Solving the PDE $u_x + u_y = 0$: $x - y$ Implies General Solution $u(x, y) = F(x - y)$? Solving $x = s - phi(s)$ for $s$ in Terms of $x$?

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Solve the PDE $u_x + u_y = 0$ in the domain $y > phi(x)$, $x in mathbbR$, given that $u = g(x)$ on the curve $y = phi(x)$, where $phi(x) = dfracx1 + $.

The characteristic equations are

$$fracdxdt = 1, fracdydt = 1$$

with solution

$$x = t + C_1, y = t + C_2$$

So $x - y$ is a constant, which implies that a general solution is of the form

$$u(x, y) = F(x - y)$$

where $F$ is an arbitrary function.

Question 1: Why does $x - y$ being a constant imply that a general solution is of the form $u(x, y) = F(x - y)?$

Given the general solution

$$u(x, y) = F(x - y)$$

with initial curve

$$Gamma = (s, phi(s), g(s))$$

we have the solution

$$u(s, phi(s)) = F(s - phi(s)) = g(s)$$

It just remains to write the solution in terms of $x$ and $y$.

Writing the solution in terms of $x$ and $y$ requires that $x = s - phi(s)$ be solved for $s$ in terms of $x$ so as to give $A(x) = g(s(x))$.

When $phi(x) = dfracx1 + $, this is accomplished by treating the cases $x ge 0$ and $x < 0$ separately.

$s(x) =
begincases
frac12(x + sqrtx^2 + 4x)&textif, xge 0\
frac12(x - sqrtx^2 - 4x)&textif, x < 0\
endcases$

Question 2: How did the author get these values for $s(x)$? What is the process?

The solution can then be written as $u(x, y) = g(s(x - y))$ for $y > phi(x)$.

I would be very thankful if someone could please help me with this.

pde vector-analysis parametric parametrization characteristics

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asked Jul 31 at 6:05

handler's handle

848

add a comment | 

up vote
2
down vote

favorite

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Solve the PDE $u_x + u_y = 0$ in the domain $y > phi(x)$, $x in mathbbR$, given that $u = g(x)$ on the curve $y = phi(x)$, where $phi(x) = dfracx1 + $.

The characteristic equations are

$$fracdxdt = 1, fracdydt = 1$$

with solution

$$x = t + C_1, y = t + C_2$$

So $x - y$ is a constant, which implies that a general solution is of the form

$$u(x, y) = F(x - y)$$

where $F$ is an arbitrary function.

Question 1: Why does $x - y$ being a constant imply that a general solution is of the form $u(x, y) = F(x - y)?$

Given the general solution

$$u(x, y) = F(x - y)$$

with initial curve

$$Gamma = (s, phi(s), g(s))$$

we have the solution

$$u(s, phi(s)) = F(s - phi(s)) = g(s)$$

It just remains to write the solution in terms of $x$ and $y$.

Writing the solution in terms of $x$ and $y$ requires that $x = s - phi(s)$ be solved for $s$ in terms of $x$ so as to give $A(x) = g(s(x))$.

When $phi(x) = dfracx1 + $, this is accomplished by treating the cases $x ge 0$ and $x < 0$ separately.

$s(x) =
begincases
frac12(x + sqrtx^2 + 4x)&textif, xge 0\
frac12(x - sqrtx^2 - 4x)&textif, x < 0\
endcases$

Question 2: How did the author get these values for $s(x)$? What is the process?

The solution can then be written as $u(x, y) = g(s(x - y))$ for $y > phi(x)$.

I would be very thankful if someone could please help me with this.

pde vector-analysis parametric parametrization characteristics

share|cite|improve this question

asked Jul 31 at 6:05

handler's handle

848

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

Solve the PDE $u_x + u_y = 0$ in the domain $y > phi(x)$, $x in mathbbR$, given that $u = g(x)$ on the curve $y = phi(x)$, where $phi(x) = dfracx1 + $.

The characteristic equations are

$$fracdxdt = 1, fracdydt = 1$$

with solution

$$x = t + C_1, y = t + C_2$$

So $x - y$ is a constant, which implies that a general solution is of the form

$$u(x, y) = F(x - y)$$

where $F$ is an arbitrary function.

