Method of characteristics example: using different change of variables to textbook, characteristic speed, and example clarification

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The following method of characteristics example is from chapter 4 of Essential Partial Differential Equations by Griffiths, Dold, and Silvester:

Example 4.2

(Half-plane problem) Solve the PDE $pu_x + qu_y = f$ (with $p$ and $q$ constant) in the domain $alpha x + beta y > 0$ given that $u = g(x)$ on the line $mathscrl: alpha x + beta y = 0$, where $q > 0$ and $beta > 0$.

The characteristic equations are readily solved to give

$$x = pt + C_1, y = qt + C_2 (4.8)$$

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.

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 $mathscrl$. 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 $C_1 = beta s$, $C_2 = -alpha s$. $s$ varies along the line $mathscrl$ and each choice $s$ selects a different characteristic:

$$x = pt + beta s, y = qt - alpha s (4.9)$$

Note that the characteristics have inherited their parameterization from that of the line $mathscrl$. Next, we integrate the ODE $dfracdudt = f$:

$$int_0^t fracdudt' dt' = int_0^t f(x(s, t'), y(s, t')) dt'$$

to give

$$u(s, t) = u(s, 0) + F(t), F(t) = int_0^t f(x(s, t'), y(s, t')) dt'$$

Here, $A(s) = u(s, 0)$ is constant on any characteristic but varies , with $s$, from one characteristic to another. The initial condition gives $u = g(x) = g(beta s)$ at $t = 0$ from which we deduce (using (4.9)) that $u(s, 0) = g(beta s)$ so that

$$u = g(beta s) + int_0^t f(x(s, t'), y(s, t')) dt' (4.10)$$

Finally, to express $u$ as a function of $x$ and $y$, the equation (4.9) have to be solved to give $s$ and $t$ in terms of $x$ and $y$:

$$t = fracalpha x + beta yalpha p + beta q, s = fracqx - pyalpha p + beta q$$

which are valid so long as $alpha p + beta q not= 0$. This condition has a geometric interpretation. Since $alpha p + beta q$ is the scalar product of the vector $(p, q)$ (which is parallel to the characteristics), and the vector $(alpha, beta)$ (which is orthogonal to the line $mathscrl$), the change of variables (4.9) is one-to-one if, and only if, (p, q) is not parallel to $(alpha, beta)$. That is the initial condition should not be specified on any line parallel to the characteristics. This is an important conclusion that applied more widely and, when it holds, we have a unique solution at a point $P$ that depends only on the initial value $g$ given at the foot $Q$ of the characteristic through $P$ and the values of $f$ along $PQ$.

Finally, note that in the simple homogeneous case the coefficients of the PDE $pu_x + qu_y = 0$ form a vector $(p, q)$, whereas the solution (4.10) depends only on $s$, that is, $qx - py$ whose coefficients form a vector $(q, -p)$ that is orthogonal to $(p, q)$. This means that boundary information is simply conveyed along characteristics. Since information travels a horizontal distance $p / q$ for each unit of vertical distance, $p/q$ is known as the characteristic speed of the equation.

When I did the change of variables for this example, I set $x(0) = s$ and $y(0) = frac-alpha sbeta$ so that we get $x(t) = pt + s$ and $y(t) = qt - fracalpha sbeta$. However, when finding the change of variables so that we get $t$ and $s$ in terms of $x$ and $y$, I get the same equations $t = fracalpha x + beta yalpha p + beta q, s = fracqx - pyalpha p + beta q$ as the example. So, despite doing the change of variables differently, is my work still correct?

And, at the end, I don't understand what is meant by this:

Finally, note that in the simple homogeneous case the coefficients of the PDE $pu_x + qu_y = 0$ form a vector $(p, q)$, whereas the solution (4.10) depends only on $s$, that is, $qx - py$ whose coefficients form a vector $(q, -p)$ that is orthogonal to $(p, q)$. This means that boundary information is simply conveyed along characteristics.

First of all, since $u = u(s, t)$ isn't (4.10) dependent on both $s$ and $t$, not just $s$?

Also, wouldn't $s$ be dependent on $fracqx - pyalpha p + beta q$, not just $qx - py$, since $s = fracqx - pyalpha p + beta q$?

And why does the last part mean that boundary information is simply conveyed along characteristics? This isn't clear to me.

Since information travels a horizontal distance $p / q$ for each unit of vertical distance, $p/q$ is known as the characteristic speed of the equation.

Why does information travels a horizontal distance $p / q$ for each unit of vertical distance? I don't see how the author got this.

