Mixing
Characteristics for nonhomogeneous wave equation $y_{tt}=y_{xx} + f$
$begingroup$
Consider the initial- and boundary-value problem
$$eqalign{
& {y_{tt}} = {y_{xx}} + f(t,x){text{ }}{text{, (t}}{text{,x)}} in {text{(0}}{text{,}}infty {text{)}} times {text{(0}}{text{,1)}} cr
& y(t,0) = y(t,1) = 0 cr
& y(0,x) = {y_0}(x),{text{ }}{y_t}(0,x) = {y_1}(x) cr} $$
To solve that, I reduced the above system to 1-d hyperbolic system by letting
$$eqalign{
& p = {y_t} - {y_x} cr
& q = {y_t} + {y_x} cr} $$
We obtain
$$eqalign{
& {p_t} = -{p_x} + f(t,x) cr
& {q_t} = {q_x} + f(t,x) cr
& p(t,0) + q(t,0) = p(t,1) + q(t,1) = 0 cr
& p(0,x) = {p_0}(x),q(0,x) = {q_0}(x) cr} $$
The characteristic lines are $x^-=t+c_1$ and $x^+=-t+c_2$ for the first and second equation respectively.
let us consider the first transport equation,
We have along x^-:
$$frac{d}{{dt}}p(t,t + {c_1}) = f(t,t + {c_1})$$
Let $f$ be defined as$$f(t,x) = left{ matrix
{f_1}(t,x),{text{ if x + t}} in (0,1),
{f_2}(t,x),{text{ if x + t}} in (1,2).
right.$$
The characteristic $x^-$ will cuts the line $x=t$ in the point $t=frac{{1 - {c_1}}}{2}$.
Integrating over $(0,t)$, we get
$$p(t,t+c_1) = {p_0}(c_1) + intlimits_0^t {f(s,s+c_1)ds} $$
How can I write the solution in function of $f_1$ and $f_2$?
Is the following expression correct?
$$
p(t,x) = {p_0}(x-t) + intlimits_0^{frac{{1 - {c_1}}}{2}} {{f_1}(s,s - t + x)ds} + intlimits_{frac{{1 - {c_1}}}{2}}^{1 - {c_1}} {{f_2}(s,s - t + x)ds} $$
Or that expression?
$$eqalign{
& p(t,t + {c_1}) = {p_0}({c_1}) + intlimits_0^t {{f_1}(s,s+c_1)ds} ,{text{ if t}} in {text{(0}}{text{,}}frac{{1 - {c_1}}}{2}) cr
& p(t,t + {c_1}) = {p_0}({c_1}) + intlimits_0^t {{f_1}(s,s + {c_1})ds} + + intlimits_{frac{{1 - {c_1}}}{2}}^t {{f_2}(s,s + {c_1})ds} ,{text{ if t}} in {text{(}}frac{{1 - {c_1}}}{2},1 - {c_1}) cr} $$
pde wave-equation characteristics hyperbolic-equations linear-pde
edited Jan 23 at 16:44
Harry49
7,44431340
asked Jan 18 at 16:08
GustaveGustave
734211
$endgroup$
add a comment |
$begingroup$
Consider the initial- and boundary-value problem
$$eqalign{
& {y_{tt}} = {y_{xx}} + f(t,x){text{ }}{text{, (t}}{text{,x)}} in {text{(0}}{text{,}}infty {text{)}} times {text{(0}}{text{,1)}} cr
& y(t,0) = y(t,1) = 0 cr
& y(0,x) = {y_0}(x),{text{ }}{y_t}(0,x) = {y_1}(x) cr} $$
To solve that, I reduced the above system to 1-d hyperbolic system by letting
$$eqalign{
& p = {y_t} - {y_x} cr
& q = {y_t} + {y_x} cr} $$
We obtain
$$eqalign{
& {p_t} = -{p_x} + f(t,x) cr
& {q_t} = {q_x} + f(t,x) cr
& p(t,0) + q(t,0) = p(t,1) + q(t,1) = 0 cr
& p(0,x) = {p_0}(x),q(0,x) = {q_0}(x) cr} $$
The characteristic lines are $x^-=t+c_1$ and $x^+=-t+c_2$ for the first and second equation respectively.
let us consider the first transport equation,
We have along x^-:
$$frac{d}{{dt}}p(t,t + {c_1}) = f(t,t + {c_1})$$
Let $f$ be defined as$$f(t,x) = left{ matrix
{f_1}(t,x),{text{ if x + t}} in (0,1),
{f_2}(t,x),{text{ if x + t}} in (1,2).
right.$$
The characteristic $x^-$ will cuts the line $x=t$ in the point $t=frac{{1 - {c_1}}}{2}$.
