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Characteristics of First-Order PDEs Example: How Was This Integration Done?
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From Essential Partial Differential Equations, by Griffiths, Dold, and Silvester:
Example 4.2 (Half-plane problem) Solve the PDE $pux + quy = f(x, y)$ (with $p$ and $q$ constant) in the domain $alpha x + beta y > 0$ given that $u = g(x)$ on the line $mathcall : alpha x + beta y = 0$, where $q > 0$ and $beta > 0$.
As in the previous example, the characteristic equations (4.4) are readily solved to give
$$x = pt + C_1, y = qt + C_2$$
where $C_1$ and $C_2$ are constant along any characteristic. The assumption $q > 0$ means that $y$ increases along a characteristic as $t$ increases -- thus the parameter $t$ may be viewed as a time-like variable. This is illustrated in Fig. 4.1 where the arrows on the characteristics indicate increasing t.
In order to determine the solution at a point $P(x, y)$, we trace the characteristic through this point backwards in $t$ until it intersects the line $mathcall$. We shall suppose that this occurs at the point $Q$ having coordinates $x = beta s$, $y = − alpha s$ when $t = 0$. Then, from (4.8), we find $C_1 = beta s$, $C_2 = − alpha s$. Here $s$ is a parameter that plays a role similar to that of $k$ in the previous example. It varies along the line l and each choice of s selects a different characteristic:
$$x = pt + beta s, y = qt − alpha s$$
Next we consider the last characteristic equation
$$fracdudt = f(x(s, t), y(s, t))$$
Integrating gives
$$u(s, t) = u(s, 0) + int_0^t f(x(s, t'), y(s, t')) dt'$$
Here $u(s, 0)$ is a constant along any characteristic but varies from one characteristic to another.
My question is, how did the author get
$$u(s, t) = u(s, 0) + int_0^t f(x(s, t'), y(s, t')) dt'$$
by integrating?
Where did the $u(s, 0)$ term come from? And how was $int_0^t f(x(s, t'), y(s, t')) dt'$ calculated by integration? For instance, I have never seen something written like $dt'$?
Thank you for any explanation you could offer.
real-analysis integration multivariable-calculus pde characteristics
asked Jul 30 at 18:34
handler's handle
848
add a comment |Â
up vote
0
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From Essential Partial Differential Equations, by Griffiths, Dold, and Silvester:
Example 4.2 (Half-plane problem) Solve the PDE $pux + quy = f(x, y)$ (with $p$ and $q$ constant) in the domain $alpha x + beta y > 0$ given that $u = g(x)$ on the line $mathcall : alpha x + beta y = 0$, where $q > 0$ and $beta > 0$.
As in the previous example, the characteristic equations (4.4) are readily solved to give
$$x = pt + C_1, y = qt + C_2$$
where $C_1$ and $C_2$ are constant along any characteristic. The assumption $q > 0$ means that $y$ increases along a characteristic as $t$ increases -- thus the parameter $t$ may be viewed as a time-like variable. This is illustrated in Fig. 4.1 where the arrows on the characteristics indicate increasing t.
In order to determine the solution at a point $P(x, y)$, we trace the characteristic through this point backwards in $t$ until it intersects the line $mathcall$. We shall suppose that this occurs at the point $Q$ having coordinates $x = beta s$, $y = − alpha s$ when $t = 0$. Then, from (4.8), we find $C_1 = beta s$, $C_2 = − alpha s$. Here $s$ is a parameter that plays a role similar to that of $k$ in the previous example. It varies along the line l and each choice of s selects a different characteristic:
$$x = pt + beta s, y = qt − alpha s$$
Next we consider the last characteristic equation
$$fracdudt = f(x(s, t), y(s, t))$$
Integrating gives
$$u(s, t) = u(s, 0) + int_0^t f(x(s, t'), y(s, t')) dt'$$
Here $u(s, 0)$ is a constant along any characteristic but varies from one characteristic to another.
My question is, how did the author get
$$u(s, t) = u(s, 0) + int_0^t f(x(s, t'), y(s, t')) dt'$$
by integrating?
Where did the $u(s, 0)$ term come from? And how was $int_0^t f(x(s, t'), y(s, t')) dt'$ calculated by integration? For instance, I have never seen something written like $dt'$?
Thank you for any explanation you could offer.
real-analysis integration multivariable-calculus pde characteristics
asked Jul 30 at 18:34
handler's handle
848
Keep in mind that $t'$ doesn't denote a derivative- it's just another variable. It might help to replace it with something like $z$. You can apply the fundamental theorem of calculus to see that this solution satisfies the differential equation. You can also check that the formula has the correct value for $u(s,t)$ at the initial time $t=0$.
– Brian Borchers
Jul 30 at 18:41
What's the derivative with respect to $t$ of $u(s,0)$? What's the derivative with respect to $t$ of the integral in the second term?