Thank you for any help you could kindly provide.

edited Aug 7 at 10:22

asked Aug 5 at 23:03

handler's handle

1508

add a commentÂ |Â

up vote
1
down vote

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The following method of characteristics example is from chapter 4 of Essential Partial Differential Equations by Griffiths, Dold, and Silvester:

Example 4.2

(Half-plane problem) Solve the PDE $pu_x + qu_y = f$ (with $p$ and $q$ constant) in the domain $alpha x + beta y > 0$ given that $u = g(x)$ on the line $mathscrl: alpha x + beta y = 0$, where $q > 0$ and $beta > 0$.

The characteristic equations are readily solved to give

$$x = pt + C_1, y = qt + C_2 (4.8)$$

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.

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 $mathscrl$. 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 $C_1 = beta s$, $C_2 = -alpha s$. $s$ varies along the line $mathscrl$ and each choice $s$ selects a different characteristic:

$$x = pt + beta s, y = qt - alpha s (4.9)$$

Note that the characteristics have inherited their parameterization from that of the line $mathscrl$. Next, we integrate the ODE $dfracdudt = f$:

$$int_0^t fracdudt' dt' = int_0^t f(x(s, t'), y(s, t')) dt'$$

to give

$$u(s, t) = u(s, 0) + F(t), F(t) = int_0^t f(x(s, t'), y(s, t')) dt'$$

Here, $A(s) = u(s, 0)$ is constant on any characteristic but varies , with $s$, from one characteristic to another. The initial condition gives $u = g(x) = g(beta s)$ at $t = 0$ from which we deduce (using (4.9)) that $u(s, 0) = g(beta s)$ so that

$$u = g(beta s) + int_0^t f(x(s, t'), y(s, t')) dt' (4.10)$$

Finally, to express $u$ as a function of $x$ and $y$, the equation (4.9) have to be solved to give $s$ and $t$ in terms of $x$ and $y$:

$$t = fracalpha x + beta yalpha p + beta q, s = fracqx - pyalpha p + beta q$$

which are valid so long as $alpha p + beta q not= 0$. This condition has a geometric interpretation. Since $alpha p + beta q$ is the scalar product of the vector $(p, q)$ (which is parallel to the characteristics), and the vector $(alpha, beta)$ (which is orthogonal to the line $mathscrl$), the change of variables (4.9) is one-to-one if, and only if, (p, q) is not parallel to $(alpha, beta)$. That is the initial condition should not be specified on any line parallel to the characteristics. This is an important conclusion that applied more widely and, when it holds, we have a unique solution at a point $P$ that depends only on the initial value $g$ given at the foot $Q$ of the characteristic through $P$ and the values of $f$ along $PQ$.

Finally, note that in the simple homogeneous case the coefficients of the PDE $pu_x + qu_y = 0$ form a vector $(p, q)$, whereas the solution (4.10) depends only on $s$, that is, $qx - py$ whose coefficients form a vector $(q, -p)$ that is orthogonal to $(p, q)$. This means that boundary information is simply conveyed along characteristics. Since information travels a horizontal distance $p / q$ for each unit of vertical distance, $p/q$ is known as the characteristic speed of the equation.

When I did the change of variables for this example, I set $x(0) = s$ and $y(0) = frac-alpha sbeta$ so that we get $x(t) = pt + s$ and $y(t) = qt - fracalpha sbeta$. However, when finding the change of variables so that we get $t$ and $s$ in terms of $x$ and $y$, I get the same equations $t = fracalpha x + beta yalpha p + beta q, s = fracqx - pyalpha p + beta q$ as the example. So, despite doing the change of variables differently, is my work still correct?

And, at the end, I don't understand what is meant by this:

Finally, note that in the simple homogeneous case the coefficients of the PDE $pu_x + qu_y = 0$ form a vector $(p, q)$, whereas the solution (4.10) depends only on $s$, that is, $qx - py$ whose coefficients form a vector $(q, -p)$ that is orthogonal to $(p, q)$. This means that boundary information is simply conveyed along characteristics.

First of all, since $u = u(s, t)$ isn't (4.10) dependent on both $s$ and $t$, not just $s$?

Also, wouldn't $s$ be dependent on $fracqx - pyalpha p + beta q$, not just $qx - py$, since $s = fracqx - pyalpha p + beta q$?

And why does the last part mean that boundary information is simply conveyed along characteristics? This isn't clear to me.

Since information travels a horizontal distance $p / q$ for each unit of vertical distance, $p/q$ is known as the characteristic speed of the equation.

Why does information travels a horizontal distance $p / q$ for each unit of vertical distance? I don't see how the author got this.