Integrating over $(0,t)$, we get
$$p(t,t+c_1) = {p_0}(c_1) + intlimits_0^t {f(s,s+c_1)ds} $$
How can I write the solution in function of $f_1$ and $f_2$?
Is the following expression correct?
$$
p(t,x) = {p_0}(x-t) + intlimits_0^{frac{{1 - {c_1}}}{2}} {{f_1}(s,s - t + x)ds} + intlimits_{frac{{1 - {c_1}}}{2}}^{1 - {c_1}} {{f_2}(s,s - t + x)ds} $$
Or that expression?
$$eqalign{
& p(t,t + {c_1}) = {p_0}({c_1}) + intlimits_0^t {{f_1}(s,s+c_1)ds} ,{text{ if t}} in {text{(0}}{text{,}}frac{{1 - {c_1}}}{2}) cr
& p(t,t + {c_1}) = {p_0}({c_1}) + intlimits_0^t {{f_1}(s,s + {c_1})ds} + + intlimits_{frac{{1 - {c_1}}}{2}}^t {{f_2}(s,s + {c_1})ds} ,{text{ if t}} in {text{(}}frac{{1 - {c_1}}}{2},1 - {c_1}) cr} $$
pde wave-equation characteristics hyperbolic-equations linear-pde
edited Jan 23 at 16:44
Harry49
7,44431340
asked Jan 18 at 16:08
GustaveGustave
734211
$endgroup$
add a comment |
$begingroup$
Consider the initial- and boundary-value problem
$$eqalign{
& {y_{tt}} = {y_{xx}} + f(t,x){text{ }}{text{, (t}}{text{,x)}} in {text{(0}}{text{,}}infty {text{)}} times {text{(0}}{text{,1)}} cr
& y(t,0) = y(t,1) = 0 cr
& y(0,x) = {y_0}(x),{text{ }}{y_t}(0,x) = {y_1}(x) cr} $$
To solve that, I reduced the above system to 1-d hyperbolic system by letting
$$eqalign{
& p = {y_t} - {y_x} cr
& q = {y_t} + {y_x} cr} $$
We obtain
$$eqalign{
& {p_t} = -{p_x} + f(t,x) cr
& {q_t} = {q_x} + f(t,x) cr
& p(t,0) + q(t,0) = p(t,1) + q(t,1) = 0 cr
& p(0,x) = {p_0}(x),q(0,x) = {q_0}(x) cr} $$
The characteristic lines are $x^-=t+c_1$ and $x^+=-t+c_2$ for the first and second equation respectively.
let us consider the first transport equation,
We have along x^-:
$$frac{d}{{dt}}p(t,t + {c_1}) = f(t,t + {c_1})$$
Let $f$ be defined as$$f(t,x) = left{ matrix
{f_1}(t,x),{text{ if x + t}} in (0,1),
{f_2}(t,x),{text{ if x + t}} in (1,2).
right.$$
The characteristic $x^-$ will cuts the line $x=t$ in the point $t=frac{{1 - {c_1}}}{2}$.
Integrating over $(0,t)$, we get
$$p(t,t+c_1) = {p_0}(c_1) + intlimits_0^t {f(s,s+c_1)ds} $$
How can I write the solution in function of $f_1$ and $f_2$?
