Question about the differentiability of solution on base characteristics curve.
$begingroup$
Let $u(x, t)$ be a function that satisfies the PDE: $u_t+uu_x = 1, x in mathbb{R}, t > 0$, and the initial condition $ubig(frac{t^2}{4}, tbig) = frac{t}{2}$. Then show that the IVP has solutions none of which is differentiable on the characteristics base curve.
My Attempt: By Lagrange's method, we have $frac{dt}{1} = frac{dx}{u}= frac{du}{1}$. On solving above, we get $u-t=a$ and $x-frac{u^2}{2} = c_2$ which further implies
$u=t+fbig(x-frac{u^2}{2}big)$. Now $ubig(frac{t^2}{4}, tbig) = frac{t}{2}$ implies that $frac{t}{2}= t+fbig(frac{t^2}{8}big) implies fbig(frac{t^2}{8}big) = -frac{t}{2} implies f(t) = mpsqrt{2t}$. This means $u = tmpsqrt{2}sqrt{x-frac{u^2}{2}}$. How to proceed further?
pde characteristics
asked Jan 24 at 16:40
Mittal GMittal G
1,301516
$endgroup$
add a comment |
$begingroup$
Let $u(x, t)$ be a function that satisfies the PDE: $u_t+uu_x = 1, x in mathbb{R}, t > 0$, and the initial condition $ubig(frac{t^2}{4}, tbig) = frac{t}{2}$. Then show that the IVP has solutions none of which is differentiable on the characteristics base curve.
My Attempt: By Lagrange's method, we have $frac{dt}{1} = frac{dx}{u}= frac{du}{1}$. On solving above, we get $u-t=a$ and $x-frac{u^2}{2} = c_2$ which further implies
$u=t+fbig(x-frac{u^2}{2}big)$. Now $ubig(frac{t^2}{4}, tbig) = frac{t}{2}$ implies that $frac{t}{2}= t+fbig(frac{t^2}{8}big) implies fbig(frac{t^2}{8}big) = -frac{t}{2} implies f(t) = mpsqrt{2t}$. This means $u = tmpsqrt{2}sqrt{x-frac{u^2}{2}}$. How to proceed further?
pde characteristics
asked Jan 24 at 16:40
Mittal GMittal G
1,301516
$endgroup$
add a comment |
$begingroup$
Let $u(x, t)$ be a function that satisfies the PDE: $u_t+uu_x = 1, x in mathbb{R}, t > 0$, and the initial condition $ubig(frac{t^2}{4}, tbig) = frac{t}{2}$. Then show that the IVP has solutions none of which is differentiable on the characteristics base curve.
My Attempt: By Lagrange's method, we have $frac{dt}{1} = frac{dx}{u}= frac{du}{1}$. On solving above, we get $u-t=a$ and $x-frac{u^2}{2} = c_2$ which further implies
$u=t+fbig(x-frac{u^2}{2}big)$. Now $ubig(frac{t^2}{4}, tbig) = frac{t}{2}$ implies that $frac{t}{2}= t+fbig(frac{t^2}{8}big) implies fbig(frac{t^2}{8}big) = -frac{t}{2} implies f(t) = mpsqrt{2t}$. This means $u = tmpsqrt{2}sqrt{x-frac{u^2}{2}}$. How to proceed further?
pde characteristics
asked Jan 24 at 16:40
Mittal GMittal G
1,301516
$endgroup$
Let $u(x, t)$ be a function that satisfies the PDE: $u_t+uu_x = 1, x in mathbb{R}, t > 0$, and the initial condition $ubig(frac{t^2}{4}, tbig) = frac{t}{2}$. Then show that the IVP has solutions none of which is differentiable on the characteristics base curve.
My Attempt: By Lagrange's method, we have $frac{dt}{1} = frac{dx}{u}= frac{du}{1}$. On solving above, we get $u-t=a$ and $x-frac{u^2}{2} = c_2$ which further implies
$u=t+fbig(x-frac{u^2}{2}big)$. Now $ubig(frac{t^2}{4}, tbig) = frac{t}{2}$ implies that $frac{t}{2}= t+fbig(frac{t^2}{8}big) implies fbig(frac{t^2}{8}big) = -frac{t}{2} implies f(t) = mpsqrt{2t}$. This means $u = tmpsqrt{2}sqrt{x-frac{u^2}{2}}$. How to proceed further?
pde characteristics
pde characteristics
asked Jan 24 at 16:40
Mittal GMittal G
1,301516
asked Jan 24 at 16:40
Mittal GMittal G
1,301516
asked Jan 24 at 16:40
Mittal GMittal G
1,301516
asked Jan 24 at 16:40
Mittal GMittal G
1,301516
asked Jan 24 at 16:40
Mittal GMittal G
1,301516
1,301516
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your calculus is correct.
$$u = tmpsqrt{2}sqrt{x-frac{u^2}{2}}.$$
Then one have to solve this equation for $u$.
$$(u-t)^2=2x-u^2$$
This is a simple quadratic equation. The solution is :
$$u(x,t)=frac{tpm sqrt{4x-t^2}}{2}$$
This function satisfies the PDE and the specified condition.
answered Jan 26 at 19:07
JJacquelinJJacquelin
44.4k21854
$endgroup$
$begingroup$
How to verify the condition for of differentability on characteristics base curve?