Thank you for any help you could kindly provide.

edited Aug 7 at 10:22

asked Aug 5 at 23:03

handler's handle

1508

add a commentÂ |Â

up vote
1
down vote

favorite

The following method of characteristics example is from chapter 4 of Essential Partial Differential Equations by Griffiths, Dold, and Silvester:

Example 4.2

(Half-plane problem) Solve the PDE $pu_x + qu_y = f$ (with $p$ and $q$ constant) in the domain $alpha x + beta y > 0$ given that $u = g(x)$ on the line $mathscrl: alpha x + beta y = 0$, where $q > 0$ and $beta > 0$.

The characteristic equations are readily solved to give

$$x = pt + C_1, y = qt + C_2 (4.8)$$

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.

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 $mathscrl$. 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 $C_1 = beta s$, $C_2 = -alpha s$. $s$ varies along the line $mathscrl$ and each choice $s$ selects a different characteristic:

$$x = pt + beta s, y = qt - alpha s (4.9)$$

Note that the characteristics have inherited their parameterization from that of the line $mathscrl$. Next, we integrate the ODE $dfracdudt = f$:

$$int_0^t fracdudt' dt' = int_0^t f(x(s, t'), y(s, t')) dt'$$

to give

$$u(s, t) = u(s, 0) + F(t), F(t) = int_0^t f(x(s, t'), y(s, t')) dt'$$

Here, $A(s) = u(s, 0)$ is constant on any characteristic but varies , with $s$, from one characteristic to another. The initial condition gives $u = g(x) = g(beta s)$ at $t = 0$ from which we deduce (using (4.9)) that $u(s, 0) = g(beta s)$ so that

$$u = g(beta s) + int_0^t f(x(s, t'), y(s, t')) dt' (4.10)$$

Finally, to express $u$ as a function of $x$ and $y$, the equation (4.9) have to be solved to give $s$ and $t$ in terms of $x$ and $y$:

$$t = fracalpha x + beta yalpha p + beta q, s = fracqx - pyalpha p + beta q$$

which are valid so long as $alpha p + beta q not= 0$. This condition has a geometric interpretation. Since $alpha p + beta q$ is the scalar product of the vector $(p, q)$ (which is parallel to the characteristics), and the vector $(alpha, beta)$ (which is orthogonal to the line $mathscrl$), the change of variables (4.9) is one-to-one if, and only if, (p, q) is not parallel to $(alpha, beta)$. That is the initial condition should not be specified on any line parallel to the characteristics. This is an important conclusion that applied more widely and, when it holds, we have a unique solution at a point $P$ that depends only on the initial value $g$ given at the foot $Q$ of the characteristic through $P$ and the values of $f$ along $PQ$.

Finally, note that in the simple homogeneous case the coefficients of the PDE $pu_x + qu_y = 0$ form a vector $(p, q)$, whereas the solution (4.10) depends only on $s$, that is, $qx - py$ whose coefficients form a vector $(q, -p)$ that is orthogonal to $(p, q)$. This means that boundary information is simply conveyed along characteristics. Since information travels a horizontal distance $p / q$ for each unit of vertical distance, $p/q$ is known as the characteristic speed of the equation.

When I did the change of variables for this example, I set $x(0) = s$ and $y(0) = frac-alpha sbeta$ so that we get $x(t) = pt + s$ and $y(t) = qt - fracalpha sbeta$. However, when finding the change of variables so that we get $t$ and $s$ in terms of $x$ and $y$, I get the same equations $t = fracalpha x + beta yalpha p + beta q, s = fracqx - pyalpha p + beta q$ as the example. So, despite doing the change of variables differently, is my work still correct?

And, at the end, I don't understand what is meant by this:

Finally, note that in the simple homogeneous case the coefficients of the PDE $pu_x + qu_y = 0$ form a vector $(p, q)$, whereas the solution (4.10) depends only on $s$, that is, $qx - py$ whose coefficients form a vector $(q, -p)$ that is orthogonal to $(p, q)$. This means that boundary information is simply conveyed along characteristics.

First of all, since $u = u(s, t)$ isn't (4.10) dependent on both $s$ and $t$, not just $s$?

Also, wouldn't $s$ be dependent on $fracqx - pyalpha p + beta q$, not just $qx - py$, since $s = fracqx - pyalpha p + beta q$?

And why does the last part mean that boundary information is simply conveyed along characteristics? This isn't clear to me.

Since information travels a horizontal distance $p / q$ for each unit of vertical distance, $p/q$ is known as the characteristic speed of the equation.

Why does information travels a horizontal distance $p / q$ for each unit of vertical distance? I don't see how the author got this.

Thank you for any help you could kindly provide.

edited Aug 7 at 10:22

asked Aug 5 at 23:03

handler's handle

1508

The following method of characteristics example is from chapter 4 of Essential Partial Differential Equations by Griffiths, Dold, and Silvester:

Example 4.2

(Half-plane problem) Solve the PDE $pu_x + qu_y = f$ (with $p$ and $q$ constant) in the domain $alpha x + beta y > 0$ given that $u = g(x)$ on the line $mathscrl: alpha x + beta y = 0$, where $q > 0$ and $beta > 0$.

The characteristic equations are readily solved to give

$$x = pt + C_1, y = qt + C_2 (4.8)$$

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.

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 $mathscrl$. 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 $C_1 = beta s$, $C_2 = -alpha s$. $s$ varies along the line $mathscrl$ and each choice $s$ selects a different characteristic:

$$x = pt + beta s, y = qt - alpha s (4.9)$$

Note that the characteristics have inherited their parameterization from that of the line $mathscrl$. Next, we integrate the ODE $dfracdudt = f$:

$$int_0^t fracdudt' dt' = int_0^t f(x(s, t'), y(s, t')) dt'$$

to give

$$u(s, t) = u(s, 0) + F(t), F(t) = int_0^t f(x(s, t'), y(s, t')) dt'$$

Here, $A(s) = u(s, 0)$ is constant on any characteristic but varies , with $s$, from one characteristic to another. The initial condition gives $u = g(x) = g(beta s)$ at $t = 0$ from which we deduce (using (4.9)) that $u(s, 0) = g(beta s)$ so that

$$u = g(beta s) + int_0^t f(x(s, t'), y(s, t')) dt' (4.10)$$

Finally, to express $u$ as a function of $x$ and $y$, the equation (4.9) have to be solved to give $s$ and $t$ in terms of $x$ and $y$:

$$t = fracalpha x + beta yalpha p + beta q, s = fracqx - pyalpha p + beta q$$

which are valid so long as $alpha p + beta q not= 0$. This condition has a geometric interpretation. Since $alpha p + beta q$ is the scalar product of the vector $(p, q)$ (which is parallel to the characteristics), and the vector $(alpha, beta)$ (which is orthogonal to the line $mathscrl$), the change of variables (4.9) is one-to-one if, and only if, (p, q) is not parallel to $(alpha, beta)$. That is the initial condition should not be specified on any line parallel to the characteristics. This is an important conclusion that applied more widely and, when it holds, we have a unique solution at a point $P$ that depends only on the initial value $g$ given at the foot $Q$ of the characteristic through $P$ and the values of $f$ along $PQ$.

Finally, note that in the simple homogeneous case the coefficients of the PDE $pu_x + qu_y = 0$ form a vector $(p, q)$, whereas the solution (4.10) depends only on $s$, that is, $qx - py$ whose coefficients form a vector $(q, -p)$ that is orthogonal to $(p, q)$. This means that boundary information is simply conveyed along characteristics. Since information travels a horizontal distance $p / q$ for each unit of vertical distance, $p/q$ is known as the characteristic speed of the equation.

When I did the change of variables for this example, I set $x(0) = s$ and $y(0) = frac-alpha sbeta$ so that we get $x(t) = pt + s$ and $y(t) = qt - fracalpha sbeta$. However, when finding the change of variables so that we get $t$ and $s$ in terms of $x$ and $y$, I get the same equations $t = fracalpha x + beta yalpha p + beta q, s = fracqx - pyalpha p + beta q$ as the example. So, despite doing the change of variables differently, is my work still correct?

And, at the end, I don't understand what is meant by this:

Finally, note that in the simple homogeneous case the coefficients of the PDE $pu_x + qu_y = 0$ form a vector $(p, q)$, whereas the solution (4.10) depends only on $s$, that is, $qx - py$ whose coefficients form a vector $(q, -p)$ that is orthogonal to $(p, q)$. This means that boundary information is simply conveyed along characteristics.

First of all, since $u = u(s, t)$ isn't (4.10) dependent on both $s$ and $t$, not just $s$?

Also, wouldn't $s$ be dependent on $fracqx - pyalpha p + beta q$, not just $qx - py$, since $s = fracqx - pyalpha p + beta q$?

And why does the last part mean that boundary information is simply conveyed along characteristics? This isn't clear to me.

Since information travels a horizontal distance $p / q$ for each unit of vertical distance, $p/q$ is known as the characteristic speed of the equation.

Why does information travels a horizontal distance $p / q$ for each unit of vertical distance? I don't see how the author got this.

Thank you for any help you could kindly provide.

edited Aug 7 at 10:22

asked Aug 5 at 23:03

handler's handle

1508

edited Aug 7 at 10:22

asked Aug 5 at 23:03

handler's handle

1508

asked Aug 5 at 23:03

handler's handle

1508

asked Aug 5 at 23:03

handler's handle

1508

add a commentÂ |Â

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