– Brian Borchers
Jul 30 at 18:53
@BrianBorchers Ohhhhhhhhhhhhhhh, the $u(s, 0)$ is from using the part 2 fundamental theorem of calculus! And $int_0^t f(x(s, t'), y(s, t')) dt'$ is from the part 1 fundamental theorem of calculus. Using the part 2 fundamental theorem of calculus, we get $u(s, t) - u(s, 0)$ for the left-hand side integration.
– handler's handle
Jul 31 at 2:05
add a comment |Â
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0
down vote
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up vote
0
down vote
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From Essential Partial Differential Equations, by Griffiths, Dold, and Silvester:
Example 4.2 (Half-plane problem) Solve the PDE $pux + quy = f(x, y)$ (with $p$ and $q$ constant) in the domain $alpha x + beta y > 0$ given that $u = g(x)$ on the line $mathcall : alpha x + beta y = 0$, where $q > 0$ and $beta > 0$.
As in the previous example, the characteristic equations (4.4) are readily solved to give
$$x = pt + C_1, y = qt + C_2$$
where $C_1$ and $C_2$ are constant along any characteristic. The assumption $q > 0$ means that $y$ increases along a characteristic as $t$ increases -- thus the parameter $t$ may be viewed as a time-like variable. This is illustrated in Fig. 4.1 where the arrows on the characteristics indicate increasing t.
In order to determine the solution at a point $P(x, y)$, we trace the characteristic through this point backwards in $t$ until it intersects the line $mathcall$. We shall suppose that this occurs at the point $Q$ having coordinates $x = beta s$, $y = − alpha s$ when $t = 0$. Then, from (4.8), we find $C_1 = beta s$, $C_2 = − alpha s$. Here $s$ is a parameter that plays a role similar to that of $k$ in the previous example. It varies along the line l and each choice of s selects a different characteristic:
$$x = pt + beta s, y = qt − alpha s$$
Next we consider the last characteristic equation
$$fracdudt = f(x(s, t), y(s, t))$$
Integrating gives
$$u(s, t) = u(s, 0) + int_0^t f(x(s, t'), y(s, t')) dt'$$
Here $u(s, 0)$ is a constant along any characteristic but varies from one characteristic to another.
My question is, how did the author get
$$u(s, t) = u(s, 0) + int_0^t f(x(s, t'), y(s, t')) dt'$$
by integrating?
Where did the $u(s, 0)$ term come from? And how was $int_0^t f(x(s, t'), y(s, t')) dt'$ calculated by integration? For instance, I have never seen something written like $dt'$?
Thank you for any explanation you could offer.
real-analysis integration multivariable-calculus pde characteristics
asked Jul 30 at 18:34
handler's handle
848
From Essential Partial Differential Equations, by Griffiths, Dold, and Silvester:
Example 4.2 (Half-plane problem) Solve the PDE $pux + quy = f(x, y)$ (with $p$ and $q$ constant) in the domain $alpha x + beta y > 0$ given that $u = g(x)$ on the line $mathcall : alpha x + beta y = 0$, where $q > 0$ and $beta > 0$.
As in the previous example, the characteristic equations (4.4) are readily solved to give
$$x = pt + C_1, y = qt + C_2$$
where $C_1$ and $C_2$ are constant along any characteristic. The assumption $q > 0$ means that $y$ increases along a characteristic as $t$ increases -- thus the parameter $t$ may be viewed as a time-like variable. This is illustrated in Fig. 4.1 where the arrows on the characteristics indicate increasing t.
In order to determine the solution at a point $P(x, y)$, we trace the characteristic through this point backwards in $t$ until it intersects the line $mathcall$. We shall suppose that this occurs at the point $Q$ having coordinates $x = beta s$, $y = − alpha s$ when $t = 0$. Then, from (4.8), we find $C_1 = beta s$, $C_2 = − alpha s$. Here $s$ is a parameter that plays a role similar to that of $k$ in the previous example. It varies along the line l and each choice of s selects a different characteristic:
$$x = pt + beta s, y = qt − alpha s$$
Next we consider the last characteristic equation
$$fracdudt = f(x(s, t), y(s, t))$$
Integrating gives
$$u(s, t) = u(s, 0) + int_0^t f(x(s, t'), y(s, t')) dt'$$
Here $u(s, 0)$ is a constant along any characteristic but varies from one characteristic to another.
My question is, how did the author get
$$u(s, t) = u(s, 0) + int_0^t f(x(s, t'), y(s, t')) dt'$$
by integrating?
Where did the $u(s, 0)$ term come from? And how was $int_0^t f(x(s, t'), y(s, t')) dt'$ calculated by integration? For instance, I have never seen something written like $dt'$?