$endgroup$
– Mittal G
Jan 27 at 15:22
$begingroup$
Why doing that ? Just check the above result $u(x,t)$ in putting it into the PDE.
$endgroup$
– JJacquelin
Jan 27 at 16:15
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Your calculus is correct.
$$u = tmpsqrt{2}sqrt{x-frac{u^2}{2}}.$$
Then one have to solve this equation for $u$.
$$(u-t)^2=2x-u^2$$
This is a simple quadratic equation. The solution is :
$$u(x,t)=frac{tpm sqrt{4x-t^2}}{2}$$
This function satisfies the PDE and the specified condition.
answered Jan 26 at 19:07
JJacquelinJJacquelin
44.4k21854
$endgroup$
$begingroup$
How to verify the condition for of differentability on characteristics base curve?
$endgroup$
– Mittal G
Jan 27 at 15:22
$begingroup$
Why doing that ? Just check the above result $u(x,t)$ in putting it into the PDE.
$endgroup$
– JJacquelin
Jan 27 at 16:15
add a comment |
$begingroup$
Your calculus is correct.
$$u = tmpsqrt{2}sqrt{x-frac{u^2}{2}}.$$
Then one have to solve this equation for $u$.
$$(u-t)^2=2x-u^2$$
This is a simple quadratic equation. The solution is :
$$u(x,t)=frac{tpm sqrt{4x-t^2}}{2}$$
This function satisfies the PDE and the specified condition.
answered Jan 26 at 19:07
JJacquelinJJacquelin
44.4k21854
$endgroup$
$begingroup$
How to verify the condition for of differentability on characteristics base curve?
$endgroup$
– Mittal G
Jan 27 at 15:22
$begingroup$
Why doing that ? Just check the above result $u(x,t)$ in putting it into the PDE.
$endgroup$
– JJacquelin
Jan 27 at 16:15
add a comment |
$begingroup$
Your calculus is correct.
$$u = tmpsqrt{2}sqrt{x-frac{u^2}{2}}.$$
Then one have to solve this equation for $u$.
$$(u-t)^2=2x-u^2$$
This is a simple quadratic equation. The solution is :
$$u(x,t)=frac{tpm sqrt{4x-t^2}}{2}$$
This function satisfies the PDE and the specified condition.
answered Jan 26 at 19:07
JJacquelinJJacquelin
44.4k21854
$endgroup$
Your calculus is correct.
$$u = tmpsqrt{2}sqrt{x-frac{u^2}{2}}.$$
Then one have to solve this equation for $u$.
$$(u-t)^2=2x-u^2$$
This is a simple quadratic equation. The solution is :
$$u(x,t)=frac{tpm sqrt{4x-t^2}}{2}$$
This function satisfies the PDE and the specified condition.
answered Jan 26 at 19:07
JJacquelinJJacquelin
44.4k21854
answered Jan 26 at 19:07
JJacquelinJJacquelin
44.4k21854
answered Jan 26 at 19:07
JJacquelinJJacquelin
44.4k21854
answered Jan 26 at 19:07
JJacquelinJJacquelin
44.4k21854
44.4k21854
$begingroup$
How to verify the condition for of differentability on characteristics base curve?
$endgroup$
– Mittal G
Jan 27 at 15:22
$begingroup$
Why doing that ? Just check the above result $u(x,t)$ in putting it into the PDE.
$endgroup$
– JJacquelin
Jan 27 at 16:15
add a comment |
$begingroup$
How to verify the condition for of differentability on characteristics base curve?
$endgroup$
– Mittal G
Jan 27 at 15:22
$begingroup$
Why doing that ? Just check the above result $u(x,t)$ in putting it into the PDE.
$endgroup$
– JJacquelin
Jan 27 at 16:15
$begingroup$
How to verify the condition for of differentability on characteristics base curve?
$endgroup$
– Mittal G
Jan 27 at 15:22
$begingroup$
How to verify the condition for of differentability on characteristics base curve?
$endgroup$
– Mittal G
Jan 27 at 15:22
$begingroup$
Why doing that ? Just check the above result $u(x,t)$ in putting it into the PDE.
$endgroup$
– JJacquelin
Jan 27 at 16:15
$begingroup$
Why doing that ? Just check the above result $u(x,t)$ in putting it into the PDE.
$endgroup$
– JJacquelin
Jan 27 at 16:15
add a comment |
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