s),ds,.$$
answered Jan 15 at 17:44
FabianFabian
20.2k3774
$endgroup$
add a comment |
$begingroup$
It is just the fundamental theorem of analysis. We have
$$int_a^b f(s),ds = F(b) - F(a) tag{1}$$
with $F'(s) =f(s)$.
Now introduce the functions $$F_L(s) = u(x+as, t+s)$$ for the left-hand side and $$f_R(s) = f(x+as, t+s)$$
for the right-hand side.
From $f_L(s)= f_R(s)$, we obtain (from (1) with $a=-t$ and $b=0$)
$$ F_L(0) - F_L(-t) = int_{-t}^0f_R(s),ds = int_{-t}^0 f(x+as, t+s),ds $$
or equivalently
$$ u(x,t) - u(x- a t, 0) = int_{-t}^0 f(x+as, t+s),ds,.$$
answered Jan 15 at 17:44
FabianFabian
20.2k3774
$endgroup$
It is just the fundamental theorem of analysis. We have
$$int_a^b f(s),ds = F(b) - F(a) tag{1}$$
with $F'(s) =f(s)$.
Now introduce the functions $$F_L(s) = u(x+as, t+s)$$ for the left-hand side and $$f_R(s) = f(x+as, t+s)$$
for the right-hand side.
From $f_L(s)= f_R(s)$, we obtain (from (1) with $a=-t$ and $b=0$)
$$ F_L(0) - F_L(-t) = int_{-t}^0f_R(s),ds = int_{-t}^0 f(x+as, t+s),ds $$
or equivalently
$$ u(x,t) - u(x- a t, 0) = int_{-t}^0 f(x+as, t+s),ds,.$$
answered Jan 15 at 17:44
FabianFabian
20.2k3774
answered Jan 15 at 17:44
FabianFabian
20.2k3774
answered Jan 15 at 17:44
FabianFabian
20.2k3774
answered Jan 15 at 17:44
FabianFabian
20.2k3774
20.2k3774
add a comment |
add a comment |
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