Question 1: Why does $x - y$ being a constant imply that a general solution is of the form $u(x, y) = F(x - y)?$

Given the general solution

$$u(x, y) = F(x - y)$$

with initial curve

$$Gamma = (s, phi(s), g(s))$$

we have the solution

$$u(s, phi(s)) = F(s - phi(s)) = g(s)$$

It just remains to write the solution in terms of $x$ and $y$.

Writing the solution in terms of $x$ and $y$ requires that $x = s - phi(s)$ be solved for $s$ in terms of $x$ so as to give $A(x) = g(s(x))$.

When $phi(x) = dfracx1 + $, this is accomplished by treating the cases $x ge 0$ and $x < 0$ separately.

$s(x) =
begincases
frac12(x + sqrtx^2 + 4x)&textif, xge 0\
frac12(x - sqrtx^2 - 4x)&textif, x < 0\
endcases$

Question 2: How did the author get these values for $s(x)$? What is the process?

The solution can then be written as $u(x, y) = g(s(x - y))$ for $y > phi(x)$.

I would be very thankful if someone could please help me with this.

pde vector-analysis parametric parametrization characteristics

share|cite|improve this question

asked Jul 31 at 6:05

handler's handle

848

Solve the PDE $u_x + u_y = 0$ in the domain $y > phi(x)$, $x in mathbbR$, given that $u = g(x)$ on the curve $y = phi(x)$, where $phi(x) = dfracx1 + $.

The characteristic equations are

$$fracdxdt = 1, fracdydt = 1$$

with solution

$$x = t + C_1, y = t + C_2$$

So $x - y$ is a constant, which implies that a general solution is of the form

$$u(x, y) = F(x - y)$$

where $F$ is an arbitrary function.

Question 1: Why does $x - y$ being a constant imply that a general solution is of the form $u(x, y) = F(x - y)?$

Given the general solution

$$u(x, y) = F(x - y)$$

with initial curve

$$Gamma = (s, phi(s), g(s))$$

we have the solution

$$u(s, phi(s)) = F(s - phi(s)) = g(s)$$

It just remains to write the solution in terms of $x$ and $y$.

Writing the solution in terms of $x$ and $y$ requires that $x = s - phi(s)$ be solved for $s$ in terms of $x$ so as to give $A(x) = g(s(x))$.

When $phi(x) = dfracx1 + $, this is accomplished by treating the cases $x ge 0$ and $x < 0$ separately.

$s(x) =
begincases
frac12(x + sqrtx^2 + 4x)&textif, xge 0\
frac12(x - sqrtx^2 - 4x)&textif, x < 0\
endcases$

Question 2: How did the author get these values for $s(x)$? What is the process?

The solution can then be written as $u(x, y) = g(s(x - y))$ for $y > phi(x)$.

I would be very thankful if someone could please help me with this.

pde vector-analysis parametric parametrization characteristics

share|cite|improve this question

asked Jul 31 at 6:05

handler's handle

848

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 6:05

handler's handle

848

asked Jul 31 at 6:05

handler's handle

848

asked Jul 31 at 6:05

handler's handle

848

848

add a comment | 

add a comment | 

1 Answer
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accepted

For the second question. Simply isolate $s$ in $x=s-phi(s)$ or in

$x=s-dfracss$. For the $sge 0$ part: $x(1+s)=s+s^2-s$ or

$s^2-xs-x=0implies s=frac12(xpmsqrtx^2 + 4x)$ We drop the solution with the minus sign because it entails that $s<0$, aganist the hypothesis for the posing of the equation. Because the same reason ($s<0$) we drop values with $x<0$. So $s=frac12(x + sqrtx^2 + 4x);textif, xge 0$. For $s<0$ we solve $x(1-s)=s-s^2-s$ or ($s^2-xs+x=0$) instead. Then, $s=frac12(xpmsqrtx^2 -4x)$ Now, we drop the plus sign because on the contrary $sge0$ and, because the same reason, it must be $x<0$. This drives to the given solution:

$$s(x) =
begincases
frac12(x + sqrtx^2 + 4x)&textif, xge 0\
frac12(x - sqrtx^2 - 4x)&textif, x < 0\
endcases$$