Thank you for any explanation you could offer.
real-analysis integration multivariable-calculus pde characteristics
asked Jul 30 at 18:34
handler's handle
848
edited Jul 31 at 1:46
asked Jul 30 at 18:34
handler's handle
848
asked Jul 30 at 18:34
handler's handle
848
asked Jul 30 at 18:34
handler's handle
848
848
Keep in mind that $t'$ doesn't denote a derivative- it's just another variable. It might help to replace it with something like $z$. You can apply the fundamental theorem of calculus to see that this solution satisfies the differential equation. You can also check that the formula has the correct value for $u(s,t)$ at the initial time $t=0$.
– Brian Borchers
Jul 30 at 18:41
What's the derivative with respect to $t$ of $u(s,0)$? What's the derivative with respect to $t$ of the integral in the second term?
– Brian Borchers
Jul 30 at 18:53
@BrianBorchers Ohhhhhhhhhhhhhhh, the $u(s, 0)$ is from using the part 2 fundamental theorem of calculus! And $int_0^t f(x(s, t'), y(s, t')) dt'$ is from the part 1 fundamental theorem of calculus. Using the part 2 fundamental theorem of calculus, we get $u(s, t) - u(s, 0)$ for the left-hand side integration.
– handler's handle
Jul 31 at 2:05
add a comment |Â
Keep in mind that $t'$ doesn't denote a derivative- it's just another variable. It might help to replace it with something like $z$. You can apply the fundamental theorem of calculus to see that this solution satisfies the differential equation. You can also check that the formula has the correct value for $u(s,t)$ at the initial time $t=0$.
– Brian Borchers
Jul 30 at 18:41
What's the derivative with respect to $t$ of $u(s,0)$? What's the derivative with respect to $t$ of the integral in the second term?
– Brian Borchers
Jul 30 at 18:53
@BrianBorchers Ohhhhhhhhhhhhhhh, the $u(s, 0)$ is from using the part 2 fundamental theorem of calculus! And $int_0^t f(x(s, t'), y(s, t')) dt'$ is from the part 1 fundamental theorem of calculus. Using the part 2 fundamental theorem of calculus, we get $u(s, t) - u(s, 0)$ for the left-hand side integration.
– handler's handle
Jul 31 at 2:05
Keep in mind that $t'$ doesn't denote a derivative- it's just another variable. It might help to replace it with something like $z$. You can apply the fundamental theorem of calculus to see that this solution satisfies the differential equation. You can also check that the formula has the correct value for $u(s,t)$ at the initial time $t=0$.
– Brian Borchers
Jul 30 at 18:41
Keep in mind that $t'$ doesn't denote a derivative- it's just another variable. It might help to replace it with something like $z$. You can apply the fundamental theorem of calculus to see that this solution satisfies the differential equation. You can also check that the formula has the correct value for $u(s,t)$ at the initial time $t=0$.
– Brian Borchers
Jul 30 at 18:41
What's the derivative with respect to $t$ of $u(s,0)$? What's the derivative with respect to $t$ of the integral in the second term?
– Brian Borchers
Jul 30 at 18:53
What's the derivative with respect to $t$ of $u(s,0)$? What's the derivative with respect to $t$ of the integral in the second term?
– Brian Borchers
Jul 30 at 18:53
@BrianBorchers Ohhhhhhhhhhhhhhh, the $u(s, 0)$ is from using the part 2 fundamental theorem of calculus! And $int_0^t f(x(s, t'), y(s, t')) dt'$ is from the part 1 fundamental theorem of calculus. Using the part 2 fundamental theorem of calculus, we get $u(s, t) - u(s, 0)$ for the left-hand side integration.
– handler's handle
Jul 31 at 2:05
@BrianBorchers Ohhhhhhhhhhhhhhh, the $u(s, 0)$ is from using the part 2 fundamental theorem of calculus! And $int_0^t f(x(s, t'), y(s, t')) dt'$ is from the part 1 fundamental theorem of calculus. Using the part 2 fundamental theorem of calculus, we get $u(s, t) - u(s, 0)$ for the left-hand side integration.
– handler's handle
Jul 31 at 2:05
add a comment |Â
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Keep in mind that $t'$ doesn't denote a derivative- it's just another variable. It might help to replace it with something like $z$. You can apply the fundamental theorem of calculus to see that this solution satisfies the differential equation. You can also check that the formula has the correct value for $u(s,t)$ at the initial time $t=0$.
– Brian Borchers
Jul 30 at 18:41
What's the derivative with respect to $t$ of $u(s,0)$? What's the derivative with respect to $t$ of the integral in the second term?
– Brian Borchers
Jul 30 at 18:53
@BrianBorchers Ohhhhhhhhhhhhhhh, the $u(s, 0)$ is from using the part 2 fundamental theorem of calculus! And $int_0^t f(x(s, t'), y(s, t')) dt'$ is from the part 1 fundamental theorem of calculus. Using the part 2 fundamental theorem of calculus, we get $u(s, t) - u(s, 0)$ for the left-hand side integration.
– handler's handle
Jul 31 at 2:05