Let u be a solution of the following PDE $u_x+xu_y=0$ and $u(x,0)=e^x$ then u(2,1) and u(-2,1)? I did this by using Cauchy method, but I got stuck in sign having square root..
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1The initial value is incompatible with the level curves from the differential equation. The problem is invalid, but could be fixed if there was a typo or information missing (domain restrictions, other boundaries etc) – Ninad Munshi Nov 17 '22 at 13:35
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That was the exact question. – Tony Nov 17 '22 at 14:11
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@NinadMunshi can you verify this answer? – Tony Nov 17 '22 at 16:15
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What do you mean? Verify what? – Ninad Munshi Nov 17 '22 at 16:53
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Wether solution by @tom is correct or not? – Tony Nov 18 '22 at 02:28
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Tom hasn't provided a solution per se, he explained in more detail that your problem is invalid. – Ninad Munshi Nov 18 '22 at 06:58
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$\frac{du}{dx} = u_x + u_y\frac{dy}{dx} = 0$ with $\frac{dy}{dx} = x$ is an equivalent formulation of the problem.
$$y = \frac{x^2}{2} + c_1, $$ $$\frac{du}{dx} = 0 \Rightarrow u = f(y-\frac{x^2}{2}) $$ At y=0: $$u(x,0) = f(-\frac{x^2}{2}) = e^{x} \Rightarrow f(a) = exp(\pm(-2a)^{\frac{1}{2}})$$
For x>0: $$u = f(y-\frac{x^2}{2}) = exp({(x^2-2y)^{\frac{1}{2}}})$$ For x<0: $$u = f(y-\frac{x^2}{2}) = exp({-(x^2-2y)^{\frac{1}{2}}})$$ The problem is ill defined there must be another initial condition stating which side of x = 0 the system starts.
tom
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