For the first question you have to consider that $u$ is constant along the characteristic curves, so is, along curves $x-y=c_1$ with $c_1$ some constant($u(x,x+c_1)=k$, $k$ constant). A solution must specify the value of $u$ for each of the values of $c_1$: $u(x,x+c_1)=F(c_1)$. So is, a particular choice of $F$ says the value of $u$ for each member of the family of characteristic curves: $u(x,y)=F(x-y)$. The function $F$ is arbitrary in the sense that by itself doesn't impose any condition on the solutions (taken apart that it must enjoy of some regularity)

share|cite|improve this answer

answered Jul 31 at 15:02

Rafa Budría

4,7661525

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for taking the time to post this. It is clear to me now.
  – handler's handle
  Jul 31 at 17:53
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

For the second question. Simply isolate $s$ in $x=s-phi(s)$ or in

$x=s-dfracss$. For the $sge 0$ part: $x(1+s)=s+s^2-s$ or

$s^2-xs-x=0implies s=frac12(xpmsqrtx^2 + 4x)$ We drop the solution with the minus sign because it entails that $s<0$, aganist the hypothesis for the posing of the equation. Because the same reason ($s<0$) we drop values with $x<0$. So $s=frac12(x + sqrtx^2 + 4x);textif, xge 0$. For $s<0$ we solve $x(1-s)=s-s^2-s$ or ($s^2-xs+x=0$) instead. Then, $s=frac12(xpmsqrtx^2 -4x)$ Now, we drop the plus sign because on the contrary $sge0$ and, because the same reason, it must be $x<0$. This drives to the given solution:

$$s(x) =
begincases
frac12(x + sqrtx^2 + 4x)&textif, xge 0\
frac12(x - sqrtx^2 - 4x)&textif, x < 0\
endcases$$

For the first question you have to consider that $u$ is constant along the characteristic curves, so is, along curves $x-y=c_1$ with $c_1$ some constant($u(x,x+c_1)=k$, $k$ constant). A solution must specify the value of $u$ for each of the values of $c_1$: $u(x,x+c_1)=F(c_1)$. So is, a particular choice of $F$ says the value of $u$ for each member of the family of characteristic curves: $u(x,y)=F(x-y)$. The function $F$ is arbitrary in the sense that by itself doesn't impose any condition on the solutions (taken apart that it must enjoy of some regularity)

share|cite|improve this answer

answered Jul 31 at 15:02

Rafa Budría

4,7661525

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for taking the time to post this. It is clear to me now.
  – handler's handle
  Jul 31 at 17:53
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

For the second question. Simply isolate $s$ in $x=s-phi(s)$ or in

$x=s-dfracss$. For the $sge 0$ part: $x(1+s)=s+s^2-s$ or

$s^2-xs-x=0implies s=frac12(xpmsqrtx^2 + 4x)$ We drop the solution with the minus sign because it entails that $s<0$, aganist the hypothesis for the posing of the equation. Because the same reason ($s<0$) we drop values with $x<0$. So $s=frac12(x + sqrtx^2 + 4x);textif, xge 0$. For $s<0$ we solve $x(1-s)=s-s^2-s$ or ($s^2-xs+x=0$) instead. Then, $s=frac12(xpmsqrtx^2 -4x)$ Now, we drop the plus sign because on the contrary $sge0$ and, because the same reason, it must be $x<0$. This drives to the given solution:

$$s(x) =
begincases
frac12(x + sqrtx^2 + 4x)&textif, xge 0\
frac12(x - sqrtx^2 - 4x)&textif, x < 0\
endcases$$

For the first question you have to consider that $u$ is constant along the characteristic curves, so is, along curves $x-y=c_1$ with $c_1$ some constant($u(x,x+c_1)=k$, $k$ constant). A solution must specify the value of $u$ for each of the values of $c_1$: $u(x,x+c_1)=F(c_1)$. So is, a particular choice of $F$ says the value of $u$ for each member of the family of characteristic curves: $u(x,y)=F(x-y)$. The function $F$ is arbitrary in the sense that by itself doesn't impose any condition on the solutions (taken apart that it must enjoy of some regularity)

share|cite|improve this answer

answered Jul 31 at 15:02

Rafa Budría

4,7661525

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for taking the time to post this. It is clear to me now.
  – handler's handle
  Jul 31 at 17:53
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