Is the following expression correct?
$$
p(t,x) = {p_0}(x-t) + intlimits_0^{frac{{1 - {c_1}}}{2}} {{f_1}(s,s - t + x)ds} + intlimits_{frac{{1 - {c_1}}}{2}}^{1 - {c_1}} {{f_2}(s,s - t + x)ds} $$
Or that expression?
$$eqalign{
& p(t,t + {c_1}) = {p_0}({c_1}) + intlimits_0^t {{f_1}(s,s+c_1)ds} ,{text{ if t}} in {text{(0}}{text{,}}frac{{1 - {c_1}}}{2}) cr
& p(t,t + {c_1}) = {p_0}({c_1}) + intlimits_0^t {{f_1}(s,s + {c_1})ds} + + intlimits_{frac{{1 - {c_1}}}{2}}^t {{f_2}(s,s + {c_1})ds} ,{text{ if t}} in {text{(}}frac{{1 - {c_1}}}{2},1 - {c_1}) cr} $$
pde wave-equation characteristics hyperbolic-equations linear-pde
edited Jan 23 at 16:44
Harry49
7,44431340
asked Jan 18 at 16:08
GustaveGustave
734211
$endgroup$
Consider the initial- and boundary-value problem
$$eqalign{
& {y_{tt}} = {y_{xx}} + f(t,x){text{ }}{text{, (t}}{text{,x)}} in {text{(0}}{text{,}}infty {text{)}} times {text{(0}}{text{,1)}} cr
& y(t,0) = y(t,1) = 0 cr
& y(0,x) = {y_0}(x),{text{ }}{y_t}(0,x) = {y_1}(x) cr} $$
To solve that, I reduced the above system to 1-d hyperbolic system by letting
$$eqalign{
& p = {y_t} - {y_x} cr
& q = {y_t} + {y_x} cr} $$
We obtain
$$eqalign{
& {p_t} = -{p_x} + f(t,x) cr
& {q_t} = {q_x} + f(t,x) cr
& p(t,0) + q(t,0) = p(t,1) + q(t,1) = 0 cr
& p(0,x) = {p_0}(x),q(0,x) = {q_0}(x) cr} $$
The characteristic lines are $x^-=t+c_1$ and $x^+=-t+c_2$ for the first and second equation respectively.
let us consider the first transport equation,
We have along x^-:
$$frac{d}{{dt}}p(t,t + {c_1}) = f(t,t + {c_1})$$
Let $f$ be defined as$$f(t,x) = left{ matrix
{f_1}(t,x),{text{ if x + t}} in (0,1),
{f_2}(t,x),{text{ if x + t}} in (1,2).
right.$$
The characteristic $x^-$ will cuts the line $x=t$ in the point $t=frac{{1 - {c_1}}}{2}$.
Integrating over $(0,t)$, we get
$$p(t,t+c_1) = {p_0}(c_1) + intlimits_0^t {f(s,s+c_1)ds} $$
How can I write the solution in function of $f_1$ and $f_2$?
Is the following expression correct?
$$
p(t,x) = {p_0}(x-t) + intlimits_0^{frac{{1 - {c_1}}}{2}} {{f_1}(s,s - t + x)ds} + intlimits_{frac{{1 - {c_1}}}{2}}^{1 - {c_1}} {{f_2}(s,s - t + x)ds} $$
Or that expression?
$$eqalign{
& p(t,t + {c_1}) = {p_0}({c_1}) + intlimits_0^t {{f_1}(s,s+c_1)ds} ,{text{ if t}} in {text{(0}}{text{,}}frac{{1 - {c_1}}}{2}) cr
& p(t,t + {c_1}) = {p_0}({c_1}) + intlimits_0^t {{f_1}(s,s + {c_1})ds} + + intlimits_{frac{{1 - {c_1}}}{2}}^t {{f_2}(s,s + {c_1})ds} ,{text{ if t}} in {text{(}}frac{{1 - {c_1}}}{2},1 - {c_1}) cr} $$
pde wave-equation characteristics hyperbolic-equations linear-pde
pde wave-equation characteristics hyperbolic-equations linear-pde
edited Jan 23 at 16:44
Harry49
7,44431340
asked Jan 18 at 16:08
GustaveGustave
734211
edited Jan 23 at 16:44
Harry49
7,44431340
asked Jan 18 at 16:08
GustaveGustave
734211
edited Jan 23 at 16:44
Harry49
7,44431340
edited Jan 23 at 16:44
Harry49
7,44431340
edited Jan 23 at 16:44
Harry49
7,44431340
7,44431340
asked Jan 18 at 16:08
GustaveGustave
734211
asked Jan 18 at 16:08
GustaveGustave
734211
asked Jan 18 at 16:08
GustaveGustave
734211
734211
add a comment |
add a comment |
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