For the second question. Simply isolate $s$ in $x=s-phi(s)$ or in

$x=s-dfracss$. For the $sge 0$ part: $x(1+s)=s+s^2-s$ or

$s^2-xs-x=0implies s=frac12(xpmsqrtx^2 + 4x)$ We drop the solution with the minus sign because it entails that $s<0$, aganist the hypothesis for the posing of the equation. Because the same reason ($s<0$) we drop values with $x<0$. So $s=frac12(x + sqrtx^2 + 4x);textif, xge 0$. For $s<0$ we solve $x(1-s)=s-s^2-s$ or ($s^2-xs+x=0$) instead. Then, $s=frac12(xpmsqrtx^2 -4x)$ Now, we drop the plus sign because on the contrary $sge0$ and, because the same reason, it must be $x<0$. This drives to the given solution:

$$s(x) =
begincases
frac12(x + sqrtx^2 + 4x)&textif, xge 0\
frac12(x - sqrtx^2 - 4x)&textif, x < 0\
endcases$$

For the first question you have to consider that $u$ is constant along the characteristic curves, so is, along curves $x-y=c_1$ with $c_1$ some constant($u(x,x+c_1)=k$, $k$ constant). A solution must specify the value of $u$ for each of the values of $c_1$: $u(x,x+c_1)=F(c_1)$. So is, a particular choice of $F$ says the value of $u$ for each member of the family of characteristic curves: $u(x,y)=F(x-y)$. The function $F$ is arbitrary in the sense that by itself doesn't impose any condition on the solutions (taken apart that it must enjoy of some regularity)

share|cite|improve this answer

answered Jul 31 at 15:02

Rafa Budría

4,7661525

For the second question. Simply isolate $s$ in $x=s-phi(s)$ or in

$x=s-dfracss$. For the $sge 0$ part: $x(1+s)=s+s^2-s$ or

$s^2-xs-x=0implies s=frac12(xpmsqrtx^2 + 4x)$ We drop the solution with the minus sign because it entails that $s<0$, aganist the hypothesis for the posing of the equation. Because the same reason ($s<0$) we drop values with $x<0$. So $s=frac12(x + sqrtx^2 + 4x);textif, xge 0$. For $s<0$ we solve $x(1-s)=s-s^2-s$ or ($s^2-xs+x=0$) instead. Then, $s=frac12(xpmsqrtx^2 -4x)$ Now, we drop the plus sign because on the contrary $sge0$ and, because the same reason, it must be $x<0$. This drives to the given solution:

$$s(x) =
begincases
frac12(x + sqrtx^2 + 4x)&textif, xge 0\
frac12(x - sqrtx^2 - 4x)&textif, x < 0\
endcases$$

For the first question you have to consider that $u$ is constant along the characteristic curves, so is, along curves $x-y=c_1$ with $c_1$ some constant($u(x,x+c_1)=k$, $k$ constant). A solution must specify the value of $u$ for each of the values of $c_1$: $u(x,x+c_1)=F(c_1)$. So is, a particular choice of $F$ says the value of $u$ for each member of the family of characteristic curves: $u(x,y)=F(x-y)$. The function $F$ is arbitrary in the sense that by itself doesn't impose any condition on the solutions (taken apart that it must enjoy of some regularity)

share|cite|improve this answer

answered Jul 31 at 15:02

Rafa Budría

4,7661525

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 15:02

Rafa Budría

4,7661525

answered Jul 31 at 15:02

Rafa Budría

4,7661525

answered Jul 31 at 15:02

Rafa Budría

4,7661525

4,7661525

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for taking the time to post this. It is clear to me now.
  – handler's handle
  Jul 31 at 17:53
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for taking the time to post this. It is clear to me now.
  – handler's handle
  Jul 31 at 17:53
  

  
  
  
  

Thank you for taking the time to post this. It is clear to me now.
– handler's handle
Jul 31 at 17:53

Thank you for taking the time to post this. It is clear to me now.
– handler's handle
Jul 31 at 17:53

add a comment | 